# Can you create an actual coherent state?

If I understood correctly, a coherent state $\lvert\alpha\rangle$ is an eigenstate of the creation and annihilation operators, meaning that adding or removing a particles does not change it. Mathematically, this is constructed by using a superposition of all possible numbers of particles. $$\lvert\alpha\rangle = \sum\limits_{n=0}^\infty c_n \lvert n\rangle$$

Is it possible to actually create a real system in such a state? If it were, would you be able to add any number of particles to that system without ever changing it? (That would make for a perfect waste bin, wouldn’t it?)

Or is this just an abstract mathematical concept?

• Coherent states are not eigenstates of the creation operator $\hat a^\dagger$, which in fact has no eigenstates at all. (It does have 'left' eigenstates, i.e. $⟨\alpha|\hat a^\dagger=⟨\alpha|\alpha^*$, but those don't help much.) The inference that 'adding' (or indeed, removing) particles does not change the state is just plain wrong. – Emilio Pisanty Jan 10 '18 at 11:41
• So they‘re not eigenstates of the creation operator, but only of the annihilation operator? If so, then at least the statement that removing particles does not change the state should hold true, shouldn’t it? – jmb Jan 10 '18 at 12:21
• Indeed, they're eigenstates of $\hat a$ but not of $\hat a^\dagger$. However, "removing particles", as a physical operation, is not normally enacted by $\hat a$, either (which can be seen by the simple fact that the physical operation would be undefined for the vacuum component of the coherent state). If you're interested, the term to google for is "photon-subtracted coherent state". – Emilio Pisanty Jan 10 '18 at 12:31

In a laser lab setting, it's very difficult to get individual photons, and laser light is typically coherent state light, unless you want to prepare it.

An answer to a similar question outlines how to a coherent state can be created by stimulated emission. Basically, we know from introductory physics that a laser works by having photons create more photons in a cascading reaction. But each time the photon has a probability of creating stimulated emission, it also has a probability of not stimulating (and then this superposition traverses the system again in a loop). The quantum superposition of all the possibilities continue to add up and form an infinite series.

There is a general principle by which coherent states can be manufactured: All types of coherent states can be produced by coupling of the system under consideration to a classical (C-number) source.

For the electromagnetic radiation, this principle was already explained by Glauber in his original work on coherent states (equations 9.16-9.21).

A more transparent derivation is given by Zhang (section $3$): When a free electromagnetic field originally at the vacuum state $|0\rangle$(no photons) is adiabatically coupled to a classical current $j^{\mu}$: $$\mathcal{L} = - \frac{1}{4} F_{\mu\nu}F^{\mu\nu} - A_{\mu}j^{\mu},$$ its state after the current application becomes: $$|0\rangle_{\mathrm{out}} = e^{-\frac{1}{2}\int d^3k\sum_{\lambda}|z_k^{\lambda}|^2} e^{\int d^3k\sum_{\lambda}z_k^{\lambda} a_k^{\lambda \dagger}}|0\rangle,$$ where: $z_k^{\lambda} = \epsilon^{\lambda}_{\mu}(k) j^{\mu}$ and $\epsilon^{\lambda}_{\mu}$ is the polarization vector.

Another example is for spin coherent states which can be manufactured by adiabatically applying a magnetic field to a spin originally at the highest weight state $|0\rangle = |j, j\rangle$.

The equations of motion $$\frac{dS_i}{dt} = \frac{i}{\hbar} \mu \epsilon_{ijk}B_j(t) S_k$$ ($\mu$ is the magnetic moment). In this case, the system will reach a state $|0\rangle_{\mathrm{out}}$ given by:

$$|0\rangle_{\mathrm{out}} = T e^{\frac{i}{\hbar} \mu \int dt \mathbf{B}(t) \cdot \mathbf{S} }|0\rangle$$ ($T$ denotes time ordering). This is a spin coherent state, because it is obtained by the action of a group element on the vacuum state.

• "the vacuum state |0⟩|0⟩(no photons) is adiabatically coupled to a classical current" Can you give an example of when this would happen? – Steven Sagona Jan 17 '18 at 21:02
• @StevenSagona I wrote that the electromagnetic field, originally at the vacuum state, is coupled to a classical source. The field is coupled, not the state. Of course the coupling is the usual minimal coupling of the current to the electromagnetic field manifested as a source terms of the Maxwell equation, (written explicitly in my answer in the Lagragian expression). I gave an explicit reference to this method in Glauber's original paper and in Zhang's work. – David Bar Moshe Jan 18 '18 at 8:35
• @StevenSagona cont. The "classicality" is achieved through the use of a source of a macroscopic number of charges. This mechanism is explained in almost every quantum optics book. Experimentally, it can be achieved for example by coupling of an optical cavity to a source driven by classical current via a waveguide; however, I concentrated on the theoretical mechanism. – David Bar Moshe Jan 18 '18 at 8:36