Physics Problems & Solutions: Operator

Showing posts with label Operator. Show all posts

Quantum Mechanics - Spin Angular Momentum

Two ions 1 and 2, at fixed separation, with spin angular momentum operators S₁ and S₂, have the interaction Hamiltonian H = −J S₁·S₂, where J > 0. The values of S₁² and S₂² are fixed at S₁(S₁+ 1) and S₂(S₂+ 1), respectively. Which of the following is the energy of the ground state of the system?

A. 0
B. –JS₁S₂
C. –J[S₁(S₁+ 1) – S₂(S₂+ 1)]
D. –(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) – S₁(S₁+1) – S₂(S₂+1)]
E. –(J/2)[(S₁(S₁+ 1) + S₂(S₂+ 1))/(S₁+ S₂)(S₁+ S₂+ 1)]

(GR9677 #77)

Solution:

S₁² ψ = S₁(S₁+ 1) ψ
S₂² ψ = S₂(S₂+ 1) ψ
S_i² ψ = S_i(S_i+ 1) ψ

H = −J S₁·S₂

Using general arithmetic equation: ab = ½ [(a + b)² − a² − b²]

H = −(J/2)[(S₁ + S₂)² − S₁² − S₂²]

Since S_i² ψ = S_i(S_i+ 1) ψ
For (S₁ + S₂)² → replace S_i with S₁ + S₂
→ (S₁ + S₂)² ψ = (S₁+ S₂)(S₁+ S₂+ 1) ψ

H = −(J/2)[(S₁+ S₂)(S₁+ S₂+ 1) − S₁(S₁+ 1) − S₂(S₂+ 1)]

Answer: D

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 - Expectation Value

If ψ is a normalized solution of the Schrodinger equation and Q is the operator corresponding to a physical observable x and the quantity ψ*Qψ may be integrated in order to obtain the

A. Normalization constant for ψ
B. Spatial overlap of Q with ψ
C. Mean value of x
D. Uncertainty in x
E. Time derivative of x

(GR8677 #56)

Solution:

Expectation value: $\langle Q\rangle=\int \psi(x)^*Q\psi(x)dx$

The predicted mean value of the result of an experiment.
The average value of the measurable quantity Q (observable x) on the state ψ

Answer: C

Quantum Mechanics - Hermitian Operator

The eigenvalues of a Hermitian operator are always

A. Real
B. Imaginary
C. Degenerate
D. Linear
E. Positive

(GR0177 #27)

Solution:

The eigenvalues are always real and hence observable.
Proof:
$\begin{aligned} A|\psi\rangle&=a|\psi\rangle \\A^\dagger|\psi\rangle&=a^*|\psi\rangle \\(A-A^\dagger)|\psi\rangle&=(a-a^*)|\psi\rangle \end{aligned}$

Hermitian operator: $A^\dagger=A$ .
So, $a-a^*=0\rightarrow a=a^* \rightarrow$ real

Answer: A

Quantum Mechanics - Expectation Value

The state $\psi=\frac{1}{\sqrt 6}\psi_{-1} +\frac{1}{\sqrt 2}\psi_1 +\frac{1}{\sqrt 3}\psi_2$ is linear combination of three orthonormal eigenstates of the operator $\hat{O}$ corresponding to eigenvalues -1,1 and 2. What is the expectation value of $\hat{O}$ for this state?

A. 2/3
B. √(7/6)
C. 1
D. 4/3
E. (√3+2√2-1)/√6

(GR0177 #29)

Solution:

The expectation value:

$\begin{aligned} \langle \hat{O}\rangle &=\langle\psi|\hat{O}|\psi\rangle \\&=O_1\left(\frac{1}{\sqrt{6}}\right )^2\langle \psi_{-1}|\psi_{-1}\rangle +O_2\left(\frac{1}{\sqrt 2}\right )^2\langle\psi_1|\psi_1\rangle +O_3\left( \frac{1}{\sqrt{3}}\right )^2\langle\psi_2|\psi_2\rangle \\&= (-1)\frac{1}{6} +(1)\frac{1}{2}+(2)\frac{1}{3}= 1\end{aligned}$

Answer: C

Quantum Mechanics - Commutation Relations

The components of the orbital angular momentum operator $\vec{L}=(L_x,L_y,L_z )$ satisfy the following commutation relations.

$\\\left[L_x,L_y \right ]=i \hbar L_z \\\left[L_y,L_z \right ]=i \hbar L_x \\\left[L_z,L_x \right ]=i \hbar L_y$

What is the value of the commutator $\left[L_x L_y,L_z \right ]$ ?

A. $2i\hbar L_x L_y$
B. $i\hbar (L_x^2+L_y^2 )$
C. $-i\hbar (L_x^2+L_y^2 )$
D. $i\hbar (L_x^2-L_y^2 )$
E. $-i\hbar (L_x^2-L_y^2 )$

(GR0177 #43)

Solution:

Commutator identities: [A, B] = − [B, A]

and, [B, AC] = A[B, C] + [B, A]C

Therefore,
$\left[\underset{A}{\underbrace{L_x L_y}},\underset{B}{\underbrace{L_z}}\right] = - \left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x L_y}}\right]$

$\left[\underset{B}{\underbrace{L_z}}, \underset{A}{\underbrace{L_x}}\underset{C}{\underbrace{L_y}}\right]=L_x [L_z,L_y ]+[L_z,L_x ] L_y$

$\begin{aligned} -[L_z,L_x L_y ]&=-L_x [L_z,L_y ]-[L_z,L_x ] L_y \\&=-L_x (-i\hbar L_x)-i\hbar L_y L_y \\&=i\hbar L_x^2-i\hbar L_y^2 \\&=i\hbar (L_x^2-L_y^2 ) \end{aligned}$

Answer: D

Quantum Mechanics - Harmonic Oscillator

Let $|n\rangle$ represent the normalized n^th energy eigenstate of the one-dimensional harmonic oscillator,

$H|n\rangle=\hbar\omega\left(n+\frac{1}{2}\right)|n\rangle$

If $|\psi\rangle$ is a normalized ensemble state that can be expanded as a linear combination

$|\psi\rangle=\frac{1}{\sqrt{14}}|1\rangle-\frac{2}{\sqrt{14}}|2\rangle+\frac{3}{\sqrt{14}}|3\rangle$

of the eigenstates, what is the expectation value of the energy operator in this ensemble state?

A. $\frac{101}{14} \hbar \omega$

B. $\frac{43}{14} \hbar \omega$

C. $\frac{23}{14} \hbar \omega$

D. $\frac{17}{\sqrt{14}} \hbar \omega$

E. $\frac{7}{\sqrt{14}} \hbar \omega$

(GR0177 #45)

Solution:

The energy eigenstates:

$\begin{aligned} H|1\rangle&=\hbar\omega\left(1+\frac{1}{2}\right)|1\rangle = \frac{3}{2}\hbar\omega|1\rangle \\H|2\rangle&=\frac{5}{2}\hbar\omega|2\rangle \\H|3\rangle&=\frac{7}{2}\hbar\omega|3\rangle \end{aligned}$

The expectation value:

$\begin{aligned} \langle H\rangle &=\langle\psi|H|\psi\rangle \\&=\frac{3}{2}\hbar\omega\left( \frac{1}{\sqrt{14}}\right )^2\langle1|1\rangle+\frac{5}{2}\hbar\omega\left(- \frac{2}{\sqrt{14}}\right )^2\langle2|2\rangle+\frac{7}{2}\hbar\omega\left( \frac{3}{\sqrt{14}}\right )^2\langle3|3\rangle \\&=\left( \frac{3}{28} +\frac{20}{28}+\frac{63}{28}\right)\hbar\omega \\&=\frac{86}{28}\hbar\omega =\frac{43}{14}\hbar\omega \end{aligned}$

Answer: B

Quantum Mechanics - Perturbation Theory

The raising and lowering operators for the quantum harmonic oscillator satisfy

$\begin{aligned} a^\dagger |n\rangle&= \sqrt{(n+1)}|n+1\rangle\\ a |n\rangle &= \sqrt{n}|n-1\rangle \end{aligned}$

for energy eigenstates $|n\rangle$ with energy E_n

. Which of the following gives the first-order shift in the n=2

energy level due to the perturbation

$\Delta H=V(a+a^\dagger )^2$

where

is constant?

A.

C. $\sqrt 2 V$
D. $2\sqrt 2 V$
E. 5 V

(GR0177 #94)

Solution:

The energy

$\Delta E=\langle n| \Delta H |n\rangle$

with n=2

and $\Delta H=V(a+a^\dagger )^2$

$\begin{aligned} \Delta E&=\langle 2| V(a+a^\dagger )^2 |2\rangle\\ &=V\langle 2| (aa+a^\dagger a^\dagger + aa^\dagger+a^\dagger a|2\rangle \\ &=V\langle 2|aa|2\rangle+V\langle 2|a^\dagger a^\dagger|2\rangle+V\langle 2|aa^\dagger|2\rangle+V\langle 2|a^\dagger a|2\rangle \end{aligned}$

Raising and lowering operators:

$\begin{aligned} a^\dagger |1\rangle &= \sqrt{1+1}|1+1\rangle=\sqrt{2}|2\rangle\\ a^\dagger |2\rangle &= \sqrt{2+1}|2+1\rangle=\sqrt{3}|3\rangle\\ a^\dagger |3\rangle &= \sqrt{3+1}|3+1\rangle=\sqrt{4}|4\rangle\\ \end{aligned}$

$\begin{aligned} a |1\rangle &= \sqrt{1}|1-1\rangle=\sqrt{1}|0\rangle\\ a |2\rangle &= \sqrt{2}|2-1\rangle=\sqrt{2}|1\rangle\\ a |3\rangle &= \sqrt{3}|3-1\rangle=\sqrt{3}|2\rangle\\ \end{aligned}$

$\langle 2|aa|2\rangle = \langle 2|a\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a|1\rangle=\sqrt{2}\langle 2|\sqrt{1}|0\rangle= \sqrt{2}\sqrt{1}\langle 2|0\rangle=0$

$\langle 2|a^\dagger a^\dagger|2\rangle = \langle 2|a^\dagger\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a^\dagger|3\rangle=\sqrt{3}\langle 2|\sqrt{4}|4\rangle= \sqrt{3}\sqrt{4}\langle 2|4\rangle=0$

$\langle 2|aa^\dagger|2\rangle = \langle 2|a\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a|3\rangle=\sqrt{3}\langle 2|\sqrt{3}|2\rangle= \sqrt{3}\sqrt{3}\langle 2|2\rangle=3$

$\langle 2|a^\dagger a|2\rangle = \langle 2|a^\dagger\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a^\dagger|1\rangle=\sqrt{2}\langle 2|\sqrt{2}|2\rangle= \sqrt{2}\sqrt{2}\langle 2|2\rangle=2$

$\Delta E=3V+2V=5V$

Answer: E