Download Chemistry 532 Problem Set 8 Fall, 2012 Solutions

Chemistry 532
Problem Set 8
Fall, 2012
Solutions
1. Evaluate the commutators [L̂i , x2 + y 2 + z 2 ] where i = x, y or z. [Hint: This problem
can be done in two lines if you think.]
Answer:
[L̂i , x2 + y 2 + z 2 ] = [L̂i , r2 ] = 0
because L̂i is independent of r for all i.
2. Let ψ be an eigenfunction of L̂z . Show that
< ψ|L̂x |ψ >=< ψ|L̂y |ψ >= 0.
Answer: Let
L̂z ψ = mh̄ψ
We have
L̂x =
Then
hψ|L̂x |ψi =
1
[L̂y , L̂z ]
ih̄
1
hψ|[L̂y , L̂z ]|ψi
ih̄
1
hψ|L̂y L̂z − L̂z L̂y |ψi
ih̄
mh̄
=
hψ|L̂y − L̂y |ψi = 0
ih̄
=
We also have
L̂y =
Then
hψ|L̂y |ψi =
1
[L̂z , L̂x ]
ih̄
1
hψ|[L̂z , L̂x ]|ψi
ih̄
1
hψ|L̂z L̂x − L̂x L̂z |ψi
ih̄
mh̄
=
hψ|L̂x − L̂x |ψi = 0
ih̄
=
1
3. Show that if any operator, Â commutes with two components of the angular momentum, it commutes with the third.
Answer: Let [Â, L̂x ] = [Â, L̂y ] = 0. We must show that [Â, L̂z ] = 0 as well. Now
1
[Â, [L̂z , L̂y ]]
ih̄
[Â, L̂z ] =
=
1
{ÂL̂x L̂y − ÂL̂y L̂x − L̂x L̂y  + L̂y L̂x Â}
ih̄
1
{ÂL̂x L̂y − ÂL̂y L̂x − ÂL̂x L̂y + ÂL̂y L̂x } = 0
ih̄
~
The same proof holds for the other components of L.
=
4. Show that the spherical harmonics are eigenfunctions of the operator L̂2x + L̂2y , and
determine the eigenvalues. This problem should take about 4 lines of math.
Answer:
L̂2 = L̂2x + L̂2y + L̂2z
L̂2x + L̂2y = L̂2 − L̂2z
(L̂2x + L̂2y )Y`m = (L̂2 − L̂2z )Y`m
= h̄2 [`(` + 1) − m2 ]Y`m
The eigenvalues are h̄2 [`(` + 1) − m2 ].
5. Calculate the first two non-vanishing terms in the expansion of the function
(
F (x) =
−1 −1 ≤ x ≤ 0
+1
0<x≤1
in Legendre polynomials. [See Table 6-3 on page 283 of the text.]
Answer:
∞
X
F (x) =
A`0 P`0 (x)
`0 =0
Z 1
−1
∞
X
F (x)P` (x) dx =
A` 0
Z 1
`0 =0
=
∞
X
A` 0
`0 =0
= A`
2`0
−1
P`0 (x)P` (x) dx
2
δ`,`0
+1
2
2` + 1
2` + 1 Z 1
A` =
F (x)P` (x) dx
2
−1
2
Z 1
1Z 1
1 Z0
A0 =
F (x) dx = [−
dx +
dx] = 0
2 −1
2
−1
0
Z 1
3
3Z 1
x dx = 3
x dx =
A1 =
2 −1
2
0
Z 1
51
(3x2 − 1)F (x) dx = 0
A2 =
2 2 −1
7Z 1 3 5 3
A3 =
x − x F (x) dx
2 −1 2 3
Z 1 3 5 3
=7
x − x dx
0 2 3
7
21 5
1
−
=−
=
2 12 2
8
Then
3
7
F (x) = P1 (x) − P3 (x) + . . .
2
8
6. Schmidt orthogonalize the set of functions {1, x, x2 , x3 , . . .} on the interval −1 ≤ x ≤ 1,
and show that the first three members of the orthogonal set at proportional to the first
three Legendre polynomials.
Answer:
v1 = 1
P0 (x) = 1
v2 = x + av1
hv1 |v2 i =
Z 1
Z 1
x dx + a
−1
dx = 0
−1
0 + 2a = 0
v2 = x
a=0
P1 (x) = x
v3 = x2 + bx + c = x2 + bv2 + cv1
hv1 |v3 i =
c=−
Z 1
2
x dx + c
−1
Z 1
x3 dx + b
−1
1
v3 = x −
3
Z 1
x2 dx = 0
−1
2
0+ b=0
3
2
dx = 0
−1
12
1
1Z 1 2
x dx = −
=−
2 −1
23
3
hv2 |v3 i =
Then
Z 1
b=0
3 2 1
3
1
P2 (x) = (3x2 − 1) =
x −
= v3
2
2
3
2
3
7. Determine uncertainty relations between the three components of the linear momentum
and the three components of the angular momentum of a system.
Answer:
1
∆px ∆Lx ≥ |h[p̂x , L̂x ]i|
2
#
"
∂
∂
∂
2
,y
−z
=0
[p̂x , L̂x ] = −h̄
∂x ∂z
∂y
Then
∆px ∆Lx = 0
1
∆px ∆Ly ≥ |h[p̂x , L̂y ]i|
2
#
"
∂
∂
∂
2
,z
−x
[p̂x , L̂y ] = −h̄
∂x ∂x
∂z
2
= h̄
∂ ∂
∂ ∂
x −x
∂x ∂z
∂z ∂x
= h̄2
!
∂
= ih̄p̂z
∂z
Then
∆px ∆Ly ≥
h̄
|hpz i|
2
Next,
∂
∂
∂
,x
−y
∂x ∂y
∂x
2
[p̂x , L̂z ] = −h̄
= −h̄2
!
∂
= −ih̄p̂y
∂y
so that
∆px ∆Lz ≥
h̄
|hpy i|
2
In a similar way
∆py ∆Lx ≥
h̄
|hpz i|
2
∆py ∆Ly = 0
∆pz ∆Lx ≥
h̄
|hpy i|
2
∆pz ∆Ly ≥
and
∆py ∆Lz ≥
h̄
|hpx i|
2
h̄
|hpx i|
2
∆pz ∆Lz = 0
8. A particle of mass m is confined to move on the surface of a sphere of radius R. The
state of the particle at time t = 0 is described by the wavefunction
Ψ(θ, φ) =
1/2
1
3
[Y11 (θ, φ) + Y10 (θ, φ) + Y1−1 (θ, φ)]
where Ylm (θ, φ) are the spherical harmonics.
4
(a) If Lz is measured for this system at time t = 0, what are the possible outcomes
of the measurement and with what relative probabilities?
Answer:
Lz = −h̄, 0, h̄
each with probability
P =
1
√
3
!2
=
1
3
(b) If L2 is measured for this system at time t = 0, what are the possible outcomes
of the measurement and with what relative probabilities?
Answer:
L2 = h̄2 1(1 + 1) = 2h̄2
P =1
(c) What is the average value of Lx over Ψ(θ, φ)?
Answer:
L̂+ = L̂x + iL̂y
L̂− = L̂x − iL̂y
so that
1
L̂x = (L̂+ + L̂− )
2
1
1
hψ|L̂x |ψi =
Y11 + Y10 + Y1−1 (L̂+ + L̂− ) Y11 + Y10 + Y1−1
3
2
√
√
√
√
h̄
= hY11 + Y10 + Y1−1 |( 2|Y11 i + 2|Y10 i + 2|Y10 i + 2|Y1−1 i)
6
√
h̄ √
2 2h̄
= 4 2=
6
3
9. The quantum mechanical Hamiltonian operator for a two-dimensional particle of mass
m confined to move on the perimeter of a ring of radius R is given by
Ĥ = −
h̄2 d2
2mR2 dφ2
√
(a) Show that the set of functions {1/ 2π exp(inφ)} form a normalized complete set
of simultaneous eigenfunctions of the Hamiltonian and the z-component of the
angular momentum. Impose boundary conditions on the set of wavefunctions to
determine the possible values of the quantum number n.
Answer:
h̄ d inφ
L̂z einφ =
e = nh̄einφ
i dφ
inφ
Ĥe
h̄2 d2 inφ
n2 h̄2 inφ
=−
e =
e
2mR2 dφ2
2mR2
einφ = ein(φ+2π) = einφ ein2π
5
Then
ein2π = 1
n = 0, ±1, ±2, . . .
Z 2π
2π
1
e−inφ einφ dφ =
=1
2π 0
2π
(b) Suppose at time t = 0 the quantum state of the particle is described by the wavefunction ψ(φ) = N cos2 (φ) where N is the normalization. If the z-component of
the angular momentum and the energy of the particle are measured simultaneously at time t = 0, what values can be obtained for each observable and with
what probabilities?
Answer:
2π
Z 2π
6π
2
4
2 3φ N
cos φ dφ = N
=1
= N2
8 0
8
0
4
N=
3π
4
3π
1/2
cos2 φ =
1/2
4
6
cn =
=
1/2
4
6
1/2
∞
X
1
cn √ einφ
2π
n=−∞
1 Z 2π
cos2 φe−inφ dφ
π 0
1 Z 2π 1 iφ
(e + e−iφ )2 e−inφ dφ
π 0 4
1/2 2
1
1
=
δn,2 + δn,−2 + δn,0
3
2
2
Then for Lz we can measure 2h̄ or −2h̄ with probabilities of 1/6 each or we can
find Lz = 0 with probability 2/3. The energy levels are
En =
n2 h̄2
2mR2
We can obtain from an energy measurement
4h̄2
E=
2mR2
i.e. n = 2
with probability 1/3 and E = 0 with probability 2/3.
10. For the 1s ground state of the hydrogen atom, find
(a) hri
Answer:
ψ1s (r) =
6
1
πa30
!1/2
e−r/a0
1 Z ∞ Z π Z 2π
hri = 3
dr
dθ
dφ r2 sin θe−r/a0 re−r/a0
πa0 0
0
0
4π Z ∞
dr r3 e−2r/a0
=
πa0 0
4
a30
=
(b) h1/ri
Answer:
1
r
3
= a0
2
=
4
a30
!
a20
4
!
=
1
a0
!
4 Z∞
dr e−2r/a0
a30 0
=
=
(d) hT i
Answer:
!
=
(c) h1/r2 i
Answer:
1
r2
6a40
24
!
4 Z∞
dr r e−2r/a0
a30 0
!
4
a30
!
a0
2
=
2
a20
h̄2 ∂ 2 ∂
L̂2
h̄2 2
r
+
− ∇ =−
2µ
2µr2 ∂r ∂r 2µr2
L̂2 ψ1s (r) = 0
d2
2 d
1 d 2d
r
=
+
2
2
r dr dr
dr
r dr
Then
!
T̂ ψ1s (r) =
Now
and
Then
d2
2 d
+
ψ1s (r)
2
dr
r dr
1
d −r/a0
= − e−r/a0
e
dr
a0
1
d2 −r/a0
e
= 2 e−r/a0
2
dr
a0
h̄2
h̄2 2 −r/a0
− ∇e
=−
2µ
2µ
7
!
1
2
−
e−r/a0
2
a0 ra0
and
h̄2
h̄2
hT i = −
hψ1s |ψ1s i +
2µa20
2µa0
2
2
h̄
h̄2
h̄
=−
+
=
2µa20 µa20
2µa20
1
r
11. An electron in a hydrogen atom occupies a 3dxy orbital. If Lz , the z-component of the
angular momentum of the electron is measured, determine what values can be obtained
and with what probabilities?
Answer:
!1/2
1
1
Ψ3dxy (r, θ, φ) = √
r2 e−r/3a0 sin2 θ sin 2φ
7
81 2π a0
Because
h̄ ∂
L̂z =
i ∂φ
we need only focus on the φ part of the wavefunction; i.e. we take
ψ(φ) = A sin 2φ
Normalizing ψ we have
A2
Z 2π
0
1 Z 4π
1 y 1
dφ sin2 2φ = A2
− sin 2y
dy sin2 y = A2
2
2 2 4
0
4π
0
2
=A π=1
or
1
ψ(φ) = √ sin 2φ
π
√
The normalized eigenfunctions of L̂z are {1/ 2πeinφ }, n = 0, ±1, ±2, . . .. Then
∞
X
1
1
√ sin 2φ =
cn √ einφ
π
2π
n=−∞
or
1 Z 2π −inφ
cn = √
e
sin 2φ dφ
π 2 0
1 Z 2π −inφ 1 i2φ
[e − e−i2φ ] dφ
= √
e
2i
π 2 0
1 Z 2π iφ(2−n)
√
=
[e
− e−iφ(2+n) ] dφ
2iπ 2 0
2π
√ [δn,2 − δn,−2 ]
=
2πi 2
1
= √ [δn,2 − δn,−2 ]
i 2
Then a measurement of Lz can find Lz = ±2h̄ each with a probability of 1/2.
8