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This integral equation from (intro) quantum mechanics confuses me

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    As you all know, ##<F> = <F>^*##, where ##F## is an observable linked with the operator ##\hat{F}##. This means ## <F> = \int \Psi^* \hat{F} \Psi d\tau = <F>^* = [\int \Psi^* (\hat{F} \Psi) d \tau]^* = \int \Psi (\hat{F} \Psi)^* d \tau \Rightarrow \int \Psi (\hat{F} \Psi)^* d \tau = \int \Psi^* \hat{F} \Psi d\tau## (btw, what theorem allows you to ignore the integral sign when using complex conjugation on a function under an integral?).

    OK. But my question is, why does that mean
    ##\int (\hat{F} \Psi_1)^* d \Psi_2\tau = \int \Psi_1^* \hat{F} \Psi_2 d\tau##, if the the wavefunctions ##\Psi_1## and ##\Psi_2## are quadratically integrable? This is what my lecture notes say, and it confuses me.

    3. The attempt at a solution

    Have no idea.
  2. jcsd
  3. Feb 6, 2014 #2


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    Of course, not all operators fulfill this condition, but only the Hermitian ones. That's why one usually says that in quantum theory the observables are represented by Hermitian operators (strictly speaking they must be even self-adjoint, but let's not deal with this subtleties now).

    Another important feature is that self-adjoint operators lead to a spectral decomposition of the squre-integrable functions. Loosely speaking in the physicists hand-waving way, you have "complete sets of (generalized) eigenfunctions" of the self-adjoint operator, i.e., functions, fulfilling
    [tex]\hat{F} u_{\lambda_n}(x)= \lambda_n u_{\lambda_n}(x),[/tex]
    and you can express all wave functions as a series of these eigenfunctions,
    [tex]\psi(x)=\sum_n \psi_n u_{\lambda_n}(x),[/tex]
    supposed the wave functions are orthonormal, i.e.,
    [tex]\int \mathrm{d} x u_{\lambda_{n_1}}^*(x) u_{\lambda_{n_2}}(x)=\delta_{n_1 n_2}.[/tex]
    Sometimes the self-adjoint operators also have continuous "eigenvalues". Then the sum becomes an integral. That's the case, e.g., for the momentum operator, which is
    [tex]\hat{p}=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x}.[/tex]
    Then the eigenvalues are all real numbers, and the eigenfunctions the plane waves
    [tex]u_{p}(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).[/tex]
    The normalization here is "to a [itex]\delta[/itex] function", i.e.,
    [tex]\int \mathrm{d} x u_{p}^*(x) u_{p'}(x)=\delta(p-p').[/tex]
    Here the expansion of a wave function into momentum-eigenfunctions is simply the Fourier transformation, i.e.,
    [tex]\psi(x)=\int \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p).[/tex]
  4. Feb 6, 2014 #3


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    Is that ``##d \Psi_2\tau##'' on the LHS a typo, or do your notes actually say that??

    It's actually hermitian conjugation in this case. But either way, it's ok because you can think of the integral as an infinite sum of infinitesimal terms. (Recall the first-principles definition of an integral...) Such conjugation acts on a sum by acting separately on each term in the sum. Strictly speaking, one should also conjugate the measure ##d\tau##, but presumably that's a real variable here(?), so the conjugation has no effect on it.
    Last edited: Feb 6, 2014
  5. Feb 7, 2014 #4
    Oops, its a horrible typo!! Sorry

    And thx for help, but can somebody answer my main concern? Ie

    "why does that mean ##\int (\hat{F} \Psi_1)^* \Psi_2d\tau = \int \Psi_1^* \hat{F} \Psi_2 d\tau##, if the the wavefunctions ##\Psi_1## and ##\Psi_2## are quadratically integrable?"
    Last edited: Feb 7, 2014
  6. Feb 7, 2014 #5


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    Well, something is not quite right with that. It should say "... if ##\hat F## is a self-adjoint operator on the space of square-integrable wave functions". (It's essentially the definition of what it means for an operator to be "self-adjoint".)
  7. Feb 8, 2014 #6
    Yes, it is also true that ##\hat{F}## is self-adjoint, sorry. But so the integral equation is just a definition? It can't be derived?

    Also, can you explain to me delta normalization? I understand it's used when the values are not discrete but continuous, but why is that the case? Why must continuous eigenvalues imply that the eigenfunctions are not normalizable in the ordinary way? And is there any physical meaning to delta-normalization? Normal normalization simply states that the probability for the particle to be present somewhere in space is 1, but what is the analogue for delta normalization?
    Last edited: Feb 8, 2014
  8. Feb 8, 2014 #7


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    It's just a particular choice of inner product on a space of functions. More abstractly, one could denote the inner product by
    $$(\Psi_2 , \Psi_1)$$or by$$\langle \Psi_2 | \Psi_1 \rangle$$
    More concretely, you're working here with the Hilbert space of square integrable functions over ##\mathbb{R}##, where the inner product is defined by
    \langle \Psi_2 | \Psi_1 \rangle ~:=~ \int \Psi^*_2(x) \Psi_1(x) \, dx

    Well, to answer to that thoroughly requires a course on Functional Analysis, and several nontrivial theorems. For physics, the basic reason is that one needs to be able to sum over the eigenvalues in somehow. For continuous spectra, the sums must become integrals, Kronecker deltas become Dirac deltas, but the trace of a Dirac delta is divergent.

    That's still the underlying motivation, but the details are nontrivial. One wishes to construct a so-called "positive operator-valued measure" over the space of possible measurement outcomes.

    But which textbook(s) are you using? The early chapters in Ballentine cover this stuff, without swallowing you in a Functional Analysis maze.
  9. Feb 9, 2014 #8
    I'm reading the notes written by a professor at my school. But I feel like I understand nothing regardless of what I read: be it griffiths or these notes. But I am able to do the problems, so I guess I'm doing OK.

    And I really don't understand delta-normalization. This dirac delta-stuff is too abstract.

    So whatever, if the main point is to just make normalization of eigenfunctions with continuous eigenvalues possible, I'm happy with that.

    thanks 4 help!
  10. Feb 9, 2014 #9


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    It's important to find a textbook that is right for you at this point in time. Don't be content with blundering through a fog.

    If you can get to the library, maybe take a look at Greiner's QM textbook. He puts more emphasis on not skipping steps.
  11. Feb 9, 2014 #10


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    I'm not sure, but from your first post it looks like you want to know how to derive ##\int (\hat{F} \Psi_1)^* \Psi_2d\tau = \int \Psi_1^* \hat{F} \Psi_2 d\tau## starting from ##\int (\hat{F} \Psi)^* \Psi d\tau = \int \Psi^* \hat{F} \Psi d\tau##.

    Try letting ##\Psi = \Psi_1+\Psi_2## in ##\int (\hat{F} \Psi)^* \Psi d\tau = \int \Psi^* \hat{F} \Psi d\tau##. Expand and simplify.

    Then do it over again with ##\Psi = \Psi_1+i\Psi_2##.
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