Physics Problems & Solutions: Commutation

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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 - 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 - Commutation Relations

Let $\hat{\mathbf{J}}$ be a quantum mechanical angular momentum operator. The commutator [ $\hat{J}_x$ $\hat{J}_y$ , $\hat{J}_x$ ] is equivalent to which of the following?

A. 0
B. iħ $\hat{J}_z$
C. iħ $\hat{J}_z$ $\hat{J}_x$
D. −iħ $\hat{J}_x$ $\hat{J}_z$
E. iħ $\hat{J}_x$ $\hat{J}_y$

(GR0877 #95)

Solution:

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

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

[ $\hat{J}_x$ $\hat{J}_y$ , $\hat{J}_x$ ] = − [ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ]

A = $\hat{J}_x$
B = $\hat{J}_x$
C = $\hat{J}_y$

[ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ] = $\hat{J}_x$ [ $\hat{J}_x$ , $\hat{J}_y$ ] + [ $\hat{J}_x$ , $\hat{J}_x$ ] $\hat{J}_y$
= $\hat{J}_x$ [ $\hat{J}_x$ , $\hat{J}_y$ ] + 0
= $\hat{J}_x$ [iħ $\hat{J}_z$ ] + 0

− [ $\hat{J}_x$ , $\hat{J}_x$ $\hat{J}_y$ ] = −iħ $\hat{J}_x$ $\hat{J}_z$

Answer: D