Yes; but both of these Hilbert spaces are subspaces of the full Hilbert space of the quantum EM field. For each mode, there will be a subspace consisting of all the Hilbert space vectors you can get by applying the creation operator for that mode only to the vacuum state an arbitrary number of times. But it's the same vacuum state in each case; and the full Hilbert space includes vectors that you get by applying two or more different creation operators to that vacuum state.
> If we want the total Hilbert space, that would be the tensor product of Hilbert space A and B
No, it would be the symmetric Fock space constructed using the direct sum of Hilbert spaces A and B ("symmetric" because photons are bosons). See here:
> a plane wave is a basis vector of the space of square-integrable functions
No, it isn't. A plane wave, i.e., a function of the form exp(ikx), is not square integrable. When we use plane waves as a basis, we have to either expand our space of functions to allow them, i.e., beyond just the square integrable functions, or restrict how linear combinations of the basis functions can be formed in order to ensure that all linear combinations we actually make use of are square integrable.
> Fock states and coherent states are basis vectors for a Hilbert space of a quantum oscillator.
Two different choices of basis, yes (and coherent states are an "overdetermined" basis--I think that's the right term--since they are not all linearly independent of each other). But Fock states, or at least the ones belonging to subspaces like your A and B above, formed by applying creation operators for only one mode to the vacuum, are plane wave states, since each "mode" is a plane wave, so any Fock state containing nonzero amplitude for only one mode will also be a plane wave.
> but both of these Hilbert spaces are subspaces of the full Hilbert space of the quantum EM field
This is definitely wrong, and it is the source of our disagreement. They are not subspaces. If they were subspaces then we would use a direct sum to build the Hilbert space of the entire EM field, and this Hilbert space would have dimensions that are the sum of the dimensions of the subspaces.
Instead, we take the tensor product of the Hilbert space for each mode. This is a drastically different operation, leading to a very different basis, different dimensions, etc.
Notice, the link you provided yourself states that the total Hilbert space is a tensor product. Only after they state that, they present a direct sum over tensor-powers of the Hilbert space of one particle, which is very different from the statement you wrote, exactly because of this tensor-power operation. I concede that if I have to be rigorous I need to specify this whole symmetrizing step that I skipped, but that is just a detail, especially when we talk about distinguishable modes.
Concerning your (justified) nitpick on square-integrable functions: substitute "square-integrable over the finite size of the oscillator".
Edit: a number of additions and extra explanations.
I'm sorry, but I disagree. You are misreading the link I gave.
> the link you provided yourself states that the total Hilbert space is a tensor product
No, it doesn't. What it says is that "n-particle states" are vectors in a symmetrized (assuming the particles are bosons) tensor product of n single-particle Hilbert spaces. Then the full Hilbert space is the (Hilbert space completion of the) direct sum of all the n-particle state Hilbert spaces, for n from zero (the n = 0 Hilbert space is just the vacuum state) to infinity.
The subtlety here is actually the proper definition of the "single particle" Hilbert space. This Hilbert space is in fact the space of all possible "modes", or more precisely the single particle Hilbert space has a basis consisting of all possible "modes", where each "mode" is a plane wave with a different momentum 3-vector (and polarization, if we include polarization, but we can ignore that complication for now). So each "single particle state" is either a single "mode", or a linear combination of multiple "modes" (which is how we can form things like wave packets).
The "n particle states" are then states containing n particles, each of which can be in any one of the single-particle states, either a single "mode" or a linear combination of them (and each particle does not have to be in the same single-particle state as any other). The full Hilbert space is then, as above, the space of all possible n-particle states, for n from zero to infinity.
It should be evident from the above that the full Hilbert space has a subspace for each single "mode", consisting of all states that only contain particles in that "mode" alone. This subspace will, as I said before, consist of the vacuum state, plus all states that can be obtained by applying the creation operator for the chosen "mode" to the vacuum state an arbitrary number of times.
> substitute "square-integrable over the finite size of the oscillator"
Doesn't help; a plane wave can't exist in a region of finite size, so with your restriction plane wave states don't exist at all.
> symmetrized __tensor product__ of n single-particle Hilbert spaces H
And clarifying your statement by copying from the wiki page (emphasis mine):
> direct sum of the symmetric or antisymmetric tensors in the __tensor powers__ of a single-particle Hilbert space
(which is very different from the direct sum you suggest in your comment)
I work with all of this in my day job and the research work I produce (which does deal a lot with quantum description of multiples modes of light) is usually perceived as correct by my academic colleagues. I am honestly trying to see how to interpret what you are saying differently, in a way that does not clash with the established knowledge on the topic, but I can not, even though I have non-negligible experience in the field. At best, I assume we are talking past each other while describing the same thing, but I am certainly uncomfortable with your use of "subspace" and "direct sum".
Going to the very first comment of yours that started this: My claim is that a mode (something that has a creation operator) can very well contain one, or two photons, or a coherent state, or anything else from its Hilbert space. There is no correspondence between plane waves and Fock states. We can use Fock states with other modes instead of plane waves, and we can have plane waves with a quantum state that is a coherent state for instance.
> which is very different from the direct sum you suggest in your comment
I don't see how. I suspect that we are interpreting ordinary language words differently and talking past each other.
> a mode (something that has a creation operator)
And a corresponding annihilation operator. Yes, that's basically the definition I've been using.
> from its Hilbert space
But, as I said before, all of these states you describe are obtained by applying the creation operator for the mode to the vacuum state some number of times (and then, in some cases, forming linear combinations of states so obtained). But this process for all modes starts with the same vacuum state. There are not different vacuum states for different modes. So ultimately, all of these states belong to the same Hilbert space, since they're all obtained starting from the same vacuum state; and this Hilbert space will also contain states that have more than one "mode" in them--states obtained by applying creation operators for different modes to the vacuum state, or linear combinations of states so obtained.
> Mode B has its own Hilbert space
Yes; but both of these Hilbert spaces are subspaces of the full Hilbert space of the quantum EM field. For each mode, there will be a subspace consisting of all the Hilbert space vectors you can get by applying the creation operator for that mode only to the vacuum state an arbitrary number of times. But it's the same vacuum state in each case; and the full Hilbert space includes vectors that you get by applying two or more different creation operators to that vacuum state.
> If we want the total Hilbert space, that would be the tensor product of Hilbert space A and B
No, it would be the symmetric Fock space constructed using the direct sum of Hilbert spaces A and B ("symmetric" because photons are bosons). See here:
https://en.wikipedia.org/wiki/Fock_space
> a plane wave is a basis vector of the space of square-integrable functions
No, it isn't. A plane wave, i.e., a function of the form exp(ikx), is not square integrable. When we use plane waves as a basis, we have to either expand our space of functions to allow them, i.e., beyond just the square integrable functions, or restrict how linear combinations of the basis functions can be formed in order to ensure that all linear combinations we actually make use of are square integrable.
> Fock states and coherent states are basis vectors for a Hilbert space of a quantum oscillator.
Two different choices of basis, yes (and coherent states are an "overdetermined" basis--I think that's the right term--since they are not all linearly independent of each other). But Fock states, or at least the ones belonging to subspaces like your A and B above, formed by applying creation operators for only one mode to the vacuum, are plane wave states, since each "mode" is a plane wave, so any Fock state containing nonzero amplitude for only one mode will also be a plane wave.