Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
13.4 Magnetoconductivity of Metals 405
We define the conductivity tensor by j = σ · E. Then, it is apparent that 237
σ =
2e
3
B
0
(2π)
3
¯h
2
c
"
∞
0
dε
−
∂f
0
∂ε
"
∞
−∞
dk
z
"
T (ε,k
z
)
0
ds v(ε, k
z
,s)e
−(
1
τ
+iω)s+iq·R(ε,k
z
,s)
×
"
s
−∞
ds
v(ε, k
z
,s
)e
(
1
τ
+iω)s
−iq·R(ε,k
z
,s
)
.
(13.76)
We assume that τ depends only on ε. Now, look at the function v(ε, k
z
,s) 238
e
iq·R(ε,k
z
,s)
. The position vector R(ε, k
z
,s) consists of two parts: 239
1. AperiodicpartR
p
(ε, k
z
,s): 240
R
p
[ε, k
z
,s+ T (ε, k
z
)] = R
p
(ε, k
z
,s) . (13.77)
2. A nonperiodic part or secular part R
s
(ε, k
z
,s): 241
R
s
(ε, k
z
,s)=v
s
(ε, k
z
)s. (13.78)
Remember that the variable s is time. The v
s
is the average value of v(ε, k
z
,s) 242
over one period, and written as 243
v
s
(ε, k
z
)=
1
T (ε, k
z
)
t+T
t
v(ε, k
z
,t
)dt
. (13.79)
Then, we have, since R = R
p
+ v
s
s, 244
v(ε, k
z
,s)e
iq·R(ε,k
z
,s)
= v(ε, k
z
,s)e
iq·v
s
(ε,k
z
)s
e
iq·R
p
(ε,k
z
,s)
. (13.80)
Thus, for σ
we can write 245
σ =
2e
3
B
0
(2π)
3
¯h
2
c
"
∞
0
dε
−
∂f
0
∂ε
"
∞
−∞
dk
z
"
T (ε,k
z
)
0
ds v(ε, k
z
,s)e
−[
1
τ
+iω+iq·v
s
(ε,k
z
)]s
× e
iq·R
p
(ε,k
z
,s)
"
s
−∞
ds
v(ε, k
z
,s
)e
[
1
τ
+iω−iq·v
s
(ε,k
z
)]s
e
−iq·R
p
(ε,k
z
,s
)
.
(13.81)
We now expand the periodic function v(ε, k
z
,s)e
iq·R
p
(ε,k
z
,s)
in Fourier series 246
in s.Letω
c
(ε, k
z
)=
2π
T (ε,k
z
)
.Then 247
v(ε, k
z
,s)e
iq·R
p
(ε,k
z
,s)
=
∞
n=−∞
v
n
(ε, k
z
)e
inω
c
s
. (13.82)
Obviously, the Fourier coefficients v
n
(ε, k
z
)aregivenby 248
v
n
(ε, k
z
)=
ω
c
(ε, k
z
)
2π
2π/ω
c
0
ds v(ε, k
z
,s)e
iq·R
p
(ε,k
z
,s)−inω
c
s
. (13.83)
We substitute the Fourier expansions, (13.82), into the expression for σ
to 249
obtain 250
σ =
2e
3
B
0
(2π)
3
¯h
2
c
"
∞
0
dε
−
∂f
0
∂ε
"
∞
−∞
dk
z
"
2π/ω
c
0
ds e
−[
1
τ
+iω+iq·v
s
(ε,k
z
)]s
×
∞
n=−∞
v
n
(ε, k
z
)e
inω
c
s
"
s
−∞
ds
e
[
1
τ
+iω−iq·v
s
(ε,k
z
)]s
×
∞
n
=−∞
v
∗
n
(ε, k
z
)e
−in
ω
c
s
.
(13.84)
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
406 13 Semiclassical Theory of Electrons
First perform the integration over s
to obtain 251
s
−∞
ds
e
[
1
τ
+iω−iq·v
s
(ε,k
z
)−iω
c
n
]s
=
e
[
1
τ
+iω−iq·v
s
−in
ω
c
]s
1
τ
+iω − iq · v
s
− in
ω
c
. (13.85)
Thus, we have
252
σ =
2e
3
B
0
(2π)
3
¯h
2
c
∞
0
dε
−
∂f
0
∂ε
∞
−∞
dk
z
×
2π
ω
c
0
ds
∞
n,n
=−∞
v
n
(ε, k
z
)v
∗
n
(ε, k
z
)e
i(n−n
)ω
c
s
1
τ
+iω − iq · v
s
− in
ω
c
. (13.86)
However,
"
2π/ω
c
0
ds e
i(n−n
)ω
c
s
=
2π
ω
c
δ
nn
.Thus,wehave 253
σ=
2e
3
B
0
(2π)
3
¯h
2
c
∞
0
dε
−
∂f
0
∂ε
∞
−∞
dk
z
2π
ω
c
(ε, k
z
)
∞
n=−∞
v
n
(ε, k
z
)v
∗
n
(ε, k
z
)
1
τ
+iω−iq ·v
s
− inω
c
.
(13.87)
This can be rewritten
254
σ =
2e
3
B
0
(2π)
3
¯h
2
c
∞
0
dε
−
∂f
0
∂ε
τ
∞
−∞
dk
z
T (ε, k
z
)
×
∞
n=−∞
v
n
(ε, k
z
)v
∗
n
(ε, k
z
)
1+iτ[ω − q · v
s
−
2πn
T (ε,k
z
)
]
. (13.88)
This expression is valid even for the case of open orbits. For closed orbits it
255
is customary to define 256
ω
c
(ε, k
z
)=
eB
0
m
∗
(ε, k
z
)c
=
2π
T (ε, k
z
)
where m
∗
is the cyclotron effective mass. Then, σ can be written 257
σ =
e
2
2π
2
¯h
2
∞
0
dε
−
∂f
0
∂ε
τ(ε)
∞
−∞
dk
z
m
∗
(ε, k
z
)
×
∞
n=−∞
v
n
(ε, k
z
)v
∗
n
(ε, k
z
)
1+iτ(ε)[ω − q · v
s
− nω
c
(ε, k
z
)]
. (13.89)
At very low temperatures the integration over energy just picks out the Fermi
258
energy because −
∂f
0
∂ε
acts just like a δ function, and we have 259
σ =
e
2
2π
2
¯h
2
τ(ε
F
)
F.S.
dk
z
m
∗
(k
z
)
∞
n=−∞
v
n
(ε
F
,k
z
)v
∗
n
(ε
F
,k
z
)
1+iτ(ε
F
)[ω − q · v
s
− nω
c
(ε
F
,k
z
)]
,
(13.90)
260
where all quantities are evaluated on the Fermi surface. 261
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
13.4 Magnetoconductivity of Metals 407
13.4.1 Free Electron Model 262
For the free electron model m
∗
(k
z
)=m is a constant independent of k
z
. 263
We shall assume that τ is also constant. The energy ε(k)andvelocityv are, 264
respectively, given by 265
ε(k)=
¯h
2
k
2
2m
,
v =
¯hk
m
, (13.91)
and hence,
266
v
z
=
¯hk
z
m
,
v
⊥
= |v
⊥
| =
v
2
− v
2
z
=
¯h
m
2mε
¯h
2
− k
2
z
1/2
.
We shall choose s, the time along the orbit, such that
267
v
x
= v
⊥
cos ω
c
s,
v
y
= v
⊥
sin ω
c
s.
(13.92)
Thus, for v(ε
F
,k
z
,s)wehave 268
v(ε, k
z
,s)=
¯h
m
9
8
2mε
¯h
2
− k
2
z
cos ω
c
s,
8
2mε
¯h
2
− k
2
z
sin ω
c
s, k
z
:
. (13.93)
The periodic part of the position vector is given by
269
R
p
(ε, k
z
,s)=
v
⊥
(ε, k
z
,s)ds,
=
v
⊥
ω
c
(sin ω
c
s, −cos ω
c
s, 0) . (13.94)
Thus, we have
270
iq ·R
p
(ε, k
z
,s)=
iv
⊥
(ε, k
z
)
ω
c
(q
x
sin ω
c
s − q
y
cos ω
c
s) . (13.95)
There is no loss in generality incurred by choosing the vector q to lie in the
271
y −z plane, i.e. q
x
=0.Thus 272
iq · R
p
(ε, k
z
,s)=−
iv
⊥
(ε, k
z
)
ω
c
q
y
cos ω
c
s. (13.96)
Now, let us evaluate the Fourier coefficients v
n
in (13.83) 273
v
n
(ε, k
z
)=
ω
c
(ε, k
z
)
2π
2π/ω
c
0
ds v(ε, k
z
,s)e
−
iv
⊥
(ε,k
z
)q
y
ω
c
cos ω
c
s−inω
c
s
. (13.97)
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
408 13 Semiclassical Theory of Electrons
Let w
=
q
y
v
⊥
ω
c
and x = ω
c
s to have 274
v
n
(ε, k
z
)=
1
2π
2π
0
dx v(ε, k
z
,x)e
−iw
cos x−inx
. (13.98)
Now, we use the relation
275
e
iw
sin x
=
∞
l=−∞
J
l
(w
)e
ilx
. (13.99)
One can easily see that
276
e
−iw
cos x
=e
iw
sin(x+
3π
2
)
=
∞
l=−∞
J
l
(w
)e
ilx
e
il
3π
2
=
∞
l=−∞
(−i)
l
J
l
(w
)e
ilx
.
(13.100)
Thus, we have
277
v
n
(ε, k
z
)=
1
2π
2π
0
dx v(ε, k
z
,x)e
−inx
∞
l=−∞
(−i)
l
J
l
(w
)e
ilx
,
=
∞
l=−∞
(−i)
l
J
l
(w
)
1
2π
2π
0
dx v(ε, k
z
,x)e
i(l−n)x
. (13.101)
Now, let us investigate the individual components of v
n
. 278
v
n
x
(ε, k
z
)=
∞
l=−∞
(−i)
l
J
l
(w
)
1
2π
v
⊥
2π
0
dx cos x e
i(l−n)x
,
=
∞
l=−∞
(−i)
l
J
l
(w
)
v
⊥
2
1
2π
2π
0
dx
e
i(l−n+1)x
+e
i(l−n−1)x
,
=
v
⊥
2
∞
l=−∞
(−i)
l
J
l
(w
)[δ
l,n−1
+ δ
l,n+1
] ,
=
(−i)
n−1
2
v
⊥
[J
n−1
(w
) − J
n+1
(w
)] . (13.102)
In an analogous way, we can obtain
279
v
n
y
(ε, k
z
)=
(−i)
n−1
2i
v
⊥
[J
n−1
(w
)+J
n+1
(w
)] , (13.103)
and
280
v
n
z
(ε, k
z
)=(−i)
n
v
z
J
n
(w
). (13.104)
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
13.4 Magnetoconductivity of Metals 409
Here, we recall some properties of Bessel functions: 281
J
n
(z)=
∞
m=0
(−1)
m
z
2
2m+n
1
m!(m + n)!
, (13.105)
J
−n
(z)=(−1)
n
J
n
(z), (13.106)
lim
z→0
J
n
(z)=
1
n!
z
2
n
, (13.107)
J
n
(z)=
d
dz
J
n
(z)=
1
2
[J
n−1
(z) − J
n+1
(z)] , (13.108)
and
282
n
z
J
n
(z)=
1
2
[J
n−1
(z)+J
n+1
(z)] . (13.109)
Using some of these equations we can write the vector v
n
as 283
v
n
(ε, k
z
)=(−i)
n
⎛
⎜
⎝
iv
⊥
J
n
(w
)
(nω
c
/q
y
) J
n
(w
)
v
z
J
n
(w
)
⎞
⎟
⎠
. (13.110)
284
If we take the zero temperature limit, we have
−
∂f
0
∂ε
= δ(ε −ζ) and, hence,
285
we have 286
σ =
e
2
τm
2π
2
¯h
2
k
F
−k
F
dk
z
∞
n=−∞
v
n
(ε
F
,k
z
)v
∗
n
(ε
F
,k
z
)
1 − iτ [nω
c
(ε
F
,k
z
)+q · v
s
− ω]
. (13.111)
287
For free electrons, the secular part of the velocity is simply v
z
,thus 288
q · v
s
= q
z
v
z
= q
z
v
F
cos θ,
dk
z
=
m
¯h
dv
z
=
mv
F
¯h
d(cos θ),
v
⊥
=
2
v
2
F
− v
2
z
= v
F
sin θ, (13.112)
w
=
q
y
v
⊥
ω
c
=
q
y
v
F
ω
c
sin θ = w sin θ.
By substituting these in (13.111), we obtain
289
σ =
3σ
0
2
∞
n=−∞
"
1
−1
d(cos θ)
×
⎛
⎜
⎜
⎝
isinθJ
n
(w sin θ)
n
w
J
n
(w sin θ)
cos θJ
n
(w sin θ)
⎞
⎟
⎟
⎠
(
−isinθJ
n
(w sin θ),
n
w
J
n
(w sin θ),cos θJ
n
(w sin θ)
)
1−iτ [nω
c
(ε
F
,k
z
)−ω+q
z
v
F
cos θ]
,
(13.113)
where J
n
(x)=dJ
n
(x)/dx and σ
0
is the dc conductivity given by σ
0
= 290
n
0
e
2
τ
m
.Inaddition,sinθJ
n
(w sin θ)=
∂
∂w
J
n
(w sin θ). Hence, we can rewrite 291
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
410 13 Semiclassical Theory of Electrons
(13.113) as 292
σ =
3σ
0
2
∞
n=−∞
"
1
−1
d(cos θ)
×
⎛
⎜
⎜
⎝
i
∂
∂w
J
n
(w sin θ)
n
w
J
n
(w sin θ)
cos θJ
n
(w sin θ)
⎞
⎟
⎟
⎠
(
−i
∂
∂w
J
n
(w sin θ),
n
w
J
n
(w sin θ),cos θJ
n
(w sin θ)
)
1−iτ [nω
c
(ε
F
,k
z
)−ω+q
z
v
F
cos θ]
.
(13.114)
293
This result was first obtained by Cohen, Harrison, and Harrison.
2
294
13.4.2 Propagation Parallel to B
0
295
To acquaint ourselves with the properties of σ, let us first evaluate it for the 296
case q B
0
, i.e. q =(0, 0,q). In the limit q
y
→ 0, w → 0 and, hence, we have 297
n
w
J
n
(w sin θ) −→
1
2
sin θ(δ
n,1
+ δ
n,−1
),
i
∂
∂w
J
n
(w sin θ) −→
i
2
sin θ(δ
n,1
− δ
n,−1
), (13.115)
cos θJ
n
(w sin θ) −→ cos θδ
n,0
.
It is easy to see that σ
xz
= σ
zx
= σ
yz
= σ
zy
= 0 because of the δ functions 298
involved. The nonvanishing components of σ are σ
xx
= σ
yy
, σ
xy
= −σ
yx
and 299
σ
zz
. They can easily be evaluated to be written 300
σ
zz
=
3
2
σ
0
1
−1
d(cos θ)cos
2
θ
1+iωτ − iq
z
v
F
τ cos θ
, (13.116)
and
301
σ
±
= σ
xx
∓ iσ
xy
=
3
4
σ
0
1
−1
d(cos θ)sin
2
θ
1+i(ω ∓ ω
c
)τ − iq
z
v
F
τ cos θ
. (13.117)
Notice that when q
z
→ 0 the integral in (13.116) becomes
"
1
−1
d(cos θ)cos
2
θ = 302
2
3
so that, in that case, we have 303
σ
zz
=
σ
0
1+iωτ
. (13.118)
The q
z
→ 0 limit of the integral in (13.117) becomes
"
1
−1
d(cos θ)sin
2
θ =
4
3
304
so that 305
σ
±
=
σ
0
1+i(ω ∓ ω
c
)τ
. (13.119)
2
M.H. Cohen, M.J. Harrison, W.A. Harrison, Phys. Rev. 117, 937 (1960).
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
13.4 Magnetoconductivity of Metals 411
13.4.3 Propagation Perpendicular to B
0
306
We consider the case q ⊥ B
0
, i.e. q =(0,q,0) to write (13.114) as 307
σ =
3σ
0
2
∞
n=−∞
"
1
−1
d(cos θ)
×
⎛
⎜
⎜
⎝
i
∂
∂w
J
n
(w sin θ)
n
w
J
n
(w sin θ)
cos θJ
n
(w sin θ)
⎞
⎟
⎟
⎠
(
−i
∂
∂w
J
n
(w sin θ),
n
w
J
n
(w sin θ),cos θJ
n
(w sin θ)
)
1−iτ [nω
c
(ε
F
,k
z
)−ω]
.
(13.120)
We define the following functions
308
c
n
(w)=
1
2
1
−1
d(cos θ)cos
2
θJ
2
n
(w sin θ),
s
n
(w)=
1
2
1
−1
d(cos θ)sin
2
θ[J
n
(w sin θ)]
2
, (13.121)
g
n
(w)=
1
2
1
−1
d(cos θ) J
2
n
(w sin θ).
It is obvious that the angular integrations appearing in σ
xz
, σ
zx
, σ
yz
,andσ
zy
309
vanish because the integrands are odd functions of cos θ. For the nonvanishing 310
components we obtain 311
σ
xx
=3σ
0
∞
n=−∞
s
n
(w)
1 − iτ (nω
c
− ω)
, (13.122)
σ
yy
=
3σ
0
w
2
∞
n=−∞
n
2
g
n
(w)
1 − iτ (nω
c
− ω)
, (13.123)
σ
xy
= −σ
yx
=
3σ
0
i
2w
∞
n=−∞
ng
n
(w)
1 − iτ (nω
c
− ω)
, (13.124)
and
312
σ
zz
=3σ
0
∞
n=−∞
c
n
(w)
1 − iτ (nω
c
− ω)
. (13.125)
13.4.4 Local vs. Nonlocal Conduction
313
What we have been studying is the nonlocal theory of the electrical conduc- 314
tivity of a solid. It is worth emphasizing once again what is meant by nonlocal 315
conduction, and in which case the nonlocal theory reduces to a local theory. 316
The result we have obtained is 317
j(q,ω)=σ(q,ω) · E(q,ω). (13.126)
318
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
412 13 Semiclassical Theory of Electrons
It is easy to show that this expression corresponds to the relation 319
j(r,t)=
K(r − r
,t− t
) ·E(r
,t
)d
3
r
dt
. (13.127)
320
Let us take the Fourier transforms of each side by multiplying by e
iq·r−iωt
and 321
integrating 322
"
j(r,t)e
iq·r−iωt
d
3
rdt
=
"
K
(r − r
,t− t
) ·E(r
,t
)e
iq·r
−iωt
e
iq·(r−r
)−iω(t−t
)
d
3
r
dt
d
3
rdt.
The left-hand side is simply j(q,ω). The right-hand side can be simplified by
323
letting r − r
= x and t − t
= s 324
j(q,ω)=
K(x,s)e
iq·x−iωs
d
3
xds
4
56 7
·
E(r
,t
)e
iq·r
−iωt
d
3
r
dt
4 56 7
.
σ
(q,ω) E(q,ω)
(13.128)
This is just the relation we gave in (13.126) if
325
σ(q,ω)=
K(x,s)e
iq·x−iωs
d
3
xds
or
326
K(r − r
,t− t
)=
1
(2π)
4
σ
(q,ω)e
iω(t−t
)−iq·(r−r
)
d
3
qdω. (13.129)
Consider for a moment what would happen if σ
(q,ω) were independent of q. 327
In that case, we have 328
K(r − r
,ω)=
1
(2π)
3
σ
(ω)e
−iq·(r−r
)
d
3
q,
= σ
(ω) δ(r − r
). (13.130)
Thus, we have
329
j(r,ω)=σ(ω) · E(r,ω). (13.131)
This is just Ohm’s law in the local theory, in which j(r) depends only on the
330
electric field at the same point r. Thus, the local theory is the special case of 331
the general nonlocal theory, in which the q dependence of σ is unimportant. 332
By looking at the expressions we derived one can see that σ is essentially 333
independent of q if 334
1. ql 1, in the absence of a magnetic field, where l = v
F
τ is the electron 335
mean free path. 336
2. q
⊥
r
c
1orq
⊥
l
0
1, and q
z
l
0
1, in the presence of a magnetic field. 337
Here r
c
is the radius of the cyclotron orbit. 338
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
13.5 Quantum Theory of Magnetoconductivity of an Electron Gas 413
13.5 Quantum Theory of Magnetoconductivity 339
of an Electron Gas 340
The evaluation of σ(q,ω,B
0
) for a quantum mechanical system is very similar 341
to our evaluation of the wave vector and frequency dependent conductivity in 342
the absence of the field B
0
. We will give a very brief summary of the technique 343
here.
3
344
The zero-order Hamiltonian for an electron in the presence of a vector 345
potential A
0
=(0,xB
0
, 0) is given by 346
H
0
=
1
2m
p
2
x
+(p
y
+
e
c
B
0
x)
2
+ p
2
z
. (13.132)
The eigenfunctions and eigenvalues of H
0
can be written as 347
|ν = |nk
y
k
z
=
1
L
e
ik
y
y+ik
z
z
u
n
x +
¯hk
y
mω
c
,
ε
ν
= ε
n,k
y
,k
z
=
¯h
2
k
2
z
2m
+¯hω
c
n +
1
2
. (13.133)
Perturbing self-consistent electromagnetic fields E(r,t)andB(r,t)are
348
assumed to be of the form e
iωt−iq·r
. These fields can be derived from the 349
potentials A(r,t)andφ(r,t): 350
E = −
1
c
˙
A −∇φ = −
iω
c
A +iqφ,
B = ∇×A = −iq × A.
(13.134)
As in the Lindhard case, the theory can be shown to be gauge invariant (we
351
will not prove it here but it is done in the references listed earlier). Therefore, 352
we can take a gauge in which the scalar potential φ = 0. Then, we write the 353
linearized (in A) Hamiltonian as 354
H = H
0
+ H
1
, (13.135)
where H
0
is given by (13.132) and H
1
is the perturbing part 355
H
1
=
e
2c
(v
0
· A + A · v
0
) . (13.136)
Here, v
0
=
1
m
p +
e
c
A
0
is the velocity operator in the presence of the field
356
A
0
. From here on, one can simply follow the steps we carried out in evaluating 357
σ(q,ω,B
0
=0).Weuse 358
H
0
|ν = ε
ν
|ν,
ρ
0
|ν = f
0
(ε
ν
)|ν.
(13.137)
3
For details one is referred to the references by J.J. Quinn, S. Rodriguez, Phys.
Rev. 128, 2480 (1962) and M.P. Greene, H.J. Lee, J.J. Quinn, S. Rodriguez, Phys.
Rev. 177, 1019 (1969).
Uncorrected Proof
BookID 160928 ChapID 13 Proof# 1 - 29/07/09
414 13 Semiclassical Theory of Electrons
The perturbation is given by (13.136) and use that 359
A(r,t)=A(q,ω)e
iωt−iq·r
.
The resulting expression for j(q,ω) can be written (for the collisionless limit
360
where τ →∞) 361
j(q,ω)=−
ω
2
p
4πc
[1
+ I(q,ω)] · A(q,ω). (13.138)
362
Here, ω
2
p
=
4πn
0
e
2
m
and n
0
=
N
V
.Thesymbol1 stands for the unit tensor and 363
I(q,ω)=
m
N
νν
f
0
(ε
ν
) − f
0
(ε
ν
)
ε
ν
− ε
ν
− ¯hω
ν
|V(q)|νν
|V(q)|ν
∗
. (13.139)
364
The operator V(q)isgivenby 365
V(q)=
1
2
v
0
e
iq·r
+
1
2
e
iq·r
v
0
, (13.140)
and v
0
=
1
m
{p +
e
c
A
0
}. The matrix elements of V(q)aregivenby 366
ν
|V
z
(q)|ν = δ(k
y
,k
y
+ q
y
)δ(k
z
,k
z
+ q
z
)
¯h
m
k
z
+
q
z
2
f
n
n
(q
y
),
ν
|V
y
(q)|ν = δ(k
y
,k
y
+q
y
)δ(k
z
,k
z
+q
z
)
(
¯hq
y
2m
f
n
n
(q
y
)+
¯hω
c
2m
1/2
X
(+)
n
n
(q
y
)
'
,
ν
|V
x
(q)|ν = δ(k
y
,k
y
+ q
y
)δ(k
z
,k
z
+ q
z
)
(
i
¯hω
c
2m
1/2
X
(−)
n
n
(q
y
)
'
(13.141)
Intheseequations,wehavetakenq
x
= 0; this can be done without loss of 367
generality. The function f
n
n
(q
y
) is the two-center harmonic oscillator integral: 368
f
n
n
(q
y
)=
∞
−∞
u
n
(x +
¯hq
y
mω
c
)u
n
(x)dx, (13.142)
and
369
X
(±)
n
n
(q
y
)=(n +1)
1/2
f
n
,n+1
(q
y
) ± n
1/2
f
n
,n−1
(q
y
). (13.143)
The function f
n
n
also appears in another useful matrix element 370
ν
|e
iq·r
|ν = δ(k
y
,k
y
+ q
y
)δ(k
z
,k
z
+ q
z
)f
n
n
(q
y
). (13.144)
The function f
n
n
can be evaluated in terms of associated Laguerre polyno- 371
mials. For n
≥ n,wehave 372
f
n
n
(q)=
n!
n
!
1/2
ξ
(n
−n)/2
e
−ξ/2
L
n
−n
n
(ξ), (13.145)