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Chapter 3
3.1
If a were to
o material would begin to behave more like an
insulator.
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increase, the bandgap energy
would decrease and the material would begin to behave less like a semiconductor and more like a metal. If a
o3.2
Schrodinger's wave equation is:
??2?2??x,t? ?V?x????x,t? 2m?x2 ?j?were to decrease, the bandgap energy would increase and the
???x,t?
?t Assume the solution is of the form:
AHA12GAGGAGAGGAFFFFAFAF
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???E???kx? ??x,t??u?x?exp?j???t???? ????????? This equation may be written as
?u?x??2u?x?2mE ?ku?x??2jk??2u?x??0
?x?x2?
2 Region I: V?x??0. Substituting the assumed solution into the wave equation,
Setting u?x??u?x? for
1region I, the equation
we
becomes:
obtain:
????2???E??????jkuxexpjkx? ??t?????? 2m?x??????????d2u1?x?dx2?2jkdu1?x??k2??2u1?x??0 dx?? where ?Q.E.D.
In Region II, V?x??V.
O ????u?x??E?????exp?j?kx???t??? ???x???????????2?2mE?2
????jE??E???????uxexpjkx? ?j?????t????? ???????????? which becomes
????2??E???2kx? ??t???jk?u?x?exp?j???? 2m?????????? ?2jk???u?x??E???exp?j?kx???t???? ?x?????????Assume the same
???2u?x??E??????? ?expjkx?t????? ??2??x?????????????E???kx? ??Eu?x?exp?j???t???? ?????????AHA12GAGGAGAGGAFFFFAFAF
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form of the solution:
???E???kx? ??x,t??u?x?exp?j???t???? ????????? Setting u?x??u?x? for
2region II, this
equation becomes
Substituting into Schrodinger's wave equation, we find:
????2??E???2?????jkuxexpjkx? ??t?????? 2m??????????d2u2?x?dx2?2jkdu2?x?
dx???u2?x??0 ?2mVO?22k??? ????2? where again
?2jk???u?x??E???exp?j?kx???t???? ?x????????????2u?x??E????? ?exp?j?kx???t??? ??2?x??????????????E???kx? ?VOu?x?exp?j???t???? ????????????E???kx? ?Eu?x?exp?j???t???? ????????? This equation can be written as:
?u?x??2u?x? ?ku?x??2jk ??x?x22mVO2mE ???ux?u?x??0
?2?22AHA12GAGGAGAGGAFFFFAFAF
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?Q.E.D.
2?du1?x?2mE ?j???k?Aexp?j???k?x?
dx?2 ?j???k?Bexp??j???k?x?
and the second derivative becomes
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d2u1?x?2dx2 ??j???k??Bexp??j???k?x?
??j???k??Aexp?j???k?x?
2 Substituting these equations into the differential
3.3
We have
equation, we find
d2u1?x?dx2?2jkdu1?x??k2??2u1?x??0 dx?? Assume the solution is of the form:
u1?x??Aexp?j???k?x?
????k?Aexp?j???k?x?
2 ????k?Bexp??j???k?x?
2 ?2jk?j???k?Aexp?j???k?x?
?j???k?Bexp??j???k?x?? ?k2??2?Aexp?j???k?x? ?Bexp??j???k?x???0
?? ?Bexp??j???k?x?
Combining terms, we obtain
??2?2?k?k2?2k???k??k2??2 ?Aexp?j???k?x?
???2?2?k?k2?2k???k??k2??2
The first derivative is
????????????AHA12GAGGAGAGGAFFFFAFAF
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?Bexp??j???k?x??0
We find that
We have the
solutions
0?0
u1?x??Aexp?j???k?x?
Q.E.D.
For the
differential equation in
u2?x? and the
?Bexp??j???k?x?
for 0?x?a and
u2?x??Cexp?j???k?x?
?Dexp??j???k?x?
for ?b?x?0.
The first boundary condition is
u1?0??u2?0?
proposed solution, the procedure is exactly the same as above. _______________________________________
which yields
A?B?C?D?0
The second boundary condition is
3.4
du1dx?x?0du2dx
x?0 which yields
AHA12GAGGAGAGGAFFFFAFAF
半导体物理与器件第四版课后习题答案3
