Matrix Elements of x and p

The ladder operators turn the otherwise hard problem of computing integrals like $\int \psi_m^*(x)\,x\,\psi_n(x)\,dx$ into simple bookkeeping. Because $\hat{x}$ and $\hat{p}$ are linear in $a$ and $a^\dagger$, every matrix element is just a pair of square roots.

Matrix elements of position

Use $\hat{x} = (a + a^\dagger)/\sqrt{2}$ and act to the right with the lowering and raising rules $a|n\rangle = \sqrt{n}|n-1\rangle$, $a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$:

Projecting onto $\langle m|$ and using orthonormality $\langle m|k\rangle = \delta_{mk}$,

$\boxed{\ \langle m|\hat{x}|n\rangle = \frac{1}{\sqrt{2}}\left(\sqrt{n}\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1}\right)\ }$

The element is nonzero only for $m = n \pm 1$: position connects a level only to its immediate neighbours. This is the harmonic selection rule.

Matrix elements of momentum

The same calculation for $\hat{p} = i(a^\dagger - a)/\sqrt{2}$ gives

Momentum obeys the same $\Delta n = \pm 1$ rule, but its elements are pure imaginary and antisymmetric in $m \leftrightarrow n$ — consistent with $\hat{p}$ being Hermitian (its matrix is $i$ times a real antisymmetric matrix, hence equal to its own conjugate transpose).

Reading the tables

Restoring physical units multiplies $\hat{x}$ by $\sqrt{\hbar/2m\omega}$ and $\hat{p}$ by $\sqrt{m\omega\hbar/2}$, but the structure is fixed by the dimensionless coefficients above. A few specific values:

The growth of the elements like $\sqrt{n}$ reflects the widening of the wavefunction at higher levels — exactly the $\langle\hat{x}^2\rangle = n + \tfrac{1}{2}$ trend found earlier.

Try it

This is a numerical exercise — your code should return a number. Build the truncated matrix of $\hat{x} = (a + a^\dagger)/\sqrt{2}$ and read off the element $\langle 2|\hat{x}|3\rangle = \sqrt{3}/\sqrt{2} = \sqrt{3/2}$.

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