34
4.1 v1
4.1 v1
4.1 v2
4.2 v1
4.2 v2
4.3 v1
1
Physics 615 Oct. 26, 2006Homework Solutions #7
1 [20 pts] Do problem 4.3 from Peskin and Schroeder. This is one of thesimpler examples of a model with spontaneous symmetry breaking.
Solution 1 (a) We are going to explore the linear sigma model, basedon N real scalar fields1 Φi(x) and their conjugate momenta Πi, with theHamiltonian
H =∫
d3x
(1
2
∑i
(Πi)2 +1
2
∑i
(~∇Φi)2 + V (Φ2)
),
with
V (Φ2) =1
2m2Φ2 +
λ
4(Φ2)2.
Note that here Φ2 means∑
i(Φi)2 and not, of course, the i = 2 component of
Φ. This Hamiltonian (and the associated Lagrangian) are clearly invariantunder global rotations of Φi in the N dimensional space.
When m2 > 0 and λ → 0, the Hamiltonian reduces to a sum of N inde-pendent pieces each depending on just one of the N real scalar fields Φi, andeach piece is simply the free Klein-Gordon Hamiltonian. Thus the interactionpicture fields can be expanded as usual in terms of N sets of creation andannihilation operators ai †
~p and ai~p, with the different fields commuting with
each other, [ai
~p, aj †~p′]
= (2π)3δijδ3(~p−~p ′).
Thus the contractions of Φi with Φj give 0 unless i = j, in which case theygive the usual free Klein Gordon propagator
DF (x−y) = φI(x)φI(y) = 〈0|TφI(x)φI(y) |0〉 =∫ d4p
(2π)4
1
p2 −m2 + iεe−ip(x−y).
1I will use the book’s notation with a superscript i, Φi, though I don’t understand whythey did not use a subscript.
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 2
The interaction hamiltonian is the λ term in V , so each factor broughtdown from the exponential gives
−iλ
4
∫d4z
∑ab
Φa(x)Φa(x)Φb(x)Φb(x).
When this factor is contracted with 4 other fields ΦiΦjΦkΦ`, (or externalstates |~p, j〉), there are 4! ways these other fields can choose among the fourin the interaction vertex. The first, Φi, has 4 equivalent choices. The second,Φj can then either choose the one Φ with the same index in the interaction,giving a δij, or it has two choices for selecting one of the two with the otherindex. In the first case the third field, Φk has two choices, and then Φ` cancouple only to the one remaining, giving 8δijδk`, while in the second case,the Φk can couple to the one field with index paired to the Φi or to the onepaired with Φj , giving 8δikδj` + 8δi`δjk. All together, therefore, the vertex is
i j
k l
= −2iλ(δijδk` + δikδj` + δi`δjk)
The differential cross section for two equal mass particles is
(dσ
dΩ
)CM
=|M|2
64π2E2cm
,
withM = −2λ(δijδk` + δikδj` + δi`δjk).
For i = k = 1, j = ` = 2, the factor in parentheses is just 1, which is also thecase for i = j = 1, k = ` = 2, while for i = j = k = ` = 1, the parenthesisgives 3, so
(dσ
dΩ
)CM
(1 + 2 → 1 + 2) =
(dσ
dΩ
)CM
(1 + 1 → 2 + 2) =λ2
16π2E2cm
,
(dσ
dΩ
)CM
(1 + 1 → 1 + 1) =9λ2
16π2E2cm
.
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 3
(b) The situation is changed if the mass term inthe potential is negative,
V (Φ2) = −1
2µ2Φ2 +
λ
4(Φ2)2.
Then expanding about Φi = 0 is an unstable equi-librium, and a lower energy state would be one for
Φ(v)
v
which we have quantum fluctuations about a configuration with V minimized,that is, with Φ2 = v2 := µ2/λ. Of course the values of Φ with this squareare the values on a hypersphere2 We choose to expand about the point Φ =(0, 0, . . . , 0, v) and define
σ(x) = ΦN (x)− v, πj(x) = Φj for j = 1, ..., N − 1.
In terms of these reexpressed fields, the Lagrangian becomes
L =∫
d3x
1
2∂νσ∂νσ +
1
2
N−1∑j=1
∂νπj∂νπj − V (σ, πj)
with
V (σ, πj) = −1
2µ2[(v + σ)2 + π2
]+
λ
4
[(v + σ)2 + π2
]2(1)
= −1
2µ2v2 − µ2vσ − 1
2µ2σ2 − 1
2µ2π2
+λ
4v4 + λv3σ +
3λ
2v2σ2 + λvσ3 +
λ
4σ4
+λ
2v2π2 + λvσπ2 +
λ
2σ2π2 +
λ
4(π2)2 (2)
−→v=µ/
√λ−1
2
µ4
λ− µ3
√λ
σ − 1
2µ2σ2 − 1
2µ2π2
+µ4
4λ+
µ3
√λ
σ +3µ2
2σ2 + µ
√λσ3 +
λ
4σ4
+µ2
2π2 + µ
√λσπ2 +
λ
2σ2π2 +
λ
4(π2)2 (3)
2I am using “sphere” as mathematicians do, for the locus of points with ~r 2 = R2, whichin freshman physics we call a spherical shell. The locus of points with ~r 2 ≤ R2 is called aball.
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 4
=λ
4(π2)2 +
λ
4σ4 +
λ
2σ2π2
+µ√
λσπ2 + µ√
λσ3
+µ2σ2
−µ4
4λ. (4)
I have written these in terms on the power of fields. The first line containsthe four-particle interactions, while the second line contains three particleinteractions. The next line is the mass term for the σ, but notice that themass associated with the fields πj, that is, the coefficient of π2, vanishes.Also notice that the term linear in σ vanishes, as it must, because we chose vas the minimum of V . Finally there is a shift in the vacuum energy density,which is now negative (relative to the energy of the state with Φj = 0).
Thus if we now do perturbation theory by expanding around interactionpicture fields evolving under the terms quadratic in σ and πj, treating thecubic and quartic terms as the interaction hamiltonian, we have a theorywith one scalar field of mass
√2µ, N − 1 massless fields πj , and interaction
terms
i j
k l
−2iλ(δijδk` +δikδj` + δi`δjk)
i j
−2iλδij
i j
−2iµ√
λδij −6iµ√
λ −6iλ
with propagators
π π: i
p2 + iε
and σ σ:
i
p2 − 2µ2 + iε
(c) There are four diagrams that contribute to the scattering amplitude
iM(πi(p1) + πj(p2) → πk(p3) + π`(p4)
)= iM1 + iM2 + iM3 + iM4, where
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 5
i j
k l
iM1 = −4iµ2λδijδk`
(p1 + p2)2 − 2µ2= 4iµ2λ
δijδk`
2µ2 − s
i j
k l
iM2 = −4iµ2λδikδj`
(p3 − p1)2 − 2µ2= 4iµ2λ
δikδj`
2µ2 − t
i j
k l
iM3 = −iµ2λδi`δjk
(p4 − p1)2 − 2µ2= 4iµ2λ
δi`δjk
2µ2 − u
i j
k l
iM4 = −2iλ(δijδk` + δikδj` + δi`δjk)
where s, t and u are the Mandelstam variables s = (p1 + p2)2, t = (p3− p1)
2,u = (p4−p1)
2, which satisfy s+t+u =∑
i m2i as a consequence of momentum
conservation. Thus in the total scattering amplitude, the term proportionalto δijδk` is
−2iλδijδk`
(1− 2µ2
2µ2 − s
)= +2iλδijδk` s
2µ2 − s.
Similiarly for the t and u terms, so all together
iM = 2iλ
(sδijδk`
2µ2 − s+
tδikδj`
2µ2 − t+
uδi`δjk
2µ2 − u
).
If s, t and u are all << µ2, this can be expanded in powers of s, t and u,
iM = iλ
µ2
(sδijδk`(1 +
s
2µ2) + tδikδj`(1 +
t
2µ2) + uδi`δjk(1 +
u
2µ2)
).
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 6
At threshold, s = t = u = 0, this vanishes. Furthermore, if i = j = k = `,all delta functions are 1, and we have
iM = iλ
µ2
(s + t + u +
s2
2µ2+
t2
2µ2+
u2
2µ2
)
= iλ
2µ4(s2 + t2 + u2),
which vanishes up to order ~p4. Notice if N = 2, then i = j = k = ` = 1, asthey all range only from 1 to N − 1.
(d) If a term ∆V = −aΦN (with N an index, not a power) is added to theHamiltonian, it breaks the O(N) symmetry of the Hamiltonian, but that hasalready been broken by the spontaneous symmetry breaking. However it alsoshifts the point in Φj space which is the minimum of the potential, fromthe point (0, . . . , 0, v0), with v0 = µ/
√λ, to a new point (0, . . . , 0, v), with
v = v0 + b. As V is now
V (ΦN , πj) = −1
2µ2(π2 + ΦN 2
) +λ
4
((π2)2 + 2π2ΦN 2
+ ΦN 4)− aΦN ,
the new minimum is at the point where
∂V
∂ΦN= 0 = −a− µ2ΦN + λΦN 3
,∂V
∂πj= 0 = πj
(−µ2 + λπ2 +
1
2λΦN 2
),
so 0 = −a− µ2v + λv3.If we treat a as a small parameter, a µ3/
√λ, b will also be small and
we can expand v = µ/λ+b to first order in b, which gives 0 = −a−µ2b+3µ2b,or b = a/2µ2.
Again, we set σ = ΦN − v. The expression (2) is still correct except formissing the −a(v + σ) term, but in substatuting its value for v, we pick upadditional terms. Using ∆v = a/2µ2, ∆v2 = a/µ
√λ, ∆v3 = 3a/2λ, and
∆v4 = 2aµ/λ3/2, we find
∆V = −a
(µ√λ
+ σ
)− aµ
2√
λ− a
2σ +
aµ
2√
λ+
3a
2σ +
3a√
λ
2µσ2
+aλ
2µ2σ3 +
a√
λ
2µπ2 +
aλ
2µ2σπ2
= − aµ√λ
+3a√
λ
2µσ2 +
a√
λ
2µπ2 +
aλ
2µ2σ3 +
aλ
2µ2σπ2.
615: Homework Sol. #7 Last Latexed: October 30, 2006 at 14:17 7
We see that the added term in V has produced small shifts in the σ mass,the σππ coupling constant, the σ3 coupling constant, and shifted the vacuumenergy density further downward.
m2σ → 2µ2 +
3a√
λ
µ, gσππ → 2µ
√λ +
aλ
µ2, gπππ → 6µ
√λ +
3aλ
µ2.
But by far the most important effect is that the π particles now have mass
m2π =
a√
λ
µ.
The scattering amplitude at threshold, s = 4m2π = 4a
√λ/µ, t = u = 0 now
has contributions
M1 = g2σππ
δijδk`
m2σ − 4m2
π
, M2 = g2σππ
δikδj`
m2σ
, M3 = g2σππ
δi`δjk
m2σ
,
M4 = −2λ(δijδk` + δikδj` + δi`δjk).
so together we have at threshold
M =
(4µ2λ + 4aλ3/2µ−1
2µ2 + 3a√
λ/µ− 4a√
λ/µ− 2λ
)δijδk`
+
(4µ2λ + 4aλ3/2µ−1
2µ2 + 3a√
λ/µ− 2λ
)(δikδj` + δi`δjk
)
≈ a
(√λ
µ
)3/2 (3δijδk` − δikδj` − δi`δjk
),
which is now non-vanishing and proportional to a.
4.4 v1
4.4 v2
