Download Quantum Mechanics Practice Problems Solutions

Chem3615 Quantum Mechanics Practice Problems These problems will not be handed in but are to help you prepare for the midterm 1.
2.
andB
x
2
on the function
0 Which of the following functions is an eigenfunction of the operator ? Give the eigenvalue where appropriate. k is a constant. a. k 0 0
∴ k is an eigenfunction with eigenvalues of 0 c. sin(kx) d.
∴ kx2 is not an eigenfunction 2
b. kx2 sin
∴ sin(kx) is not an eigenfunction ∴ k is an eigenfunction with eigenvalues of k 2
e.
3.
x
Find the results of operating with A
∴ is not an eigenfunction f.
∴ k is an eigenfunction with eigenvalues of ik What is the value of the following commutators , ̂
a.
, ̂
∴
b.
, ̂
, ̂
, ̂
2
, ̂
, ̂
∴
2
c.
, ̂
∴
4.
, ̂
0 0 Angular momentum, L, is given by the cross product r x p. So the angular momentum in the x axis, Lx, is given by ypz – zpy, what is the operator ? 5.
A quantum mechanical problem of importance is molecular vibration between two atoms of masses m1 and m2. The simplest case is a harmonic oscillator which has a potential energy of V(x) = ½kx2. The kinetic energy is given by ½v2, where . a. What is the Hamiltonian operator for the harmonic oscillator? 1
1
1
2
2
2
2
1
1
̂
→
2
2
2
2
b. Consider the operator, defined by ∗
∗ ∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
. Show that is Hermitian. ∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
c. Is an eigenfunctions of ? If so what is its eigenvalue. 1 ∴it is an eigenfunction with an eigenvalue of 1. 2
d. Is an eigenfunctions of ? If so what is its eigenvalue. 2
2
2
2
3
2
2
2
2
4
2
2
6
3 ∴ itisaneigenfunctionwithaneigenvalueof3. e. The total energy of the harmonics oscillator is Can the oscillator have zero energy? No minimum energy is when n =0 and 0, 1, 2, ⋯ f.
Classically the potential energy of the oscillator cannot be greater than the total energy of the oscillator. For the n =0 energy level what is the maximum value of x that the oscillator can have (the classical turning point)? 1
2
2
→
→
. Classically the maximum value of x is g. The wavefunction, n, for the n=0, 1 and 2 states are given below whereα
Ψ x
e
,Ψ x
xe
,Ψ x
the oscillator ever be in the classically forbidden region? Yes since (x) 0 for |x| > 2αx
1 e
also true for other value of n. see graph Is there anywhere (x) =0 beside x = ±
Yes for n=1 at x=0 and n=2 when 2x2 ‐1 = 0 or h. What is the average x value of the oscillator for the n=0 state? 〈x〉
Ψ
∗
x xΨ x dx
α
π
e
x
α
π
〈x〉
xe
dx 0 (since the integrand is an odd function) or 〈x〉
α
π
xe
dx
α
π
1
e
2α
0 e
dx Will i.
j.
What is the average x2 value of the oscillator for the n=0 state? 〈x 〉
〈x 〉
〈x 〉
Ψ
∗
x x Ψ x dx
α
π
1
x e
dx
dx
e
2√απ
α
π
e
α
π
1
1
xe
2α
1
π
2α
2√απ α
α
π
x
α
π
2 μk
e
dx 1
2α
e
dx The probability of electromagnetic radiation causing a transition from state n to state m is related to the transition dipole moment Ψ ∗ x xΨ x dx. i. What is the value of this integral for the transition from n=0 to n=1? 2α
π
1
2π
4α
π
xe
α
π
e
1
2α
x
e
2α
π
α
1
2α
e
dx
2α
π
e
1
2π
e
dx 0 Will this transition occur? Yes because the value of the integral does not equal 0 ii. What is the value of this integral for the transition from n=0 to n=2? α
4π
2αx
1 e
α
π
e
α
e
e
2
2π
Since both integrands are odd functions of x both integrals are equal to 0 so 0 Will this transition occur? No, because the value of the integral is 0 6.
Consider the particle in a one‐dimensional box of length L, the wavefunction is given by 2
Ψ
a) Show that the wavefunctions are orthonormal Ψ
i.e. Ψ∗
1 Hint: Ψ∗
Ψ
Ψ∗
,
2
2
2
0 Ψ
2
Using 2
Ψ
Ψ∗
Ψ
2
,
Ψ∗
Ψ∗
2
2
2
2
4
0
2
0
4
1 sin 0
Ψ
1
2
1
1
Ψ
4
1
2
Ψ
Ψ∗
Ψ∗
2
Ψ
∴
2
sin 2
4
∗
Ψ
1
0
0
0
sin 0
0
0 b) Using classical mechanics what is the probability of finding the particle i.
in the left hand third of the box (i.e. between 0 and ⅓L)? 33%, classical mechanics would say that the particle is equally likely to be anywhere in the box so there is a 33% chance of it being in any third of the box ii.
In the middle third of the box (i.e. between ⅓L and ⅔L)? 33% c) Using quantum mechanics what is the probability of finding the particle i.
in the left hand third of the box (i.e. between 0 and ⅓L)? Ψ∗
Ψ
2
2
2
2
6
2
4
2
0
2
3
2
2
4
0
2
1
3
4
2
3
Therefore there is ~ 33% chance (depending on n) of finding the particle in the left hand side of the box ii.
in the middle third of the box (i.e. between ⅓L and ⅔L)? Ψ∗
2
4
2
3
2
2
1
3
iii.
Ψ
2
2 2
6
2
2
2
2
3
2
4
4
4
3
6
2
4
3
What are these probabilities as n for the left hand third it is lim
for the middle third it is lim
→∞
→∞
so in both cases as n approaches the classical value 33% 33% 7.
A 1.0 g mass object is moving with a speed of 1.0 cm/s in a one‐dimensional box of length 1.0 cm. a. Find the quantum number n. 2
2 0.0010
0.010
0.010 /
→
3.0 10 2 6.63 10 b. Is this object expected to obey classical mechanics? Yes c. What is wavelength of radiation is emitted when the 1.0 g object in 1.0 cm box undergoes a transition from the n=2 to n=1 state? Note the energy of the n state is 1.65 10
8
.
2
1 .
6.63 10 8 0.0010
0.010
1.21 10 (greater than the distance to the furthest observed star) 8.
Consider the alpha decay of a radioactive nucleus. The alpha particle, 4He, must tunnel through energy barrier holding the nucleus together. a. Assume the energy of the alpha particle is 5 MeV = 8.0x10‐13 J and the barrier has a height of 100 Mev 1.6x10‐11 J and a width of 10 fm. What is the transmission probability each time the alpha particle hits the barrier? Assume the mass of the alpha particle is 4 times the mass of a proton (mp = 1.67x10‐27 kg) 1
1 4
√
Where ,
2 4 1.67 10 6.63 10 2
2 4 1.67 10 6.63
.
1
, 8.0 10
1.6 10 10 2
.
.
8.0 10
4.27 10 1
9.80 10 .
9.80 10 6.1 10
2
4.27 10 4 9.80 10 2 2 sinh
2
4.27 10 2
4.27 10 1.0 10
b. What is the velocity of the alpha particle? 1
2
→
2
2 8.0 10
4 1.67 10
1.55 10 / c. Assume the alpha particle bounces back and forth in a 1.0x10‐15 m box. What is the collision frequency of the alpha particle with the barrier? 1.55 10 /
1.55 10 1.55 10 d. What will be the decay rate, k, of the alpha particles passing through the barrier? 6.1 10
9.5 10 1.55 10 e. Given that the half‐life, ln 2
/
9.5 10 /
, what is the half‐life for the alpha decay? 7.3 10 23