# What does this state collapse into after the measurement?

1. Dec 2, 2007

### wdlang

assume that i have a photon field, which is in a coherent state.

now i detect a photon of this field, i.e., my detector absorbs a photon from this field.

my question is, what state will the field collapse into?

this question may be not so trivial as the examples most quantum mechanics textbooks mention

2. Dec 2, 2007

### cesiumfrog

I think Zapper recently referenced an experiment on this topic in his sticky thread (general forum?). Sometimes it collapses into a state with "more" photons.

Last edited: Dec 2, 2007
3. Dec 2, 2007

### strangerep

I don't think they're the same thing. A coherent state is an infinite superposition of
states of many particle numbers (poisson-distributed). I.e., a coherent state has indeterminate
photon-number. If you setup an experiment to measure "photon number", and repeat the
measurement many times, you'll get various different values 1,2,3,.... with decreasing
probabilities. Then, (if we take the Copenhagen interpretation of measurement), the
state after the measurement is an eigenstate of photon number (since that's what you
measured). So, if you measure photon number as 1, then what you have afterwards
is (theoretically) no longer a coherent state, but rather a state of definite
photon number = 1.

But this is obviously different from an apparatus which merely absorbs 1 photon from
the coherent state and lets it continue on its way. (I think of this as a filter, not a detector)
A coherent state is an eigenstate of the annihilation operator, so you'll still have a
coherent state afterwards. In other words, the act of annihilating 1 photon does not
constitute a "detection" of photon-number = 1.

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