Key 8
1. Key to Prob. 5.30
(a) Normalization requires,
1 =
Z
∞
|ψ|2 dx = C 2
−∞
= C
∞
e−2x (1 − e−x )2 dx,
0
∞
Z
2
Z
(e−2x − 2e−3x + e−4x )dx =
0
C2
,
12
The integrals were evaluated using the formula,
∞
Z
xe−ax dx = 1/a2 . (a > 0)
0
Thus, C =
√
12 nm−1/2 .
(b) The most likely place for the electron in where the probability |ψ(x)|2 is
largest. This is also where ψ itself is largest. So setting the derivative dψ/dx
equal to zero:
0=
dψ
= C{−e−x + 2e−2x } = Ce−x {2e−x − 1}.
dx
The RHS vanishes when x = ∞ (a minimum), and when 2e−x = 1 or, x = ln 2
nm. Thus the most likely position is at xp = ln 2 nm=0.693 nm.
(c) The average position is,
hxi =
Z
∞
x|ψ|2 dx,
−∞
= C
2
Z
∞
xe−2x (1 − e−x )2 dx,
0
= C
2
Z
∞
x{e−2x − 2e−3x + e−4x }dx,
0
= C 2 {13/144} = 13/12 = 1.083 nm.
(We have used the formula in part (b) to evaluate the integrals)
1
2. Key to Prob. 5.34
(a) Since there is no preference for motion in the leftward vs. rightward direction, the average momentum of the particle will be zero, i.e. < px >= 0.
(b) Average energy of the particle is,
< E >= h
p2x
hp2 i
i + hU i = x + hU i.
2m
2m
Now < E >= E0 = h̄ω/2 for the ground state. Again,
1
1
hU (x)i = mω 2 hx2 i = h̄ω,
2
4
using, < x2 >= h̄/2mω from Problem 33. Then,
1
hp2x i = 2m(E0 − hU i) = mh̄ω.
2
(c)
s
4px =
q
(hp2x i − hpx i2 ) =
1
mh̄ω.
2
3. Key to Prob. 6.1
(a) The reflection coefficient is the ratio of the reflected to the incident wave
intensity, or,
| 12 (1 − i)|2
R= 1
= 1.
| 2 (1 + i)|2
(b) To the left of the step the particle is free and the wavenumber k is,
s
k=
2mE
.
h̄2
To the right of the step, U (x) = U and there,
s
k=
2m(U − E)
.
h̄2
2
As k’s are the same on both sides we get:
s
E
= 1, ⇒ U = 2E.
(U − E)
(c) For 10 MeV protons, E = 10 MeV, and m = 938.28 MeV/c2 . Using these
we and using h̄ = 197.3 MeV.fm (1 fm = 10−15 m) we find,
δ=
1
h̄
197.3 MeV.fm
=√
=q
= 1.44 fm.
k
2mE
(2)(938.28 MeV/c2 )(10 MeV)
4. Key to Prob. 6.7
The continuity requirements from Eqs. 6.8 are:
A + B = C + D,
(1)
ik(A − B) = α(D − C),
(2)
Ce−αL + DeαL = F eikL ,
(3)
α{DeαL − Ce−αL } = ikF eikL .
(4)
To isolate the transmission amplitude F/A we first solve Eqs. (1) and (2) and
express A in terms of C and D,
1
α
1
α
1−
C+
1+
D.
2
ik
2
ik
A=
(5)
Now from Eqs. (3) and (4) we solve for C and D in terms of F .
F
C =
2
F
D =
2
!
ik ikL αL
1−
e e ,
α
!
ik ikL −αL
1+
e e
.
α
(6)
(7)
Substituing C abd D from the above Eqs. in Eq. (5) we get,
A
e−ikL
k α
k α
=
(2 + i{ − }e−αL ) + (2 − i{ − }eαL ) ,
F
4
α k
α k
"
!
#
A
i α k
= e−ikL cosh αL +
−
sinh αL
F
2 k α
"
#
3
(8)
Now,
1
T
A ∗
,
F
!2
1 α k
2
= cosh αL +
−
sinh2 αL,
4 k α
A
A
= | |2 =
F
F


1
= 1+ 1+

4
1
= 1+
4
1
= 1+
4
α k
−
k α
!2
!2 

sinh2 αL,

α2 + k 2
sinh2 αL,
2αk
!
U2
sinh2 αL.
E(U − E)
4