What I understand of Fock states so far: They describe the quantum state of a bunch of photons. A single photon can be in several different energy states, and when these photons are tensored together - the basis for this is the Fock states.

A given Fock state has a certain intensity for various frequencies of light.

However light has several other properties - for example, the polarisation states which is described by a two dimensional Hilbert space for each photon, and the spatial degree of freedom, which, say for in a beam splitter set up, can also be described by a two dimensional Hilbert space. And there may be several other properties like angular momentum, etc.

How is the polarisation state of a photon related to it's energy state or related to states in the Fock basis?

Or are they separate things - Light in any Fock state (or a superposition of Fock states) can have any polarization?

If anyone can shed light on the other properties, that would be great as well.

I come from a quantum information background, and know very little optics and electromagnetism, so if anyone could explain it from that point of view, it would be great.

  • $\begingroup$ (1) Your first paragraph does not sound correct. --- (2) What do you know about classical electromagnetic waves (or even waves in general) - do you e.g. have some physics background? -- -- But most importantly: What precisely is your question? $\endgroup$ Aug 1, 2019 at 10:30
  • $\begingroup$ @NorbertSchuch, (2) I know the equation of an electromagnetic wave, understand the double slit experiment, etc, basically high school physics. My main question: Does a description of a state in the Fock basis also tell me the polarization state of the photons in it? $\endgroup$ Aug 1, 2019 at 18:13
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    $\begingroup$ Main answer: Yes, but this is something you have to specify separately. A Fock state $|n\rangle$ describes $n$ photons in a specific mode, this is, a specific solution for the classical equations of motion -- this is, polarization, frequency, etc., is all fixed and the same for all those photons -- but you have to specify this somewhere. In many cases, there is only one such solution (mode) which is of interest, so one omits it. If you consider horizontally and vertically polarized photons (e.g. of the otherwise same state), you could e.g. write $|n\rangle_H|m\rangle_V$, corresponding ... $\endgroup$ Aug 1, 2019 at 18:17
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    $\begingroup$ ... to $n$ horizontally polarized and $m$ vertically polarized photons. $\endgroup$ Aug 1, 2019 at 18:17
  • $\begingroup$ Got it. Would it be right to sort of think of properties of a photon like frequency, polarization, spatial degree of freedom, angular momentum, etc as lying in tensored Hilbert spaces, and the state of the photon is given by some vector in this tensored space? $\endgroup$ Aug 1, 2019 at 18:21

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A photonic quantum state can consist of multiple photons, as you correctly implied. If there is a fixed number of photons in the state, then we call it a "number state" or a "Fock state." Each photon in the Fock state can carry its own set of spatiotemporal and polarization (spin) degrees of freedom. Some people say that all the photons in a Fock state must have the exact same spatiotemporal and polarization degrees of freedom, but for the sake of this discussion, we'll allow them to be different.

The polarization degrees of freedom are represented by two discreet degrees of freedom. These are also associated with the spin angular momentum. So the latter is not a separate degree of freedom.

The spatiotemporal degrees of freedom represent three continuous degrees of freedom. One can either specify them as the three components of the propagation vector $\mathbf{k}$, or as two such components and a frequency $\omega$. The fourth quantity is then related to these by the dispersion relation $\omega=c|\mathbf{k}|$. These degrees of freedom are also associated with orbital angular momentum (OAM). So OAM is not a separate degree of freedom.

As you can probably see, the total set of degree of freedom of a Fock state becomes rather complicated. To make matters worse, one can also have photonic quantum states that do not have fixed numbers of photons. So the most general state is very complicated to specify in terms of all its degrees of freedom. However, there are ways to do it ...

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    $\begingroup$ "Each photon in the Fock state can carry its own set of spatiotemporal and polarization (spin) degrees of freedom." --- This is wrong, they all carry the same quantum numbers: Otherwise, they are distinguishable. (Unless, you mean by a Fock state something like $|n_1,n_2,n_3,...\rangle$, in which case it would be important to specify that, and make clear that for each set of quantum numbers $k$, there is a separate $n_k$.) $\endgroup$ Aug 1, 2019 at 9:07
  • $\begingroup$ @NorbertSchuch: Please read the last sentence of my first paragraph $\endgroup$ Aug 1, 2019 at 9:36
  • $\begingroup$ Saying "for the sake of this discussion, we'll allow them to be different" does not make it correct. If you are talking about a Fock state $|n\rangle$, then this always describes $n$ photons in the very same mode, with no quantum number different. (And if you have a Fock state $|n_1,n_2\rangle$, you correspondingly have $n_1$ photons in a mode with all quantum numbers equal, and $n_2$ photons in a different mode with all quantum numbers equal, but different from the other mode. $\endgroup$ Aug 1, 2019 at 10:29
  • $\begingroup$ It is not a matter of right or wrong. It is simply a matter of what one means by the terminology. (Not everybody apparently agrees with your definition of the term Fock state.) If you think that a quantum state consisting of a fixed number of photon cannot have different degrees of freedom associated with different photons, then that would be wrong. $\endgroup$ Aug 1, 2019 at 10:34
  • $\begingroup$ There is perfect agreement what is meant when you write a Fock state $|n\rangle_m$: It is $n$ photons in a certain mode $m$ (which has to be specified additionally). If you have $n$ photons distributed over different modes $m_1$ and $m_2$, this would be a some state $|n_1\rangle_{m_1}|n_2\rangle_{m_2}$ (or a superposition thereof), with $n_1+n_2=n$. $\endgroup$ Aug 1, 2019 at 13:47

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