Physics 701 - Solutions 5 - University of New Hampshire

Physics 701 - Solutions 5
Department of Physics
University of New Hampshire
(1) Griffiths 2.13
a. [5] We find A by simply normalizing the wave function.
1 = hΨ(0) |Ψ(0)i
= 9 hψ0 |Ψ0 i + 12 hψ0 |Ψ1 i + 12 hψ1 |Ψ0 i + 16 hψ1 |Ψ1 i or,
Z ∞
2
= A
9|ψ0 (x)|2 + 12ψ0∗ (x)ψ1 (x) + 12ψ1∗ (x)ψ0 (x) + 16|ψ1 (x)|2 dx
−∞
2
= A (9 + 0 + 0 + 16) = 25A2
⇒ A = 1/5
b. [5] You can immediately get the coefficients: c0 = 3A and c1 = 4A, and
also using En = h
¯ ω(n + 12 ) you get
Ψ(x, t) = A 3ψ0 (x)e−iωt/2 + 4ψ1 (x)e−i3ωt/2
|Ψ(x, t)|2 = A2 9|ψ0 (x)|2 + 12ψ0∗ (x)ψ1 (x)e−iωt + 12ψ0 (x)ψ1∗ (x)eiωt + 16|ψ1 (x)|2
1 =
9ψ0 (x)2 + 24ψ0 (x)ψ1 (x) cos(ωt) + 16ψ1 (x)2
25
c. [10] In this problem you are asked to verify Ehrenfest’s theorem, so you
cannot use it to obtain the answer. To get the answer you can do a
1
straight integration:
Z ∞
2
hˆ
xi = A
x 9ψ0 (x)2 + 24ψ0 (x)ψ1 (x) cos(ωt) + 16ψ1 (x)2 dx
−∞
Z ∞
2
xψ0 (x)ψ1 (x)dx
= 24A cos(ωt)
−∞
r
Z ∞
24 mω √ h
¯
2
=
ξ 2 e−ξ dξ
2
cos(ωt)
25 π¯
h
mω
−∞
r
24
h
¯
=
cos(ωt)
25 2mω
∞
d
d
−iωt
iωt
hˆ
pi = 12A
ψ0 (x) ψ1 (x)e
+ ψ1 (x) ψ0 (x)e
dx
i −∞
dx
dx
r
Z
i
¯ ∞h
mω √ h
2 −ξ 2 −iωt
2
2 −ξ 2 iωt
2
(1 − ξ )e e
= 12A
−ξ e e
dξ
π¯
h
i −∞
r
24 m¯
hω
sin(ωt)
= −
25
2
¯
2h
Z
Alternatively, and probably much quicker, you can use the ladder operators:
r
h
¯
x
ˆ=
(ˆ
a+ − ˆ
a− )
2mω
However, we must be careful. The ladder operators only raise the eigenstate wavefunctions, not the time components! You can see this by
explicitly working it out:
r
h
¯
∂
−iEn t/¯
h
ˆ
a± ψn (x)e
=
∓
+ mωx ψn (x)e−iEn t/¯h
2mω
∂x
√
n + 1ψn+1 (x)e−iEn t/¯h
√
=
nψn−1 (x)e−iEn t/¯h
Now using this, in Dirac notation, and writing altogether too many
2
steps, we get:
hˆ
xi = hΨ(t) |ˆ
x| Ψ(t)i
r
h
¯
=
hΨ(t) |ˆ
a+ + ˆ
a− | Ψ(t)i
2mω
r
h
¯
=
(hΨ(t) |ˆ
a+ | Ψ(t)i + hΨ(t) |ˆ
a− | Ψ(t)i)
r 2mω
h
¯
=
|A|2 9 hψ0 |ˆ
a+ | ψ0 i + 12 hψ0 |ˆ
a+ | ψ1 i e−iωt +
2mω
12 hψ1 |ˆ
a+ | ψ0 i eiωt + 16 hψ0 |ˆ
a+ | ψ0 i
+ 9 hψ0 |ˆ
a− | ψ0 i + 12 hψ0 |ˆ
a− | ψ1 i e−iωt +
12 hψ1 |ˆ
a− | ψ0 i eiωt + 16 hψ0 |ˆ
a− | ψ0 i
r
h
¯
|A|2 12 hψ1 |ˆ
a+ | ψ0 i eiωt + 12 hψ0 |ˆ
a− | ψ1 i e−iωt
=
r 2mω
√
√
h
¯
=
|A|2 12 1eiωt + 12 1e−iωt
2mω
r
h
¯
24
cos(ωt)
=
25 2mω
And for the momentum:
r
h
¯ mω
hˆ
pi = i
hΨ(t) |ˆ
a+ − ˆ
a− | Ψ(t)i
2
r
√
√ −iωt h
¯ mω
2
iωt
|A| 12 1e − 12 1e
= i
2
r
24 h
¯ mω
sin(ωt)
= −
25
2
Check Ehrenfest’s theorem, Eq. [1.38]:
r
dhˆ
xi
24
h
¯
m
= −
mω sin(ωt) = hˆ
pi
dt
25 2mω
*
+
r
ˆ
dhˆ
pi
24 m¯
hω
dV
= −
ω cos(ωt) = −mω 2 hˆ
xi = −
dt
25
2
dˆ
x
Naively, with ψ2 (x) instead of ψ1 (x), the oscillating frequency for hˆ
xi
and hˆ
pi would be 2ω instead. However, hˆ
xi and hˆ
pi are actually zero
3
because the oscillating cross terms vanish (The total wave function is
even in x, or alternatively, the ladder operators don’t “reach that rung”).
The actual frequency is zero.
Z ∞
2
xψ0 (x)ψ2 (x)dx
hˆ
xi = 24A cos(2ωt)
−∞
= 0
It is also easy to see that the cross terms vanish because a
ˆ and a
ˆ† do
not couple ψ0 with ψ2 , nor would they for ψ0 and state with n > 1.
d. [5] The particle energy can be h
¯ ω/2 or 3¯
hω/2.
P (E = h
¯ ω/2) =
9/25
P (E = 3¯
hω/2) =
16/25
(2) [5] Griffiths 2.18
We use the identity: eiθ = cos(θ) + i sin(θ), then:
Aeikx + Be−ikx = A cos(kx) + iA sin(kx) + B cos(−kx) + iB sin(−kx)
= A cos(kx) + iA sin(kx) + B cos(kx) − iB sin(kx)
= (A + B) cos(kx) + (iA − iB) sin(kx) ⇒
C =A+B
A = 12 (C − iD)
⇔
D = i(A − B)
B = 12 (C + iD)
(3) Griffiths 2.22
a. [5]
Z
2
1 = |A|
2
∞
2
e−2ax dx
−∞
r
= |A|
π
2a
⇒ A=
2a
π
1/4
b. [5]
Z
∞
−(ax2 +bx)
e
−∞
Z
∞
dx =
−a(x+b/2a)2 +b2 /4a
e
−∞
4
r
dx =
π b2 /4a
e
a
The expansion coefficients are
Z ∞
A
2
φ(k) = √
e−ax e−ikx dx
2π −∞
A
2
= √ e−k /4a .
2a Z
∞
1
φ(k)eikx−iE(k)t/¯h dk
Ψ(x, t) = √
2π −∞
2
Z ∞
A
k
i¯
hk 2 t
= √
exp − −
+ ikx dk
4a
2m
4πa −∞
1/4 −ax2 /(1+2ia¯ht/m)
e
2a
p
=
π
1 + 2ia¯ht/m
c. [5] The width of the Gaussian wave packet (∼ w−1 ) increases with time.
1.0
|Ψ(x, t)|2
r
2
=
0.7
0.8
❘
0.9
|Ψ(x, t)|
0.6
r
=
0.4
0.5
t
0.2
0.3
➤
=
0.1
−10
−5
0
5
10
x
5
2
2
2a e−ax /(1−2ia¯ht/m) e−ax /(1+2ia¯ht/m)
p
p
π
1 − 2ia¯ht/m
1 + 2ia¯
ht/m
2
2
2a e−2ax /[1+(2a¯ht/m) ]
p
π
1 + (2a¯
ht/m)2
r
2 −2w2 x2
we
.
π
d. [5]
hxi =
Z
∞
x|Ψ(x, t)|2 dx = 0
(The integrand is odd.)
−∞
Z
∂Ψ(x, t)
h
¯ ∞ ∗
Ψ (x, t)
dx = 0 (The integrand is odd.)
hpi =
i −∞
∂x
Z ∞
2
hx i =
x2 |Ψ(x, t)|2 dx
r−∞ Z ∞
2
1
2 2
=
w
x2 e−2w x dx =
π
4w2
−∞
Z ∞
d2
hp2 i = −¯
h2
Ψ∗ (x, t) 2 Ψ(x, t)dx
dx
−∞
"
r Z ∞
2 # −ax2 /(1+iu)
−ax2 /(1−iu)
e
2ax
2a
2a
e
√
√
−
dx
= h
¯2
π −∞
1 + iu
1 + iu
1 − iu
1 + iu
r
2 #
Z ∞"
2a
2a h
¯2
2ax
2
2
√
=
−
e−2ax /(1+u ) dx
2
π 1 + u −∞ 1 + iu
1 + iu
"
#r
r
2
2a
2a h
¯2
1
2a
(1 + u2 )
π(1 + u2 )
√
=
−
π 1 + u2 1 + iu 2 1 + iu
2a
2a
=
a¯
h2 ,
where u = 2a¯
ht/m.
σx = 1/2w
σp =
√
a¯
h
e. [5] At t = 0 the uncertainty limit is satisfied.
σx σp =
h
¯p
h
¯
1 + (2a¯ht/m)2 ≥
2
2
6
(4) [5] Griffiths 2.42
Looking at the graph that we already
made for Quiz2, you can see that for
x > 0 we have the Harmonic Oscillator
potential, while for x < 0 the potential
is infinite. We thus have solutions that
are the same as for the H.O. potential for
x > 0, but with a boundary condition
that ψ(0) = 0. That means we can only
use the odd solutions, so En = h
¯ ω(n + 21 )
for odd n, or Em = h
¯ ω(2m + 3/2) for m
any positive integer starting at zero.
(5) Griffiths 2.45 Extra Problem
This problem is actually a bit tricky. We start with the assumption that
there are two solutions with the same energy E, and then see if this leads
to a contradiction. Writing the time-independent Schr¨odinger’s equation for
the two solutions and multiplying through with the other solution we get:
)
h2
¯
d2
− 2m ψ2 dx2 ψ1 + V ψ1 ψ2 = Eψ1 ψ2
⇒
h2
¯
d2
− 2m
ψ1 dx
= Eψ1 ψ2
2 ψ2 + V ψ1 ψ2
d2
d2
ψ2 2 ψ1 − ψ1 2 ψ2 = 0 ⇒
dx dx
d
d
d
= 0⇒
ψ2 ψ1 − ψ1 ψ2
dx
dx
dx
d
d
ψ2 ψ1 − ψ1 ψ2 = K
(1)
dx
dx
where K is some constant independent of x. However, since this represents
normalizable solutions, we must have ψ1 → 0 and ψ2 → 0 as |x| → ∞, and
thus K must be zero. We then have:
d
d
ψ2 ψ1 = ψ1 ψ2 ⇒
dx
dx
1 d
1 d
ψ1 =
ψ2 ⇒
ψ1 dx
ψ2 dx
ln(ψ1 ) = ln(ψ2 ) + C ⇒
ψ1 = eC ψ2
So the two solutions are linearly dependent and thus not separate solutions.
7
Total number of points: 60
8

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