Math Infinitum

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Intuition

The product $\hat{a}^{\dagger}\hat{a}$ counts quanta: it is the number operator, and its eigenvalues are exactly the whole numbers 0, 1, 2 and so on. The proof is a short argument about a ladder that can only go down so far. Once it is done, the whole oscillator is known: every state is the bottom rung raised some number of times, and the square roots that appear when a rung is changed are all that is needed to compute anything.

An account that is paid into and drawn from only in whole coins, and can never go below zero, always holds a whole number of coins. The energy of an oscillator above its lowest level is counted the same way, in quanta.

The number operator

$\hat{N}=\hat{a}^{\dagger}\hat{a}$ is Hermitian and , so its eigenstates are the energy eigenstates. Its eigenvalues are the non-negative whole numbers, and the ladder operators act on its eigenstates with square roots.

\begin{aligned}\hat{N}\ket{n}&=n\ket{n},\qquad n=0,1,2,\dots\\ \hat{a}\ket{n}&=\sqrt{n}\,\ket{n-1}\\ \hat{a}^{\dagger}\ket{n}&=\sqrt{n+1}\,\ket{n+1}\end{aligned}

Properties

$[\hat{N},\hat{a}]=-\hat{a}$ and $[\overset{}{}$ : lowering removes one quantum and raising adds one.

The eigenvalues of \hat{N} are whole numbers

\hat{N}\ket{\nu}=\nu\ket{\nu}\ \Longrightarrow\ \nu\in\{0,1,2,\dots\}

An eigenvalue of $\hat{N}$ is the squared length of $\hat{a}\ket{\nu}$ , so it is never negative. Lowering reduces it by one each time. A chain of lowerings would reach negative eigenvalues unless it stops, and it can only stop at a state that the lowering operator kills, whose eigenvalue is zero. So the starting eigenvalue was a whole number of steps above zero.

Proof steps

\nu=\bra{\nu}\hat{a}^{\dagger}\hat{a}\ket{\nu}=\lVert\hat{a}\ket{\nu}\rVert^{2}\ge0

For a normalised eigenstate the eigenvalue is a squared length.

\hat{N}\hat{a}\ket{\nu}=(\hat{a}\hat{N}-\hat{a})\ket{\nu}=(\nu-1)\,\hat{a}\ket{\nu}

From $[\hat{N},\hat{a}]=-\hat{a}$ : lowering reduces the eigenvalue by one, or gives zero.

\hat{a}^{k}\ket{\nu}\ \text{has eigenvalue}\ \nu-k\ \text{or is}\ 0

Lower $k$ times.

\hat{a}\ket{\nu_{0}}=0\ \ \text{for some}\ \nu_{0}=\nu-k\ge0

Negative eigenvalues are impossible by the first step, so the chain must end at a state the lowering operator kills.

\nu_{0}=\lVert\hat{a}\ket{\nu_{0}}\rVert^{2}=0\ \Rightarrow\ \nu=k

The last state has eigenvalue zero, so $\nu$ is the whole number $k$ .

Applications

Practice

Counting Quanta

The number operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ counts how many quanta of energy a state holds above the lowest level.

E_{n}=\hbar\omega\bigl(n+\tfrac12\bigr)

Try it

An oscillator state has energy $4.5\,\hbar\omega$ . How many quanta does it hold?