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I am pretty sure you are confounding the basis of functions (harmonics) that one can use to describe an EM field, with the basis of a Hilbert space used to quan

I am pretty sure you are confounding the basis of functions (harmonics) that one can use to describe an EM field, with the basis of a Hilbert space used to quantize them. You can have a mode (e.g. a plane wave) and the field of that particular mode can be quantized and you get to pick whether to use Fock states or coherent states or any other basis. In other words, "plane wave" refers to a particular mode of the field and Fock/coherent states refer to the quantum state of that particular mode. Two different plane waves are two completely different modes, e.g. two completely different Hilbert spaces, not basis vectors of a given Hilbert space.

pdonis 6y ago
> I am pretty sure you are confounding the basis of functions (harmonics) that one can use to describe an EM field, with the basis of a Hilbert space used to quantize them. No, I'm not. The same mathematical functions can appear in both classical and quantum treatments, but I'm talking about the basis of the quantum Hilbert space. > Two different plane waves are two completely different modes, e.g. two completely different Hilbert spaces No. They are different Fock states in the same Hilbert space; in the Fock state basis for that Hilbert space, they are two different, orthogonal basis vectors. And that same Hilbert space can also have other choices of basis besides the Fock state basis.
- krastanov 6y ago
  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.
knzhou 6y ago
Nah, GP is right, if he's talking about how you can decompose states into superpositions of coherent states of the whole field. These are different from coherent states of individual modes. Specifically, a coherent state of a field is a state which is an eigenvector of the positive-frequency part of the field operator, when that operator is evaluated at any position.
- krastanov 6y ago
  See the sibling comment and my answer. A plane wave can very well be a plane wave "containing" one photon or containing multiple photons or containing a coherent state. There is a big difference between saying "I can use different sets of creation operators to form a basis for the field operator" (what I think you are saying) and the confounded statement mixing up basis of a Hilbert space with basis of integrable functions of the OP.
- pdonis 6y ago
  > if he's talking about how you can decompose states into superpositions of coherent states of the whole field. Yes, that's what I was talking about.

pdonis 6y ago

> I am pretty sure you are confounding the basis of functions (harmonics) that one can use to describe an EM field, with the basis of a Hilbert space used to quantize them. No, I'm not. The same mathematical functions can appear in both classical and quantum treatments, but I'm talking about the basis of the quantum Hilbert space. > Two different plane waves are two completely different modes, e.g. two completely different Hilbert spaces No. They are different Fock states in the same Hilbert space; in the Fock state basis for that Hilbert space, they are two different, orthogonal basis vectors. And that same Hilbert space can also have other choices of basis besides the Fock state basis.

krastanov 6y ago
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.

knzhou 6y ago

Nah, GP is right, if he's talking about how you can decompose states into superpositions of coherent states of the whole field. These are different from coherent states of individual modes. Specifically, a coherent state of a field is a state which is an eigenvector of the positive-frequency part of the field operator, when that operator is evaluated at any position.

krastanov 6y ago
See the sibling comment and my answer. A plane wave can very well be a plane wave "containing" one photon or containing multiple photons or containing a coherent state. There is a big difference between saying "I can use different sets of creation operators to form a basis for the field operator" (what I think you are saying) and the confounded statement mixing up basis of a Hilbert space with basis of integrable functions of the OP.
pdonis 6y ago
> if he's talking about how you can decompose states into superpositions of coherent states of the whole field. Yes, that's what I was talking about.