# Annihilating coherent state

Starting facts

Annihilation operator $a$ removes one photon from fock state. Coherent states are eigenstates of annihilation operator and they are also superposition of different fock states.

Applying $a$ on coherent state and annihilating a photon will not change the coherent state. Therefore one can annihilate infinite number of photons from coherent state and not change state at all. This does not make much sense. Can someone point me where I am wrong?

• In a coherent state the photon number is not well-defined. There is an uncertainty relation similar to the position/momentum HUR from standard QM. See Number-Phase Uncertainty – AccidentalFourierTransform Jan 24 '16 at 12:01
• You are not wrong. But I think you have to be more precise about what you find confusing if you want clarification. – Norbert Schuch Jan 24 '16 at 12:02
• @AccidentalFourierTransform As I said it is a superposition of number states. If each number state in superposition is reduced by one, than I would expect that overall coherent state have average one less excitation – sa101 Jan 24 '16 at 12:03
• @sa101 a coherent state is defined through an infinite series. Its like taking the derivative of $\mathrm e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$: the exponent of each term is reduced by one, yet the function stays the same. – AccidentalFourierTransform Jan 24 '16 at 12:06
• @MarkMitchison I agree there is no energy conservation violation. But it is confusing that you can "remove" a photon (which has one energy quantum) and still have the same state (with the same mean energy)! -- The resolution is that $a$ does not describe the process of deterministically removing exactly one photon from a given state (in fact, there is no such process which works for any input state). – Norbert Schuch Jan 25 '16 at 9:04

What you are arguing is that you can take a coherent state $\vert\alpha\rangle$, which has energy $|\alpha|^2$ (in units of $\hbar\omega$), remove one photon which has energy $1$, and still be left with $a\vert\alpha\rangle=\vert\alpha\rangle$. Thus, you would have gained $1$ unit of energy (and you could continue doing so).

The problem is that applying $a$ is not a process you can carry out physically (i.e., deterministically on any input state). If you actually try to remove one photon from a coherent state, other things will happen.

Firstly, in order to describe the process in a clean way, you will need another system into which you transfer the energy. This could be another bosonic mode $b$ (with same energy per photon) or e.g. a two-level system. Let us first consider a bosonic mode.

One possibility to transfer the photon to $b$ would be to build a unitary doing so, such as $U=b^\dagger a$. However, you can see easily that this would not be a unitary -- for the very least, you would have to add a normalization $1/\sqrt{a^\dagger a}$, which will clearly change its effect on a coherent state.

Another approach would be to build a Hamiltonian which moves photons from $a$ to $b$, such as $H=a^\dagger b + \mathrm{h.c.}$. If you time-evolve under this Hamiltonian, what you implement is a beam splitter. It is well known that applying a beam splitter on a coherent state gives you two coherent states with the same total energy. In particular, the state $\vert\alpha\rangle$ is not preserved under the action of a beam splitter. Mathematically, this related to the fact that $\exp(iHt)$ does not only contain $a$ but all powers of $a$ and combinations with $a^\dagger$s.

Alternatively, you could try to couple your coherent state to a two-level system (described by $\sigma^+$ and $\sigma^-$ with the same energy), again using $H=\sigma^+ a + \mathrm{h.c.}$. This is a Jaynes-Cummings Hamiltonian which will induce Rabi oscillations $$\cos(\sqrt{n+1}t)\vert n+1,g\rangle + \sin(\sqrt{n+1}t)\vert n,e\rangle$$ in each of the subspaces spanned by $\{\vert n+1,g\rangle,\vert n,e\rangle\}$ independently. (The second part is the atomic state: ground and excited.) It is clear that when you act on a coherent state (or in fact any superposition of all number states), you will never end up with a factorized state between the two subsystems, since the $\sqrt{n+1}t$ are generally incommensurate. Thus, there is no simple analysis. However, you can easily convince yourself that the energy at each instant is preserved; or you can try to find times $t$ where the system is approximately excited and check that the energy is approximately conserved for the joint energy of the two subsystems.

• thanks for great answer and examples. The fact that one cant get unitary evolution with just $a$ looks like proper answer to me. Following your comments on nonunitarity I found this interesting paper describing photodetection as non unitary link. I read only part of paper so far, mostly cause dont feel very comfortable with non unitary evolution and even less with trying to intuitively understand it. Still it gives some interesting interpretation of applying just $a$. Probably I will make a separate question about this paper. – sa101 Jan 25 '16 at 20:08
• @sa101 Note on the site: (Postselected) substraction of a single photon (i.e., weak beamsplitter followed by photon-resolving detector) has been proposed as a way to implement non-Gaussian operations (e.g. Fuirasek has worked on that); maybe in this context one can find more about what photon substraction is capable of. – Norbert Schuch Jan 25 '16 at 20:24
• To complete @NorbertSchuch's side note : Photon substration has been a tool for experimental quantum optics for the last decade, since these two experiments from 2006, one in Polzik’s group (Danemark) ($$PRL/arXiv), the other in Grangier’s group (France) ($$ Science, no arXiv) – Frédéric Grosshans Dec 2 '16 at 16:12

Can someone point me where I am wrong?

Consider a state that is a superposition of number states $|n\rangle$ for all $n$.

Recalling that

$$\hat a |n\rangle = \sqrt{n}|(n-1)\rangle$$

it's clear that

$$\hat a\left(A\sum_{n=0}^\infty \frac{|n\rangle}{\sqrt{n!}}\right) = A\sum_{n=1}^\infty \frac{\sqrt{n}|(n-1)\rangle}{\sqrt{n!}}=A\sum_{n=1}^\infty \frac{|(n-1)\rangle}{\sqrt{(n-1)!}}=A\sum_{m=0}^\infty \frac{|m\rangle}{\sqrt{m!}}$$

In a certain sense, the annihilation operation removed an infinity of quanta, one from each state in the (infinite) superposition of number states.

Since there is no highest number state, the resulting state is still a superposition of all number states and, by choosing the coefficient for each number state in the superposition to be $\frac{1}{\sqrt{n!}}$, the resulting state is then identical to the original state.

• It is not correct that it removes an infinite number of quanta. Also, this does not answer the question (why energy conservation is not violated). – Norbert Schuch Jan 24 '16 at 22:51
• @NorbertSchuch: It can be correct, it depends whet is meant by “in a certain sense” ;-). More specifically, if one stresses the $\sqrt n$ prefactor in $\hat a|n\rangle=\sqrt n|n-1\rangle$, it is obvious that $\hat a$ is a non physical operation that preserves neither probabilities nor energy, and that draws probabilities out of thin air. And this is indeed the formal reason why “annihilating” a photon is like moving an infinite ladder by one step: it does not change the global state – Frédéric Grosshans Dec 2 '16 at 15:46