Physics Problems & Solutions: Eigenvalues

Showing posts with label Eigenvalues. Show all posts

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 - Eigenfunction

Which of the following is an eigenfunction of the linear momentum operator −iħ ∂/∂x with a positive eigenvalue ħk; i.e., an eigenfunction that describes a particle that is moving in free space in the direction of positive x with a precise value of linear momentum?

A. cos kx
B. sin kx
C. e^−ikx
D. e^ikx
E. e^−kx

(GR8677 #57)

Solution:

Eigenequation:

$\begin{aligned} \hat{p}\psi&=a\psi \\-i\hbar\frac{\partial \psi}{dx}&=\hbar k\psi \\\frac{\partial \psi}{dx}&=ik\psi \end{aligned}$

(A) FALSE
cos kx → ∂(cos kx)/∂x = sin kx

(B) FALSE
sin kx → ∂(sin kx)/∂x = cos kx

(C) FALSE
e^−ikx → ∂(e^−ikx)/∂x = −ik e^−ikx

(D) TRUE
e^ikx → ∂(e^ikx)/∂x = ik e^ikx

(E) FALSE
e^−kx→ ∂(e^kx)/∂x = k e^kx

Answer: D

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 - Hermitian Matrix

The matrix $A= \left| \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right|$ has 3 eigenvalues $\lambda_i$ defined by $Av_i=\lambda_i v_i$ . Which of the following statements is NOT true?

A. $\lambda_1 + \lambda_2 + \lambda_3 = 0$
B. $\lambda_1, \lambda_2,$ and $\lambda_3$ are all the real numbers
C. $\lambda_1\lambda_2 = + 1$ for some pair roots
D. $\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1= 0$
E. $\lambda_i^3 =+1, i =1,2,3$

(GR9277 #98)

Solution:

Every real symmetric matrix is Hermitian, and therefore all its eigenvalues are real.

A symmetric matrix is a square matrix that is equal to its transpose.
$A=A^T \rightarrow a_{ij} = a_{ji}$

Matrix $A = \left| \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right|$

is not symmetric, since $a_{ij} \neq a_{ji}$ . Therefore, its eigenvalues are NOT real.

Answer: B