Masujima M. Applied Mathematical Methods in Theoretical Physics
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548 10 Calculus of Variations: Applications
(a) In order to keep the ϕ
µ
orthogonal and normalized to second order in the
parameters c
σµ
,onemustwrite
ϕ
µ
= ϕ
µ
+
∞
σ>A
ϕ
σ
c
σµ
−
1
2
ν≤A
ϕ
ν
τ>A
c
∗
τν
c
τµ
.
Show that this assures that
dτϕ
∗
µ
ϕ
ν
= δ
µν
to order c
2
.
(b) Wehadtheresultthat
E ≡
φ
|
H
|
φ
=
µ≤A
t
µµ
+
1
2
µ,ν≤A
v
(a)
µν,µν
.
If φ is a solution of the Hartree–Fock variational problem, then
E
≡
φ
H
φ
=
E + δ
(2)
H
,
δ
(2)
H
being a quadratic form in the coefficients, c
σµ
, c
∗
σµ
.
Find an expression for δ
(2)
H
.Usethefactthatifint
µµ
for example, ϕ
µ
is
replaced with ϕ
µ
,onegets
t
µµ
→ t
µµ
+
σ
t
µσ
c
σµ
−
1
2
t
µλ
c
∗
τλ
c
τµ
.
With this recipe, you can easily collect all quadratic terms. The result is
simplified by using the Hartree–Fock equation,
t
αβ
+ U
αβ
≡
t
αβ
+
ν≤A
v
(a)
αν,βν
= ε
α
δ
αβ
.
Show that
δ
(2)
H
=
σµ
(ε
σ
−ε
µ
)
c
σµ
2
+
1
2
c
∗
σµ
c
∗
τν
v
(a)
στ,µν
+
c
∗
σµ
v
(a)
σν,µτ
c
τν
+
1
2
v
(a)
µν,στ
c
σµ
c
τν
.
This shows that for the lowest energy state, one wants all (ε
σ
− ε
µ
) > 0. The
requirement δ
(2)
H
≥ 0 is called the stability condition.
10.20 Problems for Chapter 10 549
10.33. Consider the degeneracy of Hartree–Fock energy. One can show that there
are now trivial variations φ
of φ,forwhichδ
(2)
H
= 0.
(a) This can be seen most easily, if you recognize that, given a determinant φ,
the function,
φ
≡ exp[iλF]φ with F =
A
i=1
f
i
,
(f
i
being a ‘‘one-body operator’’) is also a determinant where λ is a small
parameter. Prove this. Show furthermore that if F is Hermitian, φ
remains
normalized.
(b) With the above φ
, find an expression for δ
H
by a power series expansion
in λ. You get an alternative form for the condition δ
H
= 0. Use it to show
for example that the mean value of the total momentum is zero for a
Hartree–Fock wavefunction φ.
(c) Show that if F is a constant of the motion,
[H, F] = 0,
then all functions φ
= exp[iλF]φ are solutions of the Hartree–Fock problem,
with the same mean energy. Describe in which way, for F = P, φ
differs
from φ.
10.34. Consider a (not very realistic) crude model of a ‘‘nucleus,’’ represented by A
particles of mass m occupying the A lowest levels of a one-dimensional harmonic
oscillator (no spin, no isospin), with the angular frequency, ω.
(a) Let F be the total momentum,
F = P =
A
j=1
p
j
.
Using the result of the mean square value of a one particle operator, calculate
P
2
which is also
(P)
2
in terms of m, ω,andA. (Note that
P
= 0.)
(b) What is the total kinetic energy of the center of mass motion of this system?
Compare this to its total kinetic energy.
(c) Define the center of mass position as
X =
1
A
A
j=1
x
j
.
Find
(X)
2
in terms of m, ω,andA. (Note that
X
= 0.)
550 10 Calculus of Variations: Applications
(d) Show that
B
(P)
2
B
(X)
2
=
2
.
10.35. Consider the quantization of self-interacting scalar field theory with
Lagrangian density,
L =
1
2
ˆ
φ(x)K(x)
ˆ
φ(x) +L
int
(
ˆ
φ) =
1
2
ˆ
φ(x)K(x )
ˆ
φ(x) +
1
4!
ˆ
φ
4
.
(a) Following the canonical procedure, construct the S operator as
S = Texp[i
d
4
x L
int
(
ˆ
φ)].
(b) Following the normal ordering prescription, show that the normal ordered S
operator is given by
: S :=: U exp
i
d
4
x L
int
(
ˆ
φ)
: .
Here we have
U = exp
i
2
d
4
xd
4
y
δ
δ
ˆ
φ(x)
D
F
0
(x − y)
δ
δ
ˆ
φ(y)
,
with
D
F
0
(x − y) =
1
i
0
T(
ˆ
φ(x)
ˆ
φ(y))
0
J=0
.
(c) Introduce the external hook coupling, L
external
(
ˆ
φ) = J(x)
ˆ
φ(x). Making use of
the formula,
exp
i
2
d
4
xd
4
y
δ
δ
ˆ
φ(x)
D
F
0
(x − y)
δ
δ
ˆ
φ(y)
exp
i
d
4
zJ(z)
ˆ
φ(z)
ˆ
φ=0
= exp
−
i
2
d
4
xd
4
yJ(x)D
F
0
(x − y)J(y)
,
obtain the generating functional of Green’s functions in functional integral
form.
10.20 Problems for Chapter 10 551
10.36. Consider the quantization of the neutral ps-ps meson theory with the
external hook coupling, L
external
(
ˆ
φ,
ˆ
ψ,
&
ψ) = J(x)
ˆ
φ(x) +η(x)
ˆ
ψ(x) +
&
ψ(x)η(x). Show
that U operator assumes the following form:
U = exp
i
2
d
4
xd
4
y
δ
δ
ˆ
φ(x)
D
F
0
(x − y)
δ
δ
ˆ
φ(y)
+ i
d
4
xd
4
y
δ
δ
ˆ
ψ
α
(x)
S
F
0,αβ
(x − y)
δ
δ
&
ψ
β
(y)
,
with
S
F
0,αβ
(x − y) =
1
i
0
T(
ˆ
ψ
α
(x)
&
ψ
β
(y))
0
J=η=η=0
.
Obtain the generating functional of Green’s functions in functional integral form.
10.37. Extend Problem 10.36 to the most general renormalizable Lagrangian density
given by
L
tot
=
&
ψ
n
(x)(iγ
µ
D
µ
− m)
n,m
ˆ
ψ
m
(x) +
1
2
D
µ
ˆ
φ(x)
i
D
µ
ˆ
φ(x)
i
−
1
4
ˆ
F
αµν
(x)
ˆ
F
µν
α
(x) +
&
ψ
n
(x)(
i
)
n,m
ˆ
ψ
m
(x)
ˆ
φ
i
(x) + V
ˆ
φ
i
(x)
,
where the potential V(φ(x)) is a locally G invariant quartic polynomial and
D
µ
ˆ
ψ(x)
n
≡
∂
µ
δ
n,m
+ i(t
γ
)
n,m
ˆ
A
γµ
(x)
ˆ
ψ
m
(x),
D
µ
ˆ
φ(x)
i
≡
∂
µ
δ
i,j
+i(θ
γ
)
i,j
ˆ
A
γµ
(x)
ˆ
φ
j
(x),
θ
∗
α
= θ
T
α
=−θ
α
,
∂V
φ(x)
∂φ
i
(x)
(θ
α
)
i,j
φ
j
(x) = 0,
)
t
α
, γ
0
m
*
= 0,
)
t
α
, γ
0
i
*
= γ
0
j
(θ
α
)
j,i
=−(θ
α
)
i,j
γ
0
j
, V
∗
= V, m
†
= γ
0
mγ
0
,
†
i
= γ
0
i
γ
0
.
Naive application of the result of Problem 10.36 to non-Abelian gauge field theory
gives a divergent result for the generating functional (of the connected part) of
Green’s functions. Introduce the Faddeev–Popov determinant in R
ξ
-gauge in the
presence of spontaneous symmetry breaking,
∂V
˜
φ(x)
∂φ
i
(x)
˜
φ
(x)=˜v
= 0, i = 1, ..., n,
F
α
1
φ
a
(x)
2
=
9
ξ
∂
µ
A
µ
α
(x) −
1
ξ
˜η(x)i(θ
α
˜v)
, ξ>0,
F
[
{
φ
a
}
]
$
x
dg(x)
$
α,x
δ
F
α
1
φ
g
a
2
− a
α
(x)
= 1.
552 10 Calculus of Variations: Applications
Obtain the generating functional (of the connected part) of Green’s function in the
functional integral form.
10.38. In quantum mechanics, the operator ordering prescription becomes an
important issue in making the transition from canonical formalism to path integral
formalism. We examine the connection among Weyl correspondence, α-ordering,
and Well-ordered operator in some details. The parameter α connects various
operator ordering prescription continuously.
(a) Define the α-ordered Hamiltonian H
(α)
(q, p) of the quantum Hamiltonian
H
q
(
ˆ
q,
ˆ
p)by
H
(α)
(q, p) ≡
p +
1
2
−α
u
H
q
(
ˆ
q,
ˆ
p)
p −
1
2
+ α
u
exp
iqu
du.
Show that the quantum Hamiltonian H
q
(
ˆ
q,
ˆ
p)isgivenby
H
q
(
ˆ
q,
ˆ
p) =
q +
1
2
+ α
v
q −
1
2
− α
v
H
(α)
(q, p)exp
ipv
dp
2π
dqdv.
(b) Consider the operator ordering prescription specified by
E(
ˆ
p,
ˆ
q;a, b) ≡ exp
i
(a
ˆ
p + b
ˆ
q)
→ exp
i
(ap + qb)
.
With the use of Baker–Campbell–Hausdorff formula,
exp
%
ˆ
A +
ˆ
B
'
= exp
%
ˆ
A
'
exp
%
ˆ
B
'
exp
−
1
2
%
ˆ
A,
ˆ
B
'
,
%
ˆ
A,
ˆ
B
'
= c-number,
as applied to E(
ˆ
p,
ˆ
q;a, b), show that
E
ˆ
p,
ˆ
q;a, b
=
(
exp
[
ia
ˆ
p/
]
exp
[
ib
ˆ
q/
]
exp
[
−iab/2
]
,
exp
[
ib
ˆ
q/
]
exp
[
ia
ˆ
p/
]
exp
[
iab/2
]
.
(c) Compute the c-number function E
(α)
(p, q;a, b) from the definition of the
α-ordering prescription specified in part (a), and using the results of part (b).
Show that E
(α)
(p, q;a, b)isgivenby
E
(α)
(p, q;a, b) = exp[i(ap + bq)/ − iαab/].
(d) Establish the following correspondence:
exp[ia
ˆ
p/]exp[ib
ˆ
q/] → exp[
i
(ap + qb)], (α =+1/2),
exp[
i
(a
ˆ
p + b
ˆ
q)] → exp[
i
(ap + qb)], (α = 0),
exp[ib
ˆ
q/]exp[ia
ˆ
p/] → exp[
i
(ap + qb)], (α =−1/2).
10.20 Problems for Chapter 10 553
(e) Show that, in order to obtain Weyl correspondence (
ˆ
p
m
ˆ
q
n
)
W
,weonlyhaveto
differentiate E(
ˆ
p,
ˆ
q;a, b)withrespecttoa and b, m times and n times,
respectively, and set a = b = 0. For example,
ˆ
p
ˆ
q
W
=
1
2
ˆ
p
ˆ
q +
ˆ
q
ˆ
p
,
ˆ
p
2
ˆ
q
W
=
1
3
ˆ
p
2
ˆ
q +
ˆ
p
ˆ
q
ˆ
p +
ˆ
q
ˆ
p
2
.
(f) Show that, by setting α = 0, we have the Weyl correspondence.
(g) Show that α = 1/2 gives the well-ordered operator, (
ˆ
p
ˆ
q-ordering), and that
α =−1/2 gives the anti-well-ordered operator, (
ˆ
q
ˆ
p-ordering).
(h) Show that, in the α-ordering prescription, the transformation function
q
t
b
, t
b
q
t
a
, t
a
in discrete representation is given by
q
t
b
, t
b
q
t
a
, t
a
=
q
t
n
=q
t
b
q
t
0
=q
t
a
n−1
$
k=1
d
f
q
t
k
9
(2π)
f
n−1
$
k=0
d
f
p
t
k
9
(2π)
f
×exp
i
n−1
k=0
f
r=1
p
r,t
k
(q
r,t
k
+δt
− q
r,t
k
) − H
(α)
(q
(α)
t
k
+δt
, p
t
k
)δt
,
with q
(α)
t
k
+δt
defined by
q
(α)
t
k
+δt
≡
1
2
−α
q
t
k
+δt
+
1
2
+α
q
t
k
.
Note that α = 0 corresponds to the mid-point rule.
10.39. Consider a simple derivation of the one pion-exchange potential from the
symmetric ps–ps meson theory. The Lagrangian density is given by
L =
1
4
%
&
ψ
α
(x), D
αβ
(x)
ˆ
ψ
β
(x)
'
+
1
4
%
D
T
βα
(−x)
&
ψ
α
(x),
ˆ
ψ
β
(x)
'
+
1
2
&
φ(x)K(x)
&
φ(x) −iG
&
ψ
α
(x)τ (γ
5
)
αβ
ˆ
ψ
β
(x)
&
φ(x),
where τ is the isospin matrix,
τ
1
=
01
10
, τ
2
=
0 −i
i 0
, τ
3
=
10
0 −1
.
Here
ˆ
ψ(x) is the iso-doublet nucleon and
&
φ(x) is the iso-triplet pion. The kernels of
the quadratic part of the Lagrangian density are given by
554 10 Calculus of Variations: Applications
D
αβ
(x) = (iγ
µ
∂
µ
− m + iε)
αβ
,
D
T
βα
(−x) = (−iγ
T
µ
∂
µ
− m + iε)
βα
,
K(x) =−∂
2
−µ
2
+iε.
The covariant scattering matrix element M in the static limit and the potential
V(
r) are related to each other in the following manner:
M
fi
(
q) =
V(
r)exp
[
−i
q ·
r
]
d
3
r, V(
r) =
1
(2π)
3
M
fi
(
q)exp[i
q ·
r]d
3
q.
Note that the potential is nothing more than the three-dimensional Fourier
transform of the lowest-order covariant scattering matrix element.
(a) Show that the covariant scattering matrix element for the direct scattering
part is given by
M
(direct)
fi
=−
G
2
τ
(1)
·τ
(2)
u
f
1
γ
5
u
i
1
u
f
2
γ
5
u
i
2
p
f
1
− p
i
1
2
−µ
2
.
(b) In the standard representation of the Dirac γ matrices, the γ
5
is given by
γ
5
= γ
5
=
0 I
I 0
=
0010
0001
1000
0100
.
In the static limit (or nonrelativistic limit), show that
u
f
1
γ
5
u
i
1
=
χ
(s
f
1
)†
, −χ
(s
f
1
)†
σ
(1)
·
p
f
1
2m
0 I
I 0
χ
(s
i
1
)
σ
(1)
·
p
i
1
2m
χ
(s
i
1
)
= χ
(s
f
1
)†
σ
(1)
·
q
2m
χ
(s
i
1
)
,
u
f
2
γ
5
u
i
2
=−χ
(s
f
2
)†
σ
(2)
·
q
2m
χ
(s
i
2
)
.
Here
q is the three-momentum transfer,
q =
p
f
2
−
p
i
2
,
and χs are the two-component spinors.
10.20 Problems for Chapter 10 555
(c) Show that the covariant scattering matrix element for the direct scattering
part is reduced to the following form in the static limit:
M
(direct)
fi
→−
G
2
τ
(1)
·τ
(2)
σ
(1)
·
q
σ
(2)
·
q
(2m)
2
q
2
+µ
2
.
(d) By applying the inverse Fourier transform to the reduced scattering matrix
element, we obtain the one pion-exchange potential in the static limit as
V(
r) = g
2
τ
(1)
·τ
(2)
1
µ
2
σ
(1)
·
∇
σ
(2)
·
∇
exp[−µr]
r
,
where the coupling constant is redefined as
g
2
=
G
2
4π
µ
2
(2m)
2
.
10.40. Consider WKB approximation to Schr
¨
odinger equation in one spatial di-
mension with the use of matrix method.
(a) Schr
¨
odinger equation in one spatial dimension is given by
d
2
dx
2
ψ(x) = w(x)ψ(x)withw(x) =
2m
2
(V(x) − E),
where V(x) is the potential, E is the total energy, and is the Planck constant
divided by 2π. Show that this can be expressed as the first-order ordinary
differential equation with the two-component wavefunction (x), ψ(x), and
dψ(x)dx as its two components, in Dirac equation form
d
dx
(x) = Q(x)(x)withQ(x) =
01
w(x)0
. (A)
(b) Writing the two component wavefunction (x)as
(x) = S(x, x
0
)(x
0
),
the transformation function S(x, x
0
) satisfies the following composition law:
S(x, x
)S(x
, x
0
) = S(x, x
0
).
From the differential equation (A), show that the transformation function
S(x, x
0
)satisfies
S(x + x, x) = 1 +xQ(x) = exp[Q(x)dx]. (B)
556 10 Calculus of Variations: Applications
Hence show that S(x, x
0
)isgivenby
S(x, x
0
) = lim
x=0
x
$
x
0
exp[Q(x)dx] = exp
x
x
0
Q(x)dx
.
(c) We first assume that Q(x) is constant in the interval [x, x + a]. From Eq. (B),
we have
S(x + a, x) = exp[aQ].
We further assume that
w(x) =−s
2
, Q =
01
−s
2
0
.
Establish the identity,
0 s
−1
−s 0
=
s
−12
0
0 s
12
01
−10
s
12
0
0 s
−12
,
and show that Q can be expressed as
Q = s
s
−12
0
0 s
12
01
−10
s
12
0
0 s
−12
.
Show further that
S(x + a, x) =
s
−12
0
0 s
12
exp
01
−10
as
s
12
0
0 s
−12
=
cos as,(1s)sinas
−s sin as,cosas
.
(d) We next divide the interval [x
0
, x] into the subintervals [x
0
, x
1
], [x
1
, x
2
], ...,let
the values of s as s
0
in [x
0
, x
1
], s
1
in [x
1
, x
2
], ...and further let the phases
(x
1
− x
0
)s
0
,(x
2
− x
1
)s
1
, ...as F
0
, F
1
, .... Show that S(x, x
0
) can be expressed as
S(x, x
0
) =
x
$
x
0
S(x
n+1
, x
n
) =
s
12
0
0 s
−12
···exp
01
−10
F
1
×
(s
0
s
1
)
12
0
0(s
1
s
0
)
12
10.20 Problems for Chapter 10 557
× exp
01
−10
F
0
s
−12
0
0
0 s
12
0
. (C)
We now consider the general factor in the product (C),
s
n−1
s
n
14
0
0
s
n
s
n−1
14
exp
01
−10
F
n
s
n
s
n+1
14
0
0
s
n+1
s
n
14
.
We let x be infinitesimal, and represent the coordinate x
n
by x and s
n
by s
so that
s
n−1
s
n
= 1 −x(s
s), s
n+1
s
n
= 1 +x(s
s), F
n
= sx.
Here s is regarded as the continuous and differentiable function s(x)ofx with
s
= dsdx. With this representation, show that the general factor written out
above becomes
1 +x
−s
2ss
−ss
2s
. (D)
WKB approximation consists of regarding
s
2s
|
s
|
. Introducing the
local de Broglie wavelength λ(x) = 2πs(x), show that the condition
s
2s
|
s
|
reads as
λ
4π
ds
dx
s.
Equation (D) becomes
1 +x
0 s
−s 0
.
With this approximation, show that the transformation function S(x, x
0
)
becomes
S(x, x
0
) =
s
12
0
0 s
−12
exp
01
−10
x
x
0
sdx
s
−12
0
0
0 s
12
0
=
(s
0
s)
12
cos F (ss
0
)
−12
sin F
−(ss
0
)
12
sin F (ss
0
)
12
cos F
with F =
x
x
0
sdx .