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

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.

• Why do you think there should be a physical difference between a state vector and that state vector multiplied by a constant? Dec 30, 2020 at 14:48
• $\hat a$ does not simply describe photon substraction. If you substract a photon, you have to normalize the resulting state -- that is, the process of photon substraction only succeeds with a certain probability (given by the normalization). Think about how you would design an experiment to substract a photon. (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.) Dec 30, 2020 at 14:54
• @BioPhysicist I think they do different, especially when consider the measurement. For example, $|0\rangle\langle 0|+\rangle=1/2|0\rangle$, the constant means the detected probability with 50%. So if we do the projector on the coherent state, $|\alpha\rangle\langle\alpha|\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$ means there is a probability detecting it. But there is no constrain $\alpha\in[0,1]$, so I know I must have something wrong, but I don’t know what it is. Dec 30, 2020 at 15:41
• @YechaoLiu Whenever you are dealing with probabilities you always have to normalize first, so multiplying by a constant doesn't change anything. The constant factor will go away when you normalize. Dec 30, 2020 at 16:43
• @YechaoLiu You cannot just "apply the annihilation operator" -- there will always be a prefactor $\eta$ in front of $\hat a$, which e.g. in the setting of the beamsplitter will be small, so $\eta\alpha\ll 1$ (otherwise what you will have implemented will be far from photon subtraction.) Dec 30, 2020 at 18:14

$$\hat a$$ 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).
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.)
• So, for the state $x|\alpha\rangle+y|0\rangle$, after annihilation operator, we always have $|\alpha\rangle$, the difference is the probability. Do I understand this correctly? Dec 30, 2020 at 15:49