Post on 26-Dec-2014
transcript
First to Second Quantization
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I. QUANTUM MECHANICS
Quantum field theory is based on the same quantum mechanics that was invented by
Heisenberg, Born, Schrodinger, Pauli, and others in 1925-26. Therefore, for completeness
and continuity, we will review first quantization, that is, (ordinary) nonrelativistic quantum
mechanics, before going on to second quantization.
A. The postulates
In any quantum mechanical process, the object of interest to calculate is the probability
amplitude - a complex number. The square of the modulus of the amplitude is proportional
to the probability that the process will occur. The postulates of quantum mechanics are
rules on how to calculate the amplitude for any process.
For a given physical system, there is a complete set of states that describes all of the
possible configurations of the system. A state, denoted by |Ψ〉, is a vector in some Hilbert
space. c|Ψ〉, c a constant, represents the same physical state. The inner product of two
states, |Ψ1〉 and |Ψ2〉, is written as 〈Ψ2|Ψ1〉. The amplitude for the process that takes the
system from the state |Ψ1〉 to |Ψ2〉 is just 〈Ψ2|Ψ1〉. The probability for this process to occur
is |〈Ψ2|Ψ1〉|2.With every physical observable A, there is associated a Hermitian operator A that acts
on the Hilbert space containing the states. If a measurement is made of the observable A,
the result of the measurement must be one of the real eigenvalues of A - no other values
are possible. The expectation value of the physical quantity A in the state |Ψ〉 is given by
〈Ψ|A|Ψ〉, where the state vector is normalized to 〈Ψ|Ψ〉 = 1.
Last, quantum mechanics provides for time evolution by requiring the states of the system
to satisfy the Schrodinger equation,
ihd
dt|Ψ(t)〉 = H|Ψ(t)〉. (1)
H is called the Hamiltonian operator and generates infinitesimal time translations. Suppose
it is time-independent, the formal solution to the Schrodinger equation is
|Ψ(t)〉 = exp(− i
htH
)|Ψ(0)〉 ≡ U(t)|Ψ(0)〉. (2)
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U(t) is the time evolution operator. Substituting Eq.(2) into Eq.(1),
ihd
dt|Ψ(t)〉 = ih
d
dtU(t)|Ψ(0)〉
= HU(t)|Ψ(0)〉= H|Ψ(t)〉 (3)
we obtain
H = ih
[d
dtU(t)
]U †(t) = −ihU(t)
d
dtU †(t). (4)
If we have a classical Hamiltonian system (say, a particle) to quantize, we may obtain
the Hamiltonian operator H using its classical Hamiltonian H(x, p) as a guide. Since both
the position x and momentum p of a particle are observable quantities, upon quantization,
they become operators x and p. We can turn H(x, p) into H by replacing x with x and p
with p. To complete the quantization, we must specify quantum conditions for x and p:
[x, p] = ih. (5)
x and p are defined such that their commutators are equal to ih times the corresponding
classical Poisson bracket. Since x and p do not commute, potential operator ordering prob-
lems arise in constructing H from H(x, p). For example, operator products xp and px are
inequivalent while xp and px are the same in the classical system. In general, we choose a
symmetric ordering, xp → (xp + px)/2.
If the quantum mechanical system we are trying to describe has no classical analog, such
as a system with spin, then we must guess what the Hamiltonian operator is and what the
quantum conditions are.
B. Heisenberg picture
So far we have been describing the Schrodinger picture, where the states of the system
are time dependent while the operators are not. In the Heisenberg picture, the states are
time independent. The operators carry the time dependence. The transformation between
the two pictures is done with the time evolution operator U(t). We want to obtain the same
results in both representations. In particular, the expectation of any operator A should be
the same:
〈Ψ(t)|A|Ψ(t)〉 = 〈Ψ(0)|U †(t)AU(t)|Ψ(0)〉 ≡ 〈Ψ(0)|AH(t)|Ψ(0)〉 (6)
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The time dependent operators in the Heisenberg picture are related to the Schrodinger
picture operators by
AH(t) = U †(t)AU(t). (7)
Since the states are now time independent, they do not have to satisfy any Schrodinger
equation. The dynamics is locked into the time dependent operators. They must satisfy an
operator equation of motion,
d
dtAH =
(d
dtU †
)AU + U †A
d
dtU
= U †U
(d
dtU †
)AU + U †A
d
dtUU †U
=i
hU †HAU − i
hU †AHU
=i
hHHAH − i
hAHHH
=i
h[HH, AH] (8)
Note that if AH commutes with HH, it is a constant of the motion. If the operator AH has
an explicit time dependence, then we must add ∂AH/∂t to the right-hand side of Eq.(8).
Equation (8) may also be obtained from the classical system by replacing ih times the
Poisson bracket with a commutator, just as in the quantum conditions, Eq.(5).
In summary, to quantize a classical system in the Heisenberg picture, we elevate observ-
ables to operators, specify quantum conditions, choose an operator ordering, and specify
a Hilbert space containing the time independent state vectors. To compute the quantum
dynamics, we must solve the operator equation of motion. We note that since xH and pH
are time dependent, we specify the commutator at equal times,
[xH(t), pH(t)] = ih. (9)
C. Harmonic oscillator in the Heisenberg picture
The classical Hamiltonian for the harmonic oscillator is
H =1
2p2 +
1
2ω2x2 =
1
2hω
(− i√
hωp +
√ω
hx
) (i√hω
p +
√ω
hx
). (10)
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To quantize this, we make x and p the operators x and p, and demand them to satisfy the
commutation relation, [x, p] = ih. It follows that
H =1
2p2 +
1
2ω2x2, (11)
dx
dt=
i
h[H, x] =
i
2h[p2, x] = p, (12)
d2x
dt2=
dp
dt=
i
h[H, p] =
iω2
2h[x2, p] = −ω2x. (13)
We note that
x(t) = A cos ωt + B sin ωt (14)
solves Eq.(13). And,
p(t) = −ωA sin ωt + ωB cos ωt. (15)
It follows that
√ωx =
√ωA cos ωt +
√ωB sin ωt,
1√ω
p = −√ωA sin ωt +√
ωB cos ωt,
and
√ωx +
i√ω
p =√
ω(A + iB) exp(−iωt),
√ωx− i√
ωp =
√ω(A− iB) exp(iωt).
Alternatively, we could express
H = hω(a†a +
1
2
), (16)
where
a =1√2
(√ω
hx +
i√hω
p
),
a† =1√2
(√ω
hx− i√
hωp
), (17)
with
[a, a†] =1
2[
√ω
hx +
i√hω
p,
√ω
hx− i√
hωp] = 1, (18)
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and
d
dta =
i
h[H, a] = iω[a†a, a] = −iωa,
d
dta† =
i
h[H, a†] = iω[a†a, a†] = iωa†. (19)
Substituting x =√
h/2ω(a + a†) and p = −i√
hω/2(a− a†) into the Hamiltonian operator,
we find
H =1
2p2 +
1
2ω2x2
=1
4hω[−(a− a†)2 + (a + a†)2]
=1
2hω(aa† + a†a)
= hω(a†a +
1
2
)(20)
Clearly if a state |n〉 is an eigenvector of H, then it must be an eigenvector of the number
operator N = a†a and vice versa.
N |n〉 = n|n〉,H|n〉 =
(n +
1
2
)hω|n〉. (21)
Next, consider the norm of the state a|n〉,
〈n|a†a|n〉 = n〈n|n〉 ≥ 0 and 〈n|n〉 > 0 ⇒ n ≥ 0. (22)
We will normalize the states |n〉 such that 〈n|n〉 = 1. The ground state has n = 0. It is the
state of lowest energy hω/2 and |0〉 satisfies
a|0〉 = 0. (23)
The first excited state has n = 1, the second n = 2, etc. The spectrum of H is En =
(n + 1/2)hω. The energy levels are evenly spaced with separation hω. To reach the excited
states from the ground state |0〉, we observe that
[N , a†] = [a†a, a†] = a†[a, a†] = a†. (24)
Thus,
N a†|n〉 = (a†N + [N , a†])|n〉 = (n + 1)a†|n〉 = (n + 1)λ|n + 1〉. (25)
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Similarly, we have
[N , a] = [a†a, a] = [a†, a]a = −a, (26)
N a|n〉 = (aN + [N , a])|n〉 = (n− 1)a|n〉 = (n− 1)µ|n− 1〉. (27)
λ and µ are constants that are fixed by making sure the states generated are properly
normalized.
λ|n + 1〉 = a†|n〉⇒ λ2 = 〈n|aa†|n〉 = 〈n|(a†a + 1)|n〉 = n + 1
⇒ λ =√
n + 1 (28)
µ|n− 1〉 = a|n〉⇒ µ2 = 〈n|a†a|n〉 = n
⇒ µ =√
n (29)
It follows that all the eigenstates of H can be generated by applying a† consecutively to |0〉.In general,
|n〉 =1√n!
(a†)n|0〉. (30)
The operators a† and a are called the ladder operators, or raising and lowering operators,
because they make states that march up and down the ladder of excitations. As we shall soon
see, free quantum field theories reduce to a collection of independent harmonic oscillators,
one for each energy-momentum. The raising and lowering operators of the collection of
oscillators provide a particle interpretation. To find a particle interpretation of any quantum
field we look for operators similar to a† and a in the quantum field theory.
D. Coordinate representation
The position x of a particle is a physical observable. Let the associated operator be x.
The eigenvectors of x satisfy
x|x〉 = x|x〉. (31)
The eigenvectors are normalized so that
〈x′|x〉 = δ(x′ − x), (32)
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where δ is the Dirac distribution. They are assumed complete,∫
dx|x〉〈x| = 1, (33)
and form a basis for the Hilbert space of states. If we expand a state vector |Ψ(t)〉 in the
position basis |x〉, the component of |Ψ(t)〉 in the |x〉 direction is 〈x|Ψ(t)〉, which is just a
number Ψ(x, t) - the wave function representing the state |Ψ(t)〉.
|Ψ(t)〉 =(∫
dx|x〉〈x|)|Ψ(t)〉 =
∫dx|x〉〈x|Ψ(t)〉 =
∫dxΨ(x, t)|x〉 (34)
|x〉 is the state of the system where the particle is at position x. 〈x|Ψ(t)〉 = Ψ(x, t) is the
amplitude for the system in the state |Ψ(t)〉 to also be in the state |x〉. Thus, |Ψ(x, t)|2dx
is the probability that the particle will be found in (x− dx/2, x + dx/2) when the system is
in state |Ψ(t)〉. Since the particle must be found somewhere,
1 =∫
dx|Ψ(x, t)|2 =∫
dx〈Ψ(t)|x〉〈x|Ψ(t)〉 = 〈Ψ(t)|Ψ(t)〉. (35)
In the coordinate representation, the Schodinger equation (1) becomes
ih∂
∂t〈x|Ψ(t)〉 = 〈x|H|Ψ(t)〉,
ih∂
∂tΨ(x, t) =
∫dx′〈x|H|x′〉〈x′|Ψ(t)〉
=∫
dx′〈x|H|x′〉Ψ(x′, t). (36)
〈x|H|x′〉 is called the matrix element of the operator H in the position basis. To compute
〈x|H|x′〉, we must determine the matrix elements 〈x|x|x′〉 of x and 〈x|p|x′〉 of p. Since |x〉is an eigenvector of x,
〈x|x|x′〉 = x〈x|x′〉 = xδ(x− x′). (37)
Suppose p has eigenvectors |p〉, then
〈x|p|x′〉 =∫
dp〈x|p|p〉〈p|x′〉
=∫
dp p〈x|p〉〈p|x′〉
=∫
dp
(−ih
∂
∂x
)〈x|p〉〈p|x′〉
= −ih∂
∂x
∫dp〈x|p〉〈p|x′〉
= −ih∂
∂x〈x|x′〉
= −ih∂
∂xδ(x− x′) (38)
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If the Hamiltonian is
H =1
2p2 + V (x), (39)
then
〈x|H|x′〉 =
[− h2
2
∂2
∂x2+ V (x)
]δ(x− x′). (40)
Substituting Eq.(40) into Eq.(36), we recover the familiar Schrodinger equation for the wave
function.
ih∂
∂tΨ(x, t) = − h2
2
∂2
∂x2Ψ(x, t) + V (x)Ψ(x, t). (41)
We note that
ihΨ∗(x, t)∂
∂tΨ(x, t) = − h2
2Ψ∗(x, t)
∂2
∂x2Ψ(x, t) + V (x)Ψ∗(x, t)Ψ(x, t),
−ihΨ(x, t)∂
∂tΨ∗(x, t) = − h2
2Ψ(x, t)
∂2
∂x2Ψ∗(x, t) + V (x)Ψ∗(x, t)Ψ(x, t),
from which we derive
∂
∂t(Ψ∗Ψ)− ih
2
∂
∂x
(Ψ∗∂Ψ
∂x−Ψ
∂Ψ∗
∂x
)= 0. (42)
From hereon, we set h = 1 and replace Ψ(x, t) with ϕ(x, t).
II. SECOND QUANTIZATION
To second quantize the Schrodinger equation,
i∂
∂tϕ(x, t) = −1
2
∂2
∂x2ϕ(x, t) + V (x)ϕ(x, t), (43)
we want to make the first quantized wave function ϕ(x, t) an operator ϕ(x, t), a time-
dependent field operator (so we will be working in the Heisenberg picture). Equation (43)
will then become an operator equation of motion,
i∂
∂tϕ(x, t) = −1
2
∂2
∂x2ϕ(x, t) + V (x)ϕ(x, t). (44)
The normalized eigenfunctions ϕn(x), with eigenvalue en, of the first quantized Hamilto-
nian,
−1
2
∂2
∂x2+ V (x), (45)
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are assumed to form a complete set, thus any solution ϕ(x, t) to the Schrodinger equation
(43) can be expanded in terms of the ϕn’s:
ϕ(x, t) =∑n
an(t)ϕn(x). (46)
In the first quantized system, ϕ(x, t) and ϕn(x) are wave functions and an(t) is just a number
times exp(−ient). After second quantizing, ϕ(x, t) becomes an operator. Making an(t) an
operator and leaving ϕn(x) a function,
ϕ(x, t) =∑n
an(t)ϕn(x) (47)
clearly solves Eq.(44).
In quantum mechanics, operator equations of motion take the form of Eq. (8),
∂
∂tϕ(x, t) = i[H, ϕ(x, t)]. (48)
Therefore, we must find a field Hamiltonian H and quantum conditions (equal-time com-
mutation relations) such that Eq.(48) reproduces Eq.(44). The commutator is the field
equivalent of [x(t), p(t)] = i for the first quantized system. In this field theory, ϕ(x, t) plays
the role of x(t). To write down the quantum commutator involving ϕ(x, t), we must find
what the momentum field π(x, t) conjugate to ϕ(x, t) is. Our strategy will be to treat the
first quantized system as a classical field theory, find an action principle that yields the
field equation (43), and use the action as a crutch to find the mometum field π(x, t) and
Hamiltonian H.
III. CLASSICAL MECHANICS
A. Methods of Lagrange and Hamilton
A classical mechanical system is described by a set of generalized coordinates q1, · · · , qN ,
the associated velocities, q1, · · · , qN , and a Lagrangian L[qi(t), qi(t), t]. The dot denotes the
time derivative d/dt. The Lagrangian governs the dynamics and is, at most, a quadratic
function of qi. The time integral
A[qi(t)] =∫ tb
tadt L[qi(t), qi(t), t] (49)
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over the Lagrangian along an arbitrary path qi(t) is called the action of this path. The path
qi(t) that is actually chosen by the system as a function of time is called the classical orbit,
qcli (t). It has the property of extremizing the action in comparison with all neighboring paths
qi(t) = qcli (t) + δqi(t) (50)
with fixed end points: δqi(ta) = δqi(tb) = 0. To express this property formally, we consider
the variation of A[qi(t)],
δA[qi(t)] ≡ {A[qi(t) + δqi(t)]−A[qi(t)]}lin
=∫ tb
tadt {L[qi(t) + δqi(t), qi(t) + δqi(t), t]− L[qi(t), qi(t), t]}lin
=∫ tb
tadt
[∂L
∂qi
δqi(t) +∂L
∂qi
δqi(t)
]
=∫ tb
tadt
[∂L
∂qi
− d
dt
(∂L
∂qi
)]δqi(t) +
∂L
∂qi
δqi(t)
∣∣∣∣∣tb
ta
=∫ tb
tadt
[∂L
∂qi
− d
dt
(∂L
∂qi
)]δqi(t) (51)
Here, repeated indices are understood to be summed - Einstein’s summation convention. For
the classical orbit qcli (t), δA[qi(t)] = 0 (Hamilton’s principle of least action) and we obtain
the Euler-Lagrange equations:∂L
∂qi
− d
dt
(∂L
∂qi
)= 0. (52)
There is an alternative formulation of classical dynamics which is based on a Legendre
transformed function of the Lagrangian called the Hamiltonian
H[qi(t), pi(t), t] ≡ qi(t)∂
∂qi
L[qi(t), qi(t), t]− L[qi(t), qi(t), t]. (53)
Its value at any time is identified with the energy of the system. The natural variables in H
are no longer qi(t) and qi(t), but qi(t) and the canonically conjugate momenta pi(t) defined
by the equations
pi(t) ≡ ∂
∂qi
L[qi(t), qi(t), t]. (54)
In order to specify the Hamiltonian H[qi(t), pi(t), t] in terms of its proper variables qi(t) and
pi(t), the equations (54) for pi(t) have to be solved for qi(t),
qi(t) = vi[qi(t), pi(t), t], (55)
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and inserted into Eq.(53), giving
H[qi(t), pi(t), t] = vi[qi(t), pi(t), t]pi(t)− L[qi(t), vi[qi(t), pi(t), t], t]. (56)
In terms of H, we have the canonical form of the action, Eq.(49),
A[qi(t), pi(t)] =∫ tb
tadt {qi(t)pi(t)−H[qi(t), pi(t), t]}. (57)
The variation of A[qi(t), pi(t)],
δA[qi(t), pi(t)] =∫ tb
tadt
[δqi(t)pi(t) + qi(t)δpi(t)− ∂H
∂qi
δqi(t)− ∂H
∂pi
δpi(t)
]
=∫ tb
tadt
{[qi(t)− ∂H
∂pi
]δpi(t)−
[pi(t) +
∂H
∂qi
]δqi(t)
}(58)
since δqi(ta) = δqi(tb) = 0. For the classical orbit, qcli (t) and pcl
i (t), δA[qi(t), pi(t)] = 0 and
we obtain the Hamilton equations:
qi(t) =∂H
∂pi
,
pi(t) = −∂H
∂qi
. (59)
These agree with the Euler-Lagrange equations (52) via Eq.(54), as can easily be verified.
pi =∂L
∂qi
=∂L
∂vi
⇒ pi =d
dt
(∂L
∂qi
),
∂H
∂qi
=∂vj
∂qi
pj − ∂L
∂qi
− ∂L
∂vj
∂vj
∂qi
=∂vj
∂qi
pj − ∂L
∂qi
− ∂vj
∂qi
pj
= −∂L
∂qi
.
The 2N -dimensional space of all qi and pi is called a phase space. An arbitrary function
F [qi(t), pi(t), t] changes along an arbitrary path as follows
d
dtF [qi(t), pi(t), t] =
∂F
∂qi
qi +∂F
∂pi
pi +∂F
∂t. (60)
If the path is a classical orbit, we may insert Eq.(59) and find
d
dtF [qi(t), pi(t), t] =
∂F
∂qi
∂H
∂pi
− ∂F
∂pi
∂H
∂qi
+∂F
∂t
=∂F
∂qi
∂H
∂pi
− ∂H
∂qi
∂F
∂pi
+∂F
∂t
≡ {F, H}+∂F
∂t, (61)
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where we have introduced the Poisson bracket,
{A,B} ≡ ∂A
∂qi
∂B
∂pi
− ∂B
∂qi
∂A
∂pi
, (62)
again with the Einstein summation convention for the repeated index i. From Eq.(61), a
function F [qi(t), pi(t)] which has no explicit dependence on time and which commutes with
H, i.e., {F, H} = 0, is a constant of motion along the classical orbit. In particular, H itself
is often of this type, i.e., H = H[qi(t), pi(t)]. In this case, the energy is a constant of motion.
Obviously, the original Hamilton equations themselves are a special case of Eq.(61).
d
dtqi = {qi, H} =
∂qi
∂qj
∂H
∂pj
− ∂H
∂qj
∂qi
∂pj
=∂H
∂pi
, (63)
d
dtpi = {pi, H} =
∂pi
∂qj
∂H
∂pj
− ∂H
∂qj
∂pi
∂pj
= −∂H
∂qi
. (64)
We also note that at each time t,
{qi, pj} =∂qi
∂qk
∂pj
∂pk
− ∂pj
∂qk
∂qi
∂pk
= δij,
{qi, qj} =∂qi
∂qk
∂qj
∂pk
− ∂qj
∂qk
∂qi
∂pk
= 0,
{pi, pj} =∂pi
∂qk
∂pj
∂pk
− ∂pj
∂qk
∂pi
∂pk
= 0. (65)
The methods of Lagrange and Hamilton provide for an elegant and flexible description
of dynamical systems. They can be applied to any chosen set of generalized coordinates qi,
the only prerequisite being the knowledge of the Lagrangian or Hamiltonian.
B. Canonical quantization
In a quantum mechanical system we may identify a set of coordinate operators {qi}, acting
on a Hilbert space whose vectors are identified with states of the system. The eigenvectors
of the coordinate operators satisfy
qi|q1, q2, · · · , qi, · · ·〉 = qi|q1, q2, · · · , qi, · · ·〉, (66)
and are states in which cooridnate i has value qi. Operators pi corresponding to classical
conjugate momenta pi = ∂L/∂qi obey canonical commutation relations with the qi,
[qi, pj] = ihδij,
[qi, qj] = 0,
[pi, pj] = 0. (67)
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The commutation relations are closely related to the classical Poission bracket.
{A,B} → 1
ih[A, B]. (68)
In particular, we have
dA
dt= {A,H}+
∂A
∂t→ dA
dt=
i
h[H, A] +
∂A
∂t. (69)
IV. CLASSICAL FIELD THEORY
In field theory, the analog of the generalized coordinates qi, is a field ϕ(~x, t), in which
the discrete index i has been replaced by the continuous position vector ~x. The position ~x
is not a coordinate, but rather a parameter that labels the field coordinate ϕ at point ~x at
a particular time t. There may be more than one field at each point in space, in which case
the fields may carry a distinguishing subscript, as in ϕa(~x, t). To qualify as a mechanical
system, the fields must be associated with a Lagrangian
L(t) =∫
d3~xL[ϕa(~x, t), ∂µϕa(~x, t)], (70)
which determines their time development.
In Eq.(70), we employ the conventions
Aµ = (A0, ~A) = gµνAν ,
Aµ = (A0,− ~A) = gµνAν , (71)
for any vector Aµ, where the Minkowski metric
gµν =
+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
. (72)
For derivatives with respect to the coordinate vector xµ = (x0, ~x), we use the notation
∂µϕa ≡ ∂ϕa
∂xµ,
∂µϕa ≡ ∂ϕa
∂xµ
. (73)
14
Here x0 ≡ ct, where c is the speed of light, so that all the xµ have dimensions of length. From
hereon, we set c = 1. Generally, we shall use lower case Greek letters (α, β, · · · , µ, ν, · · ·) for
space-time vector indices (0, 1, 2, 3), and lower case italic letters (i, j, · · ·) for purely spatial
vector indices (1, 2, 3).
Now, let us demand that the action
A[ϕa(~x, t)] =∫ t2
t1dt
∫
Rd3~xL[ϕa(~x, t), ∂µϕa(~x, t)], (74)
within some region R of space to be extremal:
δA[ϕa(~x, t)] = δ∫ t2
t1dt
∫
Rd3~xL[ϕa(~x, t), ∂µϕa(~x, t)]
=∫ t2
t1dt
∫
Rd3~xδL[ϕa(~x, t), ∂µϕa(~x, t)]
=∫ t2
t1dt
∫
Rd3~x
[∂L∂ϕa
δϕa +∂L
∂(∂µϕa)δ∂µϕa
]
=∫ t2
t1dt
∫
Rd3~x
{∂L∂ϕa
− ∂µ
[∂L
∂(∂µϕa)
]}δϕa
+∫
Rd3~x
∂L∂(∂0ϕa)
δϕa
∣∣∣∣∣t2
t1
+∫ t2
t1dt
∫
∂Rdσi
∂L∂(∂iϕa)
δϕa
= 0, (75)
and obtain the equations of motion at every point inside R for every time between t1 and
t2,∂L∂ϕa
− ∂µ
[∂L
∂(∂µϕa)
]= 0. (76)
Here, the variation is over all possible fields ϕa(~x, t) inside R,
ϕa(~x, t) → ϕa(~x, t) + δϕa(~x, t), (77)
with the functions δϕa(~x, t) satisfying
δϕa(~x, t1) = 0 = δϕa(~x, t2) (78)
for all ~x in R, and
δϕa(~y, t) = 0 (79)
for all ~y on the surface ∂R of R, but are otherwise arbitrary. But, Eq.(76) must hold at
every point in space-time since t1, t2 and R were chosen arbitrarily.
15
The field momentum conjugate to ϕa(~x, t) is
πa(~x, t) =∂L
∂(∂0ϕa(~x, t)). (80)
The field Hamiltonian is given by the Legendre transform of the Lagrangian,
H =∫
d3~x{πa(~x, t) · ∂0ϕa(~x, t)− L[ϕa(~x, t), ∂µϕa(~x, t)]}. (81)
If we are given a Lagrangian, we may quickly find the field equations from Eq.(76), the
Hamiltonian from Eq.(81), and quantize by making ϕa(~x, t) and its conjugate momentum
πa(~x, t) operators satisfying equal-time commutation relations,
[ϕa(~x, t), πb(~x′, t)] = iδabδ
3(~x− ~x′). (82)
In the absence of difficult operator ordering ambiguities, the field Hamiltonian operator and
quantum conditions, will reproduce the field equations as operator equations of motion.
A. Noether’s theorem
If the action A[ϕa(~x, t)] is unchanged by a re-parametrization of xµ and ϕa, i.e., is in-
variant under some group of transformations on xµ and ϕa, then there exist one or more
conserved quantities, i.e., combinations of fields and their derivatives which are invariant un-
der the transformations. This crucial result is known as Noether’s theorem. It accounts for
conservation of energy, momentum, angular momentum, and any other “quantum” number
which particles happen to carry, like charge, color, isospin, etc.
To prove Noether’s theorem, consider
δA[ϕa(~x, t)] = δ∫
Rd4xL[ϕa(~x, t), ∂µϕa(~x, t)]
=∫
Rd4xδL[ϕa(~x, t), ∂µϕa(~x, t)]
=∫
Rd4x
[∂L∂ϕa
δϕa +∂L
∂(∂µϕa)δ∂µϕa + ∂µLδxµ
]
=∫
Rd4x
{∂L∂ϕa
− ∂µ
[∂L
∂(∂µϕa)
]}δϕa +
∫
∂Rdσµ
[∂L
∂(∂µϕa)δϕa + δµ
νLδxν
]
=∫
∂Rdσµ
[∂L
∂(∂µϕa)δϕa +
∂L∂(∂µϕa)
∂νϕaδxν
]−
∫
∂Rdσµ
[∂L
∂(∂µϕa)∂νϕa − δµ
νL]δxν
=∫
∂Rdσµ
[∂L
∂(∂µϕa)∆ϕa − θµ
νδxν
],
16
where the total variation
∆ϕa ≡ δϕa + ∂νϕaδxν , (83)
and the enery-momentum tensor
θµν ≡
∂L∂(∂µϕa)
∂νϕa − δµνL. (84)
We will first discuss the isospin symmetry, where the fields ϕa vary according to some
small parameter δεa.
ϕa → ϕa + δϕa = ϕa +∂ϕa
∂εb
δεb (85)
That is, ∆ϕa = δϕa = (∂ϕa/∂εb)δεb. If the action A[ϕa(~x, t)] is invariant under this trans-
formation, then
δA[ϕa] =∫
∂Rdσµ
∂L∂(∂µϕa)
∆ϕa =∫
∂RdσµJ
µa δεa = 0. (86)
Here, the current
Jµa ≡
∂L∂(∂µϕb)
∂ϕb
∂εa
. (87)
It follows from ∫
∂RdσµJ
µa δεa =
∫
Rd4xδεa∂µJ
µa = 0 (88)
that
∂µJµa = 0. (89)
From this conserved current, we can also establish a conserved charge, given by the integral
over the time-component of the current:
Qa ≡∫
d3~xJ0a . (90)
Now let us integrate the conservation equation:
0 =∫
d3~x∂µJµa
=∫
d3~x∂0J0a +
∫d3~x∂iJ
ia
=d
dt
∫d3~xJ0
a +∫
dSiJia
=d
dtQa + surface term (91)
Let us assume that the fields appearing in the surface term vanish sufficiently rapidly at
infinity so that the last term can neglected. Then,
∂µJµa = 0 ⇒ d
dtQa = 0 (92)
17
In summary, the symmetry of the action implies the conservation of a current Jµa , which in
turn implies a conservation principle:
symmetry → current conservation → conservation principle (93)
The second case, when the action is invariant under the space-time symmetry of the
Lorentz and Poincare groups, will be discussed later.
B. Schrodinger field
The Lagrangian density of the Schrodinger field
L = iϕ∗∂tϕ− 1
2∂xϕ
∗∂xϕ− V (x)ϕ∗ϕ. (94)
Here, ∂t ≡ ∂/∂t and ∂x ≡ ∂/∂x. We can determine the equations of motion by treating ϕ
and ϕ∗ as independent objects, so that for instance
∂L∂(∂tϕ∗)
= 0,
∂L∂(∂xϕ∗)
= −1
2∂xϕ,
∂L∂ϕ∗
= i∂tϕ− V (x)ϕ,
give us the Schrodinger equation (43). On the other hand,
∂L∂(∂tϕ)
= iϕ∗,
∂L∂(∂xϕ)
= −1
2∂xϕ
∗,
∂L∂ϕ
= −V (x)ϕ∗,
give us the complex-conjugate of Eq.(43).
From Eq.(94), we find that the conjugate momentum field is
π(x, t) =∂L
∂(∂tϕ)= iϕ∗(x, t). (95)
Since the field conjugate to ϕ∗ is found to vanish, there are only two independent fields,
ϕ(x, t) and π(x, t).
18
The Hamiltonian is easily computed to be
H =∫
dx[π(x, t)∂tϕ(x, t)− L]
=∫
dx1
2∂xϕ
∗∂xϕ + V (x)ϕ∗ϕ
=∫
dxϕ∗[−1
2
∂2
∂x2+ V (x)
]ϕ (96)
From Eqs.(42) and (87), we have
J0 =∂L
∂(∂0ϕ)
∂ϕ
∂ε+
∂L∂(∂0ϕ∗)
∂ϕ∗
∂ε
= iϕ∗∂ϕ
∂ε
= ϕ∗ϕ. (97)
It follows that
ϕ(ε) = ϕ(0) exp(−iε). (98)
And,
J1 =∂L
∂(∂1ϕ)
∂ϕ
∂ε+
∂L∂(∂1ϕ∗)
∂ϕ∗
∂ε
=1
2∂xϕ
∗(iϕ)− 1
2∂xϕ(iϕ∗)
= − i
2(ϕ∗∂xϕ− ϕ∂xϕ
∗). (99)
V. SECOND QUANTIZATION (CONT’)
In canonical field quantization, the classical fields ϕ(x, t) and π(x, t) are replaced by
operators ϕ(x, t) and
π(x, t) = iϕ†(x, t), (100)
where ϕ∗(x, t) is replaced by the Hermitian conjugate field operator ϕ†(x, t). We impose on
these operators the equal-time commutation relations:
[ϕ(x, t), ϕ†(x′, t)] = δ(x− x′), (101)
[ϕ(x, t), ϕ(x′, t)] = 0, (102)
[ϕ†(x, t), ϕ†(x′, t)] = 0. (103)
19
Using the Hamiltonian
H(t) =∫
dx[1
2∂xϕ
†∂xϕ + V (x)ϕ†ϕ], (104)
and the above equal-time commutation relations, we find
[H(t), ϕ(x, t)] =∫
dy[1
2∂yϕ
†(y, t)∂yϕ(y, t) + V (y)ϕ†(y, t)ϕ(y, t), ϕ(x, t)]
=∫
dy{
1
2∂y[ϕ
†(y, t), ϕ(x, t)]∂yϕ(y, t) + V (y)[ϕ†(y, t), ϕ(x, t)]ϕ(y, t)}
= −∫
dy{
1
2[∂yδ(y − x)]∂yϕ(y, t) + δ(y − x)V (y)ϕ(y, t)
}
=1
2∂2
xϕ(x, t)− V (x)ϕ(x, t) (105)
Alternatively, we could use
[H(t), ϕ(x, t)] =∫
dy[ϕ†(y, t)(−1
2∂2
y − V (y))
ϕ(y, t), ϕ(x, t)]
=∫
dy[ϕ†(y, t), ϕ(x, t)](−1
2∂2
y − V (y))
ϕ(y, t)
= −∫
dyδ(y − x)(−1
2∂2
y − V (y))
ϕ(y, t)
=1
2∂2
xϕ(x, t)− V (x)ϕ(x, t)
Therefore,∂
∂tϕ(x, t) = i[H, ϕ(x, t)]
implies
i∂
∂tϕ(x, t) = −1
2∂2
xϕ(x, t) + V (x)ϕ(x, t).
We note that we have picked a particular ordering for the operators in Eq.(104), called
normal ordering, to avoid the vacuum energy divergence.
A. Particle interpretation
Recall the expansion (47),
ϕ(x, t) =∑
i
ai(t)ϕi(x).
It follows that
ϕ†(x, t) =∑
i
a†i (t)ϕ∗i (x). (106)
20
Substitute Eq.(47) into the above equal-time commutation relations (101),
δ(x− x′) = [ϕ(x, t), ϕ†(x′, t)]
= [∑
i
ϕi(x)ai(t),∑
j
ϕ∗j(x′)a†j(t)]
=∑
i,j
ϕi(x)ϕ∗j(x′)[ai(t), a
†j(t)] (107)
only if
[ai(t), a†j(t)] = δij. (108)
This is because the first quantized energy eigenfunctions are assumed to be complete,
∑
i
ϕi(x)ϕ∗i (x′) =
∑
i
〈x|ϕi〉〈ϕi|x′〉
= 〈x|(∑
i
|ϕi〉〈ϕi|)|x′〉
= 〈x|x′〉= δ(x− x′) (109)
Alternatively, we could express ai(t) and a†i (t) in terms of ϕ(x, t) and ϕ†(x, t) respectively.
ai(t) =∫
dxϕ∗i (x)ϕ(x, t),
a†i (t) =∫
dxϕi(x)ϕ†(x, t). (110)
It follows that
[ai(t), a†j(t)] =
∫dx
∫dyϕ∗i (x)ϕj(y)[ϕ(x, t), ϕ†(y, t)]
=∫
dx∫
dyϕ∗i (x)ϕj(y)δ(x− y)
=∫
dxϕ∗i (x)ϕj(x)
=∫
dx〈ϕi|x〉〈x|ϕj〉= 〈ϕi|ϕj〉= δij (111)
Similarly,
[ai(t), aj(t)] =∫
dx∫
dyϕ∗i (x)ϕ∗j(y)[ϕ(x, t), ϕ(y, t)] = 0,
[a†i (t), a†j(t)] =
∫dx
∫dyϕi(x)ϕj(y)[ϕ†(x, t), ϕ†(y, t)] = 0. (112)
21
Next, substitute the expansion (47) into the Hamiltonian operator.
H(t) =∫
dxϕ†[−1
2
∂2
∂x2+ V (x)
]ϕ
=∫
dx∑
i
ϕ∗i (x)a†i (t)
[−1
2
∂2
∂x2+ V (x)
] ∑
j
ϕj(x)aj(t)
=∑
i,j
ej a†i (t)aj(t)
∫dxϕ∗i (x)ϕj(x)
=∑
i
eia†i (t)ai(t) (113)
The time dependence of the operators ai(t) is determined by
d
dtai(t) = i[H(t), ai(t)] =
∑
j
iej[a†j(t)aj(t), ai(t)] = −ieiai(t), (114)
which is solved by
ai(t) = exp(−ieit)ai(0) ≡ exp(−ieit)ai. (115)
It follows that
a†i (t) = exp(ieit)a†i , (116)
and H =∑
i eia†i ai. We note that the use of the eigenfunction basis {ϕi(x)} has the effect
that the time dependence of the operators ai(t) becomes trivial, being characterized by a
simple phase factor.
For fixed i, we note that a†i and ai look identical to the raising and lowering operators of
the harmonic oscillator. The field Hamiltonian is just an infinite sum of harmonic oscillator
Hamiltonians. The expansion (47) has reduced the quantum field theory to an infinite set
of harmonic oscillators. Following the discussion given on the harmonic oscillator, we can
now develop a particle interpretation. In the following we will construct the state vectors,
making use of the time-independent operators ai. This is possible within the Heisenberg
picture where the operators are time dependent while the state vectors are constant.
The lowest energy state of H, the ground state or bare vacuum, is the one that is empty,
ai|0〉 = 0. (117)
The destruction operator for any i, ai, finds no particle (excitation) to annihilate in the
empty vacuum |0〉, so the result is the null vector.
a†i |0〉 is a state of energy ei. It describes one particle of energy ei, created by a†i , the
creation operator for mode i. a†j|0〉 is also a one-particle state except that the energy of the
22
particle is ej. We have only one field in this theory, so there is only one type of particle.
The difference in energy between the two one-particle states must be due to a difference in
momentum between the two.
a†i a†j|0〉 is a two-particle state with energy ei + ej. The collection of all the states spanned
by the states formed by operating on |0〉 with any number of creation operators for any
mode i is called a Fock space. In general,
1√n1!n2! · · ·
(a†1)n1(a†2)
n2 · · · |0〉 ≡ |n1, n2, · · ·〉. (118)
These are eigenstates of ni ≡ a†i ai:
ni|n1, n2, · · · , ni, · · ·〉 = ni|n1, n2, · · · , ni, · · ·〉,
and the particle-number operator
N ≡ ∑
i
ni =∑
i
a†i ai, (119)
N |n1, n2, · · · , ni, · · ·〉 =
(∑
i
ni
)|n1, n2, · · · , ni, · · ·〉 = N |n1, n2, · · · , ni, · · ·〉,
where N is the total number of particles.
Last, we note that ϕ(x, t) is expanded in terms of ai(t) only, while ϕ†(x, t) is expanded in
terms of a†i (t) only. Thus, ϕ(x, t) is a destruction operator and ϕ†(x, t) is a creation operator.
ϕ†(x, t)|0〉 is a one-particle state where the particle is located at position x at time t.
〈0|ϕ(x, t)ϕ†(x′, t)|0〉 =∑
i,j
ϕi(x)ϕ∗j(x′)〈0|ai(t)a
†j(t)|0〉
=∑
i,j
ϕi(x)ϕ∗j(x′)〈0|{[ai(t), a
†j(t)] + a†j(t)ai(t)}|0〉
=∑
i,j
ϕi(x)ϕ∗j(x′)δij
=∑
i
ϕi(x)ϕ∗i (x′)
= δ(x− x′) (120)
So,
|x; t〉 ≡ ϕ†(x, t)|0〉. (121)
23
B. Particles in a box
Consider the case where V (x) = 0 with 0 ≤ x ≤ L. The normalized eigenfunctions,
ϕn(x) =1√L
exp(i2nπx
L
), (122)
are plane waves or momentum eigenfunctions. The expansions (47) and (106) turn into
Fourier series,
ϕ(x, t) =1√L
∑n
exp(i2nπx
L
)an(t),
ϕ†(x, t) =1√L
∑n
exp(−i
2nπx
L
)a†n(t), (123)
and Eq.(110) becomes
an(t) =1√L
∫dx exp
(−i
2nπx
L
)ϕ(x, t),
a†n(t) =1√L
∫dx exp
(i2nπx
L
)ϕ†(x, t). (124)
It follows that the state
a†n(t)|0〉 =1√L
∫dx exp
(i2nπx
L
)ϕ†(x, t)|0〉. (125)
Since we are integrating over x, a†n(t) creates a particle via ϕ†(x, t) at every point x with
amplitude exp(i2nπx/L)/√
L. In other words, the state a†n(t)|0〉 is a one-particle state with
position probability amplitude at each point given by exp(i2nπx/L)/√
L. That is, a†n(t)
creates a particle with wave function exp(i2nπx/L)/√
L. The square of this wave function is
independent of x, so a†n(t) creates a one-particle state and we do not know where the particle
is. It has equal probability of being anywhere. The particle has a definite momentum and
we have no idea what its position is, in accordance with the uncertainty principle.
Likewise, ϕ†(x, t) creates a particle at x, a definite position. Since we know its position,
we have no idea what its momentum and energy are. To create a particle definitely located
at x, we need to use a†n(t) for every n. The particle has equal probability of having any
momentum and hence n.
24
C. First quantized one-particle system
Now consider the general case where V (x) 6= 0. From above, a state containing one
particle described by wave function f1(x1, t) is given by
∫dx1 f1(x1, t)ϕ
†(x1, t)|0〉. (126)
Suppose this state has a definite energy, then it must be an eigenstate of H.
H∫
dx1 f1(x1, t)ϕ†(x1, t)|0〉
=∫
dx∫
dx1 f1(x1, t)ϕ†(x, t)
[−1
2
∂2
∂x2+ V (x)
]ϕ(x, t)ϕ†(x1, t)|0〉
=∫
dx∫
dx1 f(x1, t)ϕ†(x, t)
[−1
2
∂2
∂x2+ V (x)
]δ(x− x1)|0〉
=∫
dxϕ†(x, t)∫
dx1 f(x1, t)
[−1
2
∂2
∂x2+ V (x)
]δ(x− x1)|0〉
=∫
dx1
[−1
2
∂2
∂x21
+ V (x1)
]f(x1, t)ϕ
†(x1, t)|0〉
=∫
dx1 E1f(x1, t)ϕ†(x1, t)|0〉
= E1
∫dx1 f(x1, t)ϕ
†(x1, t)|0〉
which is true provided
[−1
2
∂2
∂x21
+ V (x1)
]f1(x1, t) = E1f1(x1, t). (127)
So, we see how the first quantized one-particle system emerges from the quantum field theory.
D. First quantized two-particle system
A two-particle state with one particle at x1 and another at x2 is ϕ†(x1, t)ϕ†(x2, t)|0〉. If
the two particles have wave function f2(x1, x2, t), then the state is
∫dx1dx2f2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉. (128)
Applying H(t) to this state, we have
∫dx
∫dx1
∫dx2f2(x1, x2, t)ϕ
†(x, t)
[−1
2
∂2
∂x2+ V (x)
]ϕ(x, t)ϕ†(x1, t)ϕ
†(x2, t)|0〉
25
=∫
dx∫
dx1
∫dx2f2(x1, x2, t)ϕ
†(x, t)
[−1
2
∂2
∂x2+ V (x)
]δ(x− x1)ϕ
†(x2, t)|0〉
+∫
dx∫
dx1
∫dx2f2(x1, x2, t)ϕ
†(x, t)
[−1
2
∂2
∂x2+ V (x)
]ϕ†(x1, t)ϕ(x, t)ϕ†(x2, t)|0〉
=∫
dx ϕ†(x, t)∫
dx2
∫dx1f2(x1, x2, t)
[−1
2
∂2
∂x2+ V (x)
]δ(x− x1)ϕ
†(x2, t)|0〉
+∫
dx ϕ†(x, t)∫
dx1
∫dx2f2(x1, x2, t)
[−1
2
∂2
∂x2+ V (x)
]δ(x− x2)ϕ
†(x1, t)|0〉
=∫
dx1dx2
[−1
2
∂2
∂x21
+ V (x1)
]f2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉
+∫
dx1dx2
[−1
2
∂2
∂x22
+ V (x2)
]f2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉
This state will be an eigenstate of H if f2 satisfies
[−1
2
∂2
∂x21
− 1
2
∂2
∂x22
+ V (x1) + V (x2)
]f2(x1, x2, t) = E2f2(x1, x2, t). (129)
This is a first quantized two-body Schrodinger equation.
E. First quantized N-particle system
If we apply the field Hamiltonian operator to its N -particle eigenstate,
∫dx1 · · · dxNfN(x1, · · · , xN , t)ϕ†(x1, t) · · · ϕ†(xN , t)|0〉, (130)
we find that the N -body wave function must satisfy
[−1
2
N∑
i=1
∂2
∂x2i
+N∑
i=1
V (xi)
]fN(x1, · · · , xN , t) = ENfN(x1, · · · , xN , t). (131)
In this way, all of the N -body first quantized systems are contained in the corresponding
quantum field theory. The operators ϕ(x, t) and ϕ†(x, t) change the number of particles
present so we can treat physical processes where the particle number is changing in a unified
manner. One way to exactly solve a quantum field theory is to solve the N -body Schrodinger
equation (131) for general N . Then any amplitude for any process will amount to a multi-
dimensional integral over the initial and final state wave functions.
26
F. Bosons and fermions
The quantum conditions we have chosen specify that ϕ†(x, t) commutes with itself. It
follows that
∫dx1dx2f2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉
=∫
dx2dx1f2(x2, x1, t)ϕ†(x2, t)ϕ
†(x1, t)|0〉
=∫
dx1dx2f2(x2, x1, t)ϕ†(x1, t)ϕ
†(x2, t)|0〉
Therefore,
f2(x1, x2, t) = f2(x2, x1, t). (132)
f2(x2, x1, t) is symmetric under the exchange of x1 and x2. The symmetric exchange is due
to the use of commutators in the quantum conditions. By the Pauli principle, this exchange
symmetry implies that we are dealing with bosons.
The wave function for two identical fermions must be antisymmetric under the exchange
of coordinates:
ψ2(x1, x2, t) = −ψ2(x2, x1, t). (133)
It follows that
∫dx1dx2ψ2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉
=∫
dx2dx1ψ2(x2, x1, t)ϕ†(x2, t)ϕ
†(x1, t)|0〉
= −∫
dx1dx2ψ2(x1, x2, t)ϕ†(x2, t)ϕ
†(x1, t)|0〉
If ϕ†(x1, t) and ϕ†(x2, t) commute, then
∫dx1dx2ψ2(x1, x2, t)ϕ
†(x1, t)ϕ†(x2, t)|0〉 = 0.
To avoid this, we demand
ϕ†(x1, t)ϕ†(x2, t) = −ϕ†(x2, t)ϕ
†(x1, t),
or
ϕ†(x1, t)ϕ†(x2, t) + ϕ†(x2, t)ϕ
†(x1, t) ≡ {ϕ†(x1, t), ϕ†(x2, t)} = 0. (134)
27
From hereon, we will use braces, {}, to denote an anticommutator. Therefore, to quantize
the field ϕ(x, t) satisfying Eq.(44) and obtain fermions, we must use anticommutators for
the quantum conditions,
{ϕ(x, t), ϕ†(x′, t)} = δ(x− x′),
{ϕ(x, t), ϕ(x′, t)} = 0,
{ϕ†(x, t), ϕ†(x′, t)} = 0. (135)
VI. LORENTZ INVARIANCE
A. Lorentz transformations
A Lorentz transformation is a linear, homogeneous change of coordinates from xµ to x′µ,
xµ → x′µ = Λµνx
ν , (136)
that preserves the interval x2 between xµ and the origin:
x2 ≡ t2 − ~x2 ≡ xµxµ = gµνxµxν . (137)
It follows that
x′2 = gµνx′µx′ν
= gµνΛµρx
ρΛνσx
σ
= gµνΛµρΛ
νσx
ρxσ
= gρσxρxσ
= x2
and the matrix (Λ)µν ≡ Λµ
ν must obey
gµνΛµρΛ
νσ = gρσ. (138)
If we take the determinant of Eq.(138), we get (det Λ)2 = 1, which implies that det Λ = ±1.
Transformations with det Λ = +1 are proper, and transformations with det Λ = −1 are
improper. Equation (138) implies
gµνΛµ0Λ
ν0
28
= Λ00Λ
00 − Λi
0Λi0
= g00
= 1
This means that (Λ00)
2 − Λi0Λ
i0 = 1; thus, either Λ0
0 ≥ +1 or Λ00 ≤ −1. Transformations
with Λ00 ≥ +1 are orthochronous, while those with Λ0
0 ≤ −1 are non-orthochronous.
B. Lorentz group
The set of all Lorentz transformations forms a group. The product of any two Lorentz
transformations is another Lorentz transformation. The product is associative. There is an
identity transformation,
Λµν = δµ
ν .
Every Lorentz transformation has an inverse:
gµνΛµρΛ
νσ = gρσ
⇒ ΛνρΛνσ = gρσ
⇒ gραΛναΛνσ = gραgασ
⇒ Λ ρν Λν
σ = δρσ
It follows that
(Λ−1)ρν = Λ ρ
ν = gνµΛµαgαρ, (139)
since, by definition, (Λ−1)ρνΛ
νσ = δρ
σ.
C. Infinitesimal Lorentz transformations
For an infinitesimal Lorentz transformation, we can write
Λµν = δµ
ν + δωµν . (140)
δω with both indices down (or up) is antisymmetric: Eq.(138) implies
gµνΛµρΛ
νσ = gµν(δ
µρ + δωµ
ρ)(δνσ + δων
σ)
≈ gµν(δµρδ
νσ + δµ
ρδωνσ + δν
σδωµρ)
29
= gρσ + δωρσ + δωσρ
= gρσ
It follows that
δωρσ = −δωσρ. (141)
Thus there are six independent infinitesimal Lorentz transformations (in four spacetime
dimensions). These can be divided into three rotations and three boosts:
δωij = −εijknkδθ (142)
for a rotation by angle δθ about the unit vector n,
δωi0 = niδη (143)
for a boost in the direction n by rapidity δη. For example,
Rz(θ) =
cos θ − sin θ 0
sin θ cos θ 0
0 0 1
and
Rz(δθ) =
1 −δθ 0
δθ 1 0
0 0 1
And,
Bz(η) =
cosh η 0 0 − sinh η
0 1 0 0
0 0 1 0
− sinh η 0 0 cosh η
,
where cosh η = γ ≡ 1/√
1− v2/c2, sinh η = γv/c, and rapidity η = tanh−1(v/c). So,
Bz(δη) =
1 0 0 −δη
0 1 0 0
0 0 1 0
−δη 0 0 1
.
30
We observe that
Rz(δθ) =
1 −δθ 0
δθ 1 0
0 0 1
=
1 0 0
0 1 0
0 0 1
− iδθ
0 −i 0
i 0 0
0 0 0
= 1− iδθ(ζ1|ζ1〉〈ζ1|+ ζ0|ζ0〉〈ζ0|+ ζ−1|ζ−1〉〈ζ−1|)= (1− iζ1δθ)|ζ1〉〈ζ1|+ (1− iζ0δθ)|ζ0〉〈ζ0|+ (1− iζ−1δθ)|ζ−1〉〈ζ−1|,
where |ζ1〉, |ζ0〉, and |ζ−1〉 are eigenvectors of the Hermitian matrix with eigenvalues ζ1 = 1,
ζ0 = 0, and ζ−1 = −1 respectively. It follows that
limn→∞[Rz(δθ)]
n = limn→∞(1− iζ1δθ)
n|ζ1〉〈ζ1|+ |ζ0〉〈ζ0|+ limn→∞(1− iζ−1δθ)
n|ζ−1〉〈ζ−1|= exp(−iθ)|ζ1〉〈ζ1|+ |ζ0〉〈ζ0|+ exp(iθ)|ζ−1〉〈ζ−1|= Rz(θ)
Here, δθ = limn→∞ θ/n.
Not all Lorentz transformations can be reached by compounding infinitesimal ones. In-
finitesimal transformations of the form Λ = 1 + δω are clearly proper and orthochronous:
det Λ = εκλµνΛκ0Λ
λ1Λ
µ2Λ
ν3
= εκλµν(δκ0 + δωκ
0)(δλ1 + δωλ
1)(δµ2 + δωµ
2)(δν3 + δων
3)
≈ εκλµν(δκ0δ
λ1δ
µ2δ
ν3 + δωκ
0δλ1δ
µ2δ
ν3 + δκ
0δωλ1δ
µ2δ
ν3 + δκ
0δλ1δω
µ2δ
ν3 + δκ
0δλ1δ
µ2δω
ν3
= 1,
Λ00 = δ0
0 + δω00 = 1.
Since the product of any two proper orthochronous Lorentz transformations is proper or-
thochronous:
det Λ = 1, det Λ′ = 1 ⇒ det(ΛΛ′) = det Λ det Λ′ = 1,
Λ00 ≥ 1, Λ′00 ≥ 1 ⇒ (ΛΛ′)0
0 = Λ0µΛ′µ0 = Λ0
0Λ′00 + Λ0
iΛ′i0 ≥ 1,
any transformation that can be reached by compounding infinitesimal ones is proper or-
thochronous. Thus, the Lorentz transformations that can be reached by compounding in-
finitesimal ones are both proper and orthochronous, and they form a subgroup. Generally,
31
when a theory is said to be Lorentz invariant, this means under the proper orthochronous
subgroup only.
Two discrete transformations that take one out of the proper orthochronous subgroup
are parity and time reversal. The parity transformation is
P µν =
+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
= (P−1)µν . (144)
It is orthochronous, but improper. The time-reversal transformation is
T µν =
−1 0 0 0
0 +1 0 0
0 0 +1 0
0 0 0 +1
= (T−1)µν . (145)
It is nonorthochronous and improper. From hereon, in this chapter, we will treat the proper
orthochronous subgroup only. Parity and time reversal will be treated separately in later
chapters.
D. Quantum Lorentz transformations
In quantum theory, symmetries are represented by unitary (or antiunitary) operators.
This means that we associate a unitary operator U(Λ) to each proper, orthochronous Lorentz
transformation Λ. These operators must obey the composition rule
U(Λ′Λ) = U(Λ′)U(Λ). (146)
For an infinitesimal transformation, we can write
U(1 + δω) = I +i
2δωµνM
µν , (147)
where Mµν = −M νµ is a set of Hermitian operators called the generators of the Lorentz
group.
Let Λ′ = 1 + δω′, then to linear order in δω′,
U(Λ)−1U(Λ′)U(Λ) = U(Λ−1Λ′Λ)
32
⇒ U(Λ)−1(I +
i
2δω′µνM
µν)
U(Λ) = U((Λ−1)µρ(δ
ρσ + δω′ρσ)Λσ
ν)
⇒ I +i
2δω′µνU(Λ)−1MµνU(Λ) = U(δµ
ν + (Λ−1)µρδω
′ρσΛσ
ν)
= U(1 + Λ−1δω′Λ)
Therefore,
I +i
2δω′µνU(Λ)−1MµνU(Λ) = I +
i
2(Λ−1δω′Λ)ρσM
ρσ
where
δω′µνU(Λ)−1MµνU(Λ) = (Λ−1)ρµδω′µνΛ
νσM
ρσ
= gµαΛαρδω
′µνΛ
αρM
ρσ
= δω′µνΛµρΛ
νσM
ρσ
Since δω′µν is arbitrary, we have
U(Λ)−1MµνU(Λ) = ΛµρΛ
νσM
ρσ (148)
We observe that each vector index on Mµν undergoes its own Lorentz transformation. This
is a general result: any operator carrying one or more vector indices should behave similarly.
For example, for the energy-momentum four-vector P µ, where P 0 is the Hamiltonian H and
P i are the components of the total three-momentum operator,
U(Λ−1)P µU(Λ) = ΛµνP
ν . (149)
Now, let Λ = 1 + δω in Eq.(148), then to linear order in δω,(I − i
2δωρσM
ρσ)
Mµν(I +
i
2δωρσM
ρσ)
= (δµρ + δωµ
ρ)(δνσ + δων
σ)Mρσ
Mµν − i
2δωρσM
ρσMµν +i
2δωρσM
µνMρσ = Mµν + δωµρδ
νσM
ρσ + δµρδω
νσM
ρσ
− i
2δωρσ[Mρσ, Mµν ] = gµαδωαρM
ρν + gναδωασMµσ
i
2δωρσ[Mµν , Mρσ] = gµρδωρσM
σν + gνσδωσρMµρ
=1
2(gµρδωρσM
σν + gµσδωσρMρν + gνσδωσρM
µρ + gνρδωρσMµσ)
=1
2δωρσ(gµρMσν − gµσMρν − gνσMµρ + gνρMµσ)
=1
2δωρσ(−gµρM νσ + gµσM νρ − gνσMµρ + gνρMµσ)
= −1
2δωρσ(gµρM νσ − gνρMµσ − gµσM νρ + gνσMµρ)
33
Therefore,
[Mµν , Mρσ] = i(gµρM νσ − gνρMµσ − gµσM νρ + gνσMµρ). (150)
These commutation relations specify the Lie algebra of the Lorentz group. We can identify
the components of the angular momentum operator as Ji ≡ −1/2εijkMjk, and the compo-
nents of the boost operator Ki ≡ M i0.
[Ji, Jj] =1
4εimnεjpq[M
mn, Mpq]
=i
4εimnεjpq(g
mpMnq − gnpMmq − gmqMnp + gnqMmp)
= − i
4(εimnεjmqM
nq − εimnεjnqMmq − εimnεjpmMnp + εimnεjpnMmp)
= − i
4(εminεmjqM
nq + εimnεjqnMmq + εinmεjpmMnp + εimnεjpnM
mp)
= −iM ij
= iεijkJk (151)
since
εijkJk = −1
2εkijεklmM lm = −1
2(δilδjm − δimδjl)M
lm = −M ij.
[Ji, Kj] = −1
2εimn[Mmn, M j0]
= − i
2εimn(gmjMn0 − gnjMm0 − gm0Mnj + gn0Mmj)
=i
2(εijnMn0 − εimjM
m0)
= iεijkMk0
= iεijkKk (152)
[Ki, Kj] = [M i0, M j0]
= i(gijM00 − g0jM i0 − gi0M0j + g00M ij)
= iM ij
= −iεijkJk (153)
Similarly, we can let Λ = 1 + δω in Eq.(149), then to linear order in δω,
(I − i
2δωρσM
ρσ)
P µ(I +
i
2δωρσM
ρσ)
= (δµν + δωµ
ν)Pν
34
P µ − i
2δωρσ[Mρσ, P µ] = P µ + gµαδωανP
ν
i
2δωρσ[P µ, Mρσ] = gµρδωρσP
σ
=1
2(gµρδωρσP
σ + gµσδωσρPρ)
=1
2δωρσ(gµρP σ − gµσP ρ)
Therefore,
[P µ, Mρσ] = i(gµσP ρ − gµρP σ). (154)
It follows that
[H, Ji] = −1
2εijk[P
0, M jk]
= − i
2εijk(g
0kP j − g0jP k)
= 0, (155)
[Pi, Jj] =1
2εjmn[P i, Mmn]
=i
2εjmn(ginPm − gimP n)
=i
2(−εjmiP
m + εjinP n)
= iεijkPk, (156)
[H, Ki] = [P 0, M i0]
= i(g00P i − g0iP 0)
= −iPi (157)
[Pi, Kj] = −[P i, M j0]
= −i(gi0P j − gijP 0)
= −iδijH (158)
Also, the components of P µ should commute with each other.
[P i, P j] = 0,
35
[P i, H] = 0. (159)
In summary,
[H, P i] = 0,
[P i, P j] = 0,
[Ji, Jj] = iεijkJk,
[Ji, Kj] = iεijkKk,
[Ki, Kj] = −iεijkJk,
[H, Ji] = 0,
[Pi, Jj] = iεijkPk,
[H, Ki] = −iPi,
[Pi, Kj] = −iδijH
form the Lie algebra of the Poincare group.
E. Noether’s theorem (cont’)
Now let us investigate the case when the action is invariant under the space-time sym-
metry of the Lorentz and Poincare groups. First, consider the translation
xµ → x′µ = xµ + aµ, (160)
where aµ is a constant. a0 represents time displacements, and ai represents space displace-
ments. From Eq.(84),
θµν ≡
∂L∂(∂µϕa)
∂νϕa − δµνL
is the energy-momentum tensor. By integrating the energy-momentum tensor, we can gen-
erate conserved quantities. Define
Pν ≡∫
d3~xθ0ν . (161)
Then
∂µθµν = 0
⇒ ∂0θ0ν + ∂iθ
iν = 0
⇒ d
dtPν = 0. (162)
36
We observe that
P0 =∫
d3~xθ00 =
∫d3~x
∂L∂(∂0ϕa)
∂0ϕa − L = H. (163)
Therefore, displacement in time (space) leads to the conservation of energy (momentum).
Here,
Pi =∫
d3~xθ0i =
∫d3~x
∂L∂(∂0ϕa)
∂iϕa. (164)
Next, consider the Lorentz transformation
xµ → x′µ = xµ + δωµνx
ν . (165)
It follows that
∂µ(θµνδω
ναxα) = 0
⇒ ∂µ(θµνδωναxα) = 0
⇒ ∂µ(θµνδωναxα + θµαδωανxν) = 0
⇒ δωνα∂µ(θµνxα − θµαxν) = 0
⇒ ∂ρ(θρµxν − θρνxµ) = 0 (166)
Define
Mµν ≡∫
d3~x(θ0µxν − θ0νxµ). (167)
Then
∂0(θ0µxν − θ0νxµ) + ∂i(θ
iµxν − θiνxµ) = 0
⇒ d
dtMµν = 0 (168)
37