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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. 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 $|\phi(p)|^2$ ?

  2. From the pairs of operators given below, identify the ones which commute. Here $l$ and $j$ correspond to the orbital angular momentum and the total angular momentum, respectively.
  3. An electromagnetic pulse has a pulse width of $10^{-3}$ s. The uncertainty in the momentum of the corresponding photon is of the order of $10^{-N}$ kg m $s^{-1}$, where $N$ is an integer. The value of $N$ is ________ (speed of light = $3 \times 10^8$ m $s^{-1}$, h = $6.6 \times 10^{-34}$ J s)
  4. In cylindrical coordinates $(s, \varphi, z)$, which of the following is a Hermitian operator?
  5. Let $|\psi_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $|\psi_2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ represent two possible states of a two-level quantum system. The state obtained by the incoherent superposition of $|\psi_1\rangle$ and $|\psi_2\rangle$ is given by a density matrix that is defined as $\rho ≡  c_1|\psi_1\rangle\langle\psi_1| + c_2|\psi_2\rangle\langle\psi_2|$. If $c_1 = 0.4$ and $c_2 = 0.6$, the matrix element $\rho_{22}$ (rounded off to one decimal place) is ________

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