Quantum Field Theory: Spring 2010 Prof. Dave Goldberg June 1, 2012 Contents 0 Expectations and Notation 4 1 Worked Example: The SHO 5 1.1 The Lagrangian and the Equations of Motion ........... 5 1.2 The Classical Solution to the SHO ................. 6 1.3 Noether’s Theorem: Part 1 ..................... 7 1.4 SHO → QHO ............................. 8 1.5 Evolution of the Free Field Solution ................ 11 1.6 The Heisenberg and Interaction Representation .......... 12 1.6.1 The Heisenberg Representation ............... 12 1.6.2 The Interaction Representation ............... 13 1.6.3 The Interaction Unitary Operator ............. 14 1.7 Example: Perturbed QHO ...................... 15 2 Classical Free Fields 17 2.1 Natural Units ............................. 17 2.2 The Lagrangian ............................ 18 2.3 Minimizing The Action ....................... 20 2.4 The Klein-Gordan Equation ..................... 21 2.5 What the Lagrangian means ..................... 22 2.6 Noether’s Theorem: Part 2 ..................... 22 2.6.1 Displacements in space/time ................ 25 2.6.2 Another Lagrangian, and another conserved current ... 26 3 Free Quantized Scalar Fields 28 3.1 From Continuous to Quantized Field: The Real-valued Scalar Free Field ............................... 28 3.2 The Creation and Annihilation Operators ............. 29 3.3 The Hamiltonian ........................... 30 3.4 The Vacuum ............................. 31 3.5 Operators and Observables ..................... 32 3.6 Normalizing the field ......................... 34 3.7 The Propagator ............................ 35 1
Transcript
Quantum Field Theory: Spring 2010
Prof. Dave Goldberg
June 1, 2012
Contents
0 Expectations and Notation 4
1 Worked Example: The SHO 51.1 The Lagrangian and the Equations of Motion . . . . . . . . . . . 51.2 The Classical Solution to the SHO . . . . . . . . . . . . . . . . . 61.3 Noether’s Theorem: Part 1 . . . . . . . . . . . . . . . . . . . . . 71.4 SHO → QHO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Evolution of the Free Field Solution . . . . . . . . . . . . . . . . 111.6 The Heisenberg and Interaction Representation . . . . . . . . . . 12
• You should be comfortable enough with the material in the first lecturethat about ∼ 90% of it should be review.
• If you are an undergrad, you should have taken (and passed with a B orbetter in both) Quantum Mechanics I & II, as well as Classical MechanicsI & II.
• You should be familiar with SR. I don’t expect you’ve taken SR, but youshould know about Lorentz boosts, for example.
Notation:
• p is an operator, p is a variable or observable.
• ~p represents a 3-vector, while p is a 4-vector (to be defined). pµ (or withany other Greek letter) are the components of the 4-vector.
• I’ll be using Einstein summation convention. That is, if you see aµbµ, itimplicitly means:
aµbµ =
3∑
µ=0
aµbµ
In other words, it sums to a scalar. Whenever you see pairs of indiceslike this, you do the sum. I’ll explain the significance of the upstairs anddownstairs indices when the time comes.
4
1 Worked Example: The SHO
• Tong 2.1, 3.1
• Gross Chapter 3.1
If you don’t understand the Simple Harmonic Oscillator (SHO), then youwill not understand this course. Further, by working through the SHO, you’llknow exactly how we’re going to introduce quantum fields.
1.1 The Lagrangian and the Equations of Motion
In freshman year, you learned about the Simple Harmonic Oscillator. We’regoing to start by describing everything we know in the classical regime. Noquantum whatsoever, at least for a little bit. You’ll know when quantum showsup, because you’ll start seeing an h or ~.
We have a system with a Kinetic and potential energy:
K =1
2mx2
U =1
2mω2x2
where we used the magical relationship:
ω =
√
k
m
This yields a Lagrangian:
L =1
2mx2 − 1
2mω2x2 (1)
We can compute the canonical momentum via:
p =∂L
∂x= mx (2)
which, of course it is.Finally, we get the Euler-Lagrange equations for a non-relativistic field with
one degree of freedom, which yields:
d
dt
(
∂L
∂x
)
=∂L
∂x(3)
x = −ω2x (4)
5
1.2 The Classical Solution to the SHO
We know that the simple Harmonic Oscillator has a solution of the form:
x(t) =1√2
[c exp(−iωt) + c∗ exp(iωt)] (5)
We have long since grown past the point where we need to talk about sines andcosines. We’re big boys and girls. However, because there is both a c and a c∗,the overall position is necessarily real.
What’s more, since c is complex, there are two variables, which uniquelygives both the amplitude of the oscillation, and the phase.
Furthermore:
v =dx
dt= − iω√
2[c exp(−iωt) − c∗ exp(iωt)]
This yields the classical energy:
K =1
2mω2cc∗ − 1
4mω2
[
c2 exp(−2iωt) + c∗2 exp(2iωt)]
U =1
2mω2cc∗ +
1
4mω2
[
c2 exp(−2iωt) + c∗2 exp(2iωt)]
so:E = mω2cc∗ (6)
You should bear this in mind. We know that in quantum mechanics, the calcu-lation of energy is very important. When we introduce the operator called theHamiltonian, it’s nothing more than the generator of the energy. We also knowthat it plays a very important role in evolving the wave-function.
If you like, you can imagine that at some point in the future – perhaps whenwe introduce quantum mechanics, c and c∗ (which are currently just numbers– albeit complex ones – might be turned into something else. So, perhaps it’smore appropriate to say:
E =mω
~× ~ω × 1
2(cc∗ + c∗c) (7)
There’s nothing wrong with what I did. I’m allowed to do it, and you can’t stopme. However... my introduction of ~, as I did, should certainly raise some redflags that we’re about to move into quantum mechanics. But clearly for nowthere’s nothing technically quantum about my choice.
If I define a dimensionless number, a, such that:
a =
√
mω
hc (8)
(which again, I’m perfectly free to do), then we get:
E =1
2~ω(aa∗ + a∗a) (9)
in full generality.
6
1.3 Noether’s Theorem: Part 1
When we get into QFT proper, I’m going to skip a proof of Noether’s theorem(but it will appear in your notes in case you’re curious). Basically, Noether’stheorem states:
If the action is invariant under some transformation, then thereis a conserved quantity for the system.
What conserved quantity?We’re going to deal (in this case) with a system of n degrees of freedom, qi.
QFT formally introduces an infinite number of degrees of freedom, which makesthings a bit complicated. Now, we know that for a properly minimized action,we get:
d
dt
(
∂L
∂qi
)
=∂L
∂qi
which is just the Euler-Lagrange equations.However, we can imagine adjusting the Lagrangian by an amount dL by
varying the degrees of freedom. That is:
δL =∑ ∂L
∂qiδqi
If we can adjust the system such that L remains fixed, then clearly the actionis fixed. Of course, the RHS of this equation looks a lot like the LHS of theEuler-Lagrange equations. Thus:
d
dt
[
∑
i
∂L
∂qi
dqidα
]
=dL
dα(10)
where α is the parameter under which the Lagrangian doesn’t explicitly depend.For example, if we’re working in Cartesian coordinates, then the transfor-
mation:x→ x+ α
represents a coordinate shift in the x-direction, and thus:
dx
dα= 1
Pretty boring.More interesting, perhaps, is a rotation around the z-axis:
x → x cosα+ y sinα
y → y cosα− x sinα
yielding:
dx
dα= y
dy
dα= −x
7
Thus, the stuff in the parentheses is a conserved quantity.For example, consider:
L =1
2m∑
i
q2i − V (q)
where the potential is assumed to be function of all possible (Cartesian) coor-dinates.
Consider, q = x. We get:
d
dt[mx] = −dV
dx
So if there is no explicit dependence of the potential in the x-direction, theconserved quantity is the x-component of momentum.
I’ll leave it as an exercise to show that if you vary, say, the azimuthal coor-dinate, φ, you get conservation of angular momentum in the z-direction.
There is a special case: conservation of the action over variations in time.In that case, it is clear that the RHS of equation (10) becomes dL/dt, and thusthe entire equation may be combined to yield:
d
dt
[
∑
i
∂L
∂qiqi − L
]
= 0 (11)
You may recognize the bit in the hard brackets as the Hamiltonian, or equiva-lently, energy.
For our sample Lagrangian this yields:
H =1
2m∑
i
q2i + V (q)
Of course.Remember: energy and time invariance are very intimately related.
1.4 SHO → QHO
In your undergraduate quantum class, when you went from classical to quantumfields, you probably did so by directly solving the Schroedinger equation. If youdid that for a SHO, you’d find the Hermite polynomials. We’re not going to dothat here.
Instead, we note that in our new coordinates:
x(t) =
√
~
mω
1√2
[a exp(−iωt) + a∗ exp(iωt)]
p(t) = −i√m~ω
1√2
[a exp(−iωt) − a∗ exp(iωt)]
8
and therefore, algebraically, we get:
a(t) =1√2
[
√
mω
~x(t) + i
√
1
mω~p(t)
]
(12)
a∗(t) =1√2
[
√
mω
~x(t) − i
√
1
mω~p(t)
]
(13)
(14)
where we’ve absorbed the exponential term in our definition of a(t), and a∗(t).Furthermore, since we’re clearly sliding into the quantum mechanical picture
anyway, let’s go whole hog. For now, we’ll choose the Schroedinger picture, sinceit’s the one you’re most familiar with. In that case, all of our observables becomeoperators on some wave-function:
p(t) → p|ψ(t)〉 = p|ψ(t)〉 (15)
where I’ve gone all the way and used the Dirac “bra” and “ket” notation ex-plicitly. Further, all operators in this system (Schroedinger Picture) are time-independent. So we get:
a =1√2
[
√
mω
~x+ i
√
1
mω~p
]
a† =1√2
[
√
mω
~x− i
√
1
mω~p
]
where I’ve done a switcheroo from ∗ to † because I’m now using operatorsrather than numbers. Further, as you will recall from your QM course, wealmost always care only about commutation relations.
You’ll recall~p = −i~∇ (16)
in general andp = −i~∂x
in particular (where ∂x = ∂∂x for shorthand).
Likewise,~x = ~x (17)
So:
[x, p]ψ = x(−i~∂x)ψ − (−i~∂x)xψ= i~ψ
or[x, p] = i~ (18)
9
Consequently:
[a, a†] =1
2(TERMS...)[x, x] +
1
2(TERMS...)[p, p] − i
2~[x, p] +
i
2~[p, x]
= 0 + 0 +1
2+
1
2= 1
or equivalently:aa† = a†a+ 1 (19)
which was TOTALLY the entire point of this exercise. Now, I know you’ve seenthis before, but since we’re going to be using very similar results for some time,it’s important that you absolutely get this.
Remember our energy (equation 9)? It now becomes an operator as well:
H =1
2~ω[
aa† + a†a]
(20)
= ~ω
[
a†a+1
2
]
(21)
(22)
What are the meanings of these operators? Well, I trust you already know that.They’re simply creation and annihilation (ladder) operators. We can even proveit.
For instance, suppose we say that the system is in state, |n〉 with Eigen-energy ~ωN (n is not necessarily an integer). Equivalently, we have:
a†a|n〉 = (N − 1
2)|n〉
Now consider:
Ha†|n〉 = ~ω
(
a†a+1
2
)
a†|n〉
= ~ω
(
a†aa† +1
2a†)
|n〉
= ~ω
(
a†[
a†a+ 1]
+1
2a†)
|n〉
= ~ω
(
a†a†a+3
2a†)
|n〉
= ~ω
(
a†(N − 1/2) +3
2a†)
|n〉
= ~ω(N + 1)a†|n〉
10
Or equivalently (and throwing in the normalization for good measure),
a†|n〉 =√n+ 1|n+ 1〉 (23)
And where similar arguments show:
a|n〉 =√n|n− 1〉 (24)
And finally,N = a†a (25)
which is the number operator. Because of the square root bit, it n can’t benegative. So it turns out that the only definition which makes sense if for n = 0to be the ground state (E0 = 1/2~ω, and all other states to be steps of integervalues such that:
En =
(
n+1
2
)
~ω (26)
This is the magic of quantum mechanics, and this is a result that I expectyou’ve seen in Quantum I or II.
These step-up and step-down operators are going to turn out to be exactlyanalogous to the creation and annihilation operators for particles in a field.
Hint of things to come:In QFT, the square of the amplitude of a field is going to be
something like the number of particles.
1.5 Evolution of the Free Field Solution
Suppose we have a simple harmonic oscillator. In principle, we can describe itat any instant by:
|ψ〉 = cn|n〉where out of sheer laziness, I’m generally going to omit explicit sums. This cnis different that the c’s we saw before.
Now the thing about quantum mechanics (and QFT) is that it is unitary,which means that the state of the system tells you everything you need to knowabout the future evolution. Or more specifically, we can define the evolutionvia:
i~ ˙|ψ〉 = H |ψ〉 (27)
The time-dependent Schroedinger equation. Please note that we are going towant a form that looks something like this for every system and field we en-counter because it will tell us how a field will evolve.
For a single mode, cn is a single value, and we get:
i~cn|n〉 = Encn|n〉
which is solved by
cn(t) = cn(0) exp
(
− iEnt~
)
11
In other words, if we’re in an energy Eigenstate, the free field solution says thatwe’re going to stay there, and only change in phase.
In practice, what we’d like to do is define a “Unitary Operator” (Evolutionoperator, if you like):
U(t) = exp
(
− iHt~
)
(28)
such that:|f〉 = U(t)|i〉
and similarly〈f | = 〈i|U †(t)
where “f” and “i” denote the final and initial states of the system, respectively.Get used to it. We’ll use that sort of shorthand a lot. Of course, both Unitarity(and inspection) guarantee that:
U † = U−1 (29)
If you’re confused about how we take the exponent of an operator, recallthat this can simply be re-written as a Taylor series:
U(t) = I − tH +1
2t2HH − ...
For a system decomposed into eigenstates of the Hamiltonian, the Unitaryoperator simply changes the phase of each of the cn coefficients.
In reality, though, QM tells us only about the Observables of a system. Forany operator, O, there is an observable O, such that:
〈O(t)〉 = 〈i|U †(t)OU(t)|i〉
Remember: by definition you only get to measure the observables and eigen-states. Anything which leaves them unchanged is fair game. Which brings usto...
1.6 The Heisenberg and Interaction Representation
1.6.1 The Heisenberg Representation
Nothing prevents us from saying that the operators change with time, and thatthe states of the system remain constant. Indeed, this is the “Heisenberg” pic-ture of quantum mechanics. The one we’ve been using so far is the Schroedingerpicture. Both are pretty simple so long as our basis states are eigen-functionsof the Hamiltonian.
Essentially:OH = U †OU (30)
And likewise:
|ψ〉H = ˆU(t)†|ψ(t)〉S (31)
12
which cancels the |ψ(t)〉 = U(t)|ψ(0)〉 from the Schroedinger picture and makesthe wave-function fixed.
ψH(t) = ψH(0)
The observables evolve exactly as the do in the Schroedinger picture.Incidentally, since the Hamiltonian necessarily commutes with itself:
HH = HS
This works all well and good if we’re dealing with Eigen-functions of theHamiltonians, but suppose we’re not?
1.6.2 The Interaction Representation
Suppose instead that we have:
H = H0 + Hint(t) (32)
We imagine that our system is in some Eigenstate of H0, and then an interactionis introduced. What happens next?
We’ve already seen (in the Schroedinger interpretation), a wave-functionevolves via:
|f〉 = U(t, t0)|i〉where I’ve introduced the extra argument in U to arbitrarily allow a wave-function to evolve from t0 (rather than 0) to t. In general, I’ll omit the explicitreference to t unless I actually need it.
Now, what happens if we define:
U ≡ U0UI (33)
Be careful! This isn’t as trivial as it seems, and U isn’t constructed simply fromHint and the reason should be clear. H0 and Hint don’t commute, and whenyou expand out an exponential, you’re going to get all sorts of combinations ofthem.
But now consider some operator, O. In the Heisenberg representation, we’dwrite it as:
O(t) = UO(t0)U†
Where the O(t) on the left is the Heisenberg version of the operator and onthe left, it’s the Schroedinger. Let’s expand the unitary operator using, equa-tion (33). We get:
O(t) = U †I U
†0 O(t0)U0UI
= U †I O0(t)UI
where O0(t) is the operator that you would have gotten from H0 alone: thefree-field version.
13
This is the Interaction Representation. Basically, this means that you findthe eigen-vectors and operators as if they were in the Heisenberg representationin the free-field limit and then see how they change with time.
Solving for UI will be the main challenge of pretty much everything we’re do-ing. Naturally, once we have it, finding the evolution of a state in the interactionpicture will be quite straightforward (if difficult to implement):
|f〉 = UI |i〉 (34)
and if the interaction is only for a finite duration, this allows us to evolve systemthrough an interaction.
1.6.3 The Interaction Unitary Operator
But given some specified interaction, how do we calculate the interaction oper-ator?
I’ll forgo the algebra and point out that because it doesn’t commute withH0, the interaction term is going to have to be defined recursively. Importantly:
U(t, t0) 6= exp(−iHint(t− t0)/~)
I’ll simply give the answer (and then explain it). First, define the following:
HI = U †0 HintU0 (35)
which is just the interaction unitary operator written in the interaction repre-sentation.
U(t, t0) = I − i
~
∫ t
t0
dt′HI(t′) +
(
− i
~
)2 ∫ t
t0
dt′∫ t′
t0
dt′′HI(t′)HI(t
′′) + ... (36)
The first term is easy to explain. No interaction means that a state is unchanged.The second term is simple as well. It’s simply the Taylor expansion over a shortperiod of time. The third is where things get confusing. The Issue is that t′′ < t′
according to the limits, so the Hamiltonians have to be applied in the order aswritten.
Naturally, we could keep writing terms forever. This is the origin of thefact that there are infinitely many Feynman diagrams to describe a process.However, the further they are to the right (in this case), the less important theyare going to be.
To keep everything tidy, we can express the entire thing as:
U(t, t0) = T exp
[
− i
~
∫ t
t0
dt′HI(t′)
]
(37)
where the T function basically means: “at any given time, expand the wholething out and sort it so that all terms with the earliest time terms go furthestto the right.”
It’s just a shorthand, but one which will be very useful, even once we moveinto QFT.
14
1.7 Example: Perturbed QHO
Let’s put this into practice. Consider for the moment a QHO. We’ll start it inthe ground state. This will be commonplace for a number of QFT calculations.Then at t0 = 0, we start applying a force, F0 for a period τ , after which westop. Classically, the work done is thus F0x, and thus:
Hint =F0√
2
√
~
mω
(
a+ a†)
All I’ve done here is expand out x from our original definition.It’s straightforward (but a bit tedious) to compute HI , so I’ll give you the
answer:
HI =F0√
2
√
~
mω
(
ae−iωt + a†eiωt)
(38)
The form shouldn’t surprise you.So what about the Unitary operator itself?The first term, of course, is I. The second term is a bit tougher, but still
straightforward:
− i
~
∫ t
t0
dt′HI(t′) =
F0√2
√
1
m~ω
1
ω
[
a(
e−iωt − 1)
− a†(
eiωt − 1)]
and so on, with the 3rd term, which I’ll leave as an exercise. I know, obnoxious,right?
What happens if we apply this force on the ground state for a relativelyshort time? Well, the time terms in the parentheses reduce, and we get:
UI(t) ≃ I +F0√
2
√
1
m~ω
1
ω
[
a(−iωt) − a†(iωt)]
= I − iF0t1√2
√
1
m~ω(a+ a†)
So what is the probability of, say, pushing the oscillator to the first excitedstate?
P10 = S210 = |〈1|UI |0〉|2
As written, this is easy! After all, only the a† operator matters. So we get:
S10 = −iF0t1√2
√
1
m~ω
or
P = F 20 t
2 1
2m~ω
What do we make of this?
15
Well, let’s think about this classically. If we assume that F0 is small, andt is much less than a period then classically. Basically, it’s as if we gave theoscillator an impulse:
p = F0t
Of course, this means that the total energy of the system is:
∆E = K =p2
2m=F 2
0 t2
2m
But this is (by construction), a small number. For example, it is only:
∆E
~ω=
F 20 t
2
2m~ω
of the excitation energy.Hot damn! This is exactly the quantum probability that we found. This is,
of course, a good thing. They should limit to the same value.
16
2 Classical Free Fields
• Gross 1.1, 1.2, 2.1, 4.1, 8.1-8.2
• Tong: 0.1, 1
We’re dealing with fields in this class, not particles. Oh sure, at the end ofthe day we’re going to have to relate what we learn about the fields to particlebehavior. But the name of the game will be to compute the properties of the fieldfirst. Typically you’ve only ever done this with photons. But why? Electrons(and quarks, and neutrinos and every other particle) should work the same way.
What is a field? A field is basically a scalar (or vector, or tensor) whathas potentially a different value at every point in space and time. A field isnot a wave-function, even though some of the same equations describe, say, arelativistic scalar field, and a single relativistic quantized particle.
For one thing, when we quantize fields we’re going to realize that talkingabout the wave-function of “an” electron is meaningless. Electrons and positronspop into and out of existence constantly. This is the curse and the beauty of aspecial relativistic theory.
Secondly, in nonrelativistic QM, the wave-function itself was never measur-able. It represented the square root of a probability, but couldn’t, itself, bemeasured. A field can be. A temperature field is a good example.
So here’s what we’re going to do. I’m going to start by describing theproperties of a field, and then we’re going to show how the dynamics fall outnaturally. To begin with, we’re going to imagine something very much like atemperature field: a real valued scalar field. But before we get into it, we’regoing to have to simplify our notation somewhat.
We’re going to take an unusual route for the next little bit. I am going toconcentrate on just giving you a flavor of how classical fields work, so that whenwe start doing quantum fields, you won’t be surprised. Because we’re doing thisin an unusual order, we’re going to skip around a bit in our texts. Don’t worry.We’ll get back to a linear progression in short order.
2.1 Natural Units
Throughout, we’re going to use “natural units,” meaning:
c = ~ = 1
This may cause a bit of confusion, not because you’re not smart enough to un-derstand units, but because I’ve already told you that we’re not using quantummechanics, and therefore you might not expect ~ to show up at all. It will, butonly when we want give some correspondence to actual particles.
Using natural units, all quantities can be expressed as energy to some power:
[m] = [E]1
(39)
17
for example. This one should be obvious, since E = mc2. Thus, for example, innatural units, the mass of a proton would be 935MeV . Don’t say, “the energy...”It’s the mass. Just in natural units.
We get a more complicated behavior when referring to length. It’s worthnoting that just like in classical mechanics (in which we define a deBrogliewavelength), we can define a “Compton wavelength”:
λC =~
mc(40)
which has the correct dimensionality. The physical interpretation of the Comp-ton wavelength is that it is the smallest scale on which a single particle can beidentified. On smaller scales, the energy goes up, and particles can be createdout of the vacuum. Moreover, it is clear that in natural units:
[L] = [E]−1
(41)
Thus large scales are low energies and vice-versa. To put things in perspective,in natural units, 1A is (1.970keV )−1. Note that this is much less than the massof an electron, and thus atoms are in the classical limit.
Finally, note that since L = cT (unit-wise), we immediately get:
[T ] = [E]−1
(42)
Of course, these units can be combined into all sorts of things. For example,energy density is in units of [E]
4, speeds are dimensionless (fractions of c), and
so on.You will need to master these units, and to assure that you do so, I’ve
included an exercise in the first homework.
2.2 The Lagrangian
I know I’ve said it before, but the wave-function below not the same thing aswe’ve seen in non-relativistic quantum mechanics. Sure, it’s an amplitude of awave, so in some way, ψ2 is going to give us some useful information, but it isnot true that |ψ2|d3~x = Pd3~x.
This is a classical field. You should be thinking about the electromagneticfield, NOT the electron wave-function. All we’re saying for the moment is thatthe wave is oscillating.
As it will turn out, the thing that we care most about when evaluatingclassical (or quantum) fields is the Lagrangian density, L, which is defined insuch a way that the action:
S ≡∫
d4xL (43)
is minimized.Let me start with the simplest possible Lagrangian density, that of a real-
valued scalar field.
18
I could just write this, or we could consider a 1-dimensional array of beads(each of mass, m), connected by springs with unstretched length, ∆ from oneanother, and where each bead is constrained to move in the y-direction. Further,we can imagine looping this string of beads into a circle so that the system isperiodic.
Thus, the kinetic energy will be:
K =∑
i
1
2my2
i
That’s the easy part. However, if we consider the vertical displacement of twobeads:
s = (yi+1 − yi)
then the potential energy stored in the ring will be:
U =1
2
∑
i
k(yi+1 − yi)2
or (just guessing)
L =1
2
∑
i
αy2i − β
(
dy
dx
)2
Replacing y with φ, the value of the field, we might even suppose the followingin the continuous (and 3-d) limit:
L =1
2φ2 − 1
2(∇φ)
2 − 1
2m2φ2 (44)
Where does this come from? Well, to some degree, I pulled it out of thinair. However, we can say a few things about it.
All Lagrangians must be real valued, and be Lorentz-invariant scalars. What’smore, in our special units, the Lagrangian density will be [E/l3] = [E]4. Sincelength has units of inverse energy, as written, our field, φ, has units of energy.
The “m” clearly stands for the mass of the particle, but I’ve implied that wehave a continuous field. Don’t worry about it. We’ll see how we get somethingkind of like particles even before formally quantizing our fields.
We want to write everything in a Lorentz invariant way. That means thateverything should be a simple scalar. In relativity, components of a vector arewritten like:
V µ
and where µ can take on the values 0,1,2,3 (t,x,y,z).Ordinary derivatives are written as:
φ,ν = ∂νφ
Either notation will do. I’m likely to use the latter. So for ν = 2, for example,this is equal to dφ/dy.
19
There are two more rules you need to know. 1st is the rule for raising orlowering indices. We can multiply by:
ηµν = ηµν =
1 0 0 00 −1 0 00 0 −1 00 0 0 −1
(45)
The second rule is that whenever you see the same dummy index “up top”and “down below” you add them all together. For example:
AµBµ = A0B0 +A1B1 +A2B2 +A3B3
= A0B0 − ~A · ~B
So this η (or the Minkowski metric, as it’s formally known) is basically a wayof taking dot products in special relativity.
Thus, our Lagrangian density becomes (and you can check my math on this):
L =1
2ηµν∂µφ∂νφ− 1
2m2φ2 (46)
This is a very general result. You’ll soon see why we’re introducing all of thisin the first place, but essentially it is guaranteed to be Lorentz invariant. Whatdoes that mean? It means that if I change my coordinates by applying a trans-form:
Λµµ =
γ vγ 0 0vγ γ 0 00 0 1 00 0 0 1
(a boost in the x-direction) or
Rµµ =
1 0 0 00 cos θ sin θ 00 − sin θ cos θ 00 0 0 1
(a rotation around the z-direction) then the Lagrangian at the rotated (orboosted) coordinate will remain the same.
I’m not going to prove this here. I will simply assert that if you multi-ply scalars or all of your ”upstairs” vector indices are contracted with all ofyour ”downstairs” ones, then you are guaranteed to have a Lorentz-invariantexpression.
Ours does.
2.3 Minimizing The Action
Suppose you have some minimum action, S (remember that we expect action tobe minimized – that’s why we introduced it), parameterized by some continuousfield, φ(x).
20
If we add some small perturbation to the field, δφ(x). We then get:
δS =
∫
d4x
[
∂L∂φ
δφ+∂L
∂(∂µφ)δ(∂µφ)
]
We can now use the equivalent of integration by parts where
δv = δ(∂µφ)
and
u =∂L
∂(∂µφ)
sov = δφ
and
δu = ∂µ
(
∂L∂(∂µφ)
)
δφ
Thus:
δS =
∫
d4x
[
∂L∂φ
δφ− ∂µ
(
∂L∂∂µφ
)]
δφ+ ∂µ
(
∂L∂(∂µφ)
δφ
)
where the last term goes away at sufficiently large times and distances. Thus,the terms in the square brackets must cancel for any parameterization of theminimized action:
∂µ
(
∂L∂(∂µφ)
)
=∂L∂φ
(47)
Hurrah! It’s the relativistic form of the Euler-Lagrange equations!
2.4 The Klein-Gordan Equation
Solving for our Lagrangian (equation 46) using the Euler-Lagrange equations(47) we get:
∂µ∂µφ+m2φ = 0
(and similarly with a star).The first term is an operator known as the “d’Alembertian,” and can more
simply be written as:φ+m2φ = 0 (48)
The equation itself is known as the “Klein-Gordan Equation,” and if youimagine the field as a rubber-sheet, you can see why the Lagrangian has the formit does. More to the point, we get a homogeneous linear equation (because westarted with only second-order terms and then took derivatives), which meansthat we can get lots of linearly independent solutions. That’s why we chosethis as our “free-field” solution in the first place. After all, we now have a waveequation, which has the solution:
φ(x) = aei(~k·~x−ωt) + a∗e−i(
~k·~x−ωt) (49)
21
which guaranteed to be real, and is satisfied if:
−ω2 + k2 +m2 = 0
Plugging in units:(~ω)2 = (~kc)2 + (mc2)2
So the “m” in the last term (particle) is offset by the ~’s elsewhere.This yields:
E2 = (pc)2 + (mc2)2
which should look familiar to you if you’ve ever done any SR.Incidentally, we will always assume that ω is positive, and should the need
arise, we’ll simply say “±ω.”
2.5 What the Lagrangian means
What does the Lagrangian mean? Just as with a single particle Lagrangian, wecan define a conjugate momentum density:
πµ =∂L
∂(∂µφ)
and for our K-G Lagrangian, we get:
πµ =∂L
∂(∂µφ)= ηµν∂νφ
but because we’re smart, we know that the time component of the momentumdensity (which we’ll simply call π) will be important. After all, we’re ultimatelyworried about time evolution.
Thus:π = φ (50)
Even better, we can define a conserved current. This is a very generic state-ment of the field generalization of Noether’s Theorem.
2.6 Noether’s Theorem: Part 2
We showed for a single classical particle that if the Lagrangian (or the action,generally) is invariant under some transformation then there is a conservedvalue. Now, we’re going to do something more general. Noether’s Theorem relystates:
If a system has a continuously symmetric action, then there is acorresponding conserved current.
22
Here’s what I mean. When I first introduced Noether’s Theorem for a single par-ticle, I said that the Lagrangian needed to be invariant under a transformation.Or, in other words:
∂L
∂α= 0
where α is the parameter of the symmetry transformation.This form isn’t quite correct. Indeed, in the special cause of a time transla-
tion, this wasn’t correct even for a single particle. After all, we’re really tryingto make the action invariant under a symmetry transformation. In that case,for a single particle, we’re allowed to have a change in the Lagrangian so longas:
L→ L+ δL
such that∫
dtδL = 0
integrated over all time.We can do a similar trick here. Let’s imagine that we do a transformation
such that:δL = ∂µF
µ (51)
Notice that this is exactly equivalent to taking a divergence (only in 3+1 di-mensions). Now consider:
δS =
∫
d4xδL =
∫
d4x∂µFµ
The right hand side looks exactly like the input to Gauss’s theorem:
∫
d3x∇ · ~v =
∫
s
d2x~v · d ~A
where we make the general assumption that we are integrating over infinity, andany relevant interactions will vanish. Thus:
∫
d4x∂µFµ =
∫
d3xFµdAµ
which also vanishes.In other words, if we can manipulate a transformation such that the change
in the Lagrangian takes the form above, we’re all set.How can we change the Lagrangian? Well, consider a Lagrangian of many
fields, φa (where each a represents a different field). In that case, suppose wemake a transformation:
δφa = Xa(φ) (52)
where, like the function, Fµ, X is to be determined by the transformation.
23
Now, let’s consider the transformation:
δL =∂L∂φa
δφa +∂L
∂(∂µφa)∂µ(δφa)
∂µFµ = ∂µ
(
∂L∂(∂µφa)
)
δφa +∂L
∂(∂µφa)∂µ(δφa)
= ∂µ
(
∂L∂(∂µφa)
δφa
)
= ∂µ
(
∂L∂(∂µφa)
Xa
)
∂µ
(
∂L∂(∂µφa)
Xa − Fµ)
= 0
Or, to put it another way, there is a conserved current such that:
jµ =∂L
∂(∂µφa)Xa(φ) − Fµ(φ) (53)
where there is an explicit sum over all the fields.I should note what I mean by a conserved current:
∂µjµ = 0 (54)
which is both Lorentz-invariant (note the matching indices!) and equivalent to:
j0 + ∇ ·~j = 0
Assume that j0 = ρ is the density of the current. This simply means that:∫
dV(
ρ+ ∇ ·~j)
= 0
over any arbitrary volume. Thus we can define a “charge” (literal or otherwise)such that:
Q =
∫
dV ρ
yielding:
Q+
∫
dV∇ ·~j = 0
Again, we exploit the divergence theorem to get:∫
dV∇ ·~j =
∫
d ~A ·~j = 0
over a sufficiently large volume that encompasses all currents.Thus:
Q = 0
That is what we mean by a conserved quantity! The total amount of “stuff” isconserved.
24
2.6.1 Displacements in space/time
Consider the small following displacement in space-time:
xν → xν − ǫν
φ(x) → φ(x) + ǫν∂νφ
where the latter term is δφ (and thus X(φ)).Likewise, the Lagrangian density transforms as:
L(x) → L(x) + ǫν∂νL(x)
and so:Fµ = ǫµL
The “current” as written before is:
jµ = =∂L
∂(∂µφa)Xa(φ) − Fµ(φ)
=∂L
∂(∂µφa)ǫν∂νφa − ǫµL
= ǫν[
∂L∂(∂µφa)
∂νφa − δµνL]
In fact, because I can choose any component of ǫν I like to be zero or otherwise,this actually represents 4 currents. That is:
T µν =∂L
∂(∂µφa)∂νφa − δµνL
Where we have to sum over all of the independent fields in the theory. Fortu-nately, we only have 1, φ.
This 4×4 matrix is known as the “stress-energy tensor” T µν , which becomesparticularly interesting as the source term in general relativity.
Note that T 00 is the thing we normally call, ρ, the density, while T i0 is ~J ,the momentum density.
So, for example, if we consider the energy-momentum 4-vector, T µ0 , we get:
πµa φa − δµ0L
and thus for the Klein-Gordan equation, this becomes:
T 00 = φ2 − L
= φ2 −(
1
2φ2 − 1
2(∇φ)2 − 1
2m2φ2
)
=1
2φ2 +
1
2(∇φ)2 +
1
2m2φ2 (55)
25
By direct inspection, we see that the kinetic energy density is:
K =1
2φ2
pretty much exactly what you’d expect by inspection.and the “potential” energy (normally broken into “gradient density” and
“potential”) is:
U =1
2(∇φ)2 +
1
2m2φ2
2.6.2 Another Lagrangian, and another conserved current
The stress-energy tensor is going to be a conserved current for almost any La-grangian. Indeed, as written (and provided that the Lagrangian has no explicitposition or time dependence) the exact relation will be as derived.
Real-valued scalar fields are kind of boring because once you’ve got the S-Etensor, you’re pretty much done with conserved quantities. Not so in general.Consider the Lagrangian density of the complex scalar field:
L = ηµν∂µψ∂νψ∗ −m2ψψ∗ (56)
By inspection, this relation is necessarily real. The trick to solving it in gener-ality is to consider ψ and ψ∗ as separate fields. Under those conditions, we gettwo Euler-Lagrange equations:
∂µ∂µψ +m2ψ = 0
∂µ∂µψ∗ +m2ψ∗ = 0
which you know the solutions to, of course.Well, besides time and space translation, what else keeps the Lagrangian
unchanged? How about a global shift in phase for our complex wave-functions?(AGAIN, this is NOT quantum mechanics). Thus:
ψ → ψe−iα
ψ∗ → ψ∗eiα
Clearly, a small change, α << 1 yields:
δψ = −iαψ
andδψ∗ = iαψ∗
The Lagrangian itself is real, so:
δL = 0
26
and henceFµ = 0
All fields need to be included, and thus we get:
jµ = iηµν [ψ∂νψ∗ − ψ∗∂νψ] (57)
This, as we will see, will be the electromagnetic current!Hurray! In very little work, we already see the difference between complex
and real fields. A complex field has charge!
27
3 Free Quantized Scalar Fields
• Gross 1.3-1.8, 4.1-4.5, 9.3
• Tong 2.1-2.7
It’s finally time to actually start doing Quantum Field Theory, but don’tworry, because of what we’ve done before, a lot of this will be completely natural.
3.1 From Continuous to Quantized Field: The Real-valuedScalar Free Field
Remember how we went from a continuous to a quantized version of the har-monic oscillator? We first wrote down everything in the continuous limit, andthen realized that momentum and position don’t commute:
[x, p] = i~
or, in natural units, and generalizing to 3-dimensions:
[xi, pj ] = iδij
Note the delta-function. That’s because for a 3-d oscillator, there are 3 degreesof freedom. For a continuous field, position is going to a coordinate, not adegree of freedom. The value of the field plays the same role as position did inthe QHO case, and the canonical momentum plays the role of momentum. Inany event, there are now an infinite number of degrees of freedom, and we canimmediately write down:
[
φa(~x, t), πb(~y, t)
]
= iδ(3)(~x− ~y)δba (58)
For us, the last (Kronicker) isn’t really relevant since there is only the 1-field,but the point is that a field and its conjugate momentum don’t commute but afield and other fields, or other conjugate momenta do commute.
You’ll no doubt recall the free-field solution to the classical Klein-Gordanequation (49):
φ(x) = a~pei(~p·~x−E~pt) + a∗~pe
−i(~p·~x−E~pt)
But all possible values of ~k are allowed! Since a) The field must be real, and b)
~ = 1, which means that ~k = ~p, we get:
φ(x) =
∫
d3p
(2π)31
√
2E~p
(
a~pei(~p·~x−E~pt) + a∗~pe
−i(~p·~x−E~pt))
(59)
The integral is done over all of ~p-space, and here I’ve decided (following Tong)to attach the (2π)3 to the complex→real transform, with no similar factor inreal→complex.
28
Likewise, we’ve defined:
E~p = +√
~p2 +m2 (60)
This is the most general real solution of φ(x). The only remaining confusionmight be where the
√
2E~p comes from. Don’t worry about it for now. We’ll getto it in good time. As written, there’s nothing to stop us from normalizing theintegral however we want.
However, everything in the parentheses in equation (59) looks exactly like theQuantum Harmonic Oscillator. Indeed! It is the same equation, and withoutdoing any more work, we can immediately write down the quantum version ofthe field:
φ(x) =
∫
d3p
(2π)31
√
2E~p
(
a~pei(~p·~x−ωt) + a†~pe
−i(~p·~x−E~pt))
(61)
It’s now an operator acting on the state of the system. The “system” for thispurpose is an infinitely long vector which lists the number of excitations of eachof the infinitely many possible momenta of the field.
What this means, of course, is that a~p annihilates a “particle” of momentum,
~p, while a†~p creates one.
Using our commutator relation (or simple inspection), we can immediatelyget the conjugate momentum operator on the field:
π(x) = −i∫
d3p
(2π)3
√
E~p2
(
a~pei(~p·~x−ωt) − a†~pe
−i(~p·~x−E~pt))
(62)
Note that I’ve written the field and momentum operators in the Heisenbergor Interaction (same thing for a free field) representation. This differs somewhatfrom Tong, who does it in the Schroedinger representation.
3.2 The Creation and Annihilation Operators
Using the commutation relation above:
[
φ(~x, t)), π(~y, t)]
= iδ(3)(~x− ~y)
we can quickly (albeit with a great deal of work) get the following commutationrelation:
[
a~p, a†~q
]
= (2π)3δ(3)(~p− ~q) (63)
These operators play exactly the role you might expect. For instance, supposingwe started with a vacuum (and more on that in a bit):
|0〉
then the operation:a†~p|0〉
29
has the effect of creating a particle of momentum, ~p. The a~p operator operationsas an annihilation term.
Likewise (appropriately normalized), the operation:
a†~pa~p ∝ N~p
counts the number of particles with a particular momentum. We’ll talk moreabout how to appropriately normalize these in a bit.
3.3 The Hamiltonian
Before we deal with the properties of individual particles, we need to deal withthe properties of the vacuum, and before we do that, we need to construct aHamiltonian. We’ve already seen the energy density from equation (55).
Thus, we might suppose (correctly) that the Hamiltonian is:
H =1
2
∫
d3xπ2 + (∇φ)2 +m2φ2
Nothing prevents us from taking a gradient of an operator. After all, the oper-ator, itself, can be expanded as a Fourier series, as we’ve seen. Thus:
∇φ =
∫
d3p
(2π)31
√
2E~p
(
i~pa~pei(~p·~x−ωt) − i~pa†~pe
−i(~p·~x−E~pt))
All of the terms in the Hamiltonian are quadratic in the various operators. Forexample, we could multiply out one of the terms as follows:
1
2m2
∫
d3x ˆφ(x) ˆφ(x) =∫ ∫ ∫
d3xd3pd3q
(2π)61
√
4E~pE~q
(
a~pe−ipµx
µ
+ a†~peipµx
µ)(
a~qe−iqµx
µ
+ a†~qeiqµx
µ)
=∫ ∫ ∫
d3xd3pd3q
(2π)61
√
4E~pE~q
[
a~pa~qe−i(pµ+qµ)xµ
+ a~pa†~qe
−i(pµ−qµ)xµ
+a†~pa~qei(pµ−qµ)xµ
+ a†~pa†~qei(pµ+qµ)xµ
]
The first and last of these terms are going to cancel with other terms, andeven if they didn’t, it is clear that they wouldn’t contribute in most situations.After all, since the states of the field are presumed to be Eigenstates of theHamiltonian, two creation or two annihilation operators are necessarily goingto produce zero contribution:
〈n|a~pa~q|n〉 = 0
The middle two terms are the most interesting. Let’s look at the second term:
1
2m2
∫ ∫ ∫
d3xd3pd3q
(2π)61
√
4E~pE~qa~pa
†~qe
−i(pµ−qµ)xµ
=
∫ ∫
d3pd3q
(2π)61
√
4E~pE~q(2π)3δ(~p− ~q)e−i(Ep−Eq)ta~pa
†~q =
1
2m2
∫
d3p
(2π)31
2E~pa~pa
†~p
30
More generally, our Hamiltonian will be of the form:
H =
∫
d3xd3pd3qa~pa†~q(...) + ...
with each term quadratic in one of the dummy variables. By construction, it isclear that ~q = ~p will be the only terms to produce non-zero contributions.
While I won’t duplicate Gross or Tong’s derivation here (as it’s surprisinglysimilar to our QHO), I will note the important relation:
∫
d3xei~k·~x = (2π)3δ(~k) (64)
This is one of the most useful and powerful relations we will see. Memorize it.It allows us to get rid of all of the arguments except for 1, ~p, yielding:
H =1
2
∫
d3p
(2π)3E~p
[
a~pa†~p + a†~pa~p
]
=
∫
d3p
(2π)3E~p
[
a†~pa~p +1
2(2π3)δ(0)
]
(65)
3.4 The Vacuum
What happens when we try to measure the energy of the vacuum, |0〉?I warned you about this disaster before! The ground-state (vacuum energy)
is formally infinite. This isn’t really a huge deal. After all, we’re integratingover an infinite volume of space:
E0 =
∫
d3p1
2E~pδ(0)
=1
2
∫ ∫
d3pd3x1
(2π)3ei~x·~p
∣
∣
~p=0E~p
=1
2
1
(2π)3V
∫
d3pE~p
E0 =1
(2π)3
∫
d3p1
2E~p
which is directly measurable via the Casimir effect.Of course, by this argument, even the energy density is infinite (since there
are an infinite number of modes). In reality, though, we’d imagine that there isa maximum momentum:
pmax =~
lp
beyond which integration would be pointless. However, this yields an energydensity of l−4
p = (1018GeV )4.
31
We can get rid of this completely (although in an ad hoc manner) via “normalordering.” That is:
: a†~pa~pa†~p := a†~pa
†~pa~p
which is to say, move all of the daggers to the left, and the lowering operatorsto the right. Under this new scheme:
: H :=
∫
d3p
(2π)3E~pa
†~pa~p (66)
which doesn’t blow up. Of course, it sets the vacuum energy to zero as well,but that’s the price we have to pay.
3.5 Operators and Observables
Number Operator
The Hamiltonian is just one operator. There are others, of course. Forinstance, we may want to find out how many “particles” (excitations, really)there are in the field. This operation is simple:
N =
∫
d3p
(2π)3a†~pa~p (67)
where the operator bit inside of the integral should look familiar. It was the op-erator that told us the state of the excitation of our simple Harmonic oscillator.
As an exercise, you can verify that this operator commutes with the Hamil-tonian. Thus, particle number is conserved. This will not generally be truein quantum fields. Our real-valued scalar field is an exception, not the rule.Even complex scalar fields will not conserve particle number.
Momentum Operator
The momentum operator is based on the momentum density from the clas-sical field. Recall the momentum density, T i0 (the element of the stress-energytensor)
T i0 = ηii∂L
∂(∂0φ)∂iφ
= −π∂iφSo, generalizing to the quantum field, we get:
P i = −∫
d3xπ∂iφ
= −∫
d3x
[
−i∫
d3p
(2π)3
√
E~p2
(
a~p(t)ei(~p·~x) − a(t)†~pe
−i(~p·~x))
]
×[
∫
d3q
(2π)3
√
1
2E~p(iqi)
(
a~q(t)ei(~q·~x) − a(t)†~qe
−i(~q·~x))
]
=
∫
d3p
(2π)3pia†~pa~p (68)
32
where the last step exploits the δ − function relation:
∫
d3xei(~q−~p)·~x = (2π)3δ(~p− ~q)
Of course, this is as you’d expect. The total momentum of the system issimply the total momentum of the particles.
But let’s work out an example. Suppose we start with the vacuum and createa particle of momentum, ~p:
|ψ〉 = a†~p|0〉The momentum operator can be written:
P i|ψ〉 =
∫
d3q
(2π)3qia†~qa~qa
†~p|0〉
We can then use the commutation relation:
a~qa†~p = a†~pa~q + (2π)3δ(~p− ~q)
and thus:
P i|ψ〉 =
∫
d3q
(2π)3qia†~qa
†~p
*
0a~q|0〉 +
∫
d3q
(2π)3qia†~q(2π)3δ(~p− ~q)|0〉
=
∫
d3qqia†~qδ(~p− ~q)|0〉
= pia†~p|0〉= pi|ψ〉
Just as you knew it must.
Angular Momentum
Finally, consider the following state:
a†~p=0|0〉
the creation of a particle with zero momentum.Classically, the angular momentum density tensor should be:
J µν = xµT 0ν − xνT 0µ
Clearly, we only care about the space-like components (and by construction,the tensor is anti-symmetric). We’ve already seen that when we quantize thesystem:
T 0i →∫
d3p
(2π)3pia†~pa~p
Doing no work at all, this is zero, and thus the intrinsic angular momentum ofour field is zero. You may know this is spin. Scalar Fields have no spin.
33
3.6 Normalizing the field
There is something non-Lorentz-invariant about what we’ve been doing. Essen-tially, all of our integrals have been over space. Consider the operation:
|~p〉 = a†~p|0〉
which generates a single meson of momentum, ~p.This object is clearly not invariant. Apply a Lorentz boost, and all of a
sudden, we’ve got an entirely different state on our hands. Likewise, the integralelement:
∫
d3p
can’t be Lorentz invariant for the same reason. However, a couple of things areclear. First, all states of the system must satisfy:
E2~p = ~p2 +m2
and thus, wherever integrals over momentum appear, the relation:
∫
d4pδ(p20 − ~p2 −m2)
must appear. I’ve used p0 rather than E~p as a variable. Obviously, the argumentof the δ-function will be zero where p0 = E~p.
I “remind” you of a standard relation for dirac delta functions.
δ(f(x)) =δ(x− x0)
f ′(x)|x=x0
where in this case:f(p0) = p2
0 − ~p2 −m2
and hence:
δ(p20 − ~p2 −m2) =
δ(p0 − E~p)
2E~p
and thus:∫
d4pδ(p20 − ~p2 −m2) =
∫
d3pdp0δ(p0 − E~p)
2E~p
=
∫
d3p1
2E~p(69)
The LHS is Lorentz invariant, and thus, so is the right.Since our eigenstates, |~p〉 form a complete basis set, we know that:
∫
d3p
(2π)3|~p〉〈~p| = 1
34
This can be re-written∫
d3p
(2π)32E~p
√
2E~p|~p〉√
2E~p〈~p| = 1
which, of course, remains true. However, now the integral element is Lorentzinvariant, and so is the wave-function.
Thus, we define:|p〉 =
√
2E~pa†~p|0〉 (70)
as the relativistically normalized wave-function.This explains where we got the factor in the original quantum operator for
φ.
3.7 The Propagator
Expectation values provide a useful connection between quantum and classicalmechanics. In particular, for some state of the system:
〈O〉 = 〈n|O|n〉
But let’s be more precise. Suppose I had a particular state (say the groundstate), and I wanted to measure the expected value of the field:
φ(x) =
∫
d3p
(2π)31
√
2E~p
(
a~pei(~p·~x−ωt) + a†~pe
−i(~p·~x−E~pt))
It takes very little algebra to show:
〈φ〉 = 〈0|φ|0〉 = 0
Not surprisingly, the expectation of the field is zero.What happens when we ask the question about the expectation of the square
of the field? We don’t expect this to vanish, since we know that in the vacuum,modes percolate constantly.
We can even make this general, by asking, what happens if we make aperturbation at some space-time coordinate, y, and then try to observe it at,x? Are the fields correlated?
〈φ(x)φ(y)〉 = 〈0|φ(x)φ(y)|0〉
Put another way, suppose we set up a source at some point in space and time,what are the odds that we’ll observe it at another? Presumably, this shouldonly be possible if the two events are time-like separated, right?
35
3.7.1 The Feynman Propagator
Well...
〈0|φ(x)φ(y)|0〉 =
∫ ∫
d3pd3p′
(2π)61
2√
E~pE~p′〈0|a~pa†~p′ |0〉e−ipµx
µ+ip′νxν
=
∫
d3p
(2π)31
2E~pe−ipµ(xµ−yµ) (71)
≡ D(x− y) (72)
Of course, in the first line, we could have included all 4 combinations of a anda†, but the only non canceling ones are the ones written.
This thing, D(x − y), is known as the propagator, and those of you who’vetaken E&M may find it oddly reminiscent of the Green’s function.
We’re going to introduce something that may seem like something of a cheat:
∆F (x− y) = 〈0|T φ(x)φ(y)|0〉 (73)
This is almost identical to our original propagator but with one subtle dif-ference. Because we’ve introduced the time ordering operator, the “FeynmanPropagator” (as it’s known) has the property that:
∆F (x− y) =
D(x − y) x0 > y0
D(y − x) y0 > x0
How do we express this propagator in a useful way? First, let’s make itLorentz invariant. I’ll claim (as so many have before me) that the Feynmanpropagator can be expressed as:
∆F (x− y) = i
∫
d4p
(2π)41
pµpµ −m2e−ipν(xν−yν) (74)
If you’ve mastered your Lorentz algebra, you’ll recognize that the stuff in thedenominator should blow up. After all, it’s really p2
0−~p2−m2, which is supposedto be zero. In fact, we can rewrite the denominator in the following way:
pµpµ −m2 = p2
0 − E2~p = (p0 − E~p)(p0 + E~p)
This sort of expression is our first hint of antiparticles. After all, we’ve got justas much problem with the negative sign on the energy as the positive sign onthe energy.
For convenience, for the rest of the analysis, I will assume that x0 > y0.To show that our propagator works, we need to integrate over p0:
∫
d3p
(2π)3ei~p(~x−~y)
∫ ∞
−∞
dp0
(2π)
i
(p0 − E~p)(p0 + E~p)e−ip0(x0−y0)
36
3.7.2 Evaluation with Complex Analysis
The integral over p0 looks impossible. We have two roots, at E~p and −E~p. Butcomplex analysis gives us a trick. First, introduce a slight change:
∫ ∞
−∞
dp0
(2π)
i
(p0 − E~p + iε)(p0 + E~p − iε)e−ip0(x0−y0)
And do a semi-infinite semi-circle integral from p0 = −∞ to ∞, and then aroundthe complex part of the plane.
In complex analysis – the Cauchy Residue Theorem to be exact – we learnthat if you do a counter-clockwise integral around a root:
∫
dz
z − z0f(z) = 2πif(z0)
Since our integral is counter-clockwise, we get a negative sign, and thus:∫ ∞
−∞
dp0
(2π)
i
(p0 − E~p + iε)(p0 + E~p − iε)e−ip0(x0−y0) = (75)
From the contour, the point, p0 = E~p − iε is inside the contour. So the entire1-d integral becomes:
1
2E~p − 2εie−i(E~p−iε)(x0−y0)
and thus, in the limit of ǫ→ 0, we get equation (71):
∆F (x − y) =
∫
d3p
(2π)31
2E~pe−ipµ(xµ−yµ) (76)
In practice, though, we can redefine our terms somewhat (but get exactlythe same result by defining):
∆F (x− y) =
∫
d4p
(2π)4i
pµpµ −m2 + iεe−ipµ(xµ−yµ) (77)
We are going to see that the Feynman propagator (77) is going to be very usefulto us in computing real calculations in the future.
3.7.3 Classical Field Relation to Green’s Function
Finally, I note that while I derived the propagator in terms of QFT, it has ap-plication in classical (relativistic) field theories as well. Consider the operation:
( +m2)∆F (x− y) =
∫
(−pµpµ +m2)d4p
(2π)4i
pµpµ −m2 + iεe−ipµ(xµ−yµ)
= −i∫
d4p
(2π)4e−ipµ(xµ−yµ)
= = −iδ(x− y)
37
To take a particularly simple example, imagine two interacting “electrons,”that are moving around at non-relativistic speeds or sitting still entirely. Sup-pose that the force between them is mediated by a massive, spinless, bosonicfield. In that case, the energy of interaction will be governed by, ∆F , such that
U ∝ ∆F
How does it scale with distance? We start with the Green’s function relation:(
+m2)
∆F = −iδ(x)where I’m putting the ”source” particle at the origin for convenience. In thetime static case, this expression simplifies to:
(
−∇2 +m2)
∆F = −iδ(~x)where by symmetry
∆F = ∆F (r)
Recall a couple of useful relations:
∇2
(
1
r
)
= −4πδ(~x)
and also
∇2f(r) =1
r
d2(rf)
dr2
Thus, multiplying both sides by r, we get:
d2(r∆F )
dr2−m2(r∆F ) = 0
which solves to:
∆F =e±mr
rwhere when you include the time propagation effects, only the minus sogn makessense. This ultimately yields:
U ∝ e−mr
r(78)
which for a massless mediator produces a standard 1/r potential, and drops offmore quickly otherwise.
This is another way of saying that if you have a source that is mediatedby a massive particle, the classical field will propagate in a way described bythe Klein-Gordan equation. Or, in other words, we have a source term in ourLagrangian:
L =1
2∂µφ∂µφ− 1
2m2φ2 − φρ
then we can determine the field generated by a particular source:
φ(x) =
∫
d4y∆F (x − y)ρ(y)
You create a current (or source) in some point in spacetime, and you observethe field oscillating elsewhere.
38
3.8 The Complex Scalar Field
I’d like to finish up this section by pointing out how our approaches to quan-tizing fields an be generalized. For the most part, we’re going to have free-fieldLagrangians which are Lorentz invariant, and which are also quadratic in thesource fields. This means that the E-L equations are going to be linear in thesource fields.
Let’s consider the complex scalar field:
L = ∂µψ∗∂µψ −m2ψψ∗ (79)
We saw this before in §2.6.2.We also saw that ψ and ψ∗ can be treated as two separate fields for the
purpose of the Euler-Lagrange equations, and that they produced two Klein-Gordan equations:
∂µ∂µψ +m2ψ = 0
and similarly with the asterisks. Thus, the solutions (for the classical field) are:
ψ(x) =
∫
d3p
(2π)31
√
2E~p
(
b~pei~p·~x + c∗~pe
−i~p·~x)
ψ∗(x) =
∫
d3p
(2π)31
√
2E~p
(
b∗~pe−i~p·~x + c~pe
i~p·~x)
Where the second solution is clearly the complex conjugate of the first. Every-thing in this otherwise looks identical to our real field.
Our conversion to operators is trivial:
ψ =
∫
d3p
(2π)31
√
2E~p
(
b~pei~p·~x + c†~pe
−i~p·~x)
(80)
ψ† =
∫
d3p
(2π)31
√
2E~p
(
b†~pe−i~p·~x + c~pe
i~p·~x)
(81)
What’s this? Well, the c† operators clearly create a particle, and b† clearlycreates the anti-particles (and without daggers, they are annihilators. Com-
binations of ψψ† terms always yield combinations of bb†, or b†c†, and so on.This makes sense. In one case, we get a scattering term. In another, we pro-duce particle-antiparticle pair. In fact, all 4 combinations produce reasonableinterpretations.
Since we’ve already done all of the work with the real-valued scalar field,I won’t repeat it here. However, I should at least write down the relevantcommutation relations:
[
cp, bq
]
=[
cp, b†q
]
=[
c†p, bq
]
=[
c†p, b†q
]
= 0 (82)
and[
cp, c†q
]
=[
bp, b†q
]
= (2π)3δ(~p− ~q) (83)
39
just like we had with the φ field.Finally, using the relation of conserved current we saw before, we can define
a total “charge”:
Q =
∫
d3p
(2π)3(c†~pc~p − b†~pb~p) = NC −Nb
3.9 Vector Fields (and beyond)
We’re not ready to do electrons yet. Spin 1/2 particles don’t have a classicalanalog. However, spin-1 particles (aka photons) do. Instead of a single scalarfield, the field is components of a vector, Aµ. I’ll spare you the suspense, andlet you know that A0 = Φ, the scalar potential of the E&M field, and Ai = ~A,the vector potential.
In general, we might imagine being able to construct a Lagrangian with thefollowing sort of terms: AµA
µ, or ∂µAµ∂νA
ν . Remember that these terms aresupposed to be quadratic in Aµ for free-field theories. As it turns out, to makeour result consistent with Maxwell’s equations, the Lagrangian density becomes:
L = −1
4FµνF
µν (84)
whereFµν ≡ ∂µAν − ∂νAµ (85)
This matrix (the Faraday matrix) has zeroes in the diagonal, and is anti-symmetric. As such, there are only 6 independent terms. Those 6 are the 3components of the E-field, and 3 of the B-field. We’ll talk about that later.
For the free classical field, we can solve the Euler-Lagrange equations as:
∂µFµν = 0 (86)
Using the definition of Fµν , this can be satisfied if:
Aµ(x) =∑
n,α
∫
d3p
(2π)31
√
2E~p
(
εα~pa~p,αe−~p·~x + εα∗~p a∗~p,αe
~p·~x)
where α represents one of two possible polarization states. Conversion to aquantized field operator should be fairly obvious, but we’ll talk about it morewhen we get to QED.
Those εα~p vectors components are important. Let’s suppose (for conve-nience), that ~p points in one of the 3-principle directions, call it i. α can takeon the values 1,2,3. What we get out of the ε is a unit vector only if i 6= α. Weuse the Levi-Civita symbols:
εijk =
+1 i, j, k = cyclic−1 i, j, k = anticyclic0 otherwise
40
In our notation, we care about the term εαiµ. Don’t worry what happens (fornow) when µ = 0.
It is clear that if we had a massive “photon” (massive so that it has a restframe), and created a single zero-momentum particle, these ε operators wouldguarantee an integer spin. (See Gross 2.7).
I should point out that, in principle, we could consider free QFT’s of anyinteger spin. The theory for a spin-2 particle, for example, would be the La-grangian of a graviton. However, these theories will suffice for now.
41
4 A Simple Scalar Yukawa Interaction
• Gross: Chapters 3, 9.1-9.2
• Tong: Chapter 3
We are finally ready to start consider nearly real physical interactions. I say,“nearly real,” because it is clear that that systems that will most interest us inthe real universe are those featuring interactions of spin 1/2 particles (fermions)with spin 1 particles (bosons). Still, we will get a very good impression ofhow cross-sections and decay rates are computed from our spin-0 view of theuniverse.
4.1 3rd Order Lagrangians
4.1.1 Significance of terms
Every Lagrangian we’ve seen so far is second order in φ or ψ. This has resulted inlinear classical equations of motion, which in turn have resulted in our free-fieldtheory. However, there is nothing to prevent perturbations to this Lagrangianof the form:
Lint = − 1
3!λ3φ
3 − 1
4!λ4φ
4 + ...
or even cross terms along the lines of:
Lint = − 1
3!λ3φψψ
∗ (87)
where I’ve included the last as a numbered equation, because we’re going to ex-amine it in great detail. It strongly resembles the interaction Lagrangian for anelectron field with a photon field. Don’t worry, for now, about what constitutesa legitimate Lagrangian. We’ll derive those from symmetry arguments in duecourse.
An analysis of the various terms indicates something:
• [λ3] = E1, which means at high energies, the term is likely to be insignif-icant. Terms like these are known as “relevant.”
• [λ4] = E0 are known as marginal, and their impact is not energy depen-dent.
• [λ5] = E−1, are known as “irrelevant”, for the simple reason that at lowenergies, they do not contribute to the Lagrangian, and only kick in athigher energies. Irrelevant terms are typically very difficult to deal with,and we won’t in this course.
42
4.1.2 The Perturbed Hamiltonian
Let’s take as our example the perturbed Hamiltonian, equation (87). It shouldbe obvious that since this is a potential energy term, we can simply read off theinteraction Hamiltonian:
Hint =1
3!λ3φψψ
† (88)
This is one of the very nice things about our Lagrangian picture, the interactionterm drops out trivially. Or, more generally:
Hint = g
∫
d3xφψψ† (89)
Note that in this case:[g] = [E]1
Note also, that I haven’t actually motivated why we might find ourselves withan interaction term like this. Don’t worry, we will, but it’ll have to wait a fewweeks.
Since the unitary operator, U contains a series of interaction Hamiltonians,it’s pretty clear that we’re going to get a bunch of φψψ†’s acting on our system.
We’ll expand everything out properly in a bit, but you should recall thateach of the three operators in the Hamiltonian is a combination of creation andannihilation operators, which means that we have the following 8 combinations:
1. a~pb~p′ b†~p′′
2. a~pb~p′ c~p′′
3. a~pc†~p′ b
†~p′′
4. a~pc†~p′ c~p′′
5. a†~pb~p′ b†~p′′
6. a†~pb~p′ c~p′′
7. a†~pc†~p′ b
†~p′′
8. a†~pc†~p′ c~p′′
Let’s give the particles names. The particle created by the a† operator is myφ-particle , the particle created by c† a ψ particle, and the one created by a b†
a ψ particle.Interaction 1 absorbs a φ into an ψ and gives it a boost. One could, if one
were so inclined, even draw a picture representing the term:
43
ψ φ
ψ
For those of you who missed it, this is a proto-Feynman diagram. For now,whenever I draw one, a solid line means a ψ particle, and a ψ “going backin time” is a ψ. A dashed line is a φ particle. Different sources use differentconventions, but I prefer my time axis running vertically. It makes it look likea space-time diagram. Be aware, however, that the x-axis is meaningless.
Note also that this particular diagram won’t represent a process that canactually happen. There is no way to absorb a single particle without a changein identity and simultaneously conserve both momentum and energy.
The diagram for 3 is particularly interesting:
φ
ψ ψ
A φ decays into a ψ + ψ pair!And you can figure out by looking at the other 6 combinations what the
relevant processes are in each. Notice that charge (N+ − N−) is conserved inevery process.
We could, of course, write down a similar set of terms for any possibleinteraction Lagrangian, and they will all have similar expressions.
How do we use these observations to do actual calculations?
4.2 Particle Decay
Rather than give you a bunch of rules to simply memorize, we’re going to startby exploring φ decay as a worked example to motivate what comes next. In theprocess, we’re going to work through a fair amount of formalism that will helpus in general.
44
4.2.1 The Interaction Hamiltonian
In general, we want to consider the reaction rate from one state, |i〉 to another|f〉, where in the case of φ-decay, for example:
|i〉 =√
2E~pa†~p|0〉
where we’re using the appropriately normalized initial state.Likewise,
|f〉 =√
4E~p′E~p′′ b†~p′ c
†~p′′ |0〉
where the order of the creation operators don’t matter because they commute.Since we are doing a quantum mechanical calculation, it’s clear that we will
almost always need to compute the amplitude:
〈f |U(t, t0)|i〉
What is the Unitary operator for this theory? Remember that we’re workingin the “Interaction Representation,” which means that operators in the theoryhave a time variability.
For our sake, let’s only consider the φ terms in the interaction Hamiltonian.because the different fields commute with each other in the free-field Hamilto-nian, we can look at them all separately.
Rather than give ourselves a headache trying to do all of the integrals overmomentum space, let’s do a much simpler example, the QHO. As a reminder:
H0 = ω
(
N +1
2
)
What happens if there’s an interaction term in the Hamiltonian (in the Schrodingerrepresentation) which looks like:
Hint = a
?Then:
HI(t) = U †0 (t)HintU0(t)
where:
U0 = e−iH0t
=
(
1 + (−i) Nωt+ (−i)2 1
2N2ω2t2 + ...
)
e−iωt/2
and similarly for U †0 . Of course, the final expontial terms in each combine
multiplicatively to form 1. Thus, to find HI , we have lots of terms of the form:
NnaNmωn+mtn+min(−i)mn!m!
45
and similarly with a dagger.It is easy to show that:
N a = a(N − 1)
andN a† = a†(N + 1)
and thus:NnaNm = aNm(N − 1)n
and similarly for the dagger.Expanding this out, we get:
U †0 aU0 = a× (1 + iωt[N − 1] − iωt[N ] + ...)
which isa(1 − iωt+ ...)
ora→ ae−iωt (90)
where it is easy to show the opposite sign:
a† → a†eiωt (91)
How does this translate to our various interaction terms in our QFT Hamil-tonian? Let’s just consider 1 term, the one most relevant to our φ decay:
HI = g
∫
d3x
∫
d3pd3p′d3p′′
(2π)91
√
8E~pE~p′E~p′′a~pb
†~p′ c
†~p′′e
−i(pµ−p′µ−p′′µ)xµ
(92)
Okay, it was an awful lot of work just to get the exponent, but that exponenthas the nice property of equalling zero if and only if energy is conserved in ourdecay. We will find a similar expression regardless of the type of interaction.
4.2.2 The S-Matrix
Now that we’ve generated an interaction Hamiltonian we can start trying tofigure out how a particular system will evolve. We need to use the Unitaryevolution operator:
U(t, t0) = T exp
(
−i∫ t
t0
dtHI
)
which, in the limit of t0 → −∞, and t → ∞ is simply known as the S-matrixamplitude:
Sfi = 〈f |S|i〉 (93)
The S-matrix will obviously be very important in doing calculations of scat-tering and decay rates. In general, we are going to work it into a form like:
Sfi = −iAfi(2π)4δ(
∑
p)
(94)
46
Since momentum (and energy, the zeroth component of the momentum 4-vector)must be conserved on all interactions, there will always be a delta-function.The normalization is arbitrary, but useful, and the i pre-factor is just standardnotation. It obviously doesn’t matter since the square of Sfi will always bepositive.
Clearly, the square of Sfi is going to be related to things like the cross sectionand the decay rate, but how?
4.2.3 Fermi’s Golden Rule
Let’s consider two different states, |m〉 and |n〉, with some energy difference,ω in an ordinary QM setting. We’re still going to use natural units, however.We’re not animals.
To first order:HI = Hinte
−iωt
so:
U(t) = I − iHint
∫
dt′e−iωt
We can ignore the I (because n 6= m), and thus:
〈m|U(t)|n〉 = −〈m|Hint|n〉e−iωt − 1
ω
The probability of a transition is simply the square of this:
This last term on the right has some nice properties. For one thing, it’seven. For another:
∫ ∞
−∞dω
(
1 − cosωt
ω2
)
= πt
It also drops off quickly with ω. Taking a look of the plot of the function:
47
As t→ ∞, this term approaches a delta function:
(
1 − cosωt
ω2
)
→ πtδ(ω)
This δ−function, unsurprisingly, suggests that energy is supposed to be con-served. Thus:
Pn→m = 2π|〈m|Hint|n〉|2Tδ(EN − EM )
orP = 2π|〈m|Hint|n〉|2δ(EN − EM ) (95)
which is the exact form that we’d hoped for. Energy conservation is guaranteed,and the pre-factor (for a 1-d system) looks awfully familiar.
4.2.4 The Decay Amplitude
Let’s stop with the generalities and start computing the actual amplitude forour φ decay. First, let’s make our initial and final states explicit:
|i〉 =√
2Epa†p|0〉
and|f〉 =
√
4Ep′Ep′′ b†p′ c
†p′′ |0〉
or in “bra” form:〈f | =
√
4Ep′Ep′′〈0|bp′ cp′′
48
Recall equation (92), and thus we get (to first order):
Sfi = −ig∫ ∞
−∞dt√
4Ep′Ep′′ 〈0|bp′ ×
cp′′
∫
d3x
∫
d3qd3q′d3q′′
(2π)91
√
8E~qE~q′E~q′′a~q b
†~q′ c
†~q′′e
−i(qµ−q′µ−q′′µ)xµ ×√
2Epa†p|0〉
= −ig∫
d4x
∫ ∫ ∫
d3qd3q′d3q′′
(2π)9
√
EpEp′Ep′′
EqEq′E′q′〈0|apa†q b†p′ bq′ c
†p′′ cq′′ |0〉e−i(qµ−q′µ−q′′µ)xµ
Each of the contributions:
〈0|apa†q|0〉 = (2π)3δ(~p− ~q)
which quickly yields:
Sfi = −ig∫
d4xe−i(pµ−p′µ−p′′µ)xµ〈0|0〉
or
Sfi = −ig∫
d4xe−i(pµ−p′µ−p′′µ)xµ
This is dramatically simpler.
A note on infinite integralsThe form of our S-matrix element still has a 4-dimensional integral in it over
all of space and time. Handling these expressions can be done in a number ofways as we’ve already seen.
Let’s start by doing this in 1-d. I’ve done the following trick a bunch oftimes, but I want to make this explicit. Consider the 1-dimensional integral:
∫ L/2
L/2
eikxdx =1
ik
(
eikL/2 − e−ikL/2)
=2
ksin(kL/2)
= Lsin(kL/2)
kL/2
= L
where the last step is under the assumption that k → 0, and L is finite.As you know:
∫
eikxdx = (2π)δ(k)
which means that in going from an infinite box to a finite box of length, L, wecan substitute:
(2π)δ(0) → L
49
or more generally:∫
d4xe−ipµxµ
= (2π)4δ(4)(p) = V T (96)
where we imagine integrating over a finite volume and time, and then extendingthose limits to infinity.
We will find this particularly useful when we get relations like the square ofa delta function. I assure you, this will ultimately be most useful for keepingtravel of terms.
Returning to the calculation of the amplitudeWe are now almost completely done computing our S-matrix. Using the just
derived results, we can convert it to:
Sfi = −ig∫
d4xe−i(pµ−p′µ−p′′µ)xµ
= −ig(2π)4δ(p− p′ − p′′)
Wait! I warned you that the S-matrix would ultimately take the form:
Sfi = −iAfi(2π)4δ(
∑
p)
and it does!In the case of a first order decay (looking at only at the first order term in
the unitary operator), we get:Afi = g (97)
Not the most surprising thing in the world.This work is not for vain. Having done all of these integrals and gotten such
a simple result is what will allow us to compute the “Feynman rules” in thenext chapter. We don’t have to do these integrals every time.
But how do we turn the amplitude or the S-matrix into an actual decayrate?
4.2.5 Calculation of the Decay Rate
We have chosen a particularly straightforward problem. 2-body decay is inmany respects much simpler than three or more body decay, or of scattering.
If we assume that the φ particle is of mass, m and is initially at rest, and thatthe ψ particles each have mass, M , then by definition, the outgoing momentamust be equal and opposite. Energy conservation yields:
m = 2√
(M2 + p2F )
where pF is the uniquely determined outgoing momentum:
pF =
√
(m
2
)2
−M2 (98)
50
Any two body decay will have a unique outgoing momentum (up to theentirely random direction), while with 3 or more particles, there is clearly goingto be some flexibility.
It should be clear that the probability of of transition between two stateswill be proportional to:
P ∝ |Sfi|2
since Sfi is simply the amplitude of the oscillation. However, to normalize thetransition, we need to compute:
P =|〈f |S|i〉|2〈f |f〉〈i|i〉
The terms in the denominator are easy. For instance,
i =√
2Epa†p|0〉
and so:
〈i|i〉 = 2Ep〈0|0〉= 2Ep(2π)3δ(3)(0)
= 2EpV
With a similar relationship for the outgoing particles. Don’t worry. The V ’swill ultimately cancel out.
Multiplying it out, we get:
P =|Afi|2 [δ (
∑
p)]2(2π)8
8EpEp′Ep′′V 3
The square of the delta function produces some difficulties, but remember thatwe can replace one of the deltas with
(2π)4δ(
∑
p)
= V T
thus:
P =|Afi|2 [δ (
∑
p)] (2π)4
8EpEp′Ep′′V 2T
or we get a rate of:
Γ =|Afi|2 [δ (
∑
p)] (2π)4
8EpEp′Ep′′V 2
or, more generally:
Γ = |Afi|2(2π)4δ(
∑
p) 1
2E~pI
∏
out states
1
2E~piV
(99)
51
All we have to do is compute the Afi coefficients (which we’ve done for thisparticular theory) and integrate over all possible outgoing momenta:
Γ =∏
outstates
(∫
d3pi(2π)3
1
2Epi
)
|Afi|2(2π)4δ(pI − pF )1
2E~pI
(100)
You’ll notice that all of the V ’s drop out. You’ll also notice that this is such astaggeringly important equation that I’ve put a box around it.
Now all that remains is to actually do the integral, where Ep = m, and (inour case):
p =
m000
Γ = g2
∫ ∫
d3p′d3p′′
(2π)2δ(p− p′ − p′′)
1
8mEp′Ep′′
= g2
∫
d3p′d3p′′
(2π)21
8mEp′Ep′′δ(m− Ep′ − Ep′′ )δ
(3)(~p′ + ~p′′)
= g2
∫
d3p′
(2π)21
8mE2p′δ(m− 2Ep′)
= g2
∫
p′2
8πmE2p′dp′δ(m− 2Ep′)
We need to be a bit careful integrating over the delta-function. In reality, sincewe’re integrating over p′, we should express it as:
δ(m− 2Ep′) = δ(m− 2√
M2 + p′2)
where we know that the value pF is the one that satisfies the dirac-delta function.Recall:
δ(f(x)) =δ(x− x0)∣
∣f ′(x)|x=x0
∣
∣
so
δ(m− 2Ep′) =δ(p′ − pF )
2pF /EpF
=mδ(p′ − pF )
4pF
Thus:
Γ = g2
∫
1
32πpF
p′2
E2p′dp′δ(p′ − pF )
= g2 1
32πpF
p2F
(m/2)2
So, we get to first order in the center of mass frame:
Γ = g2 pF8πm2
(101)
52
Inspection will assure you that this dimensionally has units of energy (or inversetime, which is appropriate for a rate).
Congratulations! We’ve done our first real QFT calculation!Note also:
• The rate of the decay is proportional to g2. Higher order terms will includehigher powers of g as we’ll see. The weaker the coupling term (in termsof g/m) the slower the decay. Also, the more that 1st order calculationsare sufficient.
• The larger the mass gap between m and 2M , the larger pF will be, andthus, the higher the rate of decay. You could generalize this, if you like,to say that all things being equal, high energy decays will occur faster.
4.2.6 Lessons Learned so Far
I want to reiterate that our decay calculation is not exact. We’ve only calculatedthe scattering amplitude to first order.
Moreover, the calculations we’ve done so far only really hold for the scalartheory we’ve been working on.
That said, we can already get an idea of how the Feynman calculus is goingto work for us. As a quick reminder, here’s how things played out. We startedby imagining an interaction as follows:
p
p′ p′′
This diagram has exactly 1 vertex, which by construction conserves charge,and, as we’ll see, momentum and energy. Recall that the whole point of thecalculation was to compute the scattering amplitude, Afi, which we then used inour decay calculation. This will be the point of all of these diagrams. Ultimately,each will produce an amplitude and if there is more relevant diagram, we’llsimply add them together.
So, how do we use this diagram to compute an amplitude WITHOUT usinga bunch of creation and annihilation operators? Using what we’ve seen so far:
1. Label all external 4-momenta with a label, pi (or with primes, as I’ve donehere).
53
2. For each vertex, write down:
−(ig)(2π)4δ
(
∑
i
qi
)
where I’ve generalized the momentum to include “internal” lines (whichwe haven’t seen yet). That’s why I call it q. In our case, we have 1 vertex,so:
(−ig)(2π)4δ(p− p′ − p′′)
As you can so, ingoing lines and outgoing lines get an opposite sign. Youneed to be consistent here, especially with internal lines that connect twovertices. A particle that is “ingoing” to one vertex will be “outgoing” toanother.
3. The final result should be reducable to:
−iAfi(2π)4δ(
∑
pi
)
and from this we can quickly read off:
Afi = g
much faster than our original approach.
4. Once we have the amplitude we compute:
Γ =∏
out states
(
V
∫
d3pi(2π)3
)
|Afi|2(2π)4δ(pI − pF )1
2E~pI
∏
out states
1
2E~piV
Of course, decays are not the only possibility. We are also very much interestedin scattering. That will be the topic of our next set of notes, and with it,
54
5 Scattering and Feynman Rules
• Gross: 9.3-9.5
Scattering is a bit more complicated than decays, but the form is nearlyidentical. At issue is the fact that the rate of interaction is related to the relativespeed of the interacting particles. Very generally, for a 2-particle scatter we maysay:
dσ =1
4E1E2
1
|~v1 − ~v2||Afi|2(2π)4δ
(
∑
p)
∏
out states
d3pi(2π)3
1
2E~pi
(102)
If you haven’t done anything with cross sections before, the idea is that therate at which a particular scattering event occurs is something like:
Γ = nσv
In this case, the reason that it’s a dσ is because it’s the cross section corre-sponding to a very specific result – two particles flying off at specified momentaand directions for example. We’ll integrate and normalize in due course.
While I haven’t derived the cross-sectional relationship by any means, itdoes seem reasonable, especially because it includes a scattering amplitude Afi,squared.
Unlike with our decay, we are going to have to introduce higher order dia-grams. For instance:
p1
p2
p′1
φ
p′2
ψ ψ
ψ ψ
This is the story of two charged ψ particles scattering off one another by “ex-changing” a φ.
5.1 The Propagator
I grant you that I basically pulled the scattering cross-section relationship outof thin air. We will get to it in due course, but for now, our biggest issue comesfrom the fact that we have an “internal line” in the diagram. This shouldn’t bea surprise.
55
Let’s consider the initial and final states of the system:
|i〉 ∝ c†p1 c†p2 |0〉
and〈f | ∝ 〈0|cp′1 cp′2
It should be clear that all 1st order terms in the Hamiltonian produce a zeroamplitude. For instance:
〈f |Hinte−iωt|i〉 = 〈0|cp′1 cp′2(c
†ca†)c†p1 c†p2 |0〉
= 0
I picked only 1 term in the Hamiltonian (and didn’t bother with the momentumsubscripts), but it’s obvious by inspection that all of them will produce zeroamplitude.
Instead, we’re going to need the 2nd order term in the unitary operator:
(−i)2∫ t
t0
Hdt′∫ t
t′Hdt′′
In this case, the creation and annihilation terms are going to look like:
(cc†a†)(cc†a)
where (roughly) the term in the left parentheses correspond to events at theleft vertex, and the right term corresponds to the events in the right vertex.Notice also that it’s completely arbitrary which one gets the creation of the φand which one gets the annihilation.
Unlike with our 1-vertex calculations, we’re going to have the possibility ofinteractions at two different points in spacetime. Writing it out explicitly yields:
Sfi = (−i)2g2〈0|cp′1 cp′2√
4Ep′1Ep2
×(
∫
d4x
∫
d3q1d3q2d
3k
(2π)91
√
8Eq1EkEq2cq1 a
†k c
†q2e
−i(q1−q2+k)µxµ
)
×(
∫
d4y
∫
d3q3d3q4d
3k′
(2π)91
√
8Eq3Ek′Eq4cq3 ak′ c
†q4e
−i(q3−q4−k′)µyµ
)
×√
4Ep1Ep2 c†p1 c
†p2 |0〉
Dear god! That looks terrifying! But really, it’s not as bad as it looks. Forone thing, the limits of the time intergation simplify.
But before I do that, I’ve already done a bit of a switcheroo on you. Nor-mally, a 2nd order Taylor series expansion looks like:
ex = 1 + x+x2
2+ ...
But I’ve neglected the 1/2 out front. That’s because the labeling of whichoutgoing particle is p′1 and which is p′2 is arbitrary.
56
p1
p2
p′1
φ
p′2
ψ ψ
ψ ψ
Of course, the labeling of which one gets a p′1 and which one gets a p′2 isarbitrary, and thus there is another nearly identical diagram with the two ofthem switched.
p1 p2
p′2
φ
p′1
ψ ψ
ψ ψ
In other words, we’re essentially doubling up on our amplitudes. If you’dlike a more rigorous argument, check out the discussion of Wick contraction inTong or Gross.
It’s clear, though, that all of those creation and annihilation operators helpus out, but not as much as you’d immediately suppose. Only the a terms cancelimmediately:
〈0|ak′ a†k|0〉 = (2π)3δ(~k − ~k′)
This doesn’t appear to simplify the expression much, but it does help a bit:
Sfi = (−i)2g2
∫
d4x
∫
d4y
∫
d3q1d3q2d
3q3d3q4
(2π)12
√
Ep′1Ep′2Ep1Ep2Eq1Eq2Eq3Eq4
×e−i(q1−q2)µxµ
e−i(q3−q4)µyµ〈0|cp′1 cp′2 cq3 cq1 c
†p1 c
†p2 c
†q2 c
†q4 |0〉
×[∫
d3k1
(2π)31
2Eke−ikµ(xµ−yµ)
]
You may recognize the last term as the propagator:
∆F (x− y) =
∫
d3k1
(2π)31
2Eke−ikµ(xµ−yµ) =
∫
d4k
(2π)4i
kµkµ −m2e−ikµ(xµ−yµ)
57
We can also work “outward” from our various creation and annihilation opera-tors, yielding:
〈0|cp′1 cp′2 cq3 cq1 c†p1 c
†p2 c
†q2 c
†q4 |0〉 = (2π)12δ(~p1 − ~q1)δ(~p2 − ~q3)δ(~p
′1 − ~q2)δ(~p
′2 − ~q4)
This simplifies things dramatically:
Sfi = −g2
∫
d4xd4y
∫
d4k
(2π)4e−i(p1−p
′1+k)µx
µ
e−i(p2−p′2−k)µy
µ i
kµkµ −m2 − iε
where I’ve put the extra ε term in the propagator because I know I’m going toneed it later.
In this form, the only place that we have space or time dependance is in theexponents, and so we quickly get:
Sfi = −g2
∫
d4k(2π)4δ(p1 − p′1 + k)δ(p2 − p′2 − k)1
kµkµ −m2 − iε
This delta-function makes things easy. I can integrate over k and noting thesecond function, I get:
k = p2 − p′2
Important note: This relationship does not necessarily guarantee that:
kµkµ 6= m2
Clearly, if the effective mass of the mediating φ is “on the mass shell” thecontribution is maximum (since the denominator goes to zero), but the effect isnot guaranteed. This reduces the expression to:
Sfi = −ig2(2π)4δ(p1 + p2 − p′1 − p′2)1
(p2 − p′2)µ(p2 − p′2)µ −m2 − iε
(103)
or, using or standard amplitude relation:
Afi =g2
(p2 − p′2)2 −m2 − iε
(104)
Wow! That was hard.But with practice, it’s clear that we don’t need to do all of these integrals
directly at all. Rather, to compute the scattering amplitude, Afi, we simplyneed to follow a bunch of rules. These will be known as “Feynman Rules” andthe Diagrams used to draw them, the Feynman diagrams.
5.2 The Feynman Rules for the Scalar Yukawa Interaction
We can use this to compute the amplitude Afi. We’ve come up with an expres-sion solve for this for the two-vertex system. To make things concrete, we’regoing to use this diagram:
58
p1
p2
p′1
φ→
p′2
ψ ψ
ψ ψ
The “Feynman Rules” will (naturally) produce the same result. Afi is theintegrated product of the following:
1. External momenta are described by four-vectors, pi. We’ve already seenthat the delta function in the overall scattering relation demands that:
∑
i
pi = 0
where by convention, outgoing momenta get a positive sign, and ingoingget a negative.
2. Label each internal line with a value ki. Don’t do anything with theintegral yet. This is just for the labeling purposes. Arrows pointing into avertex contribute a positive 4-momentum. Arrows pointing out contributea negative 4-momentum.
3. For each vertex, write down:
(−ig)(2π)4δ
(
∑
i
qi
)
(105)
where q’s include the external lines as well. In our example above, we’dhave:
δ(p1 − k − p′1)
for the left vertex, for example.
4. For each internal φ line, write the propagator (and an integral):
∫
d4ki(2π)4
i
k2i −m2 + iε
(106)
This propagator works equally well for our ψ particles, but obviously notfor vector or spin-1/2 particles.
59
The final result will be:−iAfi(2π)4δ4(
∑
i
pi)
where in our shorthand, outgoing momenta get a negative sign.This prescription will allow us to solve the scattering amplitudes and decay
rates for just about anything. Just draw all possible diagrams, compute theamplitudes, and add them together. Of course, this assumes that the result willconverge with only a few diagrams. If g is large, this may not happen simply.
5.3 Example: ψ − ψ Scattering
We’re now ready to compute our scattering amplitude for whatever interactionwe like. The first step, as we’ve seen, is to draw all possible diagrams. For each.
5.3.1 Calculating the Amplitude
We’ve already seen that there are two diagrams that contribute at 2nd order.One with p′1 coming out from the “left” and one coming out from the “right.”(The space-like component of the diagrams are kind of arbitrary). Let’s computejust one of the amplitudes:
Step 1: Nothing to do, cause it’s already labeled.Step 2: In my mind, I’ve put a “k” in the diagram.Step 3: There are two vertices, yielding:
(−ig)2(2π)8δ(p1 − k − p′1)δ(p2 + k − p′2)
Step 4: Taking the result from before, we get:
−ig2
∫
d4k(2π)41
k2 −m2 + iǫδ(p1 − k − p′1)δ(p2 + k − p′2)
The first delta-function yields:
k = p1 − p′1
of course, and thus the second delta function becomes:
δ(p2 − p′2 + p1 − p′1) = δ(∑
i
pi)
exactly as expected. Thus, we get:
〈f |S|i〉 = −ig2(2π)41
(p1 − p′1)2 −m2 + iǫ
δ
(
∑
i
pi
)
or:
Afi = g2 1
(p1 − p′1)2 −m2 + iǫ
(107)
This is exactly what we found doing it the hard way in equation (104).We’d get a second term by making p′1 ↔ p′2.
60
5.3.2 2 Particle Scattering Cross Sections (in general)
So we’ve got an expression for our amplitude, but what does that tell us aboutthe actual scattering cross section?
We can compute the differential cross section in equation (102). For a 2-2particle scattering experiment, we get:
dσ =1
16 E1E2E′1E
′2V
2
1
|~v1 − ~v2||Afi|2(2π)4δ(pF − pI)
This will be made much easier if we work exclusively in the center-of-massframe. That is:
~p1 + ~p2 = ~p′1 + ~p′2 = 0
Further, remember that:
~v =~p
E
in our coordinates, and thus:
|~v1 − ~v2| =√
(~p1/E1 − ~p2/E2)2
= |~p1|
√
(
1
E1+
1
E2
)2
= |~p1|
√
(
E1 + E2
E1E2
)2
= |~p1|E1 + E2
E1E2
Plugging this into our expression simplifies things somewhat:
dσ =1
16|~p|(E1 + E2)E′1E
′2V
2Afi|2(2π)4δ(pF − pI)
To get the “total” cross section, we need to integrate over:
σ =
∫ ∫
d3p′1(2π)3
d3p′2(2π)3
dσ
and cancel whenever possible.The δ function is going to help us simplify things, but first, let’s break up
the delta function into:
δ
(
∑
i
Ei
)
δ
(
∑
i
~pi
)
I’m going to keep this general, for now, and assume that there are potentiallytwo different particles, each with mass mj participating in the scatter. We knowthat m1 = m2, but that is specific to our problem.
61
Thus, the first delta-function becomes:
δ
(
E1 + E2 −√
m21 + ~p′21 −
√
m22 + ~p′22
)
and thus:
dσ =(2π)4
16|~p1|(E1 + E2)V 2
δ(
E1 + E2 −√
m21 + ~p′21 −
√
m22 + ~p′22
)
√
m21 + ~p′21
√
m22 + ~p′22
δ(~p′1+~p′2)|Afi|2
Thus, when we integrate to get σ, we can immediately integrate over d3p′2 andcancel out the second delta-function. Yielding:
σ =
∫
d3p′116(2π)2
1
|~p1|(E1 + E2)V 2
δ(
E1 + E2 −√
m21 + ~p′21 −
√
m22 + ~p′22
)
√
m21 + ~p′21
√
m22 + ~p′22
|Afi|2
From now on, I’m going to (for simplicity’s sake), simply call:
Pi = |~p1|
with Pf similarly defined. As a result:
d3p′1 = P 2f dPfdΩ
wheredΩ = sin θdθdφ
Of course, for our simple system, conservation of momentum and energy(and the fact that we’re scattering identical particles), means that Pi = Pf (butnever mind that).
Thus, the incremental cross section is:
dσ
dΩ=
∫
P 2f dPf
16(2π)21
Pi(E1 + E2)V 2
δ(
E1 + E2 −√
m21 + P 2
F −√
m22 + P 2
F
)
√
m21 + P 2
F
√
m22 + P 2
F
|Afi|2
Using the trick from before, where
δ(f(x)) =δ(x− x0)
f ′(x)|x0
It is a pain to do, but:
∫
dPFδ(
E1 + E2 −√
m21 + P 2
F −√
m22 + P 2
F
)
√
m21 + P 2
F
√
m22 + P 2
F
=1
PF (E1 + E2)
where I’ve re-defined PF as the solution to the delta function.Thus, we get the pleasingly simple relation:
dσ
dΩ=
|Afi|216(2π)2
PF(E1 + E2)2PI
(108)
62
5.3.3 The ψ − ψ cross section
We’ve done all of the heavy lifting. The result from the previous section isgeneral for all 2-2 particle scattering problems in the COM frame. Now, all weneed to do is show how this scales with our actual scattering amplitude. Recallthat we had:
Afi = g2
[
1
(p1 − p′1)2 −m2
+1
(p1 − p′2)2 −m2
]
where I’ve gotten rid of the ε because we don’t have a singularity, and as areminder, m is the mass of the mediator particle.
Rather than solve this in generality, let’s solve it for our particular case, andwith our incoming particle “1” moving along the x-axis. In that case:
(~p1 − ~p′1)2 = P 2
I
[
(1 − cos θ)2 + sin2 θ]
= 2P 2I (1 − cos θ)
and thus:(p1 − p′1)
2 = −2P 2I (1 − cosθ)
(Notice the sign).Since ~p′2 = −~p′1, we get:
(~p1 − ~p′2)2 = P 2
I
[
(1 + cos θ)2 + sin2 θ]
= 2P 2I (1 + cos θ)
Thus:
Afi = g2
[
2(P 2I +m2)
4P 4I sin2 θ +m2(1 + 4P 2
I )
]
(109)
We could do this in a number of limits, but supposing we want to do some-thing kind of like electron scattering, in which case, m → 0 (for a photon –remember, this isn’t real E&M, which is mediated by a vector field. It’s just anapproximation). In that case, we get:
Afi ≃g2
P 2I sin2 θ
and thus:dσ
dΩ=
1
64
(
g2
(2π)EP 2I sin2 θ
)2
(110)
Of course, a 2-body problem looks a lot like a problem in which 1 body scattersoff a stationary target (Rutherford scattering), which case, there is a sin−4(θ/2)term. Whoa! This is exactly equivalent!
63
5.4 Particle Interaction Energy
Let’s suppose you had two, ψ, particles held at rest at positions ~x, and ~y. Whatis the energy of interaction?
First, let’s consider the basic setup:
|i〉 = ψ(~x)ψ(~y)|0〉
which creates two ψ particles at positions, ~x and ~y. We’re not too worried aboutthe normalization, but it’s clear that this is just an approximation, becausewe’re implying that we can create a particle with fixed position, and quantummechanics forbids that sort of certainty.
Likewise, The final state is nearly identical. Roughly speaking, the initial(and final) state looks like:
|i〉 =
∫ ∫
d3pd3q
(2π)61
√
4E~pE~q
(
c†~pc†~qe
−i(~p·~x+~q·~y))
|0〉
I don’t want you to worry too much about the normalization or the missingcreation and annihilation operators. The upshot is that we have two creationoperators at work here.
In fact, for simplicity, I’m going to ignore the integration factor entirely.Now suppose we want to compute the interaction energy between the two
particles? Naively, you’d expect this to be:
Eint?= 〈i|Hint|i〉
Your intuition would be wrong.The interaction Hamiltonian density is:
Hint = gφψψ†
After all, working things out (and ignoring integrals), this relation yields:
〈0|cc (c† + b)(b† + c)(a† + a) c†c†|0〉
where I’ve lazily also left out the subscripts, p and the like.That (a†+a) is a killer. Since we start with no φ particles and end with none,
is automatically makes the entire expression 0. We need to be a bit cleverer.Remember that we have the evolution relation:
|f〉 = U(t, t0)|i〉
and that if we have an energy Eigenstate of the system:
U = exp
(
−i∫
HI(t′)dt′
)
≃ 1 − iEI∆t
Or, to put it another way, if we evaluate:
〈i|U(t, t0)|i〉
64
Then we get two terms: The first is a constant. We can ignore that. The secondis a measure of the energy of interaction.
So what is U?
U(t, t0) = I + (−ig)T∫
d3xdx0φ(x)ψ(x)ψ†(x)
+ (−ig)2 1
2T
∫ ∫
d3xd3ydx0dy0φ(x)ψ(x)ψ†(x)φ(t)ψ(y)ψ†(y)
The first term on the right we ignore. The second term, we’ve already showndoesn’t contribute. So we care about the second term. In particular. Taking
〈i|(U − 1)|i〉
we get the product of two terms, one relating to the ψ fields, and one relatingto the φ field:
Et ∝ 〈0|cc T ψ(x)ψ†(x)ψ(y)ψ†(y) c†c†|0〉
× g2〈0|∫ ∫
T φ(x)φ(y)dx0dy0y|0〉
The term on the top looks complicated, but all you need to get from it is thatthere are an even number of creation and annihilation operators, meaning thatthe term is non-zero, and a constant.
The second term can be treated entirely independently since φ and ψ com-mute with one another.
Since we have:∆F (x− y) = 〈0|T φ(x)φ(y)|0〉
our energy term can simplify to:
Et = Cg2
∫ ∫
dx0dy0∆F (x− y)
= Cg2
∫ ∫ ∫
dx0dy0 d4k
(2π)41
k2 −m2eik·(x−y)
= Cg2
∫
dx0eik0x0
∫
d4k
(2π)41
k2 −m2e−i
~k·(~x−~y)∫
dy0e−ik0y0
= Cg2
∫
dx0eik0x0
∫
d4k
(2π)41
(k0)2 − ~k2 −m2e−i
~k·(~x−~y) (2πδ(k0))
= Cg2
∫
dx0
∫
d3k
(2π)31
−~k2 −m2e−i
~k·(~x−~y)
= −Cg2t
∫
d3k
(2π)31
~k2 +m2e−i
~k·(~x−~y)
Our energy relation appears much simpler now.We will do our integral in spherical coordinates:
d3k = sin θdθdφk2dk
65
We can even (for simplicity) assume that ~x− ~y = rk, and define:
u = cos θ
and hence:du = − sin θdθ
and further note that∫
dφ = 2π. This yields:
E = − Cg2
(2π)2
∫
k2dk1
k2 +m2
∫ 1
−1
due−irku
= − Cg2
(2π)2
∫
k2dk1
k2 +m2
e−irk − eirk
−irk
= − Cg2
4π2r
∫ ∞
−∞dk
k
k2 +m2sin(rk)
Where a factor of 1/2 shows up because technically, the integral should only befrom 0 to ∞.
This integral is a bit tricky. The easiest way to solve it is to imagine that kcan be complex. In which case, there is a root at k = ±im:
k2 +m2 = (k + im)(k − im)
Thus, taking a clockwise integral over the top half of the complex plane, encom-pass the k = im root.
∫ ∞
−∞
dk
k − im
k
k + imsin(rk) =
1
i
∫ ∞
−∞
dk
k − im
k
k + imeikr
= 2πiim
im+ im(−i)e−mr
= πe−mr
Plugging in this integral, we get:
Eint = −Cg2
4πre−mr (111)
Energy scales as inverse distance, produces an attractive force, and drops offquickly if we have a massive mediating particle.
5.5 Example: ψ − ψ Annihilation
Let’s do one more example, and consider a massive ψ, and a massless φ.
66
p1
p′1
k
p2
p′2
ψ ψ
φ φ
Let’s just use the Feynman rules that we’ve seen before:
1. External lines are labeled.
2. Internal lines are labeled.
3. Vertices−g2(2π)8δ(p1 − p′1 − k)δ(p2 − p′2 + k)
4. Internal Lines:
−ig2
∫
d4k(2π)4δ(p1 − p′1 − k)δ(p2 − p′2 + k)1
k2 −M2 + iε
Notice that the internal line is a ψ, not a φ, so we have to use the appro-priate mass.
This, of course, intergates to:
−ig2(2π)4δ(p1 + p2 − p′1 − p′2)1
(p1 − p′1)2 −M2 + iε
and so, combining this term with p′1 ↔ p′2, we get:
Afi =g2
(p1 − p′1)2 −M2 + iε
+g2
(p1 − p′2)2 −M2 + iε
This looks almost identical to our result from φ− φ scattering from before.Let’s suppose, for convenience, that ~p1 = −~p2. Further, by energy conserva-
tion, it is clear that our “photons” will fly out with:
Ep′1 = Ep′2 =√
M2 + |~pI |2
The actual cross section is almost identical to the ψ − ψ scattering resultabove, except for the fact that the output and input don’t have the same mass.
Suppose that particle 1 is moving initially along the z-axis with momentum,pI (and in particular, we might be interested in knowing what happens if this
67
speed is much less than the speed of light). In that case, ingoing 4-momentum(of the 1st particle) is:
pµ1 =
√
M2 + P 2I
00PI
and the outgoing 4-momentum of the first “photon” in:
p′µ1 =
√
M2 + P 2I
√
M2 + P 2I sin θ cosφ
√
M2 + P 2I sin θ sinφ
√
M2 + P 2I cos θ
where pI is the momentum of the incoming particle(s).Thus:
(p1 − p′1)2 −M2 = −P 2
I − (M2 + P 2I ) + 2PI
√
M2 + P 2I cos θ −M2
= −2
(
M2 + P 2I − 2PI cos θ
√
M2 + P 2I
)
And if p′1 goes off at angle θ then the other particle goes off at π − θ, so:
(p1 − p′2)2 −M2 = −2
(
M2 + P 2I + 2PI cos θ
√
M2 + P 2I
)
With some fairly straightforward math, we can then solve:
− 1
2(E20 − PIE0 cos θ)
− 1
2(E20 + PIE0 cos θ)
= − 1
M2 + P 2I sin2 θ
where I defined E0 as the energy of the ingoing (or outgoing) particles.Thus we get:
Afi = − g2
M2 + P 2I sin2 θ
The amplitude is greatest when θ → 0, so the outgoing photons are preferentiallyemitted at the poles (the same direction as the incoming ψ’s).
As a reminder, the differential cross section for scattering is:
dσ =1
4E1E2
1
|~v1 − ~v2||Afi|2(2π)4δ
(
∑
i
pi
)
∏
out states
1
2E~pi
where for 2 outgoing particles, we get:
σ =
∫ ∫
d3p′1d3p′2
(2π)61
16E1E2E′1E
′2
1
|~v1 − ~v2||Afi|2(2π)4δ(p1 + p2 − p′1 − p′2)
68
Of course, all of our ingoing and outgoing energies are the same, E0. Likewise,because we were smart and work in the center of mass frame, the space-like partof the delta-function looks like:
δ(~p′1 + ~p′2)
and the timelike part yields:
δ(2E0 − |~p1| − |~p2|)
Combining all of this yields
σ = g4
∫
d3p′1(2π)2
1
16E40
E0
2pI
1
(M2 + P 2I sin2 θ)2
δ(2E0 − 2PF )
= g4
∫
2πp2FdpF sin θdθ
(2π)21
16E40
E0
2pI
1
(M2 + P 2I sin2 θ)2
δ(2E0 − 2PF )
= g4
∫
p2F
dpF sin θdθ
64π
1
E30PI
1
(M2 + P 2I sin2 θ)2
δ(E0 − PF )
2
= g4
∫
sin θdθ
128π
1
E0PI
1
(M2 + P 2I sin2 θ)2
The integral:∫
sin θdθ1
(M2 + P 2I sin2 θ)2
ends up being quite ugly. However, as an approximation, it is:
2
M4− 8P 2
I
3M6+ ...
In other words, for low speed collisions PI << M , this expression becomes:
σ ≃ g4
128π2M6v(112)
Note that this has units of [E]−2, exactly as we’d expect it to.
5.6 Example: Higher order corrections in φ decay
As a final example (albeit one that we’re not actually going to compute thedecay rate for), let’s consider a single higher-order correction to our φ decayproblem from earlier. Once again, we’re assuming both φ and ψ are massive.
69
p1
k k′
k′′
p2 p3
φ
ψ ψ
This is a bit tougher than before. It’s not tough to follow the Feynman rules,but solving the integral will be a challenge.
Believe it or not, that was the easy part. Our integral over k now becomes:
−2π2i
∫ ∞
0
dk1k21
(√
k21 +M2)(m+
√
k21 +M2)(m+ 2
√
k21 +M2)
Or, if we consider the contribution to the overall integral, we get:
4π2ig3δ(p1 − p2 − p3)
∫ ∞
0
dk1k21
(√
k21 +M2)(2m+
√
k21 +M2)(m+
√
k21 +M2)
yielding
Afi = − 1
(2π)2g3
∫ ∞
0
dk1k21
(√
k21 +M2)(m+
√
k21 +M2)(m+ 2
√
k21 +M2)
(113)which is super-nice, except for that integral.
The problem here is that at large values of k1, this thing diverges. After all,it’s clearly:
∫
dk1
k1
which is bad at infinity, and yields an infinity.This is a real infinity, and kind of a disaster. It is also our first introduction
to renormalization. I’m not going to this in generality. The more general ap-proaches (applied later) leave the solution as Lorentz invariant. However, you’llget a flavor for how it works with the following.
First, imagine multiplying in a fudge factor:
Λ2
Λ2 + k21
Clearly, this invokes a cutoff on a scale, Λ. On the other hand, for suitablylarge, Λ it does nothing. This is a regularization term and allows our integralto be finite.
I’m not going to do the exact integral. There are several reasons. First,this isn’t actually how you’d do the renormalizations. Secondly, if we actuallywanted to compute the exact 3rd order correction to the transition amplitude,we’d need to use all 3 vertex diagrams. Nevertheless, let me give you a sense ofhow everything comes out. I’ll write it in dimensional terms.
To make things concrete, we’ll suppose that the ψ particles are massless,M = 0.
Afi ∼ −g3
∫ ∞
0
dk1k1
(m+ k1)(m+ 2k1)
Λ2
Λ2 + k2
≃ −g3
∫ m
0
k1dk1
m2− g3
∫ Λ
m
dk1
k1
= −g3B − g3C ln
(
Λ
m
)
72
There’s a perturbation in the effective transition amplitude that is dependenton an (unknown) maximum scale – presumably the Planck scale, but really, itcould be anything.
This isn’t as big a deal as you might think. After all, in practice, we don’tever measure the “bare” term. We only observe the measured term.
5.7 Example: A Simple Mass Perturbation
Consider, for instance, the mass of a φ particle. We’ve been treating it as partof the free field solution, but there’s no reason that we couldn’t imagine it as aperturbation to the Hamiltonian:
Hint =1
2m2
∫
d3xφ(x)2
We could then ask about the amplitude of a 1-vertex diagram (just to un-derstand what it means).
φ, ~p1 = 0
φ, p2
We almost immediately get:
Afi =1
2m2
In other words, this diagram effectively measures the mass of a φ particle. Butwhat happens when we look at a perturbation?
k k′
φ, p1
ψ, p2
73
A little bit of math will get a similar result to the previous section:
Afi,2nd order = g2
[
B + C ln
(
Λ
M
)]
or to put it another way, the measured mass and the true mass are related via:
m2meas = m2
bare + 2g2
[
B + C ln
(
Λ
M
)]
where the latter term can be arbitrarily large. But that’s okay (within reason).We only care about the combination of terms.
74
6 The Dirac Equation
• Gross: 5, 7.4, 8.3-8.6
• Tong: Chapter 4, 5.1-5.2
Thus far, we’ve discussed scalar fields only. In reality, we know that particleshave spin. The fermions have spin=1/2, the bosons, spin=1. (The Higgs if/whenit is discovered will have spin=0, and the Graviton, spin=2, and these are alsobosons, but let’s ignore them for the time-being).
I introduced the Klein-Gordan equation previously as a solution to a rela-tivistic field, but it actually does double duty. It is also the evolution equationfor a single relativistic, scalar particle. It has a fatal flaw, however. It’s 2nd or-der in time, which means that the state of the system is insufficient to determinethe future evolution.
In this section, we’ll talk about the Dirac equation, the evolution equationfor spin=1/2 particles. Note that this is not field theory. The results ofthis study will be important for field theory, however, and toward the end ofthese notes, we’ll talk a bit about quantizing the Dirac field.
6.1 1st order vs. Lorentz Invariance
Ideally, we want an equation of the form:
iψ = Hψ (114)
which is the time-dependent Schroedinger equation. I know we’re not doingfield theory, but ~ still equals 1.
Now, consider a particle at rest. Since it’s a relativistic particle, we have therelationship:
(p0)2 −m2 = 0
which has the solutions:
p0 −m = 0
p0 +m = 0
either of which could be used to write an equation linear in energy (and thusproduce a differential wave equation linear in the time derivative).
Now, I know what you’re thinking: Only the first solution is physicallyviable. There’s no such thing as a negative energy (which is what p0 is, afterall). Dirac’s particular genius lay in the fact that he was willing to follow thisreasoning through to the end, regardless of the fact that it seems impossible.
But now, consider the momentum of the particle if it gets boosted. Clearly,under those circumstances, the “factoring” to solve for energy becomes a bitmore complicated. We know, of course, that:
pµpµ −m2 = 0
75
but factoring this must include terms that look like:
pµpµ −m2 = (βκpκ +m)(γλpλ −m) = 0 (115)
Where we’d want to figure out the elements of βκ and γµ.The basic idea is that we’d have a free wave-function along the lines of:
ψ ∝ eipµxµ
where the operator, pµ and the actual components of pµ can be related via:
pµψ = pµψ
and where we have our standard operators:
i∂µ = pµ (116)
which is exactly the same thing that we have for H and ~p in ordinary non-relativistic quantum mechanics (with ~ = 1, of course).
Supposing all of this works out, then the linearized version of the factoredmomentum equation can be written as a wave equation:
iγµ∂µψ −mψ = 0 (117)
This is the “Dirac Equation.” And now is as good a time as any to give you alittle shorthand for this. The Dirac notation for these γ’s allow us to write:
/p = γµpµ (118)
and similarly for any other combination of a γ and a vector, 1-form, or operator.It’s just a sum over 4 terms.
At any rate, the Dirac equation will clearly be satisfied if:
γµpµ −m = 0
which is our positive solution.What are the values of βµ and γµ? It might help if we multiply out eq. (115):
ηµνpµpν −m2 = (βκpκ +m)(γλpλ −m)
= β0γ0p0p0 + β1γ1p1p1 + ...
+(β0γ1 + β1γ0)p0p1 + ...
+(−β0 + γ0)mp0 + ...
−m2
It is clear by construction that of the four lines of expanded terms, only the firstand last don’t vanish. The third line immediately yields:
βκ = γκ
76
While the 1st and 2nd can readily be combined to yield:
γµ, γν = 2ηµν (119)
It is clear that γµ cannot be ordinary numbers. The elements of γ need tothemselves be 4×4 matrices! I’m not going to derive this, but it is easy to showthat we have the correct relation by multiplying out term-by-term.
The “Dirac Representation” version of the matrices are:
γ0 =
(
1 00 −1
)
(120)
γi =
(
0 σi
−σi 0
)
(121)
where ”1” actually means the 2 × 2 identity matrix, and σi is the ith Paulimatrix, which in case you’ve forgotten are:
σ1 =
(
0 11 0
)
(122)
σ2 =
(
0 −ii 0
)
(123)
σ3 =
(
1 00 −1
)
(124)
The “Dirac Representation” version of the matrices are:
γ0 =
(
1 00 −1
)
(125)
γi =
(
0 σi
−σi 0
)
(126)
where ”1” actually means the 2 × 2 identity matrix, and σi is the ith Paulimatrix, which in case you’ve forgotten are:
σ1 =
(
0 11 0
)
(127)
σ2 =
(
0 −ii 0
)
(128)
σ3 =
(
1 00 −1
)
(129)
The “Dirac Representation” version of the matrices are:
γ0 =
(
1 00 −1
)
(130)
77
γi =
(
0 σi
−σi 0
)
(131)
where ”1” actually means the 2 × 2 identity matrix, and σi is the ith Paulimatrix, which in case you’ve forgotten are:
σ1 =
(
0 11 0
)
(132)
σ2 =
(
0 −ii 0
)
(133)
σ3 =
(
1 00 −1
)
(134)
These matrices and the form is not unique. Tong uses what’s known as “Weyl”representation, for example, which produces a non-diagonal γ0 matrix. TheWeyl representation is usefully compact when dealing with theories that havea distinct handedness (like the weak force), but won’t be terribly useful for usnow. I’ll stick with Dirac representation for now.
One more thing. Now that we have our γ matrices, we could quickly derivea number of very useful relations with the “slash” notation, including:
tr(/a/b) = 4a · b (135)
γµ/aγµ = −2/a (136)
γµ/a/bγµ = 4a · b (137)
These will be very useful later when we’re actually trying to compute cross-sections and whatnot.
In the meanwhile, assure yourself that the slashed and γ relations work. I’llwait.
Dimensionally, all of this means that the wave-function in the Dirac equa-tion( 117) must be a vector of 4-components:
ψ =
ψ1
ψ2
ψ3
ψ4
Note: This is NOT a 4-vector. This is a “Dirac Spinor”, and it’s going to havesome interesting properties.
I want to return (briefly) to the Dirac equation itself. You’ll note that it isa first order operator in time. Noting that, we can determine the Hamiltonian(which is, after all, the time evolution operator for a wave-function), by breaking
78
things up:
iγ0∂0 = γ0H
(iγµ∂)µ−m)ψ = 0
= (iγ0∂0 + iγi∂i −m)ψ
iγ0∂0ψ = (−iγi∂i +m)ψ
γ0Hψ = (−iγi∂i +m)ψ
H0 = −iγ0γi∂i + γ0m (138)
Of course, in principle, we could add an interaction term as well:
HI = −eγ0γµAµ
which is the electromagnetic potential. For now, though, we’ll restrict our dis-cussion to free space.
6.2 Solutions to the Dirac Equation
For simplicity, let’s suppose our solution to the Dirac equation is independentof position. In that case, we have:
iγ0ψ −mψ = 0
Write it all out, and you realize that the first two terms in ψ are independentof the second two. Thus, we may think of ψ as:
ψ =
(
ψAψB
)
where each of ψA and ψB have two elements. the spatially invariant Diracequation simplifies to:
ψA = −imψAψB = +imψB
It’s clear that in the free-field case, both are going to have an energy term inthe exponential, but with a different sign.
We could go through a fair amount of derivation, but I’m simply going tostate the free-field solutions to the Dirac equation, and you can verify that a)work, and b) are independent. The four solutions are:
u(1)(p) =√E +m
10pz
E~p+mpx+ipy
E~p+m
(139)
79
u(2)(p) =√E +m
01
px−ipy
E~p+m
− pz
E~p+m
(140)
v(1)(p) =√E +m
px−ipy
E~p+m
− pz
E~p+m
01
(141)
v(2)(p) =√E +m
pz
E~p+mpx+ipy
E~p+m
10
(142)
I’ll justify both the form and the normalization out front in a moment.These forms are static, of course (Heisenberg representation). To turn theminto plane-waves (Schroedinger representation), we’d need to multiply:
u(s)e−ipµxµ
6.3 What the Dirac Solutions Mean 1: Solves the DiracEquation
Having gone through all of the effort of solving the Dirac equation, and finding4 solutions, your first question might be why we are going to all of the effort.There are at least three reasons. The first, is that these equations solve theDirac equation, which means that that superpositions of them also satisfy theDirac equation.
This may not seem like a huge deal, but hold on.
6.4 What the Dirac Solutions Mean 2: Orthogonality andCurrents
6.4.1 The Adjoint Spinor
One of the most important properties of the Dirac spinor is that, much likethe wavefunction in ordinary non-relativistic quantum mechanics, we expect aquadratic scalar combination of terms to yield something like a conserved andLorentz-invariant quantity. In our case, we will define a very important object,the “Adjoint Spinor,” defined as:
ψ = ψ†γ0 (143)
80
which is known as the “adjoint spinor.” The product:
ψψ
is Lorentz invariant. And what is it? Well, for u(p), we get:
u(1)u(1) =(
√
E~p +m)2 (
1 0 − pz
E~p+m 0)
10
+ pz
E~p+m
0
eipµxµ
e−ipµxµ
= (E~p +m)
(
1 − p2z
(E~p +m)2
)
= 2m
Without that crazy normalization up front, it’s clear that the result would bea function of E~p and therefore not Lorentz invariant. But now notice whathappens when we consider v(p)v(p). We now end up getting −2m.
6.4.2 Orthogonality
The Dirac and Adjoint spinors have a number of very important properties.Most notably:
u(r)u(s) = 2mδrs (144)
u(r)v(s) = v(r)u(s) = 0 (145)
v(r)v(s) = −2mδrs (146)
In other words, they are orthogonal. But not only that, they are complete:These solutions also have some rather nice relations to one another. For
example, if we consider the outer product:
∑
s
u(s)(p)u(s)(p) = /p+m (147)
∑
s
v(s)(p)v(s)(p) = /p−m (148)
6.4.3 The Conserved Norm
Clearly, the quantity:ψψ
is going to be important for any particular Dirac field. It represents somethinglike the total amount of energy, but not really the total number of particles. Wewill see that there are truly conserved currents as well:
jµ = ψγµψ (149)
It can be shown, for example, that for both u and v modes, j0 yields 2m.
81
More specifically, it can be shown that:
u(s)γµu(s) = v(s)γµv(s) = 2pµ (150)
(The sign is independent of u vs v). This is a very useful relation, indeed.I should also add that we will include some additional currents later when
we compute the stress-energy tensor. But first, we’ll need to write down theLagrangian.
6.5 What the Dirac Solutions Mean 3: Operators andTransforms
The third, and arguably most important property of a wavefunction is that itshould interact with operators. In particular, we’ll want to think about eigen-value relations:
Oψ = Oψand, of course, transformations. The two are not indepdendent of one another.
6.5.1 Operators: Momentum and Energy
For any wave-mechanical system, the most important oeprator is the Hamilto-nian, followed quickly by the momentum operator. As we’ve seen, the loweredindex form of this is:
pµ = i∂µ (151)
where the zero component, naturally, corresponds to energy. For the u-modes,the space-time dependent states can be written:
u(x) = ue−ipµxµ
which quickly yields an eigen-value of pµWith the v-modes, the exponent term is reversed, which means that the
eigenvalue is a bit tougher to interpret.Things become clearer if we look at the equivalent of the expectation value:
v(s)(p)pµv(s)(p) = +2mpµ
normalized, and positive exactly the same as the u’s.
6.5.2 Symmetry Operation: Charge Conjugation
One of the most important types of operations that we can perform on statesare symmetry operations. The idea is that we will have very simple eigen-values(-1 or 1, in most cases), or that an operation will exactly turn one particle intoanother.
In QFT and fundamental physics generally, there are three very importantsymmetry operations:
82
1. C Symmetry (“Charge Conjugation”) – which essentially means that if weturn all particles into antiparticles and vice-versa then all of the physicsof the universe will look identical.
2. P Symmetry (“Parity”) – which means that if we reflect all vectors:
~v → −~vthen physics will remain unchanged.
3. T Symmetry (“Time Reversal”) – which means that if we reverse the arrowof time, physics will remain inviolate.
In Gravity, E&M, and the Strong Force, all three of these symmetries are re-spected. In other words, if there is an operation, C which represents Chargeconjugation, then:
ψC−1HintCψ = ψHintψ
which can be achieved simply enough if C commutes with the Hamiltonian.As I said, this symmetry does hold for three of the forces, but it does not
hold for the weak force. In the weak force, all neutrinos are left-handed, andall anti-neutrinos are right-handed. Switching particles for anti-particles mostcertainly is noticable.
Presumably (you might think), the comination of:
CP
is respected by the weak force. It is not. Good thing, too, since otherwise, wewouldn’t have matter-antimatter asymmetry, and thus we wouldn’t exist.
Only:CP T
Seems to be symmetric for all of the fundamental forces.But what is the C operator? Simple inspection yields:
Cψ = iγ2ψ∗ (152)
We can quickly show that this gives the operations:
Cu(r)(p) → v(r)(p)
exactly as you might have hoped.
6.5.3 Symmetry Operation: Parity
Parity is even simpler. For the u states, it is clear that only the lower two termshave spation contributions. Thus, for u(1) applying a negative two the lowertwo terms yields a spatial reversal:
P√E +m
10pz
E~p+mpx+ipy
E~p+m
=√E +m
10
−pz
E~p+m−px−ipy
E~p+m
83
and similarly for u(2). Thus, it seems reasonable that:
P = γ0 (153)
Interesting things happen, especially when we consider a particle at rest. Inthat case:
Pu(s)(~p = 0) = u(s)(~p = 0)
The eigen-value of the parity operator is 1. u-states have an even symmetry.On the other hand, it’s clear that for a v-particle at rest:
P v(s)(~p = 0) = −v(s)(~p = 0)
Negative parity. Parity is not just a symmetry. It’s a conserved quantity.
6.5.4 Operator: Spin
But now, let’s consider the operator:
~S =1
2
(
~σ 00 ~σ
)
(154)
This is the spin operator, and if it looks crazy and unfamiliar, it should. It’sreally 3 operators, and you can imagine really wanting to know, for example:
σp = S · p|~p| (155)
the operator yielding the spin along the direction of motion of our particle. I’vemade things easy in this case, since the particle is moving in the z-direction, weget:
σpψ =1
2
1 0 0 00 −1 0 00 0 1 00 0 0 −1
10
− pz
E+p
0
trivially. Thus, for particles moving in the +z direction, u(1) and v(2) have aspin of 1/2, and u(2) and v(1) have a spin of -1/2.
This is not quite as strange is at it would seem. After all, much of theconvention for the v-particles is reversed. Let’s consider v(1):
v(1)(p) =√E +m
0
− pz
E~p+m
01
As we’ve already seen, the eigen-value of momentum of the v-states is:
pµv = −pµv
The spin operator yields -1/2, dotted with the momentum operator which yieldsanother -1. The spin is along the direction of motion, just for the u(1) state.
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6.5.5 Transform Operator: Boosts
We’re finally ready to consider transformation operations. In order to get arunning start, let’s discuss a simplified case, a 1+1 dimensional system.A Simple 1+1 System
It is clear that in the Galilean limit, a small boost, Ψ in the x-directionyields:
Λµµ =
(
1 ΨΨ 1
)
But it is clear that each frame is just a small Galilean boost from one another.Thus we can re-write this as:
Λµµ = (δµµ +1
NΨKµ
µ)N
which is just a shorthand way of saying that we perform the boosts many, manytimes. The matrix, K is what we call a generating matrix, and is, by inspection:
Kµµ =
(
0 11 0
)
Of course, we’ve seen familiar functions to this before:
limN→∞
(
1 +x
N
)N
= ex
Thus:Λ = exp (ΨK)
We’re going to do this a lot. We come up with a generating function thatultimately reveals a symmetry or an operator.
This is, of course, a series:
Λ = I + ΨK +1
2Ψ2I +
1
6Ψ3 + ...
I recognize this! It is:Λ = cosh ΨI + sinh ΨK
or, written out:
Λ =
(
coshΨ sinh Ψsinh Ψ coshΨ
)
This has the nice result that:
cosh2 Ψ − sinh2 Ψ = 1
Not coincidentally, we also have:
γ2 − (vγ)2 = 1
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Thus, this is the ordinary Lorentz boost with:
γ = coshΨ
andv = tanhΨ
How Dirac Spinors Behave under Lorentz TransformsOf course, we aren’t just boosting individual components of vectors. Rather,
we need to make sure that various terms in the Lagrangian will be preserved.We’ve already seen what energy density (the Hamiltonian) looks like. Thus, werealize that we’re going to have to do a transform to preserve the quantity:
ψγµψ
from one frame to another.Let’s suppose that we find an operator for a boost, which we’ll call:
ψ → S(Λ)ψ
We already know how γµ transforms. It behaves essentially like the componentsof a vector. Thus, we have the relation:
ψγµψ → ψS−1ΛµγµSψ
or equivalently:S−1(Λ)γµS(Λ) = Λµνγ
ν
Of course, the values of the γ matrices are the same in all frames. Also, note thatso far, all of my work is completely indepdendent from the sort of transformationthat I’m looking to do.
As with the previous simple example, we’re going to assume that can beapproximated as a small perturbation, and then we can take an exponent.
Note that at this point, I’m going to work the answer specifically for boosts,rather than rotations, but the approach will be similar. I am assuming a boostin the i-th direction, Ψi, and I will write out my results in terms of Bi, an as-yetunknown 4x4 matrix which will generate the boost:
S ≃ I + ΨiBi
soS−1 ≃ I − ΨiBi
to 1st order.Likewise, we already know that
Λ ≃ I + ΨiKi
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Multiplying everything out we get:
S−1γµS = (1 − ΨiBi)γµ(1 + ΨiBi) = γµ − Ψi [Bi, γ
µ]
Λµνγν = 1 + ΨiKi
[Bi, γµ] = −Kµ
i,νγν
We know the boost matrices. In the z-direction, for instance, it’s simply:
K3 =
0 0 0 10 0 0 00 0 0 01 0 0 0
so in the z direction we have:
[
B3, γ0]
= −γ3
and[
B3, γ3]
= −γ0
with all others zero. Together, these yield:
B3 =1
2γ0γ3 =
1
2
0 0 1 00 0 0 −11 0 0 00 −1 0 0
or more compactly:1
2αi ≡ Bi =
1
2
(
0 σiσi 0
)
(156)
Thus, a boost may be expressed as:
S = exp
(
1
2Ψiαi
)
whereα2i = I
so the series looks nearly identical to what we had in the 1+1 universe case:
S =
(
I cosh(
Ψi
2
)
σi sinh(
Ψi
2
)
σi sinh(
Ψi
2
)
I cosh(
Ψi
2
)
)
(157)
Simple comparison from boosting the u(1) state from rest yields:
cosh
(
Ψi
2
)
=
√
1 + γ
2
It’s not simple, but it is algebraically solvable.
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6.5.6 Transform Operator: Rotations
We can do virtually the same thing with rotations.A Simple 2-d system
As before, let’s just consider the ordinary rotation matrix. This is nearly
identical to the case with boosts, except:
Rµµ ≃(
1 θ−θ 1
)
Thus, the rotation generator is:
Jµµ =
(
0 1−1 0
)
and thus, the full form of the rotation operator is:
R = exp (θJ)
Note thatJ2 = −I
so, this forms a series:
R = I + θJ − θ2
2I − θ3
6J + ...
which combines to:
R =
(
cos θ sin θ− sin θ cos θ
)
as I’m sure you knew.How Dirac Spinors Behave under Rotations
To do rotations on spinors, we’re basically going to repeat our work onboosts. To simplify things, I’m only to going to solve for rotations around thez-axis, though the result will generalize easily.
Using an exactly analagous argument to boosts, I claim that that transfor-mation operator will be:
S ≃ I + θiRi
Again, using an exactly analogous argument to before, we get:
[Ri, γµ] = −Jµi,νγν
where, for a rotation around the z-axis:
J21 = 1 ; J1
2 = −1
and all other values are zero. Combining, we get:
[
R3, γ1]
= γ2
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and[
R3, γ2]
= −γ1
This negative sign is going to be trouble. To help us out, I’m going to introduceanother important gamma matrix, normally called, γ5
γ5 = iγ0γ1γ2γ3 =
(
0 II 0
)
My claim is that:
Ri = − i
2γ5αi
where the α matrices were defined above:
αi =
(
0 σiσi 0
)
This does, indeed, satisfy the commutation relationships above (you cancheck!)
We now have a rotation generator! Multiplying it out yields:
~R = − i
2γ5α3 = − i
2
(
σ 00 σ
)
(158)
What could be simpler!Thus, the rotation operator yields:
S(θ) = exp
(
−i θ32
Σ3
)
where I’ve been lazy and referred to Σ as the 4x4 block diagonal matrices ofPauli spin matrices. Multiplying this out yields:
S(θ3) = I − iθ32
Σ3 − I1
2
(
θ32
)2
I + i1
6
(
θ32
)2
Σ3...
or, more simply:S(θi) = cos(θi/2)I − i sin(θi/2)Σi (159)
As written, this will work for any direction. Multiplying it out explictly for z,we get:
S(θ3) =
e−iθ2 0 0 0
0 eiθ2 0 0
0 0 e−iθ2 0
0 0 0 eiθ2
(160)
This is a very exciting result! Both sets of terms produce a −1 for rotationsof θ = 2π. In other words, spin a fermion around a full rotation, and yourwave-function reverses. You need to spin it around twice to get back to whereyou started.
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6.6 The Dirac Lagrangian
Now that we’ve gone through some of the basic manipulations on Dirac particles(and more later, as it becomes necessary), we need to start turning this into thenotation that we’re accustomed to.
I’ll simply give you the free-field Lagrangian for a Dirac particle:
L = iψγµ∂µψ −mψψ (161)
which you can test (using ψ and ψ as separate fields) becomes the Dirac equa-tion.
First, a note on dimensionality. Since:
[m] = [E]
we immediately get:
[ψ] = [E]3/2
(162)
which may not have been what you were expecting.As for the Lagrangian itself, let’s just show that it works:
∂L∂(∂µψ)
= iψγµ
sod
dxµ
(
∂L∂(∂µψ)
)
= iγµ∂µψ
and∂L∂ψ
= −mψ
which combine to yield:(iγµ∂µ +m)ψ = 0 (163)
a similar exercise quickly yields:
(iγµ∂µ −m)ψ = 0
Likewise, now that we know the Lagrangian, we can immediately computethe stress-energy tensor:
T µν = iψγµ∂νψ (164)
You can also use this to show that the current we’ve used above is indeed theNoether current found from phase invariance.
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6.7 Quantizing the Dirac Field
Now that we know the free field solution to the Dirac equation, and further, nowthat we’ve gotten our propagators, we’re immediately in a position to write downthe quantized field:
ψ(~x) =∑
s
∫
d3p
(2π)31
√
2E~p
[
b(s)~p u(s)(p)ei~p·~x + c
(s)†~p v(s)(p)e−i~p·~x
]
ψ†(~x) =∑
s
∫
d3p
(2π)31
√
2E~p
[
b(s)†~p u(s)†(p)e−i~p·~x + c
(s)~p v(s)†(p)ei~p·~x
]
where s are the two possible spin states, and b† creates u particles and c† createsv particles.
6.7.1 The Hamiltonian: Part 1
Let’s continue with this line of reasoning (without yet using any of our commu-tation relations). Using our familiar stress-energy tensor, we can write down aHamiltonian:
H =
∫
d3xψ(−iγi∂i +m)ψ (165)
Through a number of beautiful relations (which are done in §5.1 of Tong), theHamiltonian can simplify to:
H =
∫
d3p
(2π)3E~p
[
b(s)†~p b
(s)~p − c
(s)~p c
(s)†~p
]
H =
∫
d3p
(2π)3E~p
[
b(s)†~p b
(s)~p − c
(s)†~p c
(s)~p + (2π)3δ(0)
]
The last term is the same infinity we’ve encountered before. We will (as always)simply choose to ignore it. However, we’ve got bigger fish to fry. The bigproblem here is that we presumably can make energy small by introducing lotsof anti-particles. In fact, we can imagine this as an infinite reservoir of energy.This is a real problem.
6.7.2 Anti-Commutator Relations
To rescue this, we need to introduce anti-commutator relations. In particular,we need:
ψ(~x), ψ†(~y) = δ(~x− ~y) (166)
and where other anti-commutators equal zero. These lead to the relations:
b(r)~p , b(s)†~q = (2π)3δrsδ(~p− ~q) (167)
c(r)~p , c(s)†~q = (2π)3δrsδ(~p− ~q) (168)
and all of the others vanish.
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6.7.3 The Hamiltonian: Part 2
Using the anti-commutator relation, we now can recast the Hamiltonian above.Noting:
c~pc†~p + c†~pc~p = (2π)3δ(0)
We now get a much more satisfying Hamiltonian (with only 1 embarrassinginfinity which has the opposite signs of the one we’ve seen before!):
H =
∫
d3p
(2π)3E~p
[
b(s)†~p b(s) + ~p− c
(s)†~p c
(s)~p − (2π)3δ(0)
]
(169)
6.8 Fermi-Dirac Statistics
I’ve been a little sloppy when listing which terms vanish and which don’t, soa quick summary of a few relevant commutator and anti-commutator relationsare in order. For our φ theory (real-valued scalar field), we had the following:
[a~p, a†~q] = (2π)3δ(~p− ~q)
but more relevant to this discussion, we had:
[a†~p, a†~q] = 0
which means that:a†~pa
†~q = a†~qa
†~p
and thus, that:|~p, ~q〉 = |~q, ~p〉
In our Fermionic fields, however, we have an anti-commutator relation (aris-ing from the need for anti-particles to contribute a positive energy), and sincemost of them vanish, including:
b†~p, b†q = 0
(Ignoring the spin), and we get:
b†~pb†~q = −b†~q b
†~p
and thus:|~p, ~q〉 = −|~q, ~p〉
This is the origin of the Fermi-Dirac statistical relations you’ve seen in your QMclass.
92
6.9 The Fermi Propagator
Finally, it is clear that we’re going to have to define a propagator, something ofthe form:
〈0|ψ(x)ψ(y)|0〉or similar.
You may recall that the Feynman propagator could be expressed as:
∆F (x− y) = [φ(x), φ(y)]
and we’re going to use our intuition to suppose that the anti-commutators aregoing to be the way to go in this case. We define the fermionic propagator as:
iS =
ψ(x), ψ(y)
Note that this is an outer product, not an inner one. The propagator is a 4× 4matrix, not a number.
Multiplying this out (as Tong does), we get:
iS(x− y) = (i/∂x +m)(D(x− y) −D(y − x))
where D is the one-sided propagator we’ve seen before. Building on what we’veseen before (in k-space), we get:
SF (x− y) = 〈0|T ψ(x)ψ(y)|0〉 =
∫
d4p
(2π)4e−ip·(x−y)
i(/p+m)
p2 −m2(170)
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7 Quantum Electrodynamics
• Gross 2.2-2.6, 10
• Tong: 6
7.1 Gauge Transformations and Symmetries
Now that we’re working in the realm of the real universe, we need to startthinking about how to generate real Lagrangians. Real, as in, “shows up in theactual universe.” In order to understand which Lagrangians are realizable, weneed to introduce the idea of Gauge symmetries.
Simply put, a gauge transformation is something that you can do to a fieldwhich doesn’t change anything observable. For example, consider the followingrelationships from E&M:
~E = −∇Φ − ~A
~B = ∇× ~A
where Φ and ~A are the vector and scalar fields, respectively.Remember that all we ever care about in E&M are the E&M fields. The
potentials are just helper functions. Now, what happens if we make the followingtransformation:
Φ → Φ + λ
~A → ~A−∇λ
where λ is an arbitrary function of space and time.Since the curl of a gradient is zero, the ~B field is unchanged. Likewise, it
should be clear that this transformation creates two canceling terms in the ~Efield, which subsequently remains unchanged. These are the important fields,of course, because they are the ones that appear in the Lorentz force law.
This is an example of a gauge transformation, and we can use it to putvarious constraints on the potential fields in a particular system.
In field theory, a gauge symmetry is assumed (and then experimentally ver-ified) for each of the fundamental forces. A Lagrangian is then found whichsatisfies the gauge symmetry. And lo and behold! With the exception of thestrength of the coupling term (which has to be added by hand) virtually all ofthe the theory of the force comes out automatically.
7.2 U(1) Gauge Symmetry
As an example (and a way of developing Quantum Electrodynamics) we’re goingto work out the implications of a particular Gauge Invariance: U(1). I’m goingto simplify things, for now, and work out electrodynamic theory for a charged
94
scalar field. Also, until further notice we’re just doing relativistic classical fieldtheory. I’ll let you know when it’s time to quantize things again.
So, U(1)... We’ve already seen the global version of this with the transfor-mation:
ψ → ψ exp(−iλ)where λ is a real constant angle. Provided we also apply the transformation:
ψ∗ → ψ∗ exp(iλ)
Suppose we know nothing about the Lagrangian except that a transformationof this form leaves the action of a system unchanged? Well, we’ve already seenthat global U(1) gauge invariance results in the Noether current:
jµ = i [ψ∂νψ∗ − ψ∗∂νψ]
Now, and here’s the tricky part, what if we assume that there is invariance(so-called “gauge invariance”) not only globally, but locally? There is no obviousreason (except for this is apparently how the universe works) that this should bethe case. The idea isn’t that our fields don’t change at all, but that by applyingsuch a transformation, our Lagrangian, and the E-L solutions to the Lagrangianstill make sense.
Applying our local gauge transformation, we get:
ψ → ψ exp(−iλ(x)) (171)
and similarly, with a + sign for ψ∗.Now things are not so simple. Let’s look at the Lagrangian:
L = ∂µψ∂µψ∗ −m2ψψ∗
The term:m2ψψ∗ → m2ψψ∗
No problem. However...
∂νψ → ∂νψ − iψ∂νλ(x)
Oh no! The U(1) Gauge transformation, introduced a new set of terms into theLagrangian!
Now it is not obvious to me (or anyone else, incidentally) why local gaugeinvariance should exist, but it seems to be a governing property of our universe.The U(1) transformation (the simplest one) is the generating symmetry forE&M, SU(2) is the one for the weak field, and SU(3) for the strong.
The point is that we can force the Lagrangian to be conserved if we replacethe derivative with:
∂ν → Dν = ∂ν + i∂νλ(x)
Note that this doesn’t mean that nothing changes if we locally change phases– just that doing so is allowed, but we need to adjust our Lagrangian accordinglyso that any possible phase distribution works in our Lagrangian.
95
Doing so necessarily adds 2 sets of terms. The first is:
Lint = −ig∂µλ(x) [ψ∂µψ∗ − ψ∗∂µψ]
where I’ve liberally expanded out my ηµν , and noted that we get a minus signin the Gauge term for the ψ∗ part of the expression, and thrown an arbitrarycoupling constant in front for good measure.
I can distribute my units out however I like, so I define:
Aµ(x) = ∂µλ (172)
I may as well ruin the surprise now and say that this is going to be the electro-magnetic 4-vector potential:
Aµ =
ΦAxAyAz
(173)
This very tidily yields a nice interaction term in our Lagrangian:
Lint = −gAµJµ (174)
It should be noted that the coupling term, g, is usually swallowed up into thedefinition of current and is, essentially, the electric charge, e. For now we’regoing to leave it out front.
7.3 The vector potential
And what of this term, A? We’ve already established that it’s the 4-vectorpotential from E&M.
However, we have a big problem. We don’t yet know what a “free” E&Mfield looks like. Consider our total Lagrangian (so far):
L = Lψ,free − gAµjµ
Can we just leave things like this?No.Our rule for gauge invariance is that we need to be able to apply a gauge
transformation to all of our fields, and find that the Lagrangian is unchanged.However, we now have two transforms:
ψ → ψe−iλ(x)
Aµ → Aµ + ∂µλ
where the second came from the fact that at first we didn’t have any Aµ fieldat all, and it was our gauge transformation that gave rise to it in the first place.
96
This has some rather profound implications. First, it is clear that the “free”Aµ Lagrangian must also appear in our total Lagrangian, and that the entirething must be locally gauge invariant. Further, it means that terms like:
M2AµAµ
can’t appear in the final Lagrangian (meaning that the final term from ouroriginal gauge transformation must somehow cancel). Why not? Because if weapply a gauge transformation then:
M2AµAµ + 2M2Aµ∂
µλ+M2∂µλ∂µλ
which is manifestly not gauge invariant. So the free-field Lagrangian for our Afield can’t have that term. This means that our A field represents a masslessparticle. Remember that the ψ2 term represented the mass term!
Likewise, if we imagine 2nd-order first derivative terms, they can be brokendown into a symmetric and anti-symmetric component:
Gµν = ∂µAν + ∂νAµ
Alas, this term can’t appear in the free-field Lagrangian because it transformsunder a local gauge transformation as:
Gµν → Gµν + 2∂µ∂νλ
Again, not gauge invariant.The only term that is is the anti-symmetric combination:
Fµ ≡ ∂µAν − ∂νAµ (175)
Thus, our Lagrangian can be expressed as:
L = Lfree ψ − gJµAµ + Const · FµνFµν (176)
Since the normalization doesn’t matter, in order to meet with normal conven-tion, we take the free-field Lagrangian for the E&M field to be:
LEM = −1
4FµνF
µν (177)
7.4 The 4-Potential and the Field
I’m going to drop the bombshell. Fµν is an anti-symmetric 4 × 4 tensor, whichmeans that the diagonal components are all zero (by definition). Also, the topsix components are just the negative of the bottom 6. In other words, there areonly 6 independent numbers. What are they?
Well, first, I’m going to let you in on a secret. That 4-vector, Aµ? We’regoing to define it as the electromagnetic potential:
Aµ =
ΦAxAyAz
(178)
97
and similarly:Aµ = (Φ,−Ax,−Ay,−Az) (179)
So consider, for example:
F01 = −Ax − ∂xΦ = Ex
It’s just the electric field! We have similar relations for F02 & F03, but we won’tderive them.
Likewise,F12 = −∂xAy + ∂yAx
which is the z-component of −∇× ~AFleshing it all out:
Fµν =
0 Ex Ey Ez−Ex 0 −Bz By−Ey Bz 0 −Bz−Ez −By Bx 0
(180)
For what it’s worth, if we raise the indices, we get:
Fµν =
0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BzEz −By Bx 0
(181)
Knowing nothing else about these fields except how they are generated, Inote:
∂λFµν + ∂µFνλ + ∂νFλµ = 0 (182)
This is known as a “Bianchi Identity” and similar expressions end up being verypowerful in GR.
You can check this if you like, but the first term simply produces:
∂λ∂µAν − ∂λ∂νAµ
and all 6 final terms end up canceling algebraically.This looks like 64 different identities (4 values each for µ, ν, λ), but in reality,
it’s far fewer, since permutations don’t matter.µ = ν = λ = 0 yields a trivial result. Zeroes all around.µ = 0, ν = 0, λ = 1 yields:
Ex − Ex = 0
which it does, of course. In other words, the only interesting identities are thosefor which all three indices are different.
µ = 0, ν = 1, λ = 2 yields:
∂yEx + Bz + ∂x(−Ey) = 0
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Re-arranging, and this is simply the z component of:
∇× ~E = − ~B (183)
The free-field version of Faraday’s law of inductance. The other two permuta-tions with µ = 0 yield the other two components.
Likewise, µ = 1, ν = 2, λ = 3 yields:
∂zBz + ∂xBx + ∂yBy = 0
or in the more familiar form:∇ · ~B = 0 (184)
Gauss’s law for magnetism.
7.5 The Dynamics of the Free-Field Potential
What do we learn from the free-field Lagrangian? Well, clearly we only have“dynamic” terms, and thus we care about:
∂L∂(∂µAν)
The easiest thing to note is that for any (asymmetric) combination of µ &ν, There are eight relevant terms in the Lagrangian, 4 from FµνFµν
, and 4from F νµFνµ. Thus, taking the derivative (and not worrying about signs), weget 8 terms, half of which have a ∂µAν , and half of which have a ∂νAµ, or 4combinations of Fµν . Thus:
∂L∂(∂µAν)
= −Fµν
Or, in the free field:−∂µFµν = 0 (185)
For ν = 0, this yields:−∇ · ~E = 0
and for ν = i, we get:
−∇× ~B + ~E = 0
Of course, not all fields are free. We can imagine what happens when weadd a current, or equivalently, when we look at the interaction term. In thatcase, the RHS of the Euler-Lagrange equation:
∂L∂Aν
= −∂AνJν∂Aν = −gjν
or∂µF
µν = gjν
99
By construction:∂µ∂νF
µν = 0
so we have∂µj
µ = 0
as, of course, we must for any good conserved current.Plugging in everything, we get our final two Maxwell equations:
∇ · ~E = ρ (186)
and
∇× ~B = ~E + ~J (187)
So from U(1) gauge invariance, we could, if we wished, derive all of classicalelectromagnetism.
7.6 Lorentz and Coulomb Gauge
We started this discussion by noting that the free-field electromagnetic potentialhas a very nice invariance, and we proceeded to show that transformations ofthe form:
Aµ → Aµ + ∂µλ
can be applied arbitrarily to yield the same physical observables.
7.6.1 Lorentz Gauge
This gives us an enormous amount of freedom. For example, for an arbitraryAµ field, I can imagine adding a gauge field such that:
∂µAµ + λ = 0
Basically, this is just a matter of solving the Poisson equation for the gaugefield, λ.
This means that we can arbitrarily find a solution for Aµ (for a time and
spatially varying ~E & ~B field) such that:
∂µAµ = 0
This is known as Lorentz Gauge.
7.6.2 Coulomb Gauge
We can take this a step further, but at the cost of breaking Lorentz invariance.By breaking up the Lorentz condition into two parts, we can solve:
∇ · ~A = 0
which equivalently yieldsA0 = 0
100
and thus we can arbitrarily set:
A0 = 0
for free fields.
7.7 Solution to the Classical Free Electromagnetic Field
You know what comes next. We have our free-field Lagrangian for photons.Let’s solve it! Applying the Euler-Lagrange equations, we immediately get:
∂µFµν = 0 (188)
or equivalently:∂µ∂µA
ν − ∂ν∂µAµ = 0
Here’s where Coulomb vs. Lorentz gauge comes into play. We’re going to gowith Lorentz, as it makes things much simpler, and the 2nd term drops outentirely. Thus, we get (in Lorentz gauge):
∂µ∂µAν = 0 (189)
I totally know how to solve this! This is just 4 copies of the massless Klein-Gordan equation. But those solutions aren’t entirely independent. After all,if I solve 3 of them the Lorentz gauge condition clearly allows me to solve theothers.
I’m going to assume the form:
Aµ(x) =
∫
d3p
(2π)31
√
2E~pε(r)µ (~p)
[
a(r)~p eip·x + a
(r)∗~p e−ip·x
]
(190)
Where ε(r) are each of 4 4-vectors representing the polarization. And where:
E~p = |~p| (191)
because, after all, we have a massless particle.Finally, note that there is an implicitly sum over 0-3 for r. Yes, I realize
that this is contrary to the normal convention (because both r’s are upstairs),but I don’t care.
For simplicity, let’s consider a radiation wave propagating in the +k direc-tion. In that case, the Lorentz condition becomes:
A0 − ∂zAz = 0
or equivalently:∑
r
ε(r)0 E~p − ε
(r)3 pz = 0
101
It is clear that there are two independent polarization modes which satisfythe Lorentz condition:
ε(2) =
0010
and ε(1) which is defined similarly.More generally, for any momentum, there will be two polarization modes
defined perpendicular to the direction of motion. More generally, we have:
ε(r)(p) · p = 0 (192)
for the transverse modes, and
ε(r) · ε(s) = ηrs (193)
generally.Note that even though only the r=1 & 2 modes make any obvious physical
sense. The particular components vary based on the direction that the momen-tum of the mode is pointing. However, for now, we’ll keep all 4 of the modes inmind.
Besides describing the field generally, we can also write down some importantquantities, like the energy density. I’ll leave it as an exercise to show that, interms of what we’ve already defined, we get:
E =1
2
∫
d3x(
~E · ~E + ~B · ~B)
(194)
exactly as you learned when you were knee-high to a grasshopper.
7.8 Quantizing the Photon Field
7.8.1 The A operator
Now that we’ve written the free-field photon expansion, it’s trivial for us toquantize the electromagnetic field. Namely, we have:
Aµ(x) =
∫
d3p
(2π)31
√
2E~pε(r)µ (~p)
[
a(r)~p eip·x + a
(r)†~p e−ip·x
]
(195)
Of course, as we saw in our discussion of fermionic fields, it isn’t quite suffi-cient to just write this down. After all, we’re going to need a few rules forcommutators and anti-commutators.
Before, we noted that the Lorentz condition, ∂µAµ = 0 meant that only two
physical polarizations were going to be relevant – the spacelike ones perpendic-ular to the direction of motion. But how do we impose this? Essentially, wemust demand that for any “good” state, |i〉, we have:
∂µA+µ |i〉 = 0
102
where the A+µ operator is the part of the photon operator with (confusingly)
the annihilation term. This further insures that:
〈f |∂µAµ|i〉
for any good state. This means that by construction, for a mode moving in thek direction, we get:
(a(3)~p − a
(0)~p )|i〉 = 0
(and we get a similar relationship for modes in any other direction).
7.8.2 The Hamiltonian
This is good because I’ve been holding off on writing down the commutationrelations:
[a(r)~p , a
(s)†~q ] = −ηrs(2π)3δ(~p− ~q) (196)
which is fine (if a bit weird), until we write out the Hamiltonian:
H =
∫
d3p
(2π)3E~p
(
a(i)†
~p a(i)~p − a
(0)†~p a
(0)~p
)
(197)
where there is an implicit sum in the space-like direction/ This looks worse thanit is. After all, it’s clear that for any particular mode (say in the z-direction),the 0 and 3 terms exactly cancel, so we get a positive definite contribution.
7.8.3 The Photon Propagator
As we’ve already seen, we need to compute the overall photon propagator. Thegeneral form is exactly what you’d expect. Indeed, the whole reason that Ichoose to do this analysis in Lorentz gauge is that the propagator has a par-ticularly simple form compared to Coulomb gauge. See the Tong notes if youdon’t believe me. In short, we get:
〈0|T Aµ(x)Aν (y)|0〉 =
∫
d4p
(2π)4−iηµνp2
e−ip·(x−y) (198)
7.8.4 EM Interaction Term
We’ve already seen that the general form of the EM interaction term in theLagrangian is:
LInt = −gjµAµWe now have a quantized version of both the electron current and the photonfield. We thus get the interaction term:
HInt = eψγµAµψ (199)
where I threw in the electron charge for good measure.
103
7.9 “Deriving” the Feynman Rules
I’m not actually going to define the Feynman rules for QED, but I will motivatethem. Consider the interaction Hamiltonian above. Expanding everything out,we get:
HInt = e
∫
d3x
[
∫
d3p
(2π)31
√
2E~p
(
b†~pu(s)(p)e−i~p·~x + c~pv
(s)(p)ei~p·~x)
]
×γµ
×[
∫
d3q
(2π)31
√
2E~q
(
b~qu(r)(q)ei~q·~x + c†~qv
(r)(q)e−i~q·~x)
]
×[
∫
d3p
(2π)31
√
2E~kε(t)µ
(
a~kei~k·~x + a†~k
e−i~k·~x)
]
Looks pretty complicated, right? Well, it is. But suppose we are looking at aparticular vertex in a Feynman diagram. Say, for example, that we’re lookingat one in which an electron is annihilated (b), and a new electron is created (b†)along with a photon (a†)?
We get a bunch of contributions. The integral over space and the exponentsgive us a delta function of the form:
δ(p+ k − q)
In addition (and forgetting about factors of 2π and integrations), we get a bunchof terms which look like:
b†~pu(s)(p)γµb~qu
(r)(q)ε(t)µ a†~k
or, collecting terms more reasonably, we get:
[u(s)(p)γµu(r)(q)]b†~pb~qa†~kε(t)µ
We’ll see how these play out in the Feynman rules, but I think you’ll find therules more sensible at this point.
7.10 QED Rules
It sure does.But now we’re in a position to actually write down the Feynman rules for
electrodynamics:
1. Labels: Label the incoming momentum and energies p1...pn, and corre-spondingly, the spins s1...sn. Similarly, label the internal lines, k1..kn,and r1..rn.
2. External Lines: We’re making an integral (to compute the scattering am-plitude). For each:
104
• Outgoing electron: u(s)(p)
• Incoming electron: u(s)(p)
• Outgoing positron: v(s)(p)
• Incoming positron: v(s)(p)
• Outgoing photon: εµ∗
• Incoming photon: εµ
In reality, the index (especially on the photons) will be related to thecorresponding index at the related vertex.
3. Vertex Factors: Each vertex gets a factor of:
igγµ(2π)4δ(∑
p)
where the index, µ relates to the photon coming out of it, and the sum isdone in the usual way. In order to simplify later calculations, you shouldput the γ factor between the incoming and outgoing fermions in the form:
[uγµu]
4. Propagators: We’ve already written these. They are:
Electrons and positrons:i(γµkµ +m)
k2 −m2
Photons:−iηµνk2
5. Integrate: We get a factor of
∫
d4k
(2π)4
for each internal line.
6. Cancel the Delta Function:
Remove the factor of (2π)4δ(∑
p), and we’re left with −iAfi.
7. Antisymmetrization: Include a minus sign between diagrams that differonly in the interchange of two incoming or outgoing electrons or positronsor of an incoming electron with an outgoing positron (or vice-versa).
Note: These factors look fairly complex, but an easy way to keep all of theterms contracted with their appropriate partners is to follow each fermionic linebackwards.
105
7.11 Example: Electron-Electron Scattering
Maybe it’s not so obvious so far. Let’s clear things up a bit by doing someexamples. Let’s begin with an obvious example: electron scattering.
First, let’s draw the requisite Feynman diagrams and do step 1 of our Feyn-man calculus:
p1, s1
p2, s2
p′1, s′1
γ, eµk
p′2, s′2
e e
e e
Yes. I really am too lazy to use proper subscripts. Note that this is one oftwo possible diagrams. The other has p3, s3 twisted with p4, s4 (and so clearlycontributes a minus sign). I’m going to write everything in shorthand. So, u(1),
really means u(s1)~p1
.Step 2& 3:
−g2(2π)8[u(3)γµu(1)][u(4)γνu(2)]δ(p3 + k − p1)δ(p4 − k − p2)
Step 4:We have 1 propagator and it’s a photon. Note that there are two indices.
That’s good because the two vertices on either end each have an index. Afterincluding it, we have:
ig2(2π)8[u(3)γµu(1)][u(4)γνu(2)]δ(p3 + k − p1)δ(p4 − k − p2)ηµνk2
Step 5:We have one internal line, so we do one integration:
ig2(2π)8∫
d4k
(2π)4[u(3)γµu(1)][u(4)γνu(2)]δ(p3 + k − p1)δ(p4 − k − p2)
What’s this? Well, as you may recall (or if you wish to quickly prove), we cansquare any of the γ matrices such that:
γµγµ = ηµµ
106
or, in other words, we can treat the matrices themselves as components of avector, and use the standard index raising and lowering operations, exactly aswe’ve seen, such that:
γµ = ηµνγν
Step 6:For the diagram we’ve drawn, we get:
Afi = − g2
(p4 − p2)2[u(3)γµu(1)][u(4)γµu(2)]
Step 7:Adding in the other diagram, we get:
Afi = − g2
(p1 − p3)2[u(3)γµu(1)][u(4)γµu(2)]+
g2
(p1 − p4)2[u(4)γµu(1)][u(3)γµu(2)]
Give me all of the spins and momenta, and I’ll give you a number!Of course, in reality, we may want to simplify this. We may want, for
example, to assume that we know nothing of spin and simply average over allpossible spin states. We’ll do more of this in a while, but for now, it’s importantto note that we’re capable of computing the amplitudes. From here, it’s mostlya lot of algebra.
7.12 Example: Electron-Positron Annihilation
As a final example, let’s consider electron-positron annihilation. In that case,the Feynman diagram looks like:
p1, s1
p3, s3
k, r
p2, s2
p4, s4
e e+
γ γ
This is different from the previous example in two ways. First, the outgoingparticles are both photons (bosons), which means that the second diagram, theone with the momenta switched, get a +1. Secondly, the mediator particle inthis case is an electron, which means that we’ll have a different propagator. So,let’s follow our Feynman rules.
Let’s make it simple and consider an electron and positron at rest. What hap-pens? Assuming that the photons are created along the z-axis, we get:
~p3 = (0, 0,m)
and the opposite for ~p4. This converts the denominator term to:
(p1 − p3)2 −m2 = −2m2
and an identical term for p1, p4.What about the rest? It’s important to remember that p1 (for example) and
ε4 (for example) are each just ordinary 4-vectors representing, respectively, theincoming momentum and energy of the electron, and the polarization of one ofthe outgoing photons.
There are many rules for combining, contracting, and re-ordering terms inFeynman’s slash notation. A few of the more useful ones (which you can verifyusing dummy vectors) are:
/a/b + /b/a = 2a · b
108
Since by our Lorentz condition, only the spacelike terms matter in the polariza-tion states (the transverse modes only), the dot product p1 · ε3 vanishes. Evenmore generally, p3 · ε3 = 0, so:
(
/p1− /p3 +m
)
/ε3u1 = /ε3(−/p1+ /p3
+m)u(1)
But wait! The Dirac equation, itself, helps us simply this:
(p1 −m)u(1) = 0
So/ε3(−/p1
+ /p3+m)u(1) = /ε3/p3
u(1)
Thus, we get the marginally simpler looking expression:
Afi = − g2
2m2v(2)
[
/ε4/ε3/p3
]
u(1)
Or, combining the other term:
Afi = − g2
2m2v(2)
[
/ε4/ε3/p3+ /ε3/ε4/p4
]
u(1) (200)
We’ve set ~p3 and ~p4, which means that we can evaluate /p3and /p4
direction.These yield:
/p3 = m(γ0 − γ3)
/p4 = m(γ0 + γ3)
and so our bracket expression reduces to:
m(
/ε4, /ε3γ0 − [/ε4, /ε3]γ3)
We’ve already seen that the anti-commutator is simply 2ε4 · ε3.What about the commutator term? Well, since we’re ignoring the timelike
component of polarization, we begin by computing /ε as:
/ε = −ε · γ =
(
0 −σ · εσ · ε 0
)
The commutation term can be thus written as:
[/ε4, /ε3] = 2i(ε3 × ε4) · Σ
where (Σ is the array of spin matrices).And thus:
Afi = −g2
mv(2)
[
ε4 · ε3γ0 + i(ε3 × ε4) · Σγ3]
u(1) (201)
109
This is as far as we can go without specifying the spin states of the electronpositron pair. But let’s suppose we’re interested in the superposition state:
(↑↓ − ↓↑)√2
which you may recognize as the singlet state.The scattering amplitude is then also a superposition of the two states. For
example, for ↑↓ (the electron, arbitrarily, is the first arrow), we get:
u(1) =√
2m
1000
andv(2) =
√2m(
0 0 1 0)
Now we can compute:v(2)γ0u(1) = 0
and (in a slightly more complicated term):
vΣγ3u(1) = −2mk
yielding:A↑↓ = −2ig2(ε3 × e4)z (202)
The direction comes into play because we specified that that photons are beingemitted in the z-direction. Naturally, the amplitude is maximized when thepolarization is normal to that direction for both photons.
It is straightforward to show that we get a negative of the same term for theother term in the singlet, yielding:
Afi = −2√
2ig2(ε3 × ε4)z (203)
I will point out that since the system is spin zero, it must remain so, and thusthere are two combinations of photon polarization state which must be writtenas a superposition of one another. This yields a net effect of −i
√2 in our final
calculation. Thus:Afi = −4g2 (204)
In case you’re wondering what all of this amounts two, I’ll tell you. Thecoupling constant, g is related to the fine structure constant via:
g =√
4πα (205)
After some very straightforward analysis, we get:
dσ
dΩ=
1
v
( α
m
)2
(206)
as the cross-section for electron-positron annihilation.
110
7.13 Averaging over Spins
You may have noticed that it’s kind of a pain to directly specify the spin ofevery particle in the system. Oftentimes, they are completely randomized. Wecould, of course, figure out the cross section for every possible spin state to everypossible spin state and then simply average accordingly:
〈|Afi|2〉 = average over initial spins, sum over final spins
However, it would be easier if we could simply average beforehand.Let’s consider our electron scattering amplitude:
Afi = − g2
(p1 − p3)2[u(3)γµu(1)][u(4)γµu(2)]+
g2
(p1 − p4)2[u(4)γµu(1)][u(3)γµu(2)]
Squaring, we get 3 terms, but all are of the approximately term:
The pre-factors aren’t so tough (and are, of course, spin independent anyway),but the spinor terms are complicated. We can imagine combining the 1st & 3rdterms with each other and the 2nd and 4th, and in each case, we’re going to geta combination which looks like:
G ≡ [u(a)Γ1u(b)][u(a)Γ2u(b)]∗
where the Γ is just a generic combination of different 4×4 γ matrices (one eachin covariant and contravariant form).
We first note that we can get rid of the complex conjugate with a transposeof the second term.
G = [u(a)Γ1u(b)][u(b)Γ2u(a)]
Now, we sum over all possible spins of b (because while the ordering above isimportant, the parentheses aren’t):
∑
sb
u(sb)(pb)u(sb)(pb) = (/pb +mb)
which an outer product calculation that you’ve actually done on your homework.So:
∑
sa
∑
sb
G =∑
a
u(a)Γ1(/pb +mb)Γ2u(a)
That middle stuff is just a well-defined (and spin independent) 4 × 4 matrix.
111
We can rewrite all of this as:
∑
sa
∑
sb
G =∑
a
u(sa)(pa)Qu(sa)(pa)
= Qij
∑
sa
u(sa)(pa)u(sa)(pa)
ji
= Qij(/p+ma)ji∑
sa
∑
sb
[u(a)Γ1u(b)][u(a)Γ2u(b)]∗ = Tr
[
Γ1(/pb +mb)Γ2(/pa +mb)]
The only complication now involves keeping track of all of the terms. Thereare no spins left at all. Typically, since we want the average over all spins,applying this trick involves a factor of 1/4 in front of the amplitude.
112
8 The Electroweak Model
• Gross 13.1-13.2 15.4
Here’s a neutron decay, just to show that I know how to draw it!
W−
u d d
u d u e νe
8.1 SU(2) Local Gauge Invariance
We’ve seen the secret recipe for figuring out gauge fields (and fundamentalforces):
1. Assume that the Lagrangian for a particle is invariant under some sort oflocal gauge transformation
2. Find the interaction Lagrangian which makes the cross terms go away.
3. Add the free-field Lagrangian for the (vector) mediator
This was such a success in E&M that we’ll want to try it with the weak field.So we’ll imagine that we have some wave-function (could be a Dirac spinor, ora complex field of some sort), on which we will apply the local gauge transfor-mation:
ψ → ψe−iθ(j)(x)σj
(207)
This is an SU(2) transformation. SU(2) stands for “special unitary” whichbasically means that all of the elements of the SU(2) group have the propertythat they are all of the 2 × 2 unitary matrices (Hermitian = inverse) with adeterminant of 1. You can verify that for any arbitrary vector of angles, ourgenerating phase term satisfies this requirement.
If that’s the case, then the “wave-function” must be a 2-component thing:
ψ =
(
ψ1
ψ2
)
113
Each component of this object may itself be, say, a bispinor, but basically, theψ1 and ψ2 states represent different isospins. Various doublets include:
(
e−
νe
)
or
(
du
)
On this transformation,ψ†ψ
is clearly unchanged.When we apply this transform, we get:
∂µψ → e−iθ(j)(x)σj
∂µψ − ie−iθ(j)(x)σj
ψσj∂µθ(j)
Or more generally, we can imagine defining a new vector field which obeys thetransformation:
W (j)µ →W (j)
µ +i
g∂µθ
(j)
and a derivative:Dµ ≡ ∂µ − ig
∑
W (j)µ σj
such that terms like:Dµψ
†Dµψ
are invariant upon the gauge transformation. This is what we did in E&M, afterall.
But there’s a problem. The exponent term doesn’t actually commute withthe covariant derivative. We actually need to define the W field gauge transformas:
W (j)µ →W (j)
µ − 1
g∂µθ
(j) + ǫjklW(k)µ θ(l) (208)
As a result, we find that the W field Lagrangian is still:
LW = −1
4F (i)µν F
µν(i) (209)
but with the crucial difference:
F (i)µν = ∂µW
(i)ν − ∂νW
(i)µ − gǫjklW
(k)µ W (l)
ν (210)
This doesn’t look too bad until you realize a) The free Lagrangian has 4-th orderterms in W , and b) There are no 2nd order terms in W with no other terms.
What a) and b) mean, respectively, are that a) We have a massless W field(and observationally, the W’s and Z’s have mass) and b) it’s non-trivial to createa “free” solution. In fact, though, we can ignore higher order W 3 and higherterms at low energy. The solution would simply be the same as for the photons.
114
8.2 Spontaneous Symmetry Breaking
So far, we’re not doing so well. The theory predicts a massless mediator particlefor the weak force (although it does correctly predict three of them), because ofthe cross terms, it clearly becomes complicated at high energies.
To make progress, we’re going to have to take a few steps back.Let’s imagine a scalar field with the Lagrangian:
L = ∂µφ∂µφ∗ − V (φ) (211)
V (φ) = −µ2|φ|2 + λ|φ|4 (212)
Note that normally, φ is reserved for real valued scalar fields. I’m using itfor a complex field in this case because the complex part will essentially be“swallowed” by what follows. Note, too, the dimensionality of the terms in thepotential. µ has units of energy, and λ is dimensionless.
This potential is known as the “Mexican Hat” potential (for reasons thatbecome clear when you plot it). It is somewhat strange because the “masslike”term (the one with the quadratic part of the Lagrangian) has the wrong sign.Don’t worry about how this Lagrangian arises. For now (and indefinitely,frankly, since we still don’t have an answer), it just is.
The ground state can be found at:
|φ|GS = u ≡√
−µ2
2λ(213)
And at the ground state:
Vmin = −µ4
4λ(214)
I know. You’re impatient to know what this has to do with the weak force.So far, we just have a scalar field.
Classically, we’d expect the scalar field to drop to it’s vacuum state. Theproblem is that the vacuum state is a completely arbitrary value, since the fieldcan be at whatever phase it likes, each with equal probability:
〈0|φ|0〉 = ueiθ
Note, that originally the system had a U(1) symmetry; it no longer does. Thisis known as spontaneous symmetry breaking.
Why is this a big deal? Well, remember before, when we first derived elec-tromagnetism? We started with a free field Lagrangian, and we made the sub-stitution:
∂µ → ∂µ + igAµ
and the interaction part came out naturally. More generally, we found:
L = (∂µ + igAµ)φ(∂µ − igAµ)φ∗ − V (φ) − 1
4FµνF
µν
where previously we used V (φ) = m2φ2, but now we use the form above.
115
But now consider writing the φ field as a perturbation around the minimum:
φ = (u +R)eiθ
This is completely general. Working through everything, we get:
L = ∂µR∂µR− 2µ2R2 − e2u2AµA
µ − 1
4FµνF
µν − V (R) (215)
From one field, we now have 2, each of which are massive, and where I’veswallowed all terms not to second order into V (R).
8.3 Electroweak Theory (But without the handedness)
Now that we’ve got an idea of how spontaneous symmetry works, let’s do itfor real. Well, almost. I’m not going to include the handedness inherent in theweak theory. The overall theory will work without it.
8.3.1 The Electroweak Fields at Rest
So let’s first imagine that we have two types of particles, 1) A Higgs particle(which, even though it will initially be a complex doublet, we’ll still denote asφ):
φ =
(
φ1
φ2
)
You may think that this means that the Higgs will be charged. You wouldbe wrong. We will generically assume that the Higgs interacts under somepotential, V (φ), which I’ll specify later.
2) A Dirac particle.Naively, we’d expect the Lagrangian for the entire universe to look like:
L = ∂µφ†∂µφ− V (φ) + iψγµ∂µψ −mψψ
But now suppose we allow a SU(2) ⊗ U(1) local phase transition in bothfields:
ψ → ψ exp
[
i
(
g′θ
2− g
~σ · ~θ2
)]
(216)
φ→ φ exp
[
i
(
−g′ θ2− g
~σ · ~θ2
)]
(217)
We get the factor of two in the coupling constant because there are two fields.Further, the dimensionless prefactors, g and g′ are inserted because both fieldscontribute to the final Lagrangian, and it’s only the combination of g±g′ whichends up mattering in the end. Likewise, since there are 3 Pauli matrices and 3relevant angles, I’m treating both as a 3-vector, and dotting them.
116
As defined then define the corresponding Gauge fields as transforming as:
A(i)µ → A(i)
µ − 1
g∂µθ
(j) + εjklA(k)µ θ(l) (218)
and
Bµ → Bµ +1
g′∂µθ (219)
This might look a bit confusing, but remember that these represent 4 vectorfields (all massless, as it turns out, at least initially):
LA,B = −1
4F (i)µν F
(i)µν − 1
4GµνG
µν (220)
F (i)µν ≡ ∂µA
(i)ν − ∂νA
(i)µ − gεijkA
(j)µ A(k)
ν (221)
Gµν = ∂µBν − ∂νBµ (222)
Even though Bµ is the field generated by the U(1) transformation, you shouldremember that it is not the photon.
We’ve seen this sort of thing before (although perhaps in not so complexa form). We get around all of this by redefining the derivative operators to“offset” the gauge contribution:
D(ψ)µ = ∂µ − i
g
2~σ · ~Aµ + i
g′
2Bµ (223)
D(φ)µ = ∂µ − i
g
2~σ · ~Aµ − i
g′
2Bµ (224)
From all of this we get the final (locally SU(2) ⊗ U(1) gauge invariant La-grangian):
L = D(φ)µ φ†D(φ)µφ−V (φ)+iψγµD(ψ)
µ ψ−mψψ− 1
4F (i)µν F
(i)µν− 1
4GµνG
µν (225)
So far, so good.Of course, multiplying it all out (including the D contributions), we get the
various interaction terms that we’ve come to know and love.However, the Higgs has a rather strange form and appears to be charged.
What’s more, we have 4 massless vector fields, not simply 1. How to resolvethis?
8.3.2 Symmetry Breaking in Electroweak theory
In order to really understand the electroweak theory, we need to introduce anexplicit potential for the Higgs field:
V (φ) = −µ2φ†φ+ λ(φ†φ)2 (226)
The φ field will relax eventually, and clearly it can lie in any direction in phasespace, and will produce an equivalent contribution to the energy regardless of
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which part of the doublet it occupies. Without a loss of generality, we can selectthe angle such that:
〈0|φ|0〉 =
(
0
v/√
2
)
where
v =
√
µ2
λ
This is a convenient rotation and completely arbitrary, but will prove usefullater in the derivation.
We can now consider perturbations away from the minimum, such that:
φ =
(
0(v+η)√
2
)
where η is the (single and real-valued) dynamical variable describing the Higgsfield around equilibrium. Because it only has this one degree of freedom, it’s areal-valued single scalar particle – at least in its simplest in its simplest incar-nation. Because it hasn’t actually been detected, it may be that the Higgs is abit more complex.
What happens when we expand the φ part of the Lagrangian in terms of vand η?
L = Lψ+1
2∂µη∂
µη−µ2η2+LA,B−v2
8
(
g2 ~Aµ · ~Aµ + g′2BµBµ + 2gg′BµA
(3)µ)
+Lint(227)
Plus additional terms that are either constant or higher order in η (which aren’trelevant at low energies) or constants (in v, for example), which don’t show upin the dynamics of the system.
Wow! Our Higgs has acquired a mass (=√
2µ)! But that’s not all.
8.3.3 The Higgs Mechanism
We’re not going to talk about how the fermions gain mass, though presumablyit’s through a direct coupling term relating the Higgs to the fermionic fieldwhich doesn’t come into play until the Higgs relaxes into the true vacuum state.However, the most important terms in the Lagrangian for now are the A’s andB’s:
LEW = −v2
8
(
g2 ~Aµ · ~Aµ + g′2BµBµ + 2gg′BµA
(3)µ)
It’s pretty clear that the A(3) and B fields are coupled to one another in theLagrangian, and that if we are clever, we can redefine them such that:
(
A′µ
Zµ
)
=
(
cos θW sin θW− sin θW cos θW
)(
Bµ
A(3)µ
)
(228)
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Where the Weinberg angle, θW , is defined as
tan θW =g′
g(229)
Plugging this in to the Electroweak Lagrangian yields:
LEW =g2v2
8
(
W+µ W
+µ +W−µ W
−µ +1
cos2 θWZµZµ
)
(230)
This is awesome! In our new rotated state, we’ve found that 3 of the vectorfields are massive, and related such that:
Mw =gv
2(231)
and
MZ =MW
cos θW(232)
The Weinberg angle must be determined experimentally, based on comparingthe following interactions:
e−
νe
e+
Z0
νe
e−
νe
νe
W−
e−
e−
νe
W−
νe
e−
Experimentally, θW ≃ 29.This is outstanding, because we immediately get other values of physical
interest. For example (and you have to follow through the algebra a bit):
e = g sin θW (233)
This can be found by either looking at the coupling of the current terms inW± to the new A′ field, or by looking at the coupling of the charged leptonsthemselves. At any rate, from the measured unit charge,e = 0.303 in naturalunits, we get
g = 0.65 ; g′ = 0.36 (234)
Of course, also find:MZ
MW= 1.14 (235)
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an observed ratio to ridiculous accuracy.It actually gets better. Consider the Neutron decay relation:
n→ p+ e− + ν
Fermi assumed that this relation was a 4-particle interaction (with no mediator)of the form:
L = −GF√2
[
ψpγµ(1 − 1.26γ5)ψn
] [
ψeγµ(1 − γ5)γν]
where the γ5 comes into play from the fact that neutrinos only occur in left-handed combinations (to be discussed in the next section). The 1.26 comes fromthe fact that neutrons and protons are composite particles (and the weak forcehas a mediator). Clearly, the neutron decay rate should be proportional to G2
F ,and thus the term is well-determined. It is:
GF = (299.3 GeV )−2
But the above interaction Lagrangian can be rewritten in terms of up and downcreation and annihilation operators. We may thus write:
MW =g2√
2
8GF= 80.03 GeV (236)
which is precisely the measured value.What’s more, since
MW =gv
2
we can solve for the electroweak symmetry breaking scale:
v =
√
µ2
λ=
2MW
g= 246 GeV
But we can finish up by noting that:
MH =√
2λv = 347√λ GeV
which is why we have some idea what the energies are for the Higgs.
8.4 Handedness in the weak force
I deliberately wanted to leave off an observationally important detail when talk-ing about symmetry breaking: the weak force has a preferred orientation. Let’sstart by defining a new composite γ matrix:
γ5 ≡ iγ0γ1γ2γ3 =
0 0 1 00 0 0 11 0 0 00 1 0 0
(237)
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This is a particularly handy object because we can break any bispinor into twoseparate components:
ψL =1
2(1 − γ5)ψ (238)
ψR =1
2(1 + γ5)ψ (239)
As an exercise, you should try to show that a massless spin up particle movingin the +z direction is right handed, and one moving in the opposite direction isleft handed.
The reason that this comes into play is that only left-handed multiplets obeythe SU(2) gauge invariance. Right-handed multiplets only obey the U(1) gaugeinvariance:
R → eig′θR
The implication of this is that in weak interactions, only left-handed neutrinosare created, and contrarily right-handed anti-neutrinos. We do not know whythis is, and the result has to be put in by hand.
Of course, since neutrinos are massive, there is a Lorentz frame in which aparticular neutrino is right-handed. They simply aren’t created that way.
8.5 Quantized Weak Fields
We’re not going to do any weak calculations in any detail, but at very least,you should be able to do so. So, there are a few corrections to our Feynmandiagrams:
1. Because they are massive, W and Z particles have 3 possible polarizationstates. Thus:
εµpµ = 0
exhausts the degrees of freedom. Otherwise, external W’s and Z’s (whichdon’t exist, typically, as they are short lived), would look the same asexternal photons.
2. The propagator for W and Z particles is:
−iηµν − qµqν/M2
q2 −M2
as you might expect. However, because these particles are so massive, wecan typically ignore the q terms and simply use
∼ iηµνM2
3. Weak vertices get a factor of:
−igW2√
2γµ(1 − γ5)
It is the appearance of the 1−γ5 which guarantees that we’re only includ-ing left-handed interactions.
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9 Renormalization, Revisited
• Gross: Chapters 11, 16
We have primarily focused on first-order (2 vertex) diagrams. We have foundthat doing the integrals to compute the scattering amplitudes were generallytrivial (even if reducing those amplitudes to simple algebraic forms were not).We have ignored higher order diagrams in part because the contributions typi-cally scale as gn, where n is the number of vertices in a diagram and (in QEDat least), g << 1. However, we’ve seen that even for “simple” calculations,like the vacuum energy density of QED or our ψ − φ theory, terms typicallyapproach infinity. We need a method of dealing with this, and that method willbe renormalizatin.
Consider the following diagram:
p
k′k
p′
On the face of it, this diagram looks absurd. After all, we start with asingle electron and end with a single electron. Clearly, p = p′, and there is noscattering amplitude to compute. Humor me anyway.
Let’s apply our Feynman rules to compute the scattering amplitude. Wehave, of course, already done rule #1. Applying Rules # 2& 3, we get:
−g2(2π)8u(s′)(p′)γµγνu(s)(p)δ(p′ − k − k′)δ(k + k′ − p)
We have two propagators, which combine (Rule #4) to yield:
−g2 (2π)8
k2u(s′)(p′)γν
/k′+m
k′2 −m2γνu(s)(p)δ(p′ − k − k′)δ(k + k′ − p)
Rule # 5 gets rid of one of the δ− functions, since k′ = p− k.So we get:
−g2
∫
d4k
k2δ(p′ − p)u(s′)(p′)γν
/p− /k +m
(p− k)2 −m2γνu(s)(p)
This means:
Afi = −ig2 1
(2π)4
∫
d4k
k2u(s′)(p′)γν
/p− /k +m
(p− k)2 −m2γνu(s)(p)
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For large k this scales as k−3, and thus k−3d4k diverges lograthmically with k.We get an infinite contribution, in other words.
You might reasonably ask what physical interpretation Afi yields in thiscase. Well, consider the electron viewed from its rest frame. In that case, wehave:
ψ(t) = ψ(0)e−imt
The mass in the exponent represents the “physical” mass, the only one measur-able by experiment. Thus, in our case, the scattering amplitude is essentiallythe term resulting from the perturbation:
HI = δmψψ
This is a perturbation on the self-energy of the electron. This, we really measurethe combination:
mphys = mbare + δm (240)
where: So let’s consider a spin-up particle at rest, in which case:
u = N
1000
and p0 = m, with all other components equal to zero. That simplifies thingsconsiderably. In particular, it allows us to compute the total perturbation tothe energy as:
δm = e2(∫
d4k
(2π)41
k2γµ
(/p− /k) +m
(p− k)2 −m2γµ
)
00
(241)
This simplifies to:
δm = e2
(
∫
d4k
(2π)41
k2(k20 − 2mk0 − ~k2)
γµ(/p− /k +m)γµ
)
00
Consider the useful relations:
γµ/aγµ = −2/a
γµγµ = 4
which reduces:
γµ(/p− /k) +m
(p− k)2 −m2γµ = −2/p+ 2/k + 4m
and thus:
(
γµ(/p− /k) +m
(p− k)2 −m2γµ
)
00
= −2p0 + 2k0 + 4m = 2(m+ k0)
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so
δm = 2e2∫
d4k
(2π)4m+ k0
k2(k20 − 2mk0 − ~k2)
(242)
If we then do a coordinate transformation:
w =k
m
we get:
δm = 2me2∫
d4w
(2π)41 + w0
w2(w20 − 2w0 − ~w2)
(243)
Recognizing that this is spherically symmetric (and defining |~w| = W , we get:
δm
m=
e2
2π3
∫ ∞
0
dW
∫ ∞
−∞dw0
W 2(1 + w0)
(w20 −W 2)(w2
0 −W 2 − 2w0)
This integral blows up, and not just because of the 4 roots in the denominator.If we integrate over w0 using a standard contour integral, we pick up a factorof π/4, and two of our terms drop out (and w0 →W ), yielding (on the high Wside):
δm
m=
e2
8π2
∫ ∞
0
dW
W
If we put a simple truncation, this integral yields, ln(Wmax)Or, to combine everything, we get:
δm
m= C ln
(
kmaxm
)
and multiplying our units out correctly, we get:
δm = m3α
2π
(
kmaxm
)
(244)
For an electron, even if we cutoff the mass at the Planck scale, this only producesa ∼ 20% error. in δm. Of course, it means that mbare ≃ 80% of m, since m isthe experimental mass.
We found that we needed to renormalize because the final integral was ulti-mately proportional to k1. This is bad thing, since we’re integrating to infinity.However, it could be worse. We could find that the integral isn’t normalizableat all.
Our criterion is, take NB as the number of bosonic lines attached to a vertex(for QED, this is 1), and NF as the number of fermionic lines attached to avertex. Because they have different factors of k in the propagator, they aregoing to contribute differently. Computing
n = NB +3
2NF (245)
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(4 for QED) tells us whether we’re going to encounter trouble. Theories withn ≤ 4 are renormalizable, and those with n > 4 are not. As it happens, ourQED and weak theories to date lie just on the edge. Even the 4W vertices ofthe weak theory just make it in under the line.
Equalivalently, if the energy dimensionality of the coupling term of the theoryis less than E<0, then we’re likely going to be able to renormalize.
There are a number of tricks for renormalization in theory, and regularizationin practice. As we are out of time, I suggest you take a good look at Gross tosee how it’s done.
