How to distinguish between no state and the state with zero photon?

Question

This question comes from my consideration of the superposition of coherent states such as $(|\alpha\rangle+|0\rangle)/\sqrt{N}$.

I know the annihilation operator has $\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$, which can be realized by single-photon subtraction. Then mathematically, we have $\hat{a}(|\alpha\rangle+|0\rangle)/\sqrt{N}=\alpha/\sqrt{N} \cdot |\alpha\rangle$. What I confused is that, as the state with trace not equal to one means the output is probabilistic, how we distinguish between (a) no state, such like the results form $\hat{a}|0\rangle$; and (b) zero photon state $|0\rangle$.

Also how to understand $\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$ physically? Is the coherent state scaled? If so, what difference between $\alpha|\alpha\rangle$ and $|\alpha\rangle$.

Or, it should be considered as $\hat{a}|\alpha\rangle=|\alpha\rangle$ and $\hat{a}(|\alpha\rangle+|0\rangle)/\sqrt{N}=|\alpha\rangle$ in the experiments.

Answer 1

$\hat a$ does not simply describe photon subtraction. If you subtract a photon, you have to normalize the resulting state -- that is, the process of photon subtraction only succeeds with a certain probability (given by the normalization).

Think about how you would design an experiment to subtract a photon: For instance, you could send the beam on a (weak) beam splitter, and install a photon counter in the other beam. If you detect exactly one photon there, then you have succeeded in subtracting a photon your state. Otherwise, your attempt to subtract one photon has failed (if you subtracted no photon, you could send the beam on another beam splitter, otherwise: bad luck).

Note, however, that this operation does not exactly implement photon subtraction, but only an approximate version thereof -- in fact, the exact operation of photon subtraction cannot be realized even probabilistically, see e.g. the introduction of https://arxiv.org/abs/1908.02207. You will, however, get a good approximation if the beam splitter is very weak so that it only reflects one photon with very small probability, i.e. with reflectivity $\eta\ll 1$ s.th. $\eta|\alpha|^2\ll1$. Then, the effective operation implemented will be approximately $\sqrt\eta \hat a\lvert\alpha\rangle = \sqrt\eta\alpha\ll 1$, and thus there is no issue that you would get a probability larger than $1$.

In the special case where you consider $\hat a|0\rangle$, the fact that the normalization is zero simply means that the probability to subtract one photon is zero. (Unsurprisingly.)