The Number Operator

We have the algebra $[a, a^\dagger] = 1$ and the factorization $\hat{H} = a^\dagger a + \tfrac{1}{2}$ . The Hermitian combination $a^\dagger a$ appears so often that it earns its own name.

Definition

The number operator is

$\hat{N} = a^\dagger a.$

It is Hermitian, since $\hat{N}^\dagger = (a^\dagger a)^\dagger = a^\dagger a = \hat{N}$ , so its eigenvalues are real and its eigenstates form an orthonormal basis. In terms of $\hat{N}$ the Hamiltonian is simply

$\hat{H} = \hbar\omega\left(\hat{N} + \tfrac{1}{2}\right),$

so $\hat{H}$ and $\hat{N}$ commute and share eigenstates. Diagonalizing the energy reduces to finding the spectrum of $\hat{N}$ — a counting problem.

Its eigenvalues are non-negative integers

Label an eigenstate by its eigenvalue: $\hat{N}|n\rangle = n\,|n\rangle$ , with $|n\rangle$ normalized. Two facts pin down what $n$ can be.

First, $n \ge 0$ . Take the expectation value:

n = \langle n|\hat{N}|n\rangle = \langle n|a^\dagger a|n\rangle = \|a|n\rangle\|^2 \ge 0,

because it is the squared norm of the vector $a|n\rangle$ . So no eigenvalue can be negative.

Second, $n$ must be an integer. From $[\hat{N}, a] = -a$ and $[\hat{N}, a^\dagger] = +a^\dagger$ (derived last lesson) one shows that $a|n\rangle$ is an eigenstate with eigenvalue $n-1$ and $a^\dagger|n\rangle$ one with eigenvalue $n+1$ . Repeatedly applying $a$ lowers the eigenvalue in integer steps. If $n$ were not an integer, this descent would eventually produce a state with negative eigenvalue, contradicting $n \ge 0$ . The only escape is that the chain terminates: at some point $a|n\rangle = 0$ , which happens precisely when $n = 0$ . Hence the allowed eigenvalues are

$n = 0, 1, 2, 3, \ldots$

The eigenstates $|n\rangle$ are called number states or Fock states, and $n$ counts the quanta of excitation in the oscillator.

The eigenvalue is the level index

Putting the two pieces together,

\hat{H}|n\rangle = \hbar\omega\left(n + \tfrac{1}{2}\right)|n\rangle,

so the integer $n$ that labels a number state is exactly the level index from the differential- equation solution. The next lessons make the raising/lowering action quantitative and pin down the ground state.

Try it

This is a numerical exercise — your code should return a number. Build $\hat{N} = a^\dagger a$ as a matrix in the number basis and confirm its diagonal entries are $0, 1, 2, \ldots$ by returning $N_{00} + N_{11} + N_{22} = 0 + 1 + 2 = 3$ .

Run your code to see the quantum state.