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Solutions Manual
Fundamentals of Quantum Mechanics: For Solid State Electronics and Optics C.L. Tang Cornell University Ithaca, N. Y.
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Chapter 2 2-1. (a) ψ(x) [in units of A]
4
2
−4
−2
0
4
2
x
6
(b) +∞
1
6
−∞
−4
1
1 = ∫ | Ψ ( x ) | 2 dx = | A | 2 [ ∫ ( 4 + x) 2 dx + ∫ ( 6 − x ) 2 dx ] =
∴ A=
(c)
1 5
3 10
= 1
250 | A |2 3
.
, by inspection.
Next, find σ2 first:
1
6
−4
1
σ 2 = | A |2 [ ∫ ( x − 1) 2 ( 4 + x) 2 dx + ∫ ( x − 1) 2 ( 6 − x) 2 dx ] =
therefore,
< x="" 2=""> = σ2 + < x="">2 =
7 2
2 - 1
.
5 2
;
(d)
The answer to this question is tricky due to the discontinuous change in the slope of the wave function at x = -4, 1, and 6. Taking this into account ,
< k.="" e.=""> = −
2-2.
2
2
h 3 3h (0 ⋅1−5 ⋅ 2+ 0 ⋅ 1) = 2m 250 50m
Given 2 sin(3 πx/a),for 0 < x="">< a="" ψ="" (x="" )="" a="" 0,="" for="" x="" ≤="" 0="" and="" x="" ≥="" a.="">
(a) < h="">=−
2
h 2 2m a
∂
a
2
∫ sin( 3πx /a) ∂x
sin( 3πx /a) dx = 2
0
2
2
9π h 2ma 2
(b)
h2 ∂2 ˆ H sin( 3πx / a ) = − sin( 3πx / a) = E sin( 3πx / a) 2 2m ∂x
∴ E =
2
2
9π h 2ma 2
.
(c) 9 π 2h
−i t 2 2 Ψ ( x , t )= sin( 3πx / a)e 2ma a
.
(d)