Masujima M. Applied Mathematical Methods in Theoretical Physics
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2.7 Problems for Chapter 2 95
Q = i
ε
|
A
ε
|
2
d
dε
ln det
G
−
(ε)G
+
0
(ε)
G
+
(ε)G
−
0
(ε)
.
We write
G
+
0
[G
+
]
−1
= 1 −G
+
0
V, G
−
0
[G
−
]
−1
= 1 −G
−
0
V; V ≡ H − K.
Then
D
G
≡ det
G
−
0
[G
−
]
−1
G
+
0
[G
+
]
−1
= det
(1 −G
−
0
V)
1
1 − G
+
0
V
.
Now,
1
1 −G
+
0
V
= 1 +G
+
0
V
1
1 −G
+
0
V
= 1 +G
+
0
T,
where
T(ε) = V
1
1 −G
+
0
(ε)V
is the scattering matrix. Show that
D
G
= det
1 +(G
+
0
− G
−
0
)V
1
1 −G
+
0
V
= det [1 −2π iδ(ε − K)T(ε)]
= det [δ
ε
,ε
− 2π iδ
ε
,ε
T
ε
,ε
(ε)].
Show further that
D
G
= 1 −2π iT
ε,ε
(ε) = S(ε),
where S(ε) is the eigenvalue of the S-matrix for the state (s, ε). Show
that
Q =−i
ε
|
A
ε
|
2
d
dε
ln S(ε).
For a wave packet of sufficiently small energy spread, show that
Q(ε) =−i
d
dε
ln S(ε). (G)
Show also that, by writing
S = exp [2iδ(ε)],
96 2 Integral Equations and Green’s Functions
where δ(ε)isaphaseshift,
Q = 2
dδ(ε)
dε
=−iS
−1
d
dε
S.
(c) Let us consider the most general conditions under which the
scattering wavefunction ψ
+
s,ε
will have the asymptotic form
ψ
+
s,ε
= [I
s,ε
−S
s
(ε)O
s,ε
], (H)
when the interacting systems are separated at great distances. Here I
s,ε
represents an incoming wave and O
s,ε
an outgoing wave. We assume
that the S-matrix is diagonal in the (s, ε) representation. We do not
necessarily suppose that ψ
+
s,ε
satisfies a Schr
¨
odinger equation,
however. The wave packet amplitude A(ε −
ε)willbeconsideredtobe
centered about a mean energy
ε andtoextendoveronlyanarrow
range of energy. Then the time-dependent wavefunction is
s
(t) =
ε
A(ε − ε)ψ
+
s,ε
exp [−iεt].
The asymptotic outgoing wave packet of this is obtained from (H) as
out
(t) =−
ε
A(ε − ε)S(ε)O
s,ε
exp [−iεt].
Show that
S(ε) = exp [ln S(ε)]
= exp
ln S(
ε) +ε
d
dε
ln S(
ε)
= S(ε)exp
ε
d
dε
ln S(
ε)
,
with ε ≡ ε − ε.Wethushave
out
(t) =−S(ε)exp[−iεt]
ε
A(ε)O
s,ε+ε
exp
−i
t + i
d
dε
ln S(
ε)
ε
.
Show that the interaction between the particles leads to a time delay,
Q =−i
d
dε
ln S(
ε), (I)
in the appearance of the asymptotic wave packet.
(d) We label S by the channel c and energy ε only. Then, the S-matrix has
the usual form
2.7 Problems for Chapter 2 97
*
c
;ε
S
c;ε
%
= δ
(c
;ε
),(c;ε)
−2πiδ(ε
− ε)T
c
c
(ε)
= δ
ε
,ε
[δ
c
,c
− 2π iT
c
c
(ε)] ≡ δ
ε
,ε
S
c
c
(ε).
Here T
c
c
is the element of the T matrix connecting channels c and c
.
Since S is unitary, it may be diagonalized. Let it then be diagonal in a
representation for which the states are labeled as (s, ε). That is,
*
s
, ε
S
s, ε
%
= δ
ε
,ε
δ
s
,s
[1 −2π iT
s
(ε)] ≡ δ
ε
,ε
δ
s
,s
S
s
(ε).
If we write U
cs
for the unitary matrix which transforms from the s to
the c representation, then
S
c
c
(ε) =
s
U
c
s
S
s
(ε)U
−1
sc
, δ
s
,s
S
s
(ε) =
c,c
U
−1
s
c
S
c
c
(ε)U
cs
.
If the scattering in the state under consideration is sufficiently larger
than that in the other states, the cross section for scattering from
channel c to channel c
is
σ
c
c
=
π
κ
2
c
2J + 1
(2S
1
+ 1)(2S
2
+1)
S
c
c
(ε) −δ
c
,c
2
.
Here κ
c
is the momentum of either of the two particles in the incident
channel c,whereasS
1
and S
2
are their respective spins.
Let us ask what will be the consequence if the lifetime Q in a particular
eigenstate (s, ε, j, ...) has a maximum value at a given energy ε = ε
0
,
but that strong scattering does not occur except near ε = ε
0
in this
state. We shall assume that Q(ε) is large and positive near ε = ε
0
,but
decreases to a value of the order of the ‘‘free flight time’’ when ε is not
close to ε
0
.
AlargevalueofQ for ε ≈ ε
0
means, of course, that the ‘‘particles stick
together’’ for an extended time. Show that Q
−1
canbeexpandedina
Taylor series of the form
1
Q
= α + β(ε − ε
0
)
2
+ γ (ε − ε
0
)
3
+···,(J)
where α, β, γ are constants and α, β>0. Normally, we might expect
that
β = O(α/ε
2
0
), γ = O(β/ε
0
), ....
98 2 Integral Equations and Green’s Functions
Our assumption of a long lifetime near the energy ε
0
will be
interpreted more precisely to mean that
α ε
2
0
β, γ = O(β/ε
0
). (K)
The conditions (K) imply that Q
−1
≈ α + β(ε − ε
0
)
2
for a range of
energies such that
|
ε − ε
0
|
O(ε
0
). Let us return to (J) and set
Z ≡ ε − ε
0
.Wethenwrite
1
Q
= i
dZ
d ln S
= α + βZ
2
+F(Z ), F(Z) ≡ γ Z
3
+···. (L)
For convenience, we shall drop the state label s on the S-matrix for the
moment. The term F(Z) will be considered as small over the range of
Z used.
Integrate (L) and show that
ln S =
i
√
αβ
tan
−1
√
α/β
ε
0
− ε
+ 2iν(ε), (M)
where
2ν(ε) =−
ε−ε
0
F(Z)dZ
(α + βZ
2
)[α + βZ
2
+ F(Z)]
,
the lower limit on the integral being chosen in a manner consistent
with (M).
It is convenient to replace α and β with two new parameters r and ,
defined by the equations
r ≡
1
2
(αβ)
−1/2
, ≡ 2
α/β.
Then (M) takes the form
S = exp
2ri tan
−1
/2
ε
0
− ε
+ 2iν(ε)
. (N)
The condition (K) implies that /ε
0
1, so the term involving
tan
−1
!
/2
ε
0
−ε
"
in (N) will be negligible except for ε ε
0
.Wehave
specified that the scattering in the state being considered is small
except (possibly) near the energy ε
0
. This implies that ν(ε)willnotgive
a very important contribution to S when ε ε
0
.Toestimatethe
contribution from ν(ε), we take r = 1, so α = /4, β = 1/.Thenfor
ε − ε
0
= /2, the series (J) is
2.7 Problems for Chapter 2 99
1
Q
=
4
+
4
+
4
2ε
0
+···. (O)
Condition (K) implies that the higher terms in the series (O) are
negligible. The scattering matrix T(ε) is defined and analytic when ε
has a sufficiently small positive imaginary part, since T was obtained
as a function of ε + iη in the limit η → 0(+). From this, we see that
S(ε) is analytic in a domain above and bounded by the real axis. If is
sufficiently small, the point ε = ε
0
+ i/2 will lie in this domain of
analyticity. In order to understand the implication of this, let us write
tan
−1
/2
ε
0
− ε
=
i
2
ln
ε − ε
0
+i
2
− ln
ε − ε
0
− i
2
.
(P)
Now, restricting ε to the domain of analyticity of S(ε), let the point ε
moveupfromtherealaxis,describeacircleaboutthepoint
ε
0
+ i(/2), and then return to its initial point on the real axis. Show
that the expression (P) does not return to its initial value, however, but
acquires an additional term,
tan
−1
/2
ε
0
− ε
→ tan
−1
/2
ε
0
− ε
+ π.
Then, moving ε around the closed contour transforms S according
to S → S exp [2πir]. This violates the condition that S(ε) be analytic
in the domain in which the contour lies, unless r is an integer.
We therefore must have r = 0, 1, 2, .... Of principal physical
interest is the case that r = 1.
Let us suppose that r = 1. From (N), writing
tan δ(ε) =
/2
ε
0
− ε
,
show that
S(ε) ≡ exp [2iδ(ε)] exp [2iν(ε)] =
1 +i tan δ
1 −i tan δ
exp [2iν(ε)]
= 1 +
i
ε
0
− ε − i(/2)
+
ε
0
− ε + i(/2)
ε
0
− ε − i(/2)
(exp [2iν] −1).
(Q)
The second term here represents ‘‘resonance scattering,’’ whereas the
third term is called ‘‘potential scattering.’’ By hypothesis, it gives only
a small contribution to S for ε ε
0
.
To discuss the scattering cross section, it is convenient to reinsert
the state label s on S(ε), writing S
s
(ε). Let the state for which S has
100 2 Integral Equations and Green’s Functions
the form (Q) be s
0
. For simplicity, let us neglect the scattering in other
states and the potential scattering in the state s
0
. Then, we may take
S
s
= 1fors = s
0
; S
s
0
= 1 +
i
ε
0
− ε − i(/2)
.
Show that S
c
c
in the channel representation is
S
c
c
= δ
c
c
+i
U
c
s
0
U
∗
cs
0
ε
0
− ε − i(/2)
.
Obtain the Breit–Wigner resonance scattering cross section as
σ
c
c
=
π
κ
2
c
2J + 1
(2S
1
+1)(2S
2
+ 1)
c
c
(ε
0
− ε)
2
+
2
/4
,
with the channel width defined by
c
≡
U
cs
0
2
,
c
≡
U
c
s
0
2
,
and the full width at half the maximum defined by
≡
c
c
.
2.30. Consider the self-interacting Schr
¨
odinger field through the nonlocal
self-interaction:
i
∂
∂t
+
2
2m
∇
2
ϕ(
x , t) =
d
x
V(
x,
x
)ϕ(
x
, t),
V(
x,
x
) =−λ
0
ρ(
x )ρ(
x
),
λ
0
> 0, ρ(
x ) = agivenc-number real function.
(a) In order to obtain the c-number normal mode, set
ϕ(
x , t) ∼ u(
x, J)exp[−iω
J
t],
and convert the differential equation into an integral equation:
u(
x, J) = u
0
(
x, J) −λ
0
d
x
G
+
(
x −
x
;ω
J
)ρ(
x
)(ρu),
with
G
+
(
x −
x
;ω
J
) ≡
2
x
1
ω
J
+ (
2
/2m)
∇
2
+ iε
x
3
,
ω
J
+
2
2m
∇
2
u
0
(
x, J) = 0, (ρu) ≡
d
x ρ(
x )u(
x , J).
2.7 Problems for Chapter 2 101
Multiplying by ρ(
x) in the integral equation from the left, and
integrating over
x,obtain
(ρu ) =
1
D
+
(ω
J
)
(ρu
0
),
where
D
+
(ω
J
) = 1 +λ
0
d
xd
x
ρ(
x )G
+
(
x −
x
;ω
J
)ρ(
x
).
Since (ρu) is proportional to (ρu
0
), the angular frequency of the system
is given by
ω
J
=
1
2
2m
k
2
≡ ω
k
.
(b) By setting
u
0
(
x , J) ≡
1
(2π)
3/2
1
√
exp [i
k ·
x ],
show that
u(
x ,
k) =
1
(2π)
3/2
1
√
d
k
δ(
k −
k
) −
λ
0
(2π)
3
G
+
(
k
;ω
k
)
ρ
∗
(
k)ρ(
k
)
D
+
(ω
k
)
exp[i
k
·
x],
where
ρ(
x ) =
1
(2π)
3
d
k exp [i
k ·
x ]ρ(
k),
G
+
(
k
;ω
k
) =
1
ω
k
− (
2
/2m)
k
2
+ iε
D
+
(ω
k
) = 1 +
λ
0
(2π)
3
d
qG
+
(
q;ω
k
)
ρ(
q)
2
.
(c) Demonstrate the orthonormality of u(
x,
k):
d
xu
∗
(
x,
k)u(
x,
k
) = δ(
k −
k
).
(d) In this model, we can have a bound state,
D(ω) = 1 +
λ
0
(2π)
3
d
q
ρ(
q)
2
ω − (
2
/2m)
q
2
= 0.
102 2 Integral Equations and Green’s Functions
Show that D(ω) is a monotonically decreasing function such that
D(−∞) = 1, D(0) = 1 −
λ
0
(2π)
3
d
q
ρ(
q)
2
(
2
/2m)
q
2
,
d
dω
D(ω) =−
λ
0
(2π)
3
d
q
ρ(
q)
2
[ω − (
2
/2m)
q
2
]
2
< 0.
If we do not have the bound state pole, D(ω) = 0, by direct
computation, we obtain
d
ku(
x,
k)u
∗
(
x
,
k) = δ(
x −
x
).
(e) Ifwehaveonepoint(onthenegativerealaxis)ω = ω
b
,whereD(ω)
vanishes, we have an extra contribution to above, which comes from
D(z/) ∼
!
z
− ω
b
"
D
(ω
b
).
Show that
d
ku(
x,
k)u
∗
(
x
,
k)
= δ(
x −
x
) +
λ
0
(2π)
3
d
q
d
q
ρ(
q)ρ
∗
(
q
)
2
(ω
b
−ω
q
)(ω
b
− ω
q
)
1
D
(ω
b
)
× exp[i
q ·
x]exp[−i
q
·
x
].
Define the bound state wavefunction by
u(
x, b) ≡
1
(2π)
3/2
−
λ
0
D
(ω
b
)
1/2
d
q
ρ(
q)
(ω
b
− ω
q
)
exp [i
q ·
x].
Show that the completeness relation becomes
d
ku(
x,
k)u
∗
(
x
,
k) + u(
x , b)u
∗
(
x
, b) = δ(
x −
x
).
(f) Show that the bound state wavefunction is normalized:
d
xu
∗
(
x , b)u(
x , b) =−
λ
0
D
(ω
b
)
d
q
ρ(
q)
2
2
(ω
b
− ω
q
)
2
= 1.
Show that the configuration space wavefunction is given by
u(
x , b) =−
m
2
(−
λ
0
D
(ω
b
)
)
1/2
d
x
ρ(
x
)
1
x −
x
exp
−
+
−
2m
2
ω
b
x −
x
.
2.7 Problems for Chapter 2 103
Show that the bound state wavefunction satisfies the homogeneous
equation:
u(
x , b) =−λ
0
d
x
G
+
(
x −
x
;ω
b
)ρ(
x
)
d
x
ρ(
x
)u(
x
, b).
(g) Expand the quantum field ϕ(
x, t)intermsofu(
x,
k)andu(
x, b):
ϕ(
x , t) =
d
kA
in
(
k)u(
x ,
k)exp[−iω
k
t]
√
+ Bu(
x , b)exp[−iω
b
t]
√
,
where A
in
(
k), A
in†
(
k
), B and B
†
are the q-number operators satisfying
the commutation relations:
[A
in
(
k), A
in†
(
k
)]
±
= δ(
k −
k
),
[A
in
(
k), A
in
(
k
)]
±
= [A
in†
(
k), A
in†
(
k
)]
±
= 0,
[B, B
†
]
±
= 1,
[B, B]
±
= [B
†
, B
†
] = 0,
[B, A
in
(
k)]
±
= [B, A
in†
(
k)]
±
= 0.
Show that ϕ(
x, t) satisfies the commutation relation with the use of the
completeness relation:
[ϕ(
x, t), ϕ
†
(
x
, t)]
±
=
d
k
d
k
u(
x ,
k)u
∗
(
x
,
k
)δ(
k −
k
) + u(
x, b)u
∗
(
x
, b)
= δ(
x −
x
).
(h) Show that the total Hamiltonian is given by
H =
d
x
2
2m
∇ϕ
†
(
x, t) ·
∇ϕ(
x , t) −λ
0
d
x
d
x
ϕ
†
(
x, t)ρ(
x )ρ(
x
)ϕ(
x
, t).
Using the equation of motion, obtain
H =
d
x
2
2m
∇ϕ
†
(
x , t) ·
∇ϕ(
x, t) + iϕ
†
(
x , t) ˙ϕ(
x , t)
+
2
2m
ϕ
†
(
x , t)
∇
2
ϕ(
x , t)
= i
d
x ϕ
†
(
x , t) ˙ϕ(
x , t).
From the normal mode expansion, we obtain
H =
d
kω
k
A
in†
(
k)A
in
(
k) +ω
b
B
†
B.
104 2 Integral Equations and Green’s Functions
This is the addition of the bound state contribution to
H
in
0
=
d
x
2
2m
∇ϕ
in†
(
x, t) ·
∇ϕ
in
(
x , t) =
d
kω
k
A
in†
(
k)A
in
(
k).
(i) Choose the state where one asymptotic field carries the momentum
k
as the initial state:
A
in†
(
k)
0
-
≡
k
-
.
Then, the Lippmann–Schwinger equation is
k
(+)
-
=
k
-
+
1
ω
k
− H
in
0
+ iε
H
int
k
(+)
-
,
where
H
int
≡−λ
0
d
x ϕ
in†
(
x ,0)ρ(
x )
d
x
ρ
∗
(
x
)ϕ
in
(
x
,0)
=−
λ
0
(2π)
3
d
kA
in†
(
k)ρ(
k)
d
k
ρ
∗
(
k
)A
in
(
k
).
Multiplying by
*
0
A
in
(
q) in the Lippmann–Schwinger equation from
the left, we obtain
d
qρ
∗
(
q)
,
q
k
(+)
-
=
ρ
∗
(
k)
D
+
(ω
k
)
,
,
q
k
(+)
-
= δ(
q −
k) −
λ
0
(2π)
3
G
+
(
q;ω
k
)
ρ(
q)ρ
∗
(
k)
D
+
(ω
k
)
.
When
k =
k
, we obtain the transition rate as
w
k
,
k
=
2π
δ(ω
k
− ω
k
)
λ
0
(2π)
3
ρ(
k
)ρ
∗
(
k)
D
+
(ω
k
)
2
= 0.
Nonvanishing transition rate implies that the scattering takes place in
this model.