4.1 v1
4.1 v2
4.2 v1
4.2 v2
4.3 v1
1
Physics 615
Oct. 26, 2006
Homework Solutions #7
1
[20 pts] Do problem 4.3 from Peskin and Schroeder. This is one of the
simpler examples of a model with spontaneous symmetry breaking.
Solution 1 (a) We are going to explore the linear sigma model, based
on N real scalar fields1 Φi (x) and their conjugate momenta Πi , with the
Hamiltonian
Z
H=
!
1X i 2 1X ~ i 2
dx
(Π ) +
(∇Φ ) + V (Φ2 ) ,
2 i
2 i
3
with
1
λ
V (Φ2 ) = m2 Φ2 + (Φ2 )2 .
2
4
P
2
i 2
Note that here Φ means i (Φ ) and not, of course, the i = 2 component of
Φ. This Hamiltonian (and the associated Lagrangian) are clearly invariant
under global rotations of Φi in the N dimensional space.
When m2 > 0 and λ → 0, the Hamiltonian reduces to a sum of N independent pieces each depending on just one of the N real scalar fields Φi , and
each piece is simply the free Klein-Gordon Hamiltonian. Thus the interaction
picture fields can be expanded as usual in terms of N sets of creation and
annihilation operators aip~† and aip~ , with the different fields commuting with
each other,
h
i
aip~, ajp~0† = (2π)3 δ ij δ 3 (~
p − ~p 0 ).
Thus the contractions of Φi with Φj give 0 unless i = j, in which case they
give the usual free Klein Gordon propagator
DF (x−y) = φI (x)φI (y) = h0| T φI (x)φI (y) |0i =
1
Z
d4 p
1
e−ip(x−y) .
4
2
2
(2π) p − m + i
I will use the book’s notation with a superscript i, Φi , though I don’t understand why
they 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 brought
down from the exponential gives
λ
−i
4
Z
d4 z
X
Φa (x)Φa (x)Φb (x)Φb (x).
ab
When this factor is contracted with 4 other fields Φi Φj Φk Φ` , (or external
states |~p, ji), there are 4! ways these other fields can choose among the four
in 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 δ ij , or it has two choices for selecting one of the two with the other
index. In the first case the third field, Φk has two choices, and then Φ` can
couple only to the one remaining, giving 8δ ij δ k` , while in the second case,
the Φk can couple to the one field with index paired to the Φi or to the one
paired with Φj , giving 8δ ik δ j` + 8δ i` δ jk . All together, therefore, the vertex is
k
l
= −2iλ(δ ij δ k` + δ ik δ j` + δ i` δ jk )
j
i
The differential cross section for two equal mass particles is
dσ
dΩ
with
!
=
CM
|M|2
,
2
64π 2 Ecm
M = −2λ(δ ij δ k` + δ ik δ j` + δ i` δ jk ).
For i = k = 1, j = ` = 2, the factor in parentheses is just 1, which is also the
case for i = j = 1, k = ` = 2, while for i = j = k = ` = 1, the parenthesis
gives 3, so
dσ
dΩ
dσ
dΩ
!
(1 + 2 → 1 + 2) =
!CM
dσ
dΩ
!
(1 + 1 → 2 + 2) =
CM
9λ2
(1 + 1 → 1 + 1) =
.
2
16π 2 Ecm
CM
λ2
,
2
16π 2 Ecm
615: Homework Sol. #7
3
Last Latexed: October 30, 2006 at 14:17
(b) The situation is changed if the mass term in
the potential is negative,
Φ(v)
1
λ
V (Φ2 ) = − µ2 Φ2 + (Φ2 )2 .
2
4
v
Then expanding about Φi = 0 is an unstable equilibrium, and a lower energy state would be one for
which we have quantum fluctuations about a configuration with V minimized,
that is, with Φ2 = v 2 := µ2 /λ. Of course the values of Φ with this square
are the values on a hypersphere2 We choose to expand about the point Φ =
(0, 0, . . . , 0, v) and define
σ(x) = ΦN (x) − v,
π j (x) = Φj
for j = 1, ..., N − 1.
In terms of these reexpressed fields, the Lagrangian becomes
Z
L =


−1
1
1 NX
ν

∂ν σ∂ σ +
dx
∂ν π j ∂ ν π j − V (σ, {π j })
2
2 j=1
3
with
V (σ, {π j })
=
=
−→√
v=µ/ λ
i
i2
1 h
λh
− µ2 (v + σ)2 + π 2 +
(v + σ)2 + π 2
2
4
1 2 2
1 2 2 1 2 2
2
− µ v − µ vσ − µ σ − µ π
2
2
2
λ 4
3λ 2 2
λ
3
+ v + λv σ + v σ + λvσ 3 + σ 4
4
2
4
λ 2 2
λ
λ
+ v π + λvσπ 2 + σ 2 π 2 + (π 2 )2
2
2
4
µ3
1 2 2 1 2 2
1 µ4
−√ σ− µ σ − µ π
−
2 λ
2
2
λ
√
µ3
3µ2 2
µ4
λ
σ + µ λσ 3 + σ 4
+ +√ σ+
4λ
2
4
λ
2
√
µ
λ
λ
+ π 2 + µ λσπ 2 + σ 2 π 2 + (π 2 )2
2
2
4
(1)
(2)
(3)
2
I am using “sphere” as mathematicians do, for the locus of points with ~r 2 = R2 , which
in freshman physics we call a spherical shell. The locus of points with ~r 2 ≤ R2 is called a
ball.
615: Homework Sol. #7
=
Last Latexed: October 30, 2006 at 14:17
λ 2 2 λ 4 λ 2 2
(π ) + σ + σ π
4
2
√ 42
√
+µ λσπ + µ λσ 3
+µ2 σ 2
µ4
− .
4λ
4
(4)
I have written these in terms on the power of fields. The first line contains
the four-particle interactions, while the second line contains three particle
interactions. The next line is the mass term for the σ, but notice that the
mass associated with the fields π j , that is, the coefficient of π 2 , vanishes.
Also notice that the term linear in σ vanishes, as it must, because we chose v
as 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 interaction
picture fields evolving under the terms quadratic in σ and π j , treating the
cubic and quartic terms as the
√ interaction hamiltonian, we have a theory
with one scalar field of mass 2 µ, N − 1 massless fields π j , and interaction
terms
k
l
j
i
−2iλ(δ ij δ k` +
δ ik δ j` + δ i` δ jk )
j i
i
−2iλδ ij
j
√
−2iµ λδ ij
√
−6iµ λ
−6iλ
with propagators
π π:
p2
i
+ i
and σ σ:
11
00
p2
i
− 2µ2 + i
(c) There
are four diagrams that contribute
to the scattering amplitude
i
j
k
`
iM π (p1 ) + π (p2 ) → π (p3 ) + π (p4 ) = iM1 + iM2 + iM3 + iM4 , where
615: Homework Sol. #7
k
l
i
j
k
l
i
k
5
Last Latexed: October 30, 2006 at 14:17
iM1 = −4iµ2 λ
δ ij δ k`
δ ij δ k`
2
=
4iµ
λ
(p1 + p2 )2 − 2µ2
2µ2 − s
iM2 = −4iµ2 λ
δ ik δ j`
δ ik δ j`
2
=
4iµ
λ
(p3 − p1 )2 − 2µ2
2µ2 − t
j
l
iM3 = −iµ2 λ
i
j
k
l
δ i` δ jk
δ i` δ jk
2
=
4iµ
λ
(p4 − p1 )2 − 2µ2
2µ2 − u
iM4 = −2iλ(δ ij δ k` + δ ik δ j` + δ i` δ jk )
i
j
where s, t and u are the Mandelstam variables s = (p1 + p2 )2 , t = (p3 − p1 )2 ,
P
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 proportional
to δ ij δ k` is
ij k`
−2iλδ δ
2µ2
1− 2
2µ − s
!
= +2iλδ ij δ k`
s
.
−s
2µ2
Similiarly for the t and u terms, so all together
!
sδ ij δ k`
tδ ik δ j`
uδ i` δ jk
iM = 2iλ
.
+
+
2µ2 − s 2µ2 − t 2µ2 − u
If s, t and u are all << µ2 , this can be expanded in powers of s, t and u,
!
λ
s
t
u
iM = i 2 sδ ij δ k` (1 + 2 ) + tδ ik δ j` (1 + 2 ) + uδ i` δ jk (1 + 2 ) .
µ
2µ
2µ
2µ
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
!
λ
s2
t2
u2
iM = i 2 s + t + u + 2 + 2 + 2
µ
2µ
2µ
2µ
λ
= i 4 (s2 + t2 + u2 ),
2µ
which vanishes up to order p~4 . Notice if N = 2, then i = j = k = ` = 1, as
they all range only from 1 to N − 1.
(d) If a term ∆V = −aΦN (with N an index, not a power) is added to the
Hamiltonian, it breaks the O(N) symmetry of the Hamiltonian, but that has
already been broken by the spontaneous symmetry breaking. However it also
shifts the point in {Φj } space which is
√ the minimum of the potential, from
the point (0, . . . , 0, v0), with v0 = µ/ λ, to a new point (0, . . . , 0, v), with
v = v0 + b. As V is now
1
λ 2 2
2
2
4
V (ΦN , π j ) = − µ2 (π 2 + ΦN ) +
(π ) + 2π 2 ΦN + ΦN − aΦN ,
2
4
the new minimum is at the point where
∂V
1 N2
∂V
2 N
N3
j
2
2
= 0 = −a − µ Φ + λΦ ,
= 0 = π −µ + λπ + λΦ
,
∂ΦN
∂π j
2
so 0 = −a − µ2 v + λv 3 .
√
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−µ2 b+3µ2 b,
or b = a/2µ2 .
Again, we set σ = ΦN − v. The expression (2) is still correct except for
missing the −a(v + σ) term, but in substatuting its√value for v, we pick up
additional terms. Using ∆v = a/2µ2 , ∆v 2 = a/µ λ, ∆v 3 = 3a/2λ, and
∆v 4 = 2aµ/λ3/2 , we find
√
!
aµ
3a λ 2
a
3a
µ
aµ
σ
∆V = −a √ + σ − √ − σ + √ + σ +
2
2µ
λ
2 λ 2
2 λ
√
aλ 3 a λ 2
aλ
+ 2σ +
π + 2 σπ 2
2µ
2µ
2µ
√
√
3a λ 2 a λ 2
aµ
aλ
aλ
σ +
π + 2 σ 3 + 2 σπ 2 .
= −√ +
2µ
2µ
2µ
2µ
λ
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 σ 3 coupling constant, and shifted the vacuum
energy density further downward.
√
√
√
3a λ
aλ
3aλ
2
2
mσ → 2µ +
gπππ → 6µ λ + 2 .
,
gσππ → 2µ λ + 2 ,
µ
µ
µ
But by far the most important effect is that the π particles now have mass
√
λ
a
m2π =
.
µ
√
The scattering amplitude at threshold, s = 4m2π = 4a λ/µ, t = u = 0 now
has contributions
2
M1 = gσππ
δ ij δ k`
,
m2σ − 4m2π
2
M2 = gσππ
δ ik δ j`
,
m2σ
2
M3 = gσππ
δ i` δ jk
,
m2σ
M4 = −2λ(δ ij δ k` + δ ik δ j` + δ i` δ jk ).
so together we have at threshold
!
4µ2 λ + 4aλ3/2 µ−1
√
√
M =
− 2λ δ ij δ k`
2µ2 + 3a λ/µ − 4a λ/µ
!
4µ2 λ + 4aλ3/2 µ−1
√
− 2λ δ ik δ j` + δ i` δ jk
+
2µ2 + 3a λ/µ
√ !3/2
λ
≈ a
3δ ij δ k` − δ ik δ j` − δ i` δ jk ,
µ
which is now non-vanishing and proportional to a.
4.4 v1
4.4 v2