固体物理学习题解答-陈长乐 khdaw

导读:80b)?(x?xn?b)dx=1,1|2=2|1=0,1|1=2|2=C?C,1|H???|1?ECC1|H|2=0.?|1?|2?EC?C2|H2|H(4.9)(4.10)?|2=1|HCC?e?ikxn??(x?xn)H?neik(xn′+b)?(x?xn′?b)dxn′=C?Cxn′?xn+b)??(x?xn)H?(x?xn′n,neik(′??b)

固体物理学习题解答-陈长乐 khdaw

80

b)?(x?xn?b)dx=1, 1|2 = 2|1 =0, 1|1 = 2|2 =C?C,

1|H ???|1 ?ECC 1|H|2 =0.

?|1 ?|2 ?EC?C 2|H 2|H

(4.9)

(4.10)

?|2 = 1|H

CC?

e

?ikxn

??(x?xn)H

?

n

eik(xn′+b)?(x?xn′?b)dx

n′

=

C?C

xn′?xn+b)

??(x?xn)H?

(x?xn′n,n

eik(′

??b)dxRm=xn′?xn,xn=0,

1|H

?|2 =C?Ceik(xm+b)

??(x)H?

m

?b)dx.

?(x?xn??(x)H?

?(x?xm?b)dx=α, 1|H

?,

xm=0,

|2 =C?Cαeikb. 2|H

?e?ikb.H

??,

|1 =C?Cα,

1|H

?|1 =C?C ??(x?xn)H?

?(x?xn′n′

?b)dx=C?Cβ1.,

2|H

?|2 =C?Cβ2.(4.10),?CC

β1

?Eiβαekb?E

2E?(β1+β2)E+(β1β2? α?e

?ikb α?α)=0.

=0,

2

E=

β1+β2

(β1?β2)2+4α?α

,

§4.4(

81

γ2).

a=b,

E(k)=ε0?β?

(n,n)

=ε0?β?[γ1(e?ikxa+eikxa)+γ2(e?ikyb+eikyb)]=ε0?β?2γ1cos(kxa)?2γ2cos(kyb).

(2)

γ1=γ2.

s

eik·Rmγ(Rm)

?kE(k)=0

E(k)

.

kx=ky=0

,E(k)

,

(3)

Emax=ε0?β+2(γ1+γ2).

Emin=ε0?β?2(γ1+γ2);

kx=π/a,ky=π/b

,E(k)

,

?E=Emax?Emin=4(γ1+γ2).

?2E

1

2

kx=ky=0

ww

γ2b2

1

2

kx=π/a,ky=π/b

γ1a0

2

w

1

2

γ1a0

2

γ2b2

1

2

γ1a0

2

0γ2b2

4.8

,

.k

.

.

,

hd

?ky?kx

γ1acos(kxa)

2

γ2b2cos(kyb)

.

aw

.

.

.c

.

??E(k)=2kk

1

2?kx

?2E

?

,

4.7

s

β?2γ[cos(kxa)+cos(kya)],

v(k)=

1

E(k)=ε0?

[sin(kxa)i+

om

82sin(kya)j].

kx=±π/a

,

v(k)=

2γa

sin(kya)j+sin(kza)k].2γa

[sin(kxa)i+

kx=±π/a

,

v(k)=

w

ww

dt.

.k

hd

答案

dt

(4.11)

a=

dv

2

d2E(k)

d2E(k) 2

π

dk,

aw

.c

om

§4.4

83

dt

dp(k)=dpdω=

2L

2

·

d2E(k)

a

π

·

mF

dk2

dk=

2Lτ

2

dE

a

2m

,V(r)

V(r)=

,V(0)=

1

,

G′

w

?2+V(r).

V(G)e

iG′·r

ww

=V(0)+

G′=0

.k

V(G′)eiG·r,

2m,

?

2

?′=H

?′,H

G′=0

V(G′)eiG·r

hd

aw

.

?0?0(r)=E0?k(r)Hkk

?

2

1S

2m

,

?0k(r)=

.c

eik·r,

,S

.

om

dk k=±π

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