Physics Problems & Solutions: Hamiltonian

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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

Classical Mechanics - Hamiltonian

A. (p²/2m) + kx⁴
B. (p²/2m) − kx⁴
C. kx⁴
D. ½mv² − kx⁴
E. ½mv²

(GR8677 #35)

Solution:

Hamiltonian: H = T + U
U = kx⁴
T = ½mv² = p²/2m
H = (p²/2m) + kx⁴

Answer: A

Classical Mechanics - Lagrangian and Hamiltonian

Question 34-36

The potential energy of a body constrained to move on a straight line is kx⁴ where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

The body moves from x₁ at time t₁ to x₂ at time t₂. Which of the following quantities is an extremum for the x-t curve corresponding to this motion, if end points are fixed?

A. $\int_{t_1}^{t_2}\bigg(\frac{1}{2}mv^2-kx^4\bigg)dt$
B. $\int_{t_1}^{t_2}\frac{1}{2}mv^2dt$
C. $\int_{t_1}^{t_2}mxv$ $dt$
D. $\int_{x_1}^{x_2}\bigg(\frac{1}{2}mv^2+kx^4\bigg)dx$
E. $\int_{x_1}^{x_2}mv$ $dx$

(GR8677 #36)

Solution:

$\\$Lagrangian: $L=T-U \\$Hamiltonian: $ H=\int_{t_1}^{t_2} L dt \\U=kx^4 \\\\T=\frac{1}{2} mv^2 \\\rightarrow H=\int_{t_1}^{t_2}\bigg(\frac{1}{2} mv^2-kx^4\bigg)dt$

Answer: A

Quantum Mechanics - Schrodinger Equation

The Hamiltonian operator in the Schrodinger equation can be formed from the classical Hamiltonian by substituting

A. Wavelength and frequency for momentum and energy
B. A differential operator for momentum
C. Transition probability for potential energy
D. Sums over discrete eigenvalues for integrals over continuous variables
E. Gaussian distributions of observables for exact values

(GR8677 #49)

Solution:

Schrodinger Equation: Hψ(x) = Eψ(x)

Hamiltonian: $H=\frac{\hat p^2}{2m}+V(x)$

Momentum operator : $\hat p=-i\hbar \frac{\partial }{\partial x}$

Answer: B

Classical Mechanics - Hamiltonian

Two small equal masses m are connected by an ideal massless spring that has equilibrium length l₀ and force constant k. The system is free to move without friction in the plane of the page. If p₁and p₂represent the magnitude of momenta of the two masses, a Hamiltonian for this system is.

A. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-2k(l-l_0)\bigg]$

B. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-2k(l-l_0)^2\bigg]$

C. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-k(l-l_0)\bigg]$

D. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}-k(l-l_0)^2\bigg]$

E. $\frac{1}{2}\bigg[\frac{p_1^2}{m}+\frac{p_2^2}{m}+k(l-l_0)^2\bigg]$

(GR0177 #92)

Solution:

Hamiltonian: H = T + U

T_i= p_i²/2m
T₁+ T₂= p₁²/2m + p₂²/2m

U = ¹/₂k(Δl)²= ¹/₂k(l − l₀)²

H = p₁²/2m + p₂²/2m + ¹/₂k(l − l₀)²
= ¹/₂[ p₁²/m + p₂²/m + k(l − l₀)²]

Answer: E

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

Quantum Mechanics – Ladder Operator

The operator, $\hat a = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x+i\frac{\hat p}{m \omega_0}\big)$ when operating on a harmonic energy eigenstate ψ_n with energy E_n, produces another energy eigenstate whose energy is E_n− ħω₀. Which of the following is true?

I. $\hat a$ commutes with the Hamiltonian.

II. $\hat a$ is a Hermitian operator and therefore an observable.

III. The adjoint operator $\hat a^{\dagger}\neq \hat a$

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

(GR9677 #100)

Solution:

I. FALSE
Commutes if [H, a] = 0
But a is a ladder operator, a raises the energy level so that [H, a] = − ħωa

II. FALSE
Hermitian operator if $\hat a=\hat a^{\dagger}$

$\hat a = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x+i\frac{\hat p}{m \omega_0}\big)$

$\hat a^{\dagger} = \sqrt{\frac{m \omega_0}{2 \hbar}} \big(\hat x-i\frac{\hat p}{m \omega_0}\big)\neq\hat a$

III. TRUE
See II.

Answer: C