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This still does not seem to make sense to me. Consider an oscillator with two modes, A and B. Mode A has a Hilbert space spanned by the Fock states {|0⟩,|1⟩,|

This still does not seem to make sense to me. Consider an oscillator with two modes, A and B. Mode A has a Hilbert space spanned by the Fock states {|0⟩,|1⟩,|2⟩...} or equivalently by the coherent states {|α⟩, α∈C}. Mode A has its own creation and annihilation operator. Mode B has its own Hilbert space for which we can use similar bases. Mode B has its own creation and annihilation operators that are different from the ones for A. Let us just say both A and B are plane waves for the examples that follow. If mode A and mode B are orthogonal (the way two different plane waves are, and this is a purely classical statement), then we have all their operators commute and their Hilbert spaces are completely different. We can have a coherent state in mode A and it will still be a plane wave. Same with having a single photon state in mode B and it will still be a plane wave. We can not use kets related to mode B to describe anything related to mode A. If we want the total Hilbert space, that would be the tensor product of Hilbert space A and B and the basis kets in it will be denoted like {|00⟩,|10⟩,|01⟩,|11⟩,|20⟩,|21⟩,|12⟩,|02⟩,|22⟩,...}. I continue to claim you are confounding the basis of harmonic functions (modes) for the classical description of the EM field, with the basis of kets describing the quantization of one single mode. Said another way, a plane wave is a basis vector of the space of square-integrable functions, while Fock states and coherent states are basis vectors for a Hilbert space of a quantum oscillator.

pdonis 6y ago
> Mode A has a Hilbert space > 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 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.
- krastanov 6y ago
  > 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.
  - pdonis 6y ago
    > This is definitely wrong 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.

pdonis 6y ago

> Mode A has a Hilbert space > 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 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.

krastanov 6y ago
> 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.
- pdonis 6y ago
  > This is definitely wrong 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.