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 $(|alpharangle+|0rangle)/sqrt{N}$.
I know the annihilation operator has $hat{a}|alpharangle=alpha|alpharangle$, which can be realized by single-photon subtraction.
Then mathematically, we have $hat{a}(|alpharangle+|0rangle)/sqrt{N}=alpha/sqrt{N} cdot |alpharangle$.
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 $hat{a}|0rangle$; and (b) zero photon state $|0rangle$.
Also how to understand $hat{a}|alpharangle=alpha|alpharangle$ physically? Is the coherent state scaled? If so, what difference between $alpha|alpharangle$ and $|alpharangle$.
Or, it should be considered as $hat{a}|alpharangle=|alpharangle$ and $hat{a}(|alpharangle+|0rangle)/sqrt{N}=|alpharangle$ in the experiments.

Norbert Schuch · Answer

$hat a$ 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 $etall 1$ s.th. $eta|alpha|^2ll1$. Then, the effective operation implemented will be approximately $sqrteta hat alvertalpharangle = sqrtetaalphall 1$, and thus there is no issue that you would get a probability larger than $1$.
In the special case where you consider $hat a|0rangle$, the fact that the normalization is zero simply means that the probability to subtract one photon is zero. (Unsurprisingly.)

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

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