Masujima M. Applied Mathematical Methods in Theoretical Physics
Подождите немного. Документ загружается.
2.7 Problems for Chapter 2 85
(c) Upon substitution of the explicit form of P
l,j
obtained above, show that
,
k
, m
s
f (θ )
k, m
s
-
=
1
k
l
{(l + 1) exp[iδ
l+
]sinδ
l+
+ l exp[iδ
l−
]sinδ
l−
}P
l
(cos θ )
,
m
s
m
s
-
+
1
k
l
{exp[iδ
l+
]sinδ
l+
−exp[iδ
l−
]sinδ
l−
}(
LP
l
(cos θ ))
,
m
s
σ
m
s
-
,
which is the spin-matrix element of the spin-space operator F(θ),
F(θ) = 1 ·f
N
(θ) +σ · (
Lf
S
(θ)).
(d) Show that the non-spin-flip amplitude f
N
(θ) and the spin-flip
amplitude f
S
(θ) are, respectively, given by
f
N
(θ) =
1
k
l
{(l + 1) exp[iδ
l+
]sinδ
l+
+ l exp[iδ
l−
]sinδ
l−
}P
l
(cos θ ),
f
S
(θ) =
1
k
l
{exp[iδ
l+
]sinδ
l+
−exp[iδ
l−
]sinδ
l−
}P
l
(cos θ ).
(e) Show that the final spin-averaged differential cross section σ(θ )is
given by, for the unpolarized beam of spin-1/2 particle,
σ (θ ) =
1
2
m
s
m
s
,
m
s
f (θ )
m
s
-
2
=
1
2
m
s
m
s
,
m
s
f (θ )
m
s
-
∗
,
m
s
f (θ )
m
s
-
=
1
2
m
s
m
s
,
m
s
f
†
(θ)
m
s
-,
m
s
f (θ )
m
s
-
=
1
2
m
s
*
m
s
f
†
(θ)f (θ)
m
s
%
=
1
2
trF
†
(θ)F(θ ),
where F(θ)andF
†
(θ) with the dagger † in spin space are given by
F(θ) = 1 ·f
N
(θ) +σ · (
Lf
S
(θ)),
F
†
(θ) = 1 ·f
∗
N
(θ) +σ · (
Lf
S
(θ))
∗
.
(f) Making use of the identity (σ ·
a)(σ ·
b) =
a ·
b + iσ ·(
a ×
b), trσ = 0,
and tr1= 2, show that
σ (θ ) =
1
2
trF
†
(θ)F(θ ) = f
∗
N
(θ)f
N
(θ) +(
Lf
S
(θ))
∗
(
Lf
S
(θ)).
86 2 Integral Equations and Green’s Functions
(g) Plugging in the explicit form of f
N
(θ)andf
S
(θ) obtained as above into
σ (θ ), show that
σ (θ ) =
1
k
2
ll
{(l + 1) exp[iδ
l+
]sinδ
l+
+ l exp[iδ
l−
]sinδ
l−
}
∗
×{(l
+1) exp[iδ
l
+
]sinδ
l
+
+l
exp[iδ
l
−
]sinδ
l
−
}P
∗
l
(cos θ )P
l
(cos θ ) +
1
k
2
ll
{exp[iδ
l+
]sinδ
l+
− exp[iδ
l−
]sinδ
l−
}
∗
×{exp[iδ
l
+
]sinδ
l
+
− exp[iδ
l
−
]sinδ
l
−
}(
LP
l
(cos θ ))
∗
(
LP
l
(cos θ )).
(h) Identities due to orthonormality of spherical harmonics are given by
dP
∗
l
(cos θ )P
l
(cos θ ) =
4π
2l + 1
δ
ll
,
d(
LP
l
(cos θ ))
∗
(
LP
l
(cos θ )) =
4π
2l + 1
δ
ll
l(l + 1).
With the use of identities stated above, show that the total cross section
σ
tot
is given by
σ
tot
=
dσ (θ ) =
4π
k
2
l
{(l + 1) sin
2
δ
l+
+ l sin
2
δ
l−
}.
(i) Show that the imaginary part of the forward scattering amplitude is
given by
Im f
N
(θ = 0) =
1
k
l
{(l + 1) sin
2
δ
l+
+l sin
2
δ
l−
}.
Show finally the optical theorem for spin-1/2 particle. Namely, show
that
σ
tot
=
4π
k
Im f
N
(θ = 0).
2.26. Consider the polarization and the density matrix of spin-1/2 particle in the
scattering process discussed in Problem 2.25. Two-state spin systems are
represented by
|
α
and
|
β
.Theysatisfy
α
|
β
= 0,
α
|
α
=
β
|
β
= 1.
The density matrix ρ in the α–β representation is given by
ρ =
|
α
p
α
α
|
+
|
β
p
β
β
|
with p
α
+ p
β
= 1,
2.7 Problems for Chapter 2 87
where p
α
(p
β
) is the probability of finding the particle in the state
|
α
(
|
β
).
(a) Show Hermiticity of the density matrix ρ. Show also that
trρ = 1, trρ
2
≤ 1.
(b) Unique parameterization of the density matrix ρ is given by
ρ =
1
2
(1 +ε(σ ·
ˆ
n)), ε ≥ 0,
ˆ
n ·
ˆ
n = 1.
Show that the size of ε is given by 0 ≤ ε ≤ 1.
(c) Wedefinetheaverageofσ by
σ
= tr(σρ).
Show that
σ
= ε
ˆ
n.
(d) Consider the scattering where the density matrix ρ
(i)
of the incident
beam is defined by
ρ
(i)
=
m
(i)
S
m
(i)
S
-
p
i,m
(i)
S
,
m
(i)
S
,
where p
i,m
(i)
S
is the probability of the incident beam in the state
m
(i)
S
-
.
Show that the density matrix ρ
(S)
(θ) for the scattered beam is given by
ρ
(S)
(θ) =
F(θ)ρ
(i)
F
†
(θ)
tr(F(θ)ρ
(i)
F
†
(θ))
.
(e) For an unpolarized incident beam,
ρ
(i)
=
1
2
1,
show that the average of σ of the scattered beam is given by
σ
Scattered
= tr(σρ
(S)
(θ)) =
2Re{f
∗
N
(θ)(
Lf
S
(θ))}
f
∗
N
(θ)f
N
(θ) +(
Lf
S
(θ))
∗
(
Lf
S
(θ))
.
(f) For an unpolarized incident beam, in the first-order Born
approximation, show that
88 2 Integral Equations and Green’s Functions
f
Born
N
(θ) =−
4π
2
m
2
d
3
r
(2π)
3
exp[−i
q ·
r]V
0
(r)
q=
k
−
k
,
Lf
Born
S
(θ) =−
4π
2
m
2
i(
k
×
k)
×
1
q
d
dq
d
3
r
(2π)
3
exp[−i
q ·
r]V
LS
(r)
q=
k
−
k
.
Namely, f
Born
N
(θ)isrealand
Lf
Born
S
(θ) is pure imaginary. Show also that
σ
Born
Scattered
= 0, σ
Born
tot
=
4π
k
Im f
Born
N
(θ = 0) = 0.
These results clearly show that the first-order Born approximation is
not good enough in the presence of the spin–orbit coupling.
2.27. Consider the one pion-exchange potential. The one pion-exchange potential
in the static limit has the form
V(
r) = g
2
(τ
(1)
·τ
(2)
)
1
µ
2
(σ
(1)
·
∇)(σ
(2)
·
∇)
exp[−µr]
r
,(A)
where g
2
is the coupling constant,
g
2
c
= 0.08,
and µ is the inverse pion Compton wavelength,
µ =
m
π
c
,orcµ = m
π
c
2
.
(a) In carrying out the derivatives on exp[−µr]/r,rewriteV(
r)as
V(
r) =
g
2
c
(m
π
c
2
)(τ
(1)
·τ
(2)
)
σ
(1)
·
∇
µ
σ
(2)
·
∇
µ
exp[−µr]
µr
,
and introduce the dimensionless variable ρ as
ρ = µ
r,
∇
µ
=
∂
∂(µ
r)
=
∇
ρ
.
Recalling the identity
∇
ρ
k
=
df
dρ
dρ
dρ
k
=
ρ
k
ρ
(
df
dρ
),
2.7 Problems for Chapter 2 89
and that exp[−ρ]/ρ satisfies the Klein–Gordon equation with the
source term, 4πδ
3
( ρ),
(−
∇
2
ρ
+1)
exp[−ρ]
ρ
= 4πδ
3
( ρ),
show that, for all ρ,wehave
(σ
(1)
·
∇
ρ
)(σ
(2)
·
∇
ρ
)
exp[−ρ]
ρ
=
1
3
(σ
(1)
·σ
(2)
) + (3(σ
(1)
· ˆρ)(σ
(2)
· ˆρ) − (σ
(1)
·σ
(2)
))
1 +
3
ρ
+
3
ρ
2
×
exp[−ρ]
ρ
−
1
3
(σ
(1)
·σ
(2)
) ·4πδ
3
( ρ).
(b) Writing the tensor operator S
12
as S
12
≡ 3(σ
(1)
· ˆρ)(σ
(2)
· ˆρ) −
(σ
(1)
·σ
(2)
), show that
V(
r) =
g
2
c
(m
π
c
2
)
1
3
(τ
(1)
·τ
(2)
)
(σ
(1)
·σ
(2)
) +S
12
1 +
3
µr
+
3
(µr)
2
×
exp[−µr]
µr
(B)
−
g
2
c
(m
π
c
2
)
1
3
(τ
(1)
·τ
(2)
)(σ
(1)
·σ
(2)
) · 4πδ
3
(µ
r). (C)
(c) The scattering amplitude, in the first-order Born approximation, is
given by the Fourier transform
f (
q) =−
4π
2
m
red
2
d
3
r
(2π)
3
exp[−i
q ·
r]{V(
r) −(C)},
where
q is the momentum transfer given by
q =
k
−
k and m
red
is the
reduced mass of the two-nucleon system, m
red
= m/2. Show that the
term (C) is to be subtracted from V(
r). Obtain the scattering amplitude as
f (
q) =+
g
2
c
1
µ
2
!
m
π
c
"
m
m
π
τ
(1)
·τ
(2)
×
(σ
(1)
·
q)(σ
(2)
·
q)
µ
2
+
q
2
−
1
3
(σ
(1)
·σ
(2)
)
.
Setting
S =
1
2
(σ
(1)
+σ
(2)
),
12
(
q) ≡ 2
(
S ·
q)
2
−
1
3
S
2
·
q
2
= (σ
(1)
·
q)(σ
(2)
·
q) −
1
3
q
2
(σ
(1)
·σ
(2)
)
=
1
3
S
12
(
ˆ
q)q
2
,
90 2 Integral Equations and Green’s Functions
obtain the final form of the scattering amplitude as
f (
q) =+
g
2
c
m
π
c
m
m
π
1
µ
2
+
q
2
(τ
(1)
·τ
(2)
)
× (
12
(
q) −
1
3
µ
2
(σ
(1)
·σ
(2)
)).
(d) Consider the isospin factor, (τ
(1)
·τ
(2)
). By the Pauli principle, we have
Spin-triplet:
S = 1or(σ
(1)
·σ
(2)
) = 1: (τ
(1)
·τ
(2)
) =
−3forl even,
+1forl odd,
Spin-singlet:
S = 0or(σ
(1)
·σ
(2)
) =−3: (τ
(1)
·τ
(2)
) =
+1forl even,
−3forl odd.
Show that we can split any Born amplitude into even-parity and
odd-parity parts,
f
even
=
1
2
(f (
q) +f (
q
)), f
odd
=
1
2
(f (
q) −f (
q
));
q
= (−
k
) −
k =−(
k
+
k).
Obtain the spin-triplet state (S = 1) N–N scattering amplitude as
f
S=1
(
q,
q
) = f
even
+ f
odd
=
1
2
(f (
q) +f (
q
)
iso-singlet
) +
1
2
(f (
q) −f (
q
)
iso-triplet
)
=−
g
2
c
m
π
c
m
m
π
1
µ
2
+
q
2
12
(
q) −
1
3
µ
2
+
2
µ
2
+
q
2
12
(
q
) −
1
3
µ
2
,
and the spin-singlet state (S = 0) N–N scattering amplitude as
f
S=0
(
q,
q
) = f
even
+f
odd
=
1
2
(f (
q) + f (
q
)
iso-triplet
) +
1
2
(f (
q) −f (
q
)
iso-singlet
)
=−
g
2
c
m
π
c
m
m
π
µ
2
µ
2
+
q
2
−
2µ
2
µ
2
+
q
2
.
(e) In order to obtain an expression for the n–p scattering
differential cross section, show first that
2.7 Problems for Chapter 2 91
(
12
(
q))
2
=
8
9
q
4
−
2
3
q
2
12
(
q),
*
12
(
q)
%
= 0,
*
(
12
(
q))
2
%
=
8
9
q
4
,
,
12
(
q)
12
(
q
)
-
=−
4
9
q
2
q
2
.
Obtain an explicit expression for the n–p scattering differential
cross section,
σ
n-p
=
3
4
*
f
2
%
S=1
+
1
4
*
f
2
%
S=0
.
(f) In the case of the p–p scattering, show that
f
S=1
= 2f
odd
, f
S=0
= 2f
even
.
On this basis, obtain an explicit expression for the p–p scattering
differential cross section, σ
p-p
.
Hint: To Fourier transform the potential, the form (A) is preferable
to the form (B). The derivation of one pion-exchange potential from
the pseudoscalar meson theory is given as a problem in Chapter 10.
2.28. Consider a scattering of two identical particles.
(a) Consider a scattering between two identical spin 0 bosons in the
center-of-mass frame. Show that a scattering can be described by the
symmetrized scattering amplitude,
f (
k
,
k) = f (
k
,
k) + f (−
k
,
k),
and hence the differential cross section is given by
dσ
d
=
f (
k
,
k) + f (−
k
,
k)
2
.
In terms of the wave number k and the scattering angle θ , show that
dσ
d
=
f (k, θ ) +f (k, π − θ )
2
=
f (k, θ )
2
+
f (k, π − θ )
2
+ 2Ref (k, θ)f
∗
(k, π − θ ). (B-B)
Consider the partial wave expansion of f (k, θ). Show that
dσ
d
=
4
k
2
l=0,2,4,...
4π(2l + 1) exp[iδ
l
(k)] sin δ
l
(k)Y
l0
(θ)
2
.
Namely, the symmetry requirement has eliminated all partial waves of
odd angular momentum.
92 2 Integral Equations and Green’s Functions
(b) Consider now the scattering of two identical fermions with spin 1/2.
The scattering amplitude is a matrix in the four-dimensional spin
space of the two identical particles,
M(
k
,
k) = M
s
(
k
,
k) + M
t
(
k
,
k),
where s and t refer to singlet and triplet spin states, respectively. Show
that the correctly antisymmetrized amplitude is given by
M(
k
,
k) = [M
s
(
k
,
k) +M
s
(−
k
,
k)] +[M
t
(
k
,
k) −M
t
(−
k
,
k)].
When the initial state has the spin-space density matrix ρ
i
, show that
dσ
d
= trρ
i
M
†
M.
If neither the beam nor the target is polarized, ρ
i
= 1/4. Show that the
two pieces M
s
and M
t
can be obtained by use of the projection
operators P
s
and P
t
that project onto the singlet and triplet states,
respectively,
P
s
= (1/4)(1 −σ
1
·σ
2
), P
t
= (1/4)(3 +σ
1
·σ
2
),
with the properties, P
s
P
t
= 0, P
s
+ P
t
= 1; trP
s
= 1, and trP
t
= 3. Show
that the triplet and singlet scattering amplitudes are given by
M
t
= P
t
MP
t
, M
s
= P
s
MP
s
.
Consider the simple situation when the potential has the form
V = V
1
(r) +σ
1
·σ
2
V
2
(r).
Show that M can be expressed as
M = f
s
(k, θ )P
s
+ f
t
(k, θ )P
t
,
where f
s,t
are scalars in the spin space. Show that the differential cross
section becomes
dσ
d
=
f
s
(k, θ ) +f
s
(k, π − θ )
2
trρ
i
P
s
+
f
t
(k, θ ) − f
t
(k, π − θ )
2
trρ
i
P
t
.
For a completely unpolarized initial state (ρ
i
= 1/4), show that
dσ
d
=
1
4
f
s
(k, θ ) + f
s
(k, π − θ )
2
+
3
4
f
t
(k, θ ) − f
t
(k, π − θ )
2
.
2.7 Problems for Chapter 2 93
If the force potentials are completely spin independent, set V
2
= 0,
and f
s
= f
t
= f . Show that the differential cross section is given by
dσ
d
=
f (k, θ )
2
+
f (k, π − θ )
2
− Re f
∗
(k, θ ) f (k, π −θ ). (F-F)
(c) Consider the Coulomb scattering of two identical particles. Applying
the differential cross section formulas, (B-B) and (F-F), for the
Coulomb scattering, obtain the differential cross section for each case.
Comment on Problem 2.28: In order to avoid the double counting of
the scattered particles at the detector, restrict the scattering angle θ to
0 ≤ θ ≤ π/2.
2.29. Consider the Breit–Wigner resonance scattering cross section.
(a) The eigenfunctions for the noninteracting particles will be written as
χ
s,ε
;withK as the free Hamiltonian, these are assumed to satisfy the
Schr
¨
odinger equation,
Kχ
s,ε
= εχ
s,ε
.
The energy ε is chosen as one of the quantum variables labeling the
state, the remaining state labels being written as s. We suppose that
the operators whose eigenvalues are the quantities s commute with the
full Hamiltonian H = K +V. Then the scattering eigenfunctions are
written as
H ψ
+
s,ε
= εψ
+
s,ε
.
The initial state wave packet is written as
φ
s
(t) =
ε
A
ε
χ
s,ε
exp [−iεt], (A)
where A
ε
is a suitable wave packet amplitude. The wavefunction ϕ
s
is
normalized to unity,
(ϕ
s
, ϕ
s
) = 1,
ε
|
A
ε
|
2
= 1.
The time-dependent wavefunction for the system
s
(t)ischosento
coincide with ϕ
s
(t)atthetimet =−T at which the experiment is
begun. This is
s
(t) =
ε
A
ε
ψ
s,ε
exp [−iεt], (
s
(t),
s
(t)) = 1. (B)
94 2 Integral Equations and Green’s Functions
Finally, we make the special assumption that the representation s is
such that in it the S-matrix is diagonal. Then our experiment is one in
which the incident wave is an incoming spherical packet collapsing at
a point where the interaction is to occur. The scattered wave will be an
outgoing spherical packet expanding away from the local region of
interaction.
Consider now a large volume S
R
in the configuration space, which is
characterized by a radius R and is sufficiently large that the particles
do not interact when lying outside S
R
. T is so large that the wave
packets lie outside S
R
with arbitrary precision when t < −T or t > T.
Then, the ‘‘lifetime’’ Q of the scattering state is defined by the equation
Q = lim
R→∞
lim
T→∞
T
−T
dt
S
R
dτ [
∗
s
(t)
s
(t) −ϕ
∗
s
(t)ϕ
s
(t)]. (C)
Here, dτ is a volume element in the configuration space and the
integral is confined to the domain S
R
. The limit R →∞is understood
to imply that in the limit the integral extends over the entire
configuration space. Q represents the additional time, over and above
the ‘‘free flight time,’’ that the particles spend in the range of their
interaction. If Q > 0, the particles tend to ‘‘stick together,’’ whereas if
Q < 0, the interaction tends to ‘‘force them apart.’’
(b) Using the wave packet expansions, (A) and (B), show that
Q = lim
R→∞
2π
×
ε
|
A
ε
|
2
S
R
dτ
ε
[ψ
+∗
s,ε
δ(ε − H)ψ
+
s,ε
− χ
∗
s,ε
δ(ε − K)χ
s,ε
]
.
(D)
We have the relations
δ(ε − K) =
i
2π
[G
+
0
(ε) − G
−
0
(ε)], δ(ε − H)
=
i
2π
G
+
(ε) −G
−
(ε)
,(E)
G
±
0
(ε) =
1
ε ± iη − K
, G
±
(ε) =
1
ε ± iη − H
.
When the relations (E) are substituted into (D), show that
Q = i
ε
|
A
ε
|
2
tr[G
+
(ε) −G
−
(ε) − G
+
0
(ε) +G
−
0
(ε)]. (F)
By the symbol tr{···}here, we mean only a partial trace since the
variables s are held constant and the sum runs over only the energy ε
.
Simplify (F) to the form