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Q. about continuous spectra

  1. Jan 4, 2006 #1
    In Sakurai:
    The state ket for an arbitrary physical state can be expanded in terms of the |x'>
    [tex]|\alpha>=\int dx'^3 |\mathbf{x'}><\mathbf{x'}|\alpha>[/tex]
    (where the |x'> are the eigenkets of the position operator, [itex]\hat{x}|x'>=x'|x'>[/itex]).
    (Sakurai 1.6.4, p. 42)

    My question is about how this integral is mathematically defined - since the integrand is not a complex number, the Lebesgue integral as defined in analysis1 doesn't work directly. I thought about extending the defintion to kets, defining an ordering |a> <= |b> iff every <x'|a> <= <x'|b>, is this how it works? I also noticed the equation looks very much like a Fourier series (except the basis is uncountable) - it's the same idea, an expansion in a basis set... does this work formally?

    Since the explanation may be very involved, a reference to a book/section would be more than sufficient. Thanks!

    (secondary question - is there a better way to TeX the above equation?)
     
    Last edited by a moderator: Jan 4, 2006
  2. jcsd
  3. Jan 9, 2006 #2

    dextercioby

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    Well, formally, everything works due to some bright ideas by Dirac. The mathematics behind this is contained (a great part of it) in the 5 volumes by Gelfand and especially in the IV-the (written with Vilenkin) and entitled: "Generalized functions: Applications of Harmonic Analysis".

    Daniel.
     
  4. Jan 11, 2006 #3
    Consider the action of both sides of the equation on an arbitrary bra <[itex]\beta[/itex]|:

    [tex]<\beta|\alpha> = \int {dx'}^3 <\beta|x'><x'|\alpha>[/tex]

    Now everything in sight is a number and so Lebesque integration works.
     
  5. Jan 11, 2006 #4

    Hurkyl

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    If you want a more overarching reason, you can appeal to the spectral resolution theorem, Stiljes integrals and functional calculus. (This approach can even manage to avoid delta "funcions", if you find that appealing)


    Here's a quick sketch of how it works in the one-dimensional case.

    The position operator X has the spectral resolution:

    [tex]
    X := \int_{-\infty}^{+\infty} x |x\rangle\langle x| \, dx
    [/tex]

    which merely means that

    [tex]
    \langle \psi | X | \psi \rangle = \int_{-\infty}^{+\infty} x \langle \psi| x\rangle\langle x|\psi\rangle \, dx
    [/tex]

    If you want to avoid the delta functions, the "right" way to describe the spectral resolution is in terms of projection operators: if [itex]\phi[/itex] is a wave function (in the position basis), then we define:

    [tex]
    (E_x \phi)(y) := \left\{
    \begin{array}{ll}
    \phi(y) \qquad & y < x \\
    0 & y > x
    [/tex]

    I don't want to work out what you're supposed to do when x = y, but I'm pretty sure it's irrelevant to the discussion.

    Now, the "right" way to view the spectral resolution is via:

    [tex]
    \langle \psi | X | \psi \rangle = \int_{-\infty}^{+\infty} x \, d \langle \psi | E_x | \psi \rangle
    [/tex]

    This is a Stiljes integral, and it's well-defined because the thing after the d is a nondecreasing function. This leads us to say

    [tex]
    X = \int_{-\infty}^{+\infty} x \, dE_x
    [/tex]

    The reason we can write [itex]dE_x = |x\rangle\langle x| \, dx[/itex] is because the [itex]E_x[/itex] forms a continuous "distribution function", so it can be differentiated to get a "density function".


    Then, if we write f(z) := 1, we can express the identity operator functionally in terms of the position operator: [itex]\hat{1} = f(X)[/itex]. So, the functional calculus tells us:

    [tex]
    \hat{1} = f(X) = \int_{-\infty}^{+\infty} f(x) \, dE_x
    = \int_{-\infty}^{+\infty} dE_x
    = \int_{-\infty}^{+\infty} |x\rangle\langle x| \, dx
    [/tex]

    which is a "resolution of unity".

    Then, you just apply this resolution of unity to [itex]|\alpha\rangle[/itex], and you get
    [tex]
    |\alpha \rangle = \hat{1} |\alpha \rangle
    = \int_{-\infty}^{+\infty} |x\rangle\langle x|\alpha\rangle \, dx
    [/tex]


    None of this was needed to actually answer your question (since the integral of a linear operator is defined pointwise), but I thought you might find it appealing.


    Incidentally, Real Analysis by Royden has a section on the Stiljes integral. The spectral resolution, etc, I took from the Hilbert space section at Modern Physics for Mathematicians.

    You can get to that latter link, incidentally, by following the "links" button off of this forum!
     
    Last edited: Jan 11, 2006
  6. Jan 23, 2006 #5
    In my opinion you can always appeal to the simpler formalism, which resembles the one used in statistical theory. Product of functions, integral of product of functions, no bra and ket, just functions (complex ones) and its conjugates.

    Matrix formalism never showed itself as an absolutelly necessary tool.

    Best wishes

    DaTario
     
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