# Coherent States

The general states of a harmonic oscillator can be expressed as a superpostion of the energy eigenstates $|n\rangle.\!$ A class of states that is of particular importance are the eigenstates of the (non-Hermitian) lowering operator $\hat{a},\!$

$a|\alpha\rangle=\alpha|\alpha\rangle,\!$

where $\alpha\!$ can be any complex number.

These states are known as coherent states. The term, "coherent", reflects their important role in optics and quantum electronics.

Note that it is not possible to construct an eigenstate of the raising operator $\hat{a}^{\dagger}$ because $a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle;$ this fact means that application of $\hat{a}^{\dagger}$ to any superposition of harmonic oscillator eigenstates eliminates the lowest-energy state that was present in the superposition.

The following are some properties of coherent states.

## Construction of Coherent States

The coherent state with eigenvalue $\alpha\!$ is given by

$|\alpha\rangle=\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle=e^{\alpha\hat{a}^\dagger}|0\rangle.$

We may see that this is a coherent state with the given eigenvalue as follows:

$\hat{a}|\alpha\rangle=\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\hat{a}|n\rangle=\sum_{n=1}^{\infty} \frac{\alpha^n}{\sqrt{n!}}\sqrt{n}|n-1\rangle=\sum_{n=1}^{\infty}\frac{\alpha^n}{\sqrt{(n-1)!}}|n-1\rangle= \alpha\left(\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle\right)=\alpha|\alpha\rangle$

This state, however, is not normalized, so we will now normalize it. Let us introduce a normalization constant, $N,\!$ into the coherent state:

$|\alpha\rangle=Ne^{\alpha a^{\dagger}} |0\rangle$

We now determine what value of $N\!$ yields a normalized state:

$1=\langle\alpha|\alpha\rangle=\langle 0|Ne^{\alpha^*a} Ne^{\alpha a^{\dagger}} |0\rangle = N^2\langle 0|e^{\alpha^*a} e^{\alpha a^{\dagger}} |0\rangle$

We now use the fact that, for any two operators $\hat{A}$ and $\hat{B}$ that both commute with their commutator, the following formula, known as the Campbell-Baker-Hausdorff formula, holds:

$e^\hat{A} e^\hat{B} = e^{\hat{A}+\hat{B}} e^{[\hat{A},\hat{B}]/2}\!$

Similarly,

$e^\hat{B} e^\hat{A} = e^{\hat{B}+\hat{A}} e^{[\hat{B},\hat{A}]/2} = e^{\hat{A}+\hat{B}} e^{-[\hat{A},\hat{B}]/2}.$

Combining the above two formulas, we obtain

$e^\hat{A} e^\hat{B} = e^\hat{B} e^\hat{A} e^{[\hat{A},\hat{B}]}.\!$

This result applies for $\hat{A}=\alpha^\ast \hat{a}\!$ and $\hat{B}=\alpha \hat{a}^{\dagger}\!$ because the commutator for these two operators is $[\hat{A},\hat{B}]=|\alpha|^2,\!$, which is a constant. We thus obtain

\begin{align} \langle\alpha|\alpha\rangle &= N^2\langle 0|e^{\alpha^\ast\hat{a}}e^{\alpha\hat{a}^{\dagger}}|0\rangle=N^2\langle 0|e^{\alpha\hat{a}^{\dagger}}e^{\alpha^\ast\hat{a}}e^{[\alpha^\ast\hat{a},\alpha\hat{a}^{\dagger}]} |0\rangle \\ &=N^2e^{|\alpha|^2}\langle 0|e^{\alpha\hat{a}^{\dagger}}e^{\alpha^\ast\hat{a}}|0\rangle=N^2e^{|\alpha|^2}\langle 0|e^{\alpha\hat{a}^{\dagger}} |0\rangle \\ &=N^2e^{|\alpha|^2}\langle 0|0\rangle=N^2e^{|\alpha|^2} \end{align}

We have thus determined the normalization constant,

$\rightarrow N=e^{-\frac{1}{2}|\alpha|^2}.$

The normalized coherent state $|\alpha\rangle$ is therefore

$|\alpha\rangle=e^{-|\alpha |^2/2}e^{\alpha\hat{a}^{\dagger}}|0\rangle.$

## Inner Product of Two Coherent States

We have shown that, for any complex number $\alpha,\!$ there is an eigenstate $|\alpha\rangle\!$ of the lowering operator $\hat{a}.$ Therefore, we have a complete set of coherent states. However, this is not an orthogonal set. Indeed, the inner product of two coherent states $|\alpha\rangle\!$ and $|\beta\rangle\!$ can be calculated as follows:

\begin{align} \langle \beta|\alpha \rangle &= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}\langle 0|e^{\beta^\ast\hat{a}} e^{\alpha\hat{a}^\dagger} |0\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}\langle 0|e^{\alpha\hat{a}^\dagger} e^{\beta^\ast\hat{a}} e^{[\beta^\ast\hat{a},\alpha\hat{a}^\dagger]}|0\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*}\langle 0|e^{\alpha\hat{a}^\dagger} e^{\beta^\ast\hat{a}}|0\rangle \\ &= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*} \end{align}

$|\langle\beta|\alpha\rangle |^2 = e^{-|\alpha-\beta|^2}.$

Hence, the set of coherent states is not orthogonal and the distance $|\alpha-\beta|\!$ in the complex plane measures the degree to which the two eigenstates are "approximately orthogonal".