Coherent creation operator: unitary or not?

Question

In Quantum Mechanics, for coherent states $|z\rangle$ it can be prooved that if $|0\rangle$ is the vacuum state for an harmonic oscillator, therefore: \begin{equation} |z\rangle=e^{za^{\dagger}-z^*a}|0\rangle\equiv \hat{D}(z)|0\rangle \end{equation} with displacement operator $\hat{D}(z)$ that is unitary: $\hat{D}^{\dagger}(z)\hat{D}(z)=\hat{D}(z)\hat{D}^{\dagger}(z)=1$.

However, with some direct manipulations from the above stuff it can be shown that: \begin{equation} |z\rangle=e^{za^{\dagger}-z^*a}|0\rangle=e^{-\frac{|z|^2}{2}}e^{za^{\dagger}}|0\rangle \end{equation} where $D(z)\equiv e^{-\frac{|z|^2}{2}}e^{za^{\dagger}}$ isn't unitary.

Now the question is: how can be possible that in one case the operator is unitary and in the another not? Is there a "bug" in the proof in someway? I suppose the answer has to do with the fact that the Hilbert space in question is infinite dimensional but I would like a more rigorous answer mathematically and on what kind of physical consequences one has in either case.

Answer 1

Unitarity is not defined by the action on a single specific state any more than hermiticity is defined by the action on a single state. $L_x$ is hermitian but $L_x\vert \ell,\ell\rangle=\frac12 L_-\vert \ell,\ell\rangle$ acts like a non-hermitian operator because $L_+$ acting on the specific state $\vert \ell,\ell\rangle=0$. Indeed since $\hat x\propto \hat a +\hat a^\dagger$ you could conclude nonsensically that $\hat x$ is not hermitian if you consider only the action $\hat x\vert 0\rangle$

To talk about unitarity or hermiticity, you need an inner product somewhere so deducing something from the action on a single state is dangerous.

Unitarity and hermiticity are defined via the action on arbitrary states. For instance in finite dimensions $\hat A$ is hermitian if we have $$ \langle \psi\vert \hat A\phi\rangle= \langle\hat A \psi\vert \phi\rangle $$ for arbitrary $\vert \phi\rangle$ and $\langle \psi\vert$. There’s no way that get you out of using an inner product as there are realizations of $L_x$ which are not hermitian (albeit equivalent to hermitian realizations).

The same general argument holds for unitarity: you need an inner product and the action on arbitrary states.

Answer 2

Using the Baker-Campbell-Hausdorff formula, the unitary operator $$\hat{D}(z):= e^{za^\dagger-z^\ast a}, \qquad \quad [a,a^\dagger]=\mathbf{1},$$ can also be written as $$\hat{D}(z)=\underbrace{e^{-|z|^2/2} e^{z a^\dagger}}_{:=D(z)} \; e^{-z^\ast a}. $$ Therefore, the fact that $\hat{D}(z)|0\rangle$ and $D(z)|0\rangle$ coincide, being a simple consequence of $$a|0\rangle=0 \;\Rightarrow \;e^{-z^\ast a}|0\rangle=|0\rangle,$$ should not come as a big surprise. Note that already for the vector $|1\rangle:=a^\dagger |0\rangle$ we have $$\hat{D}(z) |1\rangle = D(z) |1\rangle-z^\ast D(z) |0\rangle \ne D(z) |1\rangle. $$