Position and Momentum in Terms of a, a†

The ladder operators were built from $\hat{x}$ and $\hat{p}$ . Inverting that relationship lets us express position and momentum back in terms of $a$ and $a^\dagger$ — and that is what makes the algebra so practical, because $a$ and $a^\dagger$ have such simple matrix elements.

The inversion

Adding and subtracting the definitions $a = (\hat{x} + i\hat{p})/\sqrt{2}$ and $a^\dagger = (\hat{x} - i\hat{p})/\sqrt{2}$ isolates $\hat{x}$ and $\hat{p}$ :

a + a^\dagger = \frac{2\hat{x}}{\sqrt{2}} = \sqrt{2}\,\hat{x}, \qquad a - a^\dagger = \frac{2i\hat{p}}{\sqrt{2}} = i\sqrt{2}\,\hat{p}.

Solving,

$\boxed{\ \hat{x} = \frac{a + a^\dagger}{\sqrt{2}}, \qquad \hat{p} = \frac{a - a^\dagger}{i\sqrt{2}} = \frac{i(a^\dagger - a)}{\sqrt{2}}\ }$

(in natural units; restoring constants gives $\hat{x} = \sqrt{\hbar/2m\omega}\,(a + a^\dagger)$ and $\hat{p} = i\sqrt{m\omega\hbar/2}\,(a^\dagger - a)$ ). Both $\hat{x}$ and $\hat{p}$ are Hermitian, as they must be: $\hat{x}^\dagger = (a^\dagger + a)/\sqrt{2} = \hat{x}$ , and similarly for $\hat{p}$ .

Why this is so useful

The operators $a$ and $a^\dagger$ shift the number by exactly one rung. Therefore $\hat{x}$ and $\hat{p}$ , being their sum and difference, connect only neighbouring number states:

\langle m|\hat{x}|n\rangle \neq 0 \ \text{only when}\ m = n \pm 1.

This selection rule — that $\hat{x}$ has matrix elements only between adjacent levels — is the algebraic origin of the dipole selection rule $\Delta n = \pm 1$ for harmonic transitions. We quantify these matrix elements in the next lesson.

Mean square of position

A clean application: the spread of position in a number state. Square $\hat{x}$ ,

\hat{x}^2 = \tfrac{1}{2}(a + a^\dagger)^2 = \tfrac{1}{2}\left(a^2 + (a^\dagger)^2 + a a^\dagger + a^\dagger a\right).

Sandwiched between $\langle n|$ and $|n\rangle$ , the $a^2$ and $(a^\dagger)^2$ pieces shift the level by $\pm 2$ and so are orthogonal to $|n\rangle$ ; only the number-diagonal terms survive:

\langle n|\hat{x}^2|n\rangle = \tfrac{1}{2}\left(\langle n|a a^\dagger|n\rangle + \langle n|a^\dagger a|n\rangle\right) = \tfrac{1}{2}\left((n+1) + n\right) = n + \tfrac{1}{2}.

Since $\langle\hat{x}\rangle = 0$ in any number state, this is also the variance: $(\Delta x)^2 = n + \tfrac{1}{2}$ . By symmetry $(\Delta p)^2 = n + \tfrac{1}{2}$ as well, so excited states are progressively more spread out in both position and momentum.

Try it

This is a numerical exercise — your code should return a number. Build $\hat{x} = (a + a^\dagger)/\sqrt{2}$ as a truncated matrix, square it, and return the first-excited-state spread $\langle 1|\hat{x}^2|1\rangle = n + \tfrac{1}{2} = \tfrac{3}{2}$ .

Run your code to see the quantum state.