The Commutator [a, a†]

A single equation governs the entire ladder structure of the oscillator:

$\boxed{\ [a, a^\dagger] = 1\ }.$

Every result in this module — the spectrum, the normalization factors, the matrix elements — is a consequence of this one relation together with the Hamiltonian $\hat{H} = a^\dagger a + \tfrac{1}{2}$. Let us derive it.

The derivation

Recall $a = (\hat{x} + i\hat{p})/\sqrt{2}$ and $a^\dagger = (\hat{x} - i\hat{p})/\sqrt{2}$. The commutator is bilinear, so

Expand using linearity. The pure $[\hat{x}, \hat{x}]$ and $[\hat{p}, \hat{p}]$ terms vanish (every operator commutes with itself), leaving only the cross terms:

where we used the antisymmetry $[\hat{p}, \hat{x}] = -[\hat{x}, \hat{p}]$. Substituting the canonical relation $[\hat{x}, \hat{p}] = i$,

So $[a, a^\dagger] = 1$, equivalently $a a^\dagger = a^\dagger a + 1$. The annihilation operator and the creation operator do not commute; pushing one past the other costs (or refunds) a single unit.

Why this one relation is enough

Compare with the canonical commutator $[\hat{x}, \hat{p}] = i\hbar$. The ladder algebra is its dimensionless cousin: $[a, a^\dagger] = 1$ carries exactly the same physical content (the non-commutativity of position and momentum) but in a form tailored to the oscillator. Because the number operator is $\hat{N} = a^\dagger a$, the relation $a a^\dagger = a^\dagger a + 1$ immediately gives

which are precisely the statements "$a^\dagger$ raises the quantum number by one" and "$a$ lowers it by one" — the subject of the next lessons.

Try it

This is a numerical exercise — your code should return a number. Represent $a$ and $a^\dagger$ as matrices in the number basis and confirm that $[a, a^\dagger] = aa^\dagger - a^\dagger a$ acts as the identity, returning the $(1,1)$ diagonal entry (which equals $1$).

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