What is the position wavefunction of coherent states?

Question

Consider a coherent state $|\alpha\rangle$, $\alpha\in\mathbb C$. In the context of a quantum harmonic oscillator, this is defined as the eigenvector of the annihilation operator $a$: $a|\alpha\rangle=\alpha|\alpha\rangle$. In the Fock basis, this can be decomposed as $$|\alpha\rangle = e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle = e^{-|\alpha|^2/2}\sum_{k=0}^\infty \frac{\alpha^k}{\sqrt{k!}}|k\rangle.\tag A$$

An alternative way to express a coherent state is via its wavefunction in the position representation, denote it with $\psi_\alpha(x)$ (it's worth stressing that here "position" is to be understood as an abstract parameter, not necessarily related to a physical position, of which there is no notion of in this context).

How is $\psi_\alpha(x)$ derived? This is currently already to some degree discussed in the Wikipedia page, but I'm looking for the expression in dimensionless quantities and in the time-independent case. Moreover, I am also interested in the derivation of $\psi_\alpha(x)$ starting from (A), passing through the position representation of the Fock states.

Answer 1

Derivation from the eigenvalue condition

The most straightforward approach is to start from the position representation of the annihilation operator $a$.

I'll use the convention $a=\frac{1}{\sqrt2}(x+ip)$ and $a^\dagger=\frac{1}{\sqrt2}(x-ip)$, corresponding to which the canonical commutation relations read $[a,a^\dagger]=1$ and $[x,p]=i$. In the position representation, this corresponds to $a=\frac{1}{\sqrt2}(x+\partial_x)$.

The eigenvalue condition $a|\alpha\rangle=\alpha|\alpha\rangle$ then corresponds to $$x\psi_\alpha(x)+\psi_\alpha'(x)=\sqrt2\alpha\psi_\alpha(x),$$ which we can rewrite as $$\psi_\alpha'(x) = (\sqrt2\alpha-x)\psi_\alpha(x).$$ A natural ansatz to solve this is $\psi_\alpha(x)=C e^{f(x)}$ for some constant $C$. Using this we get $$f'(x) = \sqrt2\alpha-x \Longrightarrow f(x) = \sqrt2\alpha x - \frac{x^2}{2} + C'.$$ Upon some simple rearranging of the terms we get $\psi_\alpha(x) \propto \exp[-\frac12(x - \sqrt2\alpha)^2]$, which ensuring normalisation finally leads to $$\psi_\alpha(x) = \frac{e^{-\alpha_2^2}}{\pi^{1/4}} \exp\left[-\frac12(x - \sqrt2\alpha)^2\right],\tag{$R_1$}$$ where $\alpha=\alpha_1+i\alpha_2$.

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Derivation from (A)

Consider the wavefunctions $\psi_n$ of $|n\rangle$, which have the form $$\psi_n(x) = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} e^{-x^2/2}H_n(x),\tag B$$ where $H_n$ are the Hermite polynomials, defined here as $H_n(x)=(2x-\partial_x)^n \cdot 1$. This expression for $\psi_n$ comes from $$\psi_n(x) = \langle x|\frac{a^{\dagger n}}{\sqrt{n!}}|0\rangle = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} (x-\partial_x)^n e^{-x^2/2} = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} e^{-x^2/2}H_n(x).$$ Using (B) in (A), we get $$ \psi_\alpha(x) = e^{-|\alpha|^2/2} \frac{e^{-x^2/2}}{\pi^{1/4}} \sum_{k=0}^\infty \frac{\alpha^k}{\sqrt{k!}} \frac{1}{\sqrt{2^k k!}} H_k(x). $$ We now consider the identity $$\sum_{k=0}^\infty H_k(x) \frac{t^k}{k!} = e^{2xt - t^2}.$$ Using this with $t=\alpha/\sqrt2$ we get $$\psi_\alpha(x) = e^{-|\alpha|^2/2} \frac{e^{-x^2/2}}{\pi^{1/4}} e^{\sqrt2\alpha x - \alpha^2/2} = \frac{1}{\pi^{1/4}}e^{\frac12(\alpha^2-|\alpha|^2)}\exp\left[ -\frac12(x-\sqrt2\alpha)^2 \right]. \tag{$R_2$} $$ Observing that $\alpha^2-|\alpha|^2=-2\alpha_2^2 + 2i\alpha_1\alpha_2$, we see that ($R_2$) is consistent with ($R_1$), up to an irrelevant global phase. An analogous derivation is also given in Gerry, Knight (2004), section 3.3.

Answer 2

Here's an answer that takes a scenic route to get to the result. We start by finding "position eigenstates;" i.e., eigenstates of the $x$-quadrature operator $$ \hat{x}|x\rangle_x=x|x\rangle_x, $$ where the subscripts are explicitly telling us that the states refer to the $x$-quadrature basis that resolves the identity through $\int dx|x\rangle_x \vphantom{a}_x\langle x|=\mathbb{I}$. Using the translation between quadrature operators, we can rewrite this expression as $$ \frac{\hat{a}+\hat{a}^\dagger}{\sqrt{2}}|x\rangle_x=x|x\rangle_x. $$ States satisfying this kind of eigenvalue equation were studied in the early days of coherent states, now falling under the umbrella of "intelligent states" (e.g., here). It is now well known that these states must be squeezed coherent states in the limit of infinite squeezing. For example, if $x=0$, we can consider the following set of equalities, using squeeze operators $\hat{S}(z)=\exp\left[\left(z^*\hat{a}-z\hat{a}^{\dagger\,2}\right)/2\right]$: \begin{align} 0=\left(\hat{a}+\hat{a}^\dagger\right)|0\rangle_x=&\hat{S}(z)\left(\hat{a}+\hat{a}^\dagger\right)|0\rangle_x =\hat{S}(z)\lim_{r\to\infty}\frac{\hat{S}(-r)\hat{a}\hat{S}(-r)}{\cosh r}|0\rangle_x. \end{align} We can choose $z=-r$ at all points in the limit, and require the right-hand side to vanish, so this (plus some mathematical rigour) leads to the requirement $$0=\lim_{r\to\infty}\hat{a}\hat{S}(-r)|0\rangle_x.$$ We know that the zero eigenstate of $\hat{a}$ is $|0\rangle$ (in either the Fock basis or the coherent-state basis, but I'll use the lack of a label to imply the coherent-state basis), so we discover $$\lim_{r\to\infty}\hat{S}(-r)|0\rangle_x=|0\rangle\quad\Leftrightarrow\quad |0\rangle_x=\lim_{r\to\infty}\hat{S}(r)|0\rangle.$$ (I'm sure you can convince yourself of this with a sketch of a quasiprobability distribution.) The rest of the $x$-quadrature eigenstates can be found by displacing this squeezed state along the $x$ quadrature with the usual displacement operators $D(\alpha)$, which enact $D(\alpha)^\dagger \hat{x}D(\alpha)=\hat{x}+\frac{\alpha+\alpha^*}{\sqrt{2}}$: $$|x\rangle_x=D(x/\sqrt{2})|0\rangle_x=\lim_{r\to\infty}D(x/\sqrt{2})\hat{S}(r)|0\rangle.$$

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With this result, all we need to do is calculate one inner product to find the wavefunction: \begin{align} \psi_\alpha(x)=\langle x|_x |\alpha\rangle&=\lim_{r\to\infty}\langle 0|\hat{S}(r)^\dagger D(x/\sqrt{2})^\dagger |\alpha\rangle\\ &=\lim_{r\to\infty}\langle 0|\hat{S}(r)^\dagger |\alpha-x/\sqrt{2}\rangle\\ &=\lim_{r\to\infty}\frac{\exp\left[-\frac{|\alpha-x/\sqrt{2}|^2+\left(\alpha-x/\sqrt{2}\right)^{2}\tanh r}{2}\right]}{\sqrt{\cosh r}}, \end{align} where the final equality uses the overlap between a coherent state and a squeezed state [eg Gerry and Knight (7.79)]. Taking the limit yields $\tanh r\to 1$ and the denominator blows up, but normalization ensures the result \begin{align} \psi_\alpha(x)\propto \exp\left[-\frac{|\alpha-x/\sqrt{2}|^2+\left(\alpha-x/\sqrt{2}\right)^{2}}{2}\right]\propto\exp\left[-\frac{\left(\sqrt{2}\mathrm{Re}\alpha-x\right)^{2}}{2}\right]e^{ix\mathrm{Im} \alpha/\sqrt{2}}, \end{align} where the second proportionality constant discards a global phase. The overall normalization is a function of $\alpha$ that can be computed using Gaussian integration.

Answer 3

Let me first remark, that the answer depends on the field that is represented by the coherent state in question. Wikipedia assumes that they are the coherent states for a particle in a harmonic oscillator potential. However, coherent states are given by the same expressions for any bosonic field, e.g., for electromagnetic field, where the position dependence is not directly linked to the number of quanta.

Making the Wikipedia assumption, we can get the answer by taking projection to the position representation $$ \psi_\alpha(x) = \langle x|\alpha\rangle, $$ and keeping in mind that $$ \phi_n(x) = \langle x|n\rangle $$ is the n-th eigenstate of the corresponding Harmonic oscillator.