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From many quantum optics textbooks, I see that the annihilation operator $\hat{a}$ removes the particle (or photon) excitation in the quantised electromagnetic field. This operator is used in various contexts in quantum optics as an operator that removes an excitation of quantum energy for electromagnetic fields.

For example, for a given energy eigen state $\vert n \rangle$ (called a photon number state), acting $\hat{a}$ on it produces the state $\vert n-1 \rangle$, from which we understand the operation by $\hat{a}$ annihilates one photon from the state of light. For another example, when we have the Jaynes-Cummings Hamiltonian ($\propto \hat{a}\sigma^{\dagger}+\hat{a}^{\dagger}\sigma$), we understand that the first term remove an excitation of field with creating an excitation of an atom, and vice versa for the second term.

Now, it is so natural to apply the above interpretation of $\hat{a}$ to the case of coherent state $\vert \alpha \rangle$. That is, $\hat{a}\vert \alpha\rangle$ can be seen as removing one photon from the state of light. While keeping the consistent interpretation of $\hat{a}$ from the above arguments, I do not understand why the state $\vert \alpha \rangle $ can remain the same although we apply the operator $\hat{a}$? Can this be physically explained and understood?

Please do not just say that it is an eigenstate of $\hat{a}$. I know this, but do not fully understand the physics behind. It really looks to me like the operator $\hat{a}$ physically removes one photon from $\vert\alpha \rangle$, same as how we interpret it in other cases.

I know this has been asked in Annihilating coherent state, but I still do not understand why we should not say that the operation $\hat{a}$ removes photons in the case of coherent state $\vert \alpha \rangle$. I would really appreciate if anyone can help me get the right answer.

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    $\begingroup$ Possible duplicate of Annihilating coherent state $\endgroup$ – Elliot Yu Nov 4 '16 at 16:16
  • $\begingroup$ @ElliotYu I already mentioned in my original question that I have seen that link, but there I feel that most explanations are a little vague and abstract. Looks like people say the annihilation operator removes a photon when they talk about the energy eigen state and atom-field interaction model, but just don't say the annihilation operator is really an operator that remove a photon from the coherent state. Such inconsistent explanation doesn't help me completely understand. Is there any type of consistent interpretation? $\endgroup$ – Veteran Nov 4 '16 at 16:53
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    $\begingroup$ I think the point here is that the interpretation of $\hat a$ as "physically" removing a photon is just wrong. It is true in an handwavy way because it does send a Fock state into another with one less photon in it, but this cannot and should not be seen as an actual physical operation. $\hat a$, as $\hat a^\dagger$, is not an hermitian operator and as such does not represent any physical quantity of any system. Also, if it did really represent the removing of a photon, the scaling factor $\sqrt n$ would not have been there. $\endgroup$ – glS Nov 4 '16 at 19:18
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    $\begingroup$ If you have seen the duplicate, you should elaborate what about the answers you do not understand. Just saying "I don't like the answers" is not very helpful. -- On a different note, what do you mean by "really" remove a photon? If "really" means "in the sense of a physical - unitary - operation", the answer is "no". $\endgroup$ – Norbert Schuch Nov 5 '16 at 14:50
  • $\begingroup$ @NorbertSchuch;glS Many thanks for your answers. I'm fine with saying the operator $\hat{a}$ is not a physical operation as it's not Hermitian. but it still makes me feel that if we say so, we shouldn't say, for the consistent interpretation, that it annihilates a quantum of energy in the case of number states or Jaynes-Cummings model, or elsewhere. I'm a little uncomfortable with such inconsistency of using the term "annihilation". $\endgroup$ – Veteran Nov 7 '16 at 12:52
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The coherent state is a superposition of number eigenstates ($|n\rangle$). We can only say there is exactly a certain number of photons in the system for number eigenstates. The coherent state does not represent a state with a well defined number of photons, but a state that we can measure and find a given number of photons with a certain probability. The expected value of particle number in a coherent state is $$\langle \hat n\rangle = \langle \alpha|\hat{a}^\dagger \hat{a}|\alpha\rangle = \langle \alpha|\overline{\alpha}\alpha|\alpha\rangle = |\alpha|^2 \>, $$ but this is not the value that we will get from hitting the state by an $\hat{n}$.

Now when you hit a coherent state $|\alpha\rangle$ by $\hat{a}$, you are not taking one photon from a state with a definite number of them, but one from each of the (infinitely many) number states composing the coherent state and adding the result back up. As this answer from the question you cited illustrates, due to the way coherent states are constructed, after you apply the annihilation, the expected particle number stays the same. This happens because the each number state is annihilated to a lower number state, but the annihilated result from a higher number state "fills in the blank."

Note that you cannot say there is an infinite number of particles removed from the system, since each of the component number states are not normalised the way a pure number state would be, and as I said above, it makes no sense to speak of particle numbers in a superposition of number states anyway.

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A physical removal of a photon from one system would usually result in creation of a photon (with the same energy) in some other system. E.g. when a dissipative cavity loses a photon, it means that the photon leaks out from the cavity to the environment, and if the cavity field is prepared in a coherent state then the loss of photons due to dissipation leads to exponantial decay in the amplitude of the coherent state, so the state changes. This photon removal is a physical process (the two systems have to interact to make this exchange possible) and can be described by a Hermitian operator acting on both systems.

On the other hand, the annihilation operator $\hat{a}$ is not a Hermitian operator (it's not an observable), so it doesn't physically remove photons from systems. By acting with it on a state that has a well defined number of photons (i.e. states that are eigenstates of the number operator – Fock states) you end up with a state with one less photon (multiplied by a constant), but this "removal" is not accompanied by creation of a photon somewhere else and therefore is not physical, but rather an abstract mathematical operation. You can think of it like removing a photon from a system and also from the universe – indeed a true annihilation.

You have to combine the action of $\hat{a}$ with other operators to get a physically meaningful operation, e.g. $\hat{a}^{\dagger}\hat{a}$, or $\hat{a}^{\dagger}\hat{b}+\hat{a}\hat{b}^{\dagger}$ for two-mode states.

A coherent state or any other superposition of the Fock states are not eigenstates of the number operator – they don't have a well defined number of photons, so by acting on them by the annihilation operator you don't end up with a state which has one photon less. You simply end up with the same or a different state (also without a well defined number of photons). The coherent state just has a special structure (a special infinite sum of Fock states), such that it is an eigenstate of the annihilation operator. You remove abstractly (not physically) infinitely many photons, one from each Fock state that compose the coherent state, but because of this special structure you end up with the same state multiplied by a constant, and the removed photons do not exist physically (so there is no paradox with energy conservation violation).

(Notice also that by acting with $\hat{a}^{\dagger}$ on a coherent state changes it to a different state)

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