Coherent States as a measurable eigenstate of the Annihilation Operator

Question

We were discussing coherent states for the harmonic oscillator and their observability and I still can't seem to resolve this apparent contradiction

"A very important property of a coherent state is that it 's an eigenvector of the annihilation operator a."

The annihilation operator is anti-Hermitian, so its eigenvalues are imaginary, so the corresponding eigenvectors will at best be complex, right? So that means a isn't an observable, which we understand from last week, but aren't coherent states observable? They're the states which most closely resemble the dynamics of the classical harmonic oscillator, so surely we can measure them; otherwise, what is the point in distinguishing them? So how can it be an eigenvector of a non-observable? How can an observable state be an eigenvector of a non-observable operator? What subtleties are at play here that we're blind to?

Adding to that, we know that coherent states are an eigenstate of the annihilation operator, does that imply that all anti-Hermitian operators do have some states with which they obey the eigenvalue equation?

$a|\alpha> = \alpha |\alpha>$

Answer 1

The annihilation operator is anti-Hermitian, so its eigenvalues are imaginary

no. the annihilation operator is not Hermitian, but also not anti-Hermitian. If it were anti-Hermitian then it would maintain $a^{\dagger} = -a$ which is obviously not the case. Its eigenvalues are in general complex numbers, but they can have an imaginary part which is zero. As you see from the coherent state, $a|\alpha\rangle = \alpha |\alpha\rangle$, the eigenvalues cover all of $\mathbb{C}$.

but aren't coherent states observable?

states are not "observable" or "unobservable". This is not a terminology we use with respect to states, but rather with respect to operators. The annihilation operator by itself is not an observable, but it is comprised of two (non-commuting) observable operators - $x$ and $p$. The coherent state has a physical meaning, as you write "They're the states which most closely resemble the dynamics of the classical harmonic oscillator" which is true. But you don't measure a state but rather you measure $x$ or $p$. And indeed when you measure the behavior of $\langle x\rangle$ or $\langle p\rangle$ for the coherent state you get a classical behavior. Note that these observables are always real-valued.

Answer 2

States are not observable, and neither are the real or imaginary. For instance the state \begin{align} \vert\psi\rangle =\frac{1}{\sqrt{2}}\vert +\rangle + \frac{i}{\sqrt{2}}\vert -\rangle \end{align} is an eigenstate of $\sigma_y$ in the basis where $\sigma_z$ is diagonal. It is clearly neither real nor imaginary.

$\hat a$ or $\hat a^\dagger$ aren’t hermitian or anti hermitian either. In fact, thinking of a coherent state as an eigenstate of $\hat a$ is not terribly useful physically precisely because $\hat a$ isn’t hermitian (and thus not an observable). Greater physical insight (and mathematical flexibility) is gained by thinking of coherent states harmonic oscillator ground states that have been translated in the $x-p$ plane so that $\langle x\rangle =\text{Re}({\alpha})$ and $\langle p\rangle=\text{Im}(\alpha)$. Of course they satisfy the uncertainty relation $\Delta x\Delta p=\hbar$ so systems described by a coherent state do not have definite position or momentum. (The idea of coherent states for operators other than $\hat x$ and $\hat p$, and thus for operators other than $\hat a$ and $\hat a^\dagger$, is based on the definition of coherent states as translates of a suitable “ground state”.)

Coherent states can certainly be “easily” produced in the lab: if $\vert n\rangle$ denotes a state containing $n$ photons, then the probability distribution of photons in a coherent state is precisely the probability distribution of photons in a laser (up to technicalities on operations of the laser). Of course both $\vert n\rangle$ and $e^{i\varphi_n}\vert n \rangle$ have the same probability distributions but are distinct states if the phase $\varphi_n$ depends on $n$, so an argument about the probability distribution is not complete, but still because of their classical properties coherent states are an every day experimental tool in quantum optics.