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Question

Let $|m\rangle$ and $|n\rangle$ denote the energy eigenstates of a one-dimensional simple harmonic oscillator. The position and momentum operators are $\hat{X}$ and $\hat{P}$, respectively. The matrix element $\langle m|\hat{P}\hat{X}|n\rangle$ is non-zero when

The correct answer is
$m = n$ or $m = n \pm 2$

SHO Operator Expressions

The position ($\hat{X}$) and momentum ($\hat{P}$) operators for a Simple Harmonic Oscillator (SHO) can be expressed using annihilation ($a$) and creation ($a^\dagger$) operators:

  • $\hat{X} = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$
  • $\hat{P} = i\sqrt{\frac{m\omega\hbar}{2}}(a^\dagger - a)$

The product operator $\hat{P}\hat{X}$ is then:

$\hat{P}\hat{X} = \left(i\sqrt{\frac{m\omega\hbar}{2}}(a^\dagger - a)\right) \left(\sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)\right) = i\frac{\hbar}{2} (a^\dagger - a)(a + a^\dagger)$

Expanding and using the commutation relation $a a^\dagger = a^\dagger a + 1$ yields:

$\hat{P}\hat{X} = i\frac{\hbar}{2} (a^\dagger a + a^\dagger a^\dagger - a a - a a^\dagger) = i\frac{\hbar}{2} (2a^\dagger a + a^\dagger a^\dagger - a a - 1)$

Matrix Element P X Calculation

We evaluate the matrix element $\langle m|\hat{P}\hat{X}|n\rangle$ using the properties of ladder operators on SHO energy eigenstates $|n\rangle$:

  • $a|n\rangle = \sqrt{n}|n-1\rangle$
  • $a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$

The matrix elements $\langle m|\hat{O}|n\rangle$ involving the terms in $\hat{P}\hat{X}$ have specific selection rules:

  • $\langle m|a^\dagger a|n\rangle \neq 0$ only if $m = n$.
  • $\langle m|a^\dagger a^\dagger|n\rangle \neq 0$ only if $m = n+2$.
  • $\langle m|a a|n\rangle \neq 0$ only if $m = n-2$.
  • $\langle m|1|n\rangle \neq 0$ only if $m = n$.

Non-Zero Conditions Derived

The matrix element $\langle m|\hat{P}\hat{X}|n\rangle$ is a sum of terms derived from $\hat{P}\hat{X} = i\frac{\hbar}{2} (2a^\dagger a + a^\dagger a^\dagger - a a - 1)$. For the total matrix element to be non-zero, at least one of its constituent terms must be non-zero.

This requires the quantum number $n$ to satisfy one of the following conditions derived from the operators:

  • $m = n$ (from $2a^\dagger a$ and $-1$ terms)
  • $m = n+2$ (from $a^\dagger a^\dagger$ term)
  • $m = n-2$ (from $-a a$ term)

Therefore, the matrix element $\langle m|\hat{P}\hat{X}|n\rangle$ is non-zero when $m = n$ or $m = n \pm 2$.

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Important Questions from Operators Commutators Heisenberg Picture

  1. Consider an operator $\hat{A}$ which is not Hermitian. Find the possible values of $c$ and $d$ such that the operator $(c\hat{A} - d\hat{A}^\dagger)$ is Hermitian.
  2. Which of the following operators is/are self-adjoint?
  3. Consider operators $\hat{A}$, $\hat{B}$, and $\hat{C}$ for three observables of a quantum system satisfying $[\hat{A}, \hat{B}] = 0$, $[\hat{B}, \hat{C}] = 0$, and $[\hat{A}, \hat{C}] \neq 0$, with uncertainties $\Delta A, \Delta B, \Delta C$, respectively. From the options given below, which is/are implied by the commutation relations among $\hat{A}, \hat{B}$, and $\hat{C}$?
  4. The wavefunction of a particle in one dimension is given by 
    $\psi(x) = \begin{cases} M, & -a < x < a \\ 0, & \text{otherwise.} \end{cases}$ 
    Here $M$ and $a$ are positive constants. If $\phi(p)$ is the corresponding momentum space wavefunction, which one of the following plots best represents $|\phi(p)|^2$ ?

  5. If $H$ is the Hamiltonian for a free particle with mass $m$, the commutator $[x, [x, H]]$ is
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