Masujima M. Applied Mathematical Methods in Theoretical Physics
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538 10 Calculus of Variations: Applications
K(x) =−∂
2
− κ
2
+ iε,
with
{γ
µ
, γ
ν
}=2η
µν
,(γ
µ
)
†
= γ
0
γ
µ
γ
0
,
&
ψ
α
(x) = (
ˆ
ψ
†
(x)γ
0
)
α
.
Hint: Introduce the proper self-energy parts,
∗
(x, y)and
∗
(x, y), and the vertex
operator (x, y, z), and mimic the discussion in Section 10.4.
10.21. Consider the bound state problem of zero total momentum
P = 0forthe
system of identical two fermions of the mass m exchanging a spinless and massless
boson whose Lagrangian density is given by
L =
&
ψ(x)(iγ
µ
∂
µ
− m + iε)
ˆ
ψ(x)
+
1
2
∂
µ
ˆ
φ(x)∂
µ
ˆ
φ(x) −g
&
ψ(x)
ˆ
ψ(x)
ˆ
φ(x).
Define the bound state wavefunction of the two fermions in configuration space by
[U
P
(x)]
αβ
=
0
T[
ˆ
ψ
α
x
2
ˆ
ψ
β
−
x
2
]
B
,
and the bound state wavefunction in momentum space by
[u
P
(p)]
αβ
=
d
4
x exp[ipx][U
P
(x)]
αβ
.
Also define the zero total momentum
P = 0 bound state wavefunction χ(p)in
momentum space by
[u
P=0
(p)]
αβ
=
δ
αβ
χ(p)
p
2
−m
2
+ iε
.
(a) Show that the Bethe–Salpeter equation for the bound state in the ladder
approximation is given by
χ(p) = ig
2
d
4
q
(2π)
4
1
(p − q)
2
+iε
−
1
(p + q)
2
+ iε
χ(q)
q
2
− m
2
+iε
.
(b) Solve this eigenvalue problem by dropping the antisymmetrizing term in the
kernel above. The antisymmetrizing term originates from the spin-statistics
relation for the fermions. This approximation provides the solution in terms
of the Gauss hypergeometric function.
(c) Our discussion so far is covariant. Discuss the covariant normalization of the
bound state wavefunction.
Hint: This problem is discussed in the following article.
Goldstein, J.; Phys. Rev. 91, 1516, (1953).
10.20 Problems for Chapter 10 539
10.22. In Wick–Cutkosky model, we consider the bound state problem for the
system of distinguishable two spinless bosons of equal mass m exchanging a
spinless and massless boson. The Lagrangian density of this model is given by
L =
2
i=1
1
2
∂
µ
ˆ
φ
i
(x)∂
µ
ˆ
φ
i
(x) −
1
2
m
2
ˆ
φ
2
i
(x)
+
1
2
∂
µ
ˆ
φ(x)∂
µ
ˆ
φ(x)
−g
ˆ
φ
†
1
(x)
ˆ
φ
1
(x)
ˆ
φ(x) −g
ˆ
φ
†
2
(x)
ˆ
φ
2
(x)
ˆ
φ(x).
(a) Show that the Bethe–Salpeter equation for the bound state of the two bosons
ˆ
φ
1
(x
1
)and
ˆ
φ
2
(x
2
)isgivenby
S
F
(x
1
, x
2
;B) =
d
4
x
3
d
4
x
4
F
(x
1
−x
3
)
F
(x
2
− x
4
)
×(−g
2
)D
F
(x
3
− x
4
)S
F
(x
3
, x
4
;B),
where
F
(x)andD
F
(x)aregivenby
F
(x) =
d
4
k
(2π)
4
exp[ikx]
k
2
− m
2
+ iε
,
and
D
F
(x) =
d
4
k
(2π)
4
exp[ikx]
k
2
+ iε
.
(b) Transform the coordinates x
1
and x
2
to the center-of-mass coordinate X and
the relative coordinate x by
X =
1
2
(x
1
+x
2
), x = x
1
− x
2
,
and the momentum p
1
,andp
2
to the center-of-mass momentum P and the
relative momentum p by
P = p
1
+ p
2
, p =
1
2
(p
1
− p
2
).
Define the Fourier transform (p)ofS
F
(x
1
, x
2
;B )by
S
F
(x
1
, x
2
;B) = exp[−iPX]
d
4
p exp[−ipx](p).
Show that the above Bethe–Salpeter equation assumes the following form in
momentum space:
P
2
+ p
2
−m
2
P
2
−p
2
− m
2
(p) = ig
2
d
4
q
(2π)
4
(q)
(p − q)
2
+iε
.
540 10 Calculus of Variations: Applications
(c) Assume that (p) can be expressed as
(p) =−
1
−1
g(z)dz
[p
2
+ zpP − m
2
+ (P
2
4) + iε]
3
.
Substitute this expression into the Bethe–Salpeter equation in momentum
space above. Carry out the q integration using the formula,
d
4
q
1
(p − q)
2
+ iε
·
1
[q
2
+ zqP − m
2
+ (P
2
4) + iε]
3
=
iπ
2
2[−m
2
+ (P
2
4) − z
2
(P
2
4)]
·
1
[p
2
+ zpP − m
2
+ (P
2
4) + iε]
,
and compare the result with the original expression for (p)toobtainthe
integral equation for g(z)as
g(z) =
1
0
ςdς
1
−1
dy
1
−1
dx
λg(x)
2(1 − η
2
+ η
2
x
2
)
δ
z −
1
ςy + (1 −ς )x
2
.
The dimensionless coupling constant λ is given by λ = (g4πm)
2
and the
squared mass of the bound state is given by M
2
= P
2
= 4m
2
η
2
,0<η<1.
(d) Performing the ς integration, we obtain the integral equation for g(z)as
g(z) = λ
1
z
dx
1 +z
1 + x
g(x)
2(1 − η
2
+ η
2
x
2
)
+λ
z
−1
dx
1 −z
1 − x
g(x)
2(1 − η
2
+ η
2
x
2
)
.
Observe that g(z) satisfies the boundary conditions
g(±1) = 0.
(e) Differentiating the integral equation for g(z) obtained above twice, reduce it
to the second-order ordinary differential equation for g(z),
d
2
dz
2
g(z) =−
λ
1 − z
2
g(z)
1 −η
2
+ η
2
z
2
with g(±1) = 0.
This is the eigenvalue problem.
(f) In the limit, 1 1 −η>0, the function
1
1 −η
2
+ η
2
z
2
,
has a sharp peak at z = 0. Approximated this function by
1
1 −η
2
+ η
2
z
2
≈ δ(z)
1
−1
dz
1 −η
2
+ η
2
z
2
≈
π
(1 −η
2
)
12
δ(z),
10.20 Problems for Chapter 10 541
and obtain the approximate differential equation for g(z)as
d
2
dz
2
g(z) =−
λπ
(1 − η
2
)
12
g(0)δ(z).
Show that the solution for this differential equation satisfying the boundary
condition g(±1) = 0isgivenby
g(z) =
π
2
1
(1 −η
2
)
12
g(0)λ(1 −
|
z
|
),
which is nodeless and hence represents the lowest eigenfunction. Show that
the lowest approximate eigenvalue is given by
λ ≈ (2π)
9
1 −η
2
.
Hint: We cite the following articles and the following books for the Wick–Cutkosky
model (scalar meson theory), which is a solvable case of the Bethe–Salpeter
equation.
Wick, G.C.; Phys. Rev. 96, 1124, (1954).
Cutkosky, R.E.; Phys. Rev. 96, 1135, (1954).
Nishijima, K.; Fields and Particles: Field Theory and Dispersion Relations,Benjamin
Cummings, Massachusetts, 1969 and 1974, Chapter 7.
Itzykson, C., and Zuber, J.B. ; Quantum Field Theory, McGraw-Hill, New York,
1985, Chapter 10.
10.23. Solve the integro-differential equation,
d
2
X(t)
dt
2
= 2C
β
0
[X(t) −X(s)] exp
[
−w
|
t − s
|
]
ds − f (t),
with the boundary conditions
X(0) = X(β) = 0andf (t) = ik
)
δ(t − τ ) −δ(t − σ )
*
.
(a) Setting
Z (t) ≡
w
2
β
0
exp
[
−w
|
t − s
|
]
X(s)ds,
show that
d
2
Z (t)
dt
2
= w
2
[Z(t) −X(t)],
and that the original integro-differential equation becomes
d
2
X(t)
dt
2
=
4C
w
[X(t) −Z(t)] −f (t).
542 10 Calculus of Variations: Applications
(b) Derive the second-order ordinary differential equation for X(t) −Z(t)and
solve for X(t) −Z(t).
(c) Determine X(t)inthelimitofβ →∞.
10.24. Repeat the analysis of the polaron problem discussed in Section 10.6 with
the following trial action functional:
I
1
=−
1
2
d
q(τ )
dτ
2
dτ −
1
2
C
q(τ ) −
q(σ )
2
exp[−w
|
τ − σ
|
]dτ dσ ,
with w and v defined below as the variational parameters,
v
2
= w
2
+
4C
w
.
Hint for Problems 10.23 and 10.24: the polaron problems are discussed in the
following article:
Feynman, R.P.: Phys. Rev. 97., 660, (1955).
10.25. Solve a system of nonlinear differential equations in QED,
1
d
A
2
d
dς
(d
A
2
) =
4
3
g
2
16π
2
2
h
2
ψ
ψ
d
A
2
,
1
h
ψψ
d
dς
(h
ψψ
) =−
g
2
16π
2
2
h
2
ψ
ψ
d
A
2
,
1
d
dς
() =−
g
2
16π
2
2
h
2
ψ
ψ
d
A
2
with d
A
2
(0) = h
ψψ
(0) = (0) = 1.
10.26. Solve a system of nonlinear differential equations in QCD,
1
d
A
2
d
dς
(d
A
2
) =−
19
4
C
2
(G)
3
g
2
16π
2
2
A
3
d
3
A
2
−
C
2
(G)
12
g
2
16π
2
2
c
cA
h
2
c
c
d
A
2
+
8T(R)
6
g
2
16π
2
2
ψ
ψA
h
2
ψ
ψ
d
A
2
,
1
ccA
d
dς
(
ccA
) =−
C
2
(G)
8
g
2
16π
2
2
c
cA
h
2
c
c
d
A
2
−
3C
2
(G)
8
g
2
16π
2
ccA
A
3
d
2
A
2
h
cc
,
1
h
c
c
d
dς
(h
c
c
) =−
C
2
(G)
2
g
2
16π
2
2
c
cA
h
2
c
c
d
A
2
,
1
h
ψψ
d
dς
(h
ψψ
) =−
C
2
(G)
2
g
2
16π
2
2
ψ
ψA
h
2
ψ
ψ
d
A
2
,
A
3
d
A
2
=
c
c
A
h
c
c
,
A
3
d
A
2
=
ψψA
h
ψψ
,
d
A
2
(0) = h
c
c
(0) = h
ψψ
(0) =
c
c
A
(0) = 1. (10.20.10)
Hint for Problems 10.25 and 10.26: Many unknowns satisfy the identical nonlinear
differential equation.
10.27. In the Glashow–Weinberg–Salam model, there exist two options to intro-
duce the heavy leptons to eliminate Z
νν coupling.
10.20 Problems for Chapter 10 543
(a) We introduce the charged heavy lepton E
+
and form the left-handed triplet
given by
left-handed :
L =
E
+
ν
e
−
L
, Y
weak hypercharge
= 0,
and the right-handed SU(2)
weak isospin
singlets,
e
−
R
, Y
weak hypercharge
=−2andE
+
R
, Y
weak hypercharge
=+2.
Show that the neutral current is given by
j
3
µ
= E
+
L
γ
µ
E
+
L
−e
−
L
γ
µ
e
−
L
,
which contains no νγ
µ
ν term and hence neither A
µ
nor Z
µ
couple to the
neutrinos.
(b) We can further introduce the neutral heavy lepton E
0
. The right-handed
singlet E
0
R
has Y
weak hypercharge
= 0. The two left-handed doublets are given by
(ν + E
0
)
√
2
e
−
L
, Y
weak hypercharge
=−1,
and
E
+
(ν − E
0
)
√
2
L
, Y
weak hypercharge
=+1.
Show that neither the hypercharge current j
weak hypercharge
µ
nor the neutral
current j
3
µ
contains νγ
µ
ν term.
(c) Incorporate the hadrons with the SU(3)
color
quark model in the models in
parts (a) and (b).
We note that the model in part (a) is known as LPZ model and the model in
part (b) is known as PZ II model.
Hint for Problem 10.27: This problem is discussed in the following article.
Prentki, J. and Zumino, B. : Nucl. Phys. B47, 99, (1972).
10.28. One model of the elementary particle interactions which avoids the neutral
current entirely aside from the electromagnetic current was proposed by Georgi
and Glashow. The gauge group is O(3).
(a) Show that with a triplet Higgs scalar field
φ, the Higgs–Kibble mechanism
gives a mass to the charged bosons, but leaves the neutral boson massless.
544 10 Calculus of Variations: Applications
(b) In order to incorporate the leptons, the following scheme is suggested:
left-handed :
L =
E
+
ν sin β + E
0
cos β
e
−
L
,right-handed:
R =
E
+
E
0
e
−
R
,
(E
0
sin β − ν cos β)
L
is a singlet.
Show that the neutral vector boson couples in a parity-conserving manner.
Calculate the effective weak coupling constant G
F
in terms of e, β,andM
W
.
Derive an upper bound for M
W
,
M
W
≤
e
2
√
2
4G
F
.
(c) Write down the most general mass matrix that can arise from the explicit
mass terms as well as from the Yukawa couplings to the Higgs scalar field of
the form
−L
mass
= m
0
(
L
R +
R
L) +G
1
[
L
T
φ
R + h.c.]
+ G
2
[(E
0
sin β − ν cos β)
L
φ
R + h.c.].
By diagonalizing the mass matrix, derive the relation
M
E
+
+ M
e
−
= 2cosβM
E
0
.
Hint for Problem 10.28: This problem is discussed in the following article.
Georgi, H. and Glashow, S.L. : Phys. Rev. Lett. 28, 1494, (1972).
The Georgi–Glashow O(3) model was ruled out experimentally due to the
discovery of the neutral current. There exist many models of the unification of
weak and electromagnetic interactions which were ruled out.
10.29. One important feature of the Glashow–Weinberg–Salam model is the exis-
tence of the weak neutral leptonic current. To expose some striking consequences of
the existence of weak neutral leptonic currents, consider the elastic e + ν → e + ν
cross sections to the lowest order in the Glashow–Weinberg–Salam model. In
the limit where the incident neutrino energy E
ν
is small as compared with the
masses of the W and Z bosons, m
W
±
and m
Z
0
, show that the effective interaction
Lagrangian density is given by
L
eff
int
=−
1
√
2
G
F
1
ν
e
γ
ρ
(1 −γ
5
)e][eγ
ρ
(1 −γ
5
)ν
e
] + (ν
µ
γ
ρ
ν
µ
+ν
e
γ
ρ
ν
e
)(2 sin
2
θ
W
e
R
γ
ρ
e
R
− cos 2θ
W
e
L
γ
ρ
e
L
)
2
,
10.20 Problems for Chapter 10 545
where the first term and the second term represent the W
±
contribution and the
Z
0
contribution, respectively.
After a Fierz transformation on the first term, show that the effective interaction
Lagrangian density is
L
eff
int
=−
√
2G
F
ν
e
γ
ρ
1 −γ
5
2
ν
e
eγ
ρ
1 −γ
5
2
+2sin
2
θ
W
e
+
ν
µ
γ
ρ
1 −γ
5
2
ν
µ
[
eγ
ρ
−
1 − γ
5
2
+ 2sin
2
θ
W
e]
.
Show that a general effective interaction Lagrangian density,
L
eff
int
=−
√
2G
F
νγ
ρ
1 −γ
5
2
ν
eγ
ρ
C
L
1 − γ
5
2
+ C
R
1 +γ
5
2
e
,
where C
L
and C
R
are real coefficients, leads to a differential cross-section formula
in the center of mass frame,
dσ
d
=
G
2
F
E
2
ν
(2π)
2
)
C
2
L
(p · q)
2
+ C
2
R
(p · q
)
2
− C
L
C
R
m
2
e
(q · q
)
2
*
,
for the elastic scattering process, e(p) +ν(q) → e(p
) +ν(q
). Show that, for the elas-
tic scattering process, e(p) +ν(q) → e(p
) + ν(q
), the expression for the differential
cross section is obtained by interchanging the coefficients, C
L
and C
R
.
When the incident neutrino energy is much larger than the electron mass in the
center of mass frame, show that we have
(p · q)
2
= m
2
e
E
2
ν
,(p · q
)
2
= m
2
e
E
2
ν
(1 −
E
e
E
ν
)
2
,
m
2
e
q · q
m
2
e
p · p
= m
2
e
E
2
ν
E
e
E
ν
m
e
E
ν
,
and the last term above may be neglected. Obtain the differential cross sections,
dσ
d
(
ν
e
e → ν
e
e) =
G
2
F
E
2
ν
(2π)
2
4sin
4
θ
W
+ (1 +2sin
2
θ
W
)
2
1 +cos θ
2
2
,
dσ
d
(ν
e
e → ν
e
e) =
G
2
F
E
2
ν
(2π)
2
(1 +2sin
2
θ
W
)
2
+4sin
4
θ
W
1 +cos θ
2
2
,
dσ
d
(
ν
µ
e → ν
µ
e) =
G
2
F
E
2
ν
(2π)
2
4sin
4
θ
W
+ (1 −2sin
2
θ
W
)
2
1 +cos θ
2
2
,
dσ
d
(ν
µ
e → ν
µ
e) =
G
2
F
E
2
ν
(2π)
2
(1 −2sin
2
θ
W
)
2
+4sin
4
θ
W
1 +cos θ
2
2
,
546 10 Calculus of Variations: Applications
and the total cross sections,
σ (ν
e
e → ν
e
e) =
G
2
F
E
2
ν
π
4sin
4
θ
W
+
1
3
(1 +2sin
2
θ
W
)
2
,
σ (ν
e
e → ν
e
e) =
G
2
F
E
2
ν
π
(1 +2sin
2
θ
W
)
2
+
4
3
sin
4
θ
W
,
σ (
ν
µ
e → ν
µ
e) =
G
2
F
E
2
ν
π
4sin
4
θ
W
+
1
3
(1 −2sin
2
θ
W
)
2
,
σ (ν
µ
e → ν
µ
e) =
G
2
F
E
2
ν
π
(1 −2sin
2
θ
W
)
2
+
4
3
sin
4
θ
W
.
10.30. Let
1, 2, ..., A
|
be an antisymmetric wavefunction for A identical parti-
cles. ‘‘1’’ stands for all coordinates (space, spin, isospin) of particle 1.
One then defines the density matrices ρ
1
and ρ
12
as
ρ
1
:(1
|
ρ
1
|
1
) = A
dτ
2
· dτ
A
1
,2,..., A
1
,2,..., A
,
ρ
12
:
(1
,2
|
ρ
12
|
1
,2
) = A(A − 1)
dτ
3
· dτ
A
1
,2
,3,..., A
1
,2
,3,..., A
.
(a) Show that for any one-body operator
F =
A
i=1
f
i
,
one has
|
F
|
= tr(ρ
1
f
1
) =
dτ
1
dτ
1
1
ρ
1
1
f
1
.
(b) Show that for any two-body operator
F =
A
i<j1
f
ij
,
one has
|
F
|
=
1
2
tr(ρ
12
f
12
) =
1
2
dτ
1
..dτ
2
1, 2
ρ
1
,2
1
,2
f
1, 2
.
(c) The Hamiltonian of A identical particles is
H = T + V =
i
t
i
+
i<j
v
ij
.
10.20 Problems for Chapter 10 547
In terms of ρ
1
and ρ
12
, we get the expression for
E =
|
H
|
.
10.31. Consider a complete and orthogonal set of one particle wavefunction
i
|
α
= ϕ
α
(
x
i
, s
i
, t
i
),
dτ
i
α
|
i
i
|
β
= δ
αβ
, orthogonality of the state,
α
i
|
α
α
|
j
= δ(i, j), completeness of the state.
The simplest A-particle wavefunction is the Slater determinant,
1, 2, ..., A
|
φ
=
1
√
A!
det
i
|
µ
,
(
i = 1, 2, ..., a,
µ = α
1
, α
2
, ..., α
A
.
Prove the following statements, which are valid for φ:
(a)
1
ρ
1
=
A
µ=1
1
µ
µ
1
,
(b)
1, 2
ρ
1
,2
=
1
ρ
1
2
ρ
2
−
1
ρ
2
2
ρ
1
,
(c)
(ρ
1
)
2
= ρ
1
,
(d)
tr(ρ
1
) = A,
(e)
tr
2
(ρ
12
) = (A − 1)ρ
1
.
10.32. Consider the stability of Hartree–Fock solution. The variation of the de-
terminantal trial function φ is achieved by varying the individual ‘‘orbitals’’ (≡
single-particle functions), ϕ
µ
(µ = 1, 2, ..., A);
ϕ
µ
→ ϕ
µ
= ϕ
µ
+
∞
σ>A
ϕ
σ
c
σµ
.