Physics Problems & Solutions: Wave Function

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Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤ 0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ²= 2² + 3²= 13

Total probability (normalized probability):
ψ²= 1² + 1² + 2² + 3² + 1²= 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Quantum Mechanics - Wave Function

If a freely moving electron is localized in space to within ∆x₀ of x₀, its wave function can be described by a wave packet

$\psi(x,t)=\int_{-\infty}^{\infty}e^{i(kx-\omega t)} f(k)dk$

where f(k) is peaked around a central value k₀. Which of the following is most nearly the width of the peak in k?

A. $\Delta k=\frac{1}{x_0}$

B. $\Delta k=\frac{1}{\Delta x_0}$

C. $\Delta k=\frac{\Delta x_0}{x_0^2}$

D. $\Delta k=\left (\frac{\Delta x_0}{x_0} \right )k_0$

E. $\Delta k=\sqrt{k_0^2+\left (\frac{1}{x_0} \right )^2}$

(GR9277 #27)

Solution:

In quantum mechanics, the momentum p = ħk and position x wave functions are Fourier transform pairs and the relation between p and x representations forms the Heisenberg uncertainty relation:

∆x∆k ≥ 1 → ∆k ≥ 1/∆x

Or, since k and x are fourier variables, their localization would vary inversely.

Answer: B

Quantum Mechanics, Particle in a box

An attractive, one-dimensional square well has depth V₀ as shown above. Which of the following best shows a possible wave function for a bound state?

(GR9277 #29)

Solution:

1-D square well with depth V₀ → finite potential well.

Wave function of a particle in finite potential well should have these following properties:

ψ →0 as x → ± ∞ or get further into regions where the classical particle cannot penetrate at all due to its inadequate energy
ψ is always a decaying exponential function
ψ must be continuous and differentiable

Therefore:

A. FALSE. ψ does not go to zero as x → ±∞

C. FALSE. ψ is not continuous

D and E are FALSE. ψ are not decaying exponential functions

Answer: B

Electromagnetism - Fourier Series

If n is an integer ranging from 1 to infinity, is an angular frequency, and t is time, then the Fourier series for a square wave, as shown above, is given by which of the following?

A. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\sin(n\omega t)$

B. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\sin(\left(2n+1\right)\omega t)$

C. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\cos(n\omega t)$

D. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\cos(\left(2n+1\right)\omega t)$

E. $V(t)=-\frac{4}{\pi}+\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n^2}\cos(n\omega t)$

(GR9277 #39)

Solution:

The sine function is an odd function.
The cosine function is an even function.
The function in this problem is an odd square wave function.
→ C, D, E are FALSE.

Both A and B are the same, thus the right solutions. However, A is more generalized and B is the simplified answer.

For Answer A, the sine term vanishes when n is even (n = 0, 2, 4, ...) → rewrite n for odd term as 2n +1 with n starts at 0 → the answer can be simplified to B.

Answer: B

Quantum Mechanics - Wave Function

The state of a quantum mechanical system is described by a wave function. Consider two physical observables that have discrete eigenvalues: observable A with eigenvalues {α}, and observable B with eigenvalues {β}. Under what circumstances can all wave functions be expanded in a set of basis states, each of which is a simultaneous eigenfunction of both A and B?

A. Only if the values {α} and {β} are nondegenerate
B. Only if A and B commute
C. Only if A commutes with the Hamiltonian of the system
D. Only if B commutes with the Hamiltonian of the system
E. Under all circumstances

(GR9277 #50)

Solution:

For two physical quantities to be simultaneously observable, their operator representations must commute, [A, B] = 0.

Answer: B

Proof:

Wave function, ψ = ∑_{_i} c_{_i} |v_{_i}>

A |v_{_i}> = α |v_{_i}>

B |v_{_i}> = β |v_{_i}>

BA |v_{_i}> = Bα |v_{_i}> = α B |v_{_i}> = αβ |v_{_i}>

AB |v_{_i}> = Aβ |v_{_i}>= β A |v_{_i}> = βα |v_{_i}>

(BA − AB) |v_{_i}> = (αβ − βα) |v_{_i}> = 0

[A, B] = 0

Quantum Mechanics - Infinite Potential Well

_{See Problem 51

The eigenfunctions satisfy the condition

_₀∫^{^a}ψ_{_n}*(x) ψ_{_l}(x) dx = δ_nl

δ_nl = 1 if n = l, otherwise δ_nl = 0. This is a statement that the eigenfunctions are

A. Solutions to the Schrodinger equation

B. Orthonormal
C. Bounded

D. Linearly dependent
E. Symmetric

(GR9277 #52)

Solution:

Orthonormality = orthogonal and normal

_₀∫^{^a}ψ_{_n}*(x) ψ_{_l}(x) dx = 〈ψ_{_n}|ψ_{_l}〉= δ_nl

Orthogonal, δ_nl = 0

Normal, δ_nl = 1 if n = l

Answer: B}

Classical Mechanics - Wave equation

The equation $y=A \sin2\pi\left(\frac{1}{T}-\frac{x}{\lambda}\right)$ where A, T, and λ are positive constants, represents a wave whose

A. Amplitude is 2A
B. Velocity is in the negative x–direction
C. Period is T/λ
D. Speed is x/t
E. Speed is λ/T

(GR8677 #04)

Solution:

(A) FALSE.
Amplitude = maximum displacement = y_max = A.

(B) FALSE.
For a wave traveling to the right (positive x-direction): y = ƒ(x − vt)
(x − vt) = constant.
As t increases, x must increases to keep (x − vt) = constant

For a wave traveling to the left (negative x-direction): y = ƒ(x + vt)
As t increases, x decreases to keep (x + vt) = constant.

The problem gives: y = ƒ(vt − x)
As t increases, x must increases to keep (x − vt) = constant.
The waves is traveling to the right.

(C) FALSE.
The unit of T/λ (second/meter) does not match with the unit is period (second).

(D) FALSE.
The speed of the wave is λ/T.

(E) TRUE.
See D.

Answer: E

Quantum Mechanics - Normalization of Wavefunction

Eigenfunction for a rigid dumbbell rotating about its center have a φ dependence of the form ψ(φ) = Ae^imφ, where m is a quantum number and A is a constant. Which of the following values of A will properly normalize the eigenfunction?

A. √(2π)
B. 2π
C. (2π)²
D. 1/√(2π)
E. 1/(2π)

(GR8677 #28)

Solution:
Normalized Wavefunction: $\int |\psi(x)|^2dx=\int \psi(x)\psi^*(x)dx=1$
$\\\psi(\phi)=Ae^{im\phi} \\\psi^*(\phi)=Ae^{-im\phi}$

$\\\int_{0}^{2\pi} |\psi(\phi)|^2d\phi=\int_{0}^{2\pi} |A|^2d\phi=1 \\|A|^2\int_{0}^{2\pi}d\phi=|A|^2 2\pi=1\rightarrow A = \frac{1}{\sqrt{2\pi}}$

Answer: D

Nuclear & Particle Physics - Selection Rules

A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from ψ₁ to ψ₂ is forbidden if ψ₁ and ψ₂

A. have opposite parity
B. are orthogonal to each other
C. are zero at the center of the atomic nucleus
D. are both spherically symmetrical
E. are associated with different angular momenta

(GR8677 #48)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆m_l = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

A. FALSE
It’s not related to the selection rules.

B. FALSE
In any transition, eigenstates should always be mutually orthogonal.

C. FALSE.
Eigenstates zero at the center → l

0 could change, for example from 3 to 2. This is allowed.

D. TRUE.
If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. l = 0 → l = 0 is forbidden.

E. FALSE.
The selection rules require l to change.

Answer: D

Classical Mechanics - Hooke's Law

A particle is constrains to move along the x-axis under the influence of the net force F = − kx with amplitude A and frequency f, where k is a positive constant. When x = A/2, the particle speed is

A. 2πfA
B. √3πfA
C. √2πfA
D. πfA
E. (1/3) πfA

(GR8677 #77)

Solution:

$F=-kx=m\ddot{x}\rightarrow m\ddot{x}+kx=0$

Equation of motion: $\ddot{x}+\omega^2x=0$
with angular velocity: ω = √(k/m) = 2πf

Solution to the equation of motion, wave function: x = A sin ωt
Velocity: v = dx/dt = Aω cos ωt

x = A/2 → A sin ωt = A/2
sin ωt = 1/2 → ωt = 30^o
cos 30^o = ½√3

v = Aω cos ωt = A 2πf ½√3 = √3πfA

Answer: B

Quantum Mechanics - Wave Function

A system containing two identical particles is described by a wave function of the form

$\psi= \frac{1}{\sqrt{2}} \left[ \psi_\alpha(x_1)\psi_\beta(x_2)+ \psi_\beta(x_1)\psi_\alpha(x_2)\right ]$

Where x₁ and x₂ represent the spatial coordinates of the particles and α and β represent all the quantum numbers, including spin, of the states that they occupy. The particles might be

A. Electrons
B. Positrons
C. Protons
D. Neutrons
E. Deuterons

(GR8677 #89)

Solution:

Wave function: $\psi= \frac{1}{\sqrt{2}} \left[ \psi_\alpha(x_1)\psi_\beta(x_2)\pm \psi_\beta(x_1)\psi_\alpha(x_2)\right ]$

Symmetric function (+ sign): $\psi_{\alpha\beta}=\psi_{\beta\alpha}$

→ obey Bose-Einstein statistics → bosons

Anti-symmetric function (−sign): $\psi_{\alpha\beta}=-\psi_{\beta\alpha}$
→ obey Fermi-Dirac statistics → fermion

Electrons and positron (anti-elecron) → fermion

Protons and neutron → fermionic hadrons

composite particles (hadron)
composed of 3 fermionic quarks

Deuterons → bosons

nucleus of deuterium, an isotope of hydrogen
composed of a neutron (spin 1/2) and a proton (spin 1/2), total spin = 1

Answer: E

Quantum Mechanics- Wave Function

A. 10
B. 5
C. 0
D. −5
E. −10

(GR0177 #28)

Quantum Mechanics - Wave Function

Which of the following functions could represent the radial wave function for an electron in an atom? (r is the distance of the electron from the nucleus; A and b are constants.)
I. Ae^−br
II. A sin (br)
III. A/r

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

(GR0177 #30)

Solution:

Property of a wave function: Ψ is finite.
Ψ(r → ∞) and Ψ(r → 0) cannot be ∞.

Ψ(r → ∞) = Ae^−∞ → 0
Ψ(r → 0) = Ae⁻⁰ → A
I. TRUE

Ψ(r → ∞) = A sin ∞↛ 0.
II. FALSE

Ψ(r → 0) = A/0 → ∞.
III. FALSE

Answer: A

Quantum Mechanics - Spin

Let $|\alpha\rangle$ represent the state of an electron with spin up and $|\beta\rangle$ the state of an electron with spin down. Valid spin eigenfunctions for a triplet state (^3S)

of a two-electron atom include which of the following?

$\\$I. $ \hspace{5 mm} |\alpha\rangle_1|\alpha\rangle_2 \\\\$II. $ \hspace{4 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) \\\\$III. $ \hspace{3 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right )$

A. I only
B. II only
C. III only
D. I and III
E. II and III

GR0177 #82)

Solution:

Singlet state is anti-symmetric: ψ(1,2) = −ψ(2,1)
Triplet state is symmetric: ψ(1,2) = ψ(2,1)

$$I. $ \hspace{5 mm} |\alpha\rangle_1|\alpha\rangle_2$ is symmetric, because

$\\ \psi(1,2)=|\alpha\rangle_1|\alpha\rangle_2 \\ \psi(2,1)=|\alpha\rangle_2|\alpha\rangle_1=|\alpha\rangle_1|\alpha\rangle_2=\psi(1,2)$

$$II. $ \hspace{4 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right )$ is antisymmetric, because
$\begin{aligned} \\ \psi(1,2)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) \\ \psi(2,1)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_2|\beta\rangle_1-|\alpha\rangle_1|\beta\rangle_2\right )\\&=-\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2-|\alpha\rangle_2|\beta\rangle_1\right ) =-\psi(1,2) \end{aligned}$

$$III. $ \hspace{3 mm} \frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right )$ is symmetric, because
$\begin{aligned} \\ \psi(1,2)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right ) \\ \psi(2,1)&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_2|\beta\rangle_1+|\alpha\rangle_1|\beta\rangle_2\right )\\&=\frac{1}{\sqrt{2}} \left( |\alpha\rangle_1|\beta\rangle_2+|\alpha\rangle_2|\beta\rangle_1\right ) =\psi(1,2) \end{aligned}$

Thus, II is singlet state, I and III are triplet state.

Answer: D

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