How does particle detection affect a coherent state?

[snoopies622]

". . . a coherent state remains unchanged by the detection (or annihilation) of field excitation or, say, a particle."

— Wikipedia's entry on "Coherent States"


I do not understand this. Is it saying that if a system in a coherent state, like a quantum harmonic oscillator, emits a particle, like a photon, it remains in the same state? How can any system lose energy yet remain in the same state? I know that a coherent state is not a particle number (energy) eigenstate, but like any state it has an energy expection value. Shouldn't this value change when energy is emitted?

Thanks.

 

[Cthugha]

It is not about emitting a particle, rather about subtracting a particle from what is already there. The question what unchanged means is not trivial. You can express that in terms of conditional probabilities. If I detect a photon now and the probability to detect a second photon afterwards does not change, the state remains unchanged.

This conditional probability is given by the equal time second order correlation function g^2 which is defined as:
$$g^{(2)}(0)=\frac{\langle: n^2 :\rangle}{\langle n \rangle^2},$$
where the double stops denote normal ordering.
This results in:
$$g^{(2)}(0)=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}.$$
The n-1 expresses that the detection of one photon destroys it and changes the light field.
Now you can express the instantaneous photon number n as composed of the mean photon number and the deviation from that. In that case you get:
$$g^{(2)}(0)=\frac{\langle n+\delta n (n+\delta n-1) \rangle}{\langle n \rangle^2}.$$
As the expectation value of the deviation is 0, just 3 terms survive:
$$g^{(2)}(0)=\frac{\langle n^2 +(\delta n)^2-n) \rangle}{\langle n \rangle^2},$$
or equivalently:
$$g^{(2)}(0)=1+\frac{\langle(\delta n)^2 \rangle}{\langle n \rangle^2}-\frac{1}{\langle n \rangle}.$$

Now you can imagine three cases:

a) The photon number distribution is very noisy. If that happens, the second term which gives the variance of the photon number distribution will be very large. In that case the conditional probablity to detect another photon if you already detected one will be pretty large. The photons arrive in bursts. Many at a time and then few in between, so if you detect a photon at some time, the probability that there are more around is large. This increase takes place when the second term in the g2 dominates.

b) You have a well defined photon number state. In that case, the conditional probability to detect a second photon if you already detected one will go down. If you have a single photon state and detect a photon, the probability to detect another one will of course be zero. That decrease takes place when the third term in the g2 dominates.

c) The second and the third term have equal magnitude. The reduction in photon number is canceled by the noise in the photon number. This is achieved if:
$$\frac{(\langle \delta n)^2 \rangle}{\langle n \rangle^2}=\frac{1}{\langle n \rangle}.$$
That happens to be the case for a Poissonian photon number distribution. The intrinsic noise cancels the photon number reduction (on average) and the probability to detect another photon if you just detected one remains unchanged.

 

[snoopies622]

Thanks Cthugha, much to think about here.

 

[DrDu]

You should also take in mind that coherent states are eigenstates of the particle anihilation operator.