Math Infinitum

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Intuition

Of all the oscillator’s states, which moves most like a classical particle? The answer is a Gaussian with the ground state’s width, displaced from the centre: it swings back and forth at the classical frequency and never spreads. These coherent states are eigenstates of the lowering operator, their numbers of quanta are Poisson-distributed, and the light of a laser is one.

A crowd of runners on a circular track, released together and all running at exactly the same speed, stays a tight bunch for ever. A coherent state is a bunch like that: every part of it moves at the one frequency the oscillator allows, so nothing drifts apart.

The density of the coherent state with $\alpha=2$ at three moments, in units of $x_{0}$ : at $t=0$ centred at $2\sqrt2$ , a quarter period later passing the centre, half a period later at $-2\sqrt2$ . The packet keeps the ground state’s shape the whole time.

Coherent states

For any complex number $\alpha$ , the state below is an eigenstate of the lowering operator with eigenvalue $\alpha$ . It is a Gaussian of the ground state’s width, centred and moving according to $\alpha$ .

\begin{aligned}&\hat{a}\ket{\alpha}=\alpha\ket{\alpha}\\ &\ket{\alpha}=e^{-|\alpha|^{2}/2}\sum_{n=0}^{\infty}\frac{\alpha^{n}}{\sqrt{n!}}\ket{n}\end{aligned}

Properties

$\ket{0}$ is the coherent state with $\alpha=0$ . Lowering the sum term by term gives $\alpha$ times the same sum, which is the eigenvalue equation.
$\langle x\rangle=\sqrt2\,x_{0}\operatorname{Re}\alpha$ and , with and , as in the ground state.

A coherent state stays coherent

e^{-i\hat{H}t/\hbar}\ket{\alpha}=e^{-i\omega t/2}\,\ket{\alpha e^{-i\omega t}}

Each number state in the sum only acquires the phase of its energy. Taking out the half quantum’s phase, the rest multiplies $\alpha^{n}$ by $e^{-in\omega t}$ , which is $(\alpha e^{-i\omega t})^{n}$ — the same sum with turned, and the same normalisation, since turning does not change .

Proof steps

\ket{\alpha}=e^{-|\alpha|^{2}/2}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\ket{n}

The coherent state in number states.

e^{-i\hat{H}t/\hbar}\ket{n}=e^{-i\omega(n+1/2)t}\ket{n}

Each number state is stationary and acquires its own phase.

e^{-i\hat{H}t/\hbar}\ket{\alpha}=e^{-i\omega t/2}e^{-|\alpha|^{2}/2}\sum_{n}\frac{(\alpha e^{-i\omega t})^{n}}{\sqrt{n!}}\ket{n}

Take the half quantum’s phase out of the sum and join $e^{-in\omega t}$ to $\alpha^{n}$ .

\bigl|\alpha e^{-i\omega t}\bigr|=|\alpha|

Turning $\alpha$ keeps its modulus, so the normalising factor is the right one.

e^{-i\hat{H}t/\hbar}\ket{\alpha}=e^{-i\omega t/2}\,\ket{\alpha e^{-i\omega t}}

A coherent state again, with $\alpha$ turned clockwise at the rate $\omega$ .

Applications

Practice

Poisson Counts

The number of quanta in a coherent state is Poisson-distributed, with mean $|\alpha|^{2}$ — the squared modulus of the eigenvalue.

\langle\hat{N}\rangle=|\alpha|^{2}

Try it

A coherent state has $\alpha=3i$ . What is its mean number of quanta?