# I Beam in specific photon number state?

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1. Oct 8, 2016

### maxverywell

Can a beam of light be objectively in a specific photon number state (Fock state), let's say $|2\rangle$? Or is it (the specification of the state) detector/observer dependent? I.e. we can only say that a beam of light is in a Fock state, but which exactly is detector dependent (detection time interval dependent).

2. Oct 8, 2016

### vanhees71

A "beam of light" is not a Fock state but a coherent state (if you consider laser light) or (more common) thermal light, described by a statistical operator (e.g., black-body radiation for the radiation in a hot cavity).

3. Oct 8, 2016

### maxverywell

Coherent light is a poissonian light and thermal light is super-poissonian. However there is also sub-poissonian light, e.g. photon number states. Thant's what I was refering to.

The thing is that the mean photon number depends on the detection time, and thus the exact state of the light beam is dependent on the detection time (or equivalently on the length of the beam piece that we are considering), either it is coherent, thermal or a Fock state.

4. Oct 9, 2016

### A. Neumaier

How can a pair of photons make a beam? A beam is always a stream of photons.

However, one can make (approximate) Fock states that follow (roughly) a particular path.
It is quite difficult to do so.

5. Oct 9, 2016

### maxverywell

A beam of light containing a stream of photons with a fixed time spacing $\Delta t$ between them (whereas for a coherent light beam this time interval is random) is a photon number state.
If the detection time of a photo-counting detector is $T=n \Delta t$, you will always count n photon and thus the state of beam is $|n\rangle$. In other words, a beam segment of length $n c\Delta t$ (where $c$ is the speed of light) is in the photon number state $|n\rangle$. So this is objectively defined state, without refering to a specific detetor/detection time. My confusion was because of the ambiguity of the word beam without specifying it's length or detection time.

Last edited: Oct 9, 2016
6. Oct 9, 2016

### houlahound

Just trying to follow along here, how does one create a beam of two photons a temporal distance delta t apart?

What is the resolution of the detector?

7. Oct 9, 2016

### A. Neumaier

Well, this gives states of single prepared photons, not states of the beam. The beam is in a far more complicated time-dependent state oscillating between Fock states and the vacuum state.

Such single-photon states can be produced using ion traps and laser excitation; these are called photons on demand. See my slides Classical and quantum field aspects of light and references therein. For the preparation of specific multiphoton Fock states see, e.g., https://www.researchgate.net/profile/Brian_Smith52/publication/236038468_Experimental_generation_of_multi-photon_Fock_states/links/0deec53c650a715fe2000000.pdf [Broken].

This is not true. You get two independent $|1\rangle$ photons, not a 2-photon state.

Last edited by a moderator: May 8, 2017
8. Oct 9, 2016

### maxverywell

Hmm, it seems that you are right. So a segment $n c \Delta t$ of this beam is in state $|1\rangle \cdots|1\rangle$ ($n$ times), rather than $|n\rangle$.
So this beam is a stream of single-photon Fock states $|1\rangle$.
In general, we could have a beam which is a stream of m-photon Fock states $|m\rangle$, and a segment $n c \Delta t$ of such beam is in state $|m\rangle\cdots |m\rangle$ ($n$ times).
The difference is that the stream of single-photon Fock states is antibunched, whereas m-photon Fock states are bunched.

E.g.
$......$ (beam segment of six $|1\rangle$ photon states)
$..\hspace{0.2cm}..\hspace{0.2cm}..$ (beam segment of three $|2\rangle$ photon states)