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I Do I have the correct form for the number of states, the |n>?

  1. Feb 28, 2017 #1

    rocdoc

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    I cannot find any explanation of the mathematical form of the single-mode photon number states, i.e. the |n>.

    I take them to be functions with domain {0,1,2,3, …} and appropriate associated outputs.

    So |3> I take to have outputs {0,0,0,1, 0, …} , |0> to have outputs {1,0,0,0, 0, …} and similarly for other |j>.

    Am I correct?
     
  2. jcsd
  3. Feb 28, 2017 #2

    DrClaude

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    What do you mean by outputs?
     
  4. Feb 28, 2017 #3

    PeterDonis

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    You're off to a bad start. States are vectors in a Hilbert space. If you want to express them as wave functions, they have to be functions that are members of an appropriate Hilbert space (such as the space of square integrable functions on ##\mathbb{R}^n##). It's not clear that you have correctly (or even precisely) specified the functions you are describing; I would recommend doing that first, before considering anything else.

    Btw, the usual Hilbert space that is used to describe the photon number states is a Fock space of which those states form a basis:

    https://en.wikipedia.org/wiki/Fock_space
     
  5. Mar 1, 2017 #4

    rocdoc

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    A function has a set of inputs (a domain) and a set of outputs. It maps inputs to outputs. An output is a function value.

    The | n> are proper mathematical functions just like the spin functions α and β are. α and β could be described as functions with domain {-1/2, +1/2} with α having outputs {0, 1} and β outputs {1, 0}. Think “function of the eigenvalues of some observable”, I take the |n> to be functions with domain, the eigenvalues of the number operator for photons in a mode of the electromagnetic field.

    Please also see the attachment.
     

    Attached Files:

    Last edited: Mar 1, 2017
  6. Mar 1, 2017 #5

    dextercioby

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    If n is the number of photons in a quantum state, then the Dirac ket ## |n\rangle ## can be seen as a mapping ##\rho :\mathbb{N} \rightarrow \mathcal{F} ##, with the latter being the photonic Fock space. This Fock space can be built over any complex separable one-particle Hilbert space, for example ##L^2 \left(\mathbb{R}^3, d^3 x\right) ##.
     
  7. Mar 1, 2017 #6

    rocdoc

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    So I have some agreement that the domain of |n> is {0,1,2,3, …}.
     
  8. Mar 1, 2017 #7

    dextercioby

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    There's no domain. |n> is a symbol, for this means that the Fock space is abstract. The mapping I mentioned is an assignment of one vector to any natural number which shows that the eigenvectors of the Number operator in the Fock space form a countable basis.
     
  9. Mar 1, 2017 #8

    rocdoc

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    I don't follow!

    How about my first post, is the material in it correct?

    I know |n> is a symbol so are α and β !
     
    Last edited: Mar 1, 2017
  10. Mar 1, 2017 #9

    rocdoc

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    How about with the |n> viewed as Eigen functions of the photon number operator for some mode?
     
  11. Mar 1, 2017 #10

    rocdoc

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    The ket |n> stands for a concrete physical state, one with n photons in, in some mode.

    It must be associated with definite sets of numbers, i.e. be a definite function. How can its representatives in some basis be evaluated, by a scalar product say, if it isn’t itself, some definite function?
     
  12. Mar 1, 2017 #11

    PeterDonis

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    No, they can't. The Hilbert space of a qubit, which is what you are describing, has an infinite number of elements, not just two.

    This is the wrong way to think; it is not what a quantum state is.

    This is not correct.

    What textbooks or papers on quantum mechanics have you read? Do you know what a Hilbert space is? Do you know what a Fock space is? You don't seem to have the requisite background for an "I" level thread on this topic.
     
  13. Mar 1, 2017 #12

    PeterDonis

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    This is a much better point to start from. But now you need to ask: what exactly are these eigenfunctions?

    The general equation for an eigenfunction of an operator is:

    $$
    O f = \lambda f
    $$

    where ##O## is an operator, ##f## is a function, and ##\lambda## is a scalar, the eigenvalue. So if the states ##\vert n \rangle## are eigenfunctions of the photon number operator, which we will call ##N##, with eigenvalues equal to the integers, then they must satisfy the equation

    $$
    N \vert n \rangle = n \vert n \rangle
    $$

    Notice that the function ##\vert n \rangle## appears on both sides of the equation. In other words, ##\vert n \rangle## is not a function that takes an eigenvalue as input and spits out a number. It is a function that, when acted on by the operator ##N##, which is an operator that maps functions to functions, goes to the same function, multiplied by the scalar ##n## (the eigenvalue).

    So in order to find such a function, you need to know, first, what Hilbert space it belongs to, and second, what form the operator ##N## takes when expressed in that Hilbert space.
     
  14. Mar 1, 2017 #13

    rocdoc

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    I say again

    "α and β could be described as functions with domain {-1/2, +1/2}"

    These are the inputs for the spin functions α and β, and for many other spin functions as well.
     
  15. Mar 1, 2017 #14

    rocdoc

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    The function of the eigenvalues idea , I think is in Dirac 4th edition.
     
  16. Mar 1, 2017 #15

    PeterDonis

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    You are going to need to give an explicit reference (specific chapter/page/section, and give an explicit quote), because what you are saying looks obviously false, not to mention obviously inconsistent with the correct formulation I gave in post #12. If you have such a reference, PM me. Until I see a valid reference, this thread is closed.
     
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