Physics Problems & Solutions: Expectation Value

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Quantum Mechanics - Infinite Potential Well

Questions 51-53

A particle of mass m is confined to an infinitely deep square-well potential:

V(x) = ∞, x ≤ 0, x ≥ a
V(x) = 0, 0

a

The normalized eigenfunction, labeled by the quantum number n, are

$\psi_n=\sqrt{\frac{2}{a}} \sin \frac{n\pi x}{a}$

For any state n, the expectation value of the momentum of the particle is

A. 0

B. $\frac{\hbar n \pi}{a}$

C. $\frac{2\hbar n \pi}{a}$

D. $\frac{\hbar n \pi}{a} \left ( \cos n\pi-1 \right )$

E. $\frac{-i\hbar n \pi}{a} \left ( \cos n\pi-1 \right )$

(GR9277 #51)

Solution:

Infinitely deep square-well potential
→ there is zero probability for particle to be outside the well
→ 〈 $\hat p$ 〉= 0

If 〈 $\hat p$ 〉 ≠ 0 the particle would tend to go to the right or left and leave the well, which is impossible for infinitely deep square-well potential.

Answer: A

Alternative Answer #1:

Eigen function, $\hat p$ ψ_n = λψ_n
The eigenvalue, λ is associated with expectation value:〈Â〉= 〈ψ | Â | ψ〉

Since $\hat p$ is imaginer and ψ_n is real → the eigenvalue, λ is imaginer = not real = not observable
→ 〈 $\hat p$ 〉= 0

Alternative Answer #2:

since sine and cosine are orthogonal the whole period.

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 - 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 - 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