Creation and Annihilation Operators

The breakthrough of the operator method is to factor the Hamiltonian. A sum of two squares like $p^2 + (m\omega x)^2$ does not factor over the reals, but it nearly does over the complex numbers — and the small correction it leaves behind is exactly the zero-point energy.

Natural units

To keep the algebra clean we set $\hbar = m = \omega = 1$ for the rest of the module; physical factors can be restored at the end by dimensional analysis. In these units the Hamiltonian is

$\hat{H} = \frac{1}{2}\left(\hat{p}^2 + \hat{x}^2\right), \qquad [\hat{x}, \hat{p}] = i.$

Defining the ladder operators

We introduce two operators built from $\hat{x}$ and $\hat{p}$ :

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

The operator $a$ is the annihilation (or lowering) operator and $a^\dagger$ the creation (or raising) operator. Because $\hat{x}$ and $\hat{p}$ are Hermitian, $a^\dagger$ is genuinely the adjoint of $a$ — taking the dagger swaps the sign of the $i\hat{p}$ term, as shown. Note that $a$ is not Hermitian: it is not an observable but a tool for moving between states.

Inverting the definitions expresses position and momentum back in terms of the ladder operators,

\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}},

a substitution we will use repeatedly.

Why this factors the Hamiltonian

Compute the product $a^\dagger a$ directly, being careful that $\hat{x}$ and $\hat{p}$ do not commute:

a^\dagger a = \frac{(\hat{x} - i\hat{p})(\hat{x} + i\hat{p})}{2} = \frac{\hat{x}^2 + \hat{p}^2 + i(\hat{x}\hat{p} - \hat{p}\hat{x})}{2} = \frac{\hat{x}^2 + \hat{p}^2 + i[\hat{x}, \hat{p}]}{2}.

Using $[\hat{x}, \hat{p}] = i$ , the cross term becomes $i \cdot i = -1$ , so

a^\dagger a = \frac{\hat{x}^2 + \hat{p}^2}{2} - \frac{1}{2} = \hat{H} - \frac{1}{2}.

Rearranging gives the central identity of the whole module:

\boxed{\ \hat{H} = a^\dagger a + \frac{1}{2}\ } \quad(\text{in units of }\hbar\omega).

The leftover $\tfrac{1}{2}$ is the zero-point energy, and it arose entirely from the non-commutativity of $\hat{x}$ and $\hat{p}$ . Had the operators commuted (the classical case), the factorization would have been exact and the ground-state energy zero.

Try it

This is a numerical exercise — your code should return a number. Using the number-basis action $a|n\rangle = \sqrt{n}\,|n-1\rangle$ and $a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$ , build small matrices for $a$ and $a^\dagger$ , form $a^\dagger a + \tfrac{1}{2}$ , and confirm its $(0,0)$ entry is the ground-state energy $E_0 = \tfrac{1}{2}$ .

Run your code to see the quantum state.