Please turn on javascript in your browser to play chess.
Science Forum

Science Forum

  1. 22 Feb '10 19:54
    I'm having a hard time proving that A is hermitian where A acting on f gives the integral from a to x of f(x'dx'. Could anyone help me out? I can't seem to force hermiticity after doing <g|Af> by parts, but I'm not sure if I've gone as far as I can. It's kind of hard to post the math I have tried here, but if anyone could give me some advice, I'd appreciate it.
  2. Subscriber AThousandYoung
    Do ya think?
    22 Feb '10 22:14
    Originally posted by amolv06
    I'm having a hard time proving that A is hermitian where A acting on f gives the integral from a to x of f(x'dx'. Could anyone help me out? I can't seem to force hermiticity after doing <g|Af> by parts, but I'm not sure if I've gone as far as I can. It's kind of hard to post the math I have tried here, but if anyone could give me some advice, I'd appreciate it.
    I barely passed linear algebra and differential equations...sorry!
  3. 22 Feb '10 22:37
    How do you define your inner product? If you use the L2 definition of an inner product, then A is not hermitian at all if I'm not mistaken... (take e.g. g(x) = x, f(x) = x^2, a = 0)

    I could be wrong though, I'm a bit rusty on this subject.
  4. 23 Feb '10 00:08
    Originally posted by KazetNagorra
    How do you define your inner product? If you use the L2 definition of an inner product, then A is not hermitian at all if I'm not mistaken... (take e.g. g(x) = x, f(x) = x^2, a = 0)

    I could be wrong though, I'm a bit rusty on this subject.
    "How do you define your inner product?"

    Correct, using the L2 definition. Both f and g lie in Hilbert space, so they are complex. I must show that A is either hermitian or not hermitian more generally. I do not think the professor would allow us to select specific vectors. I think I've got it, though. A is non-hermitian, as <f|Ag>=-<Af|g>.
  5. 23 Feb '10 07:16
    Originally posted by amolv06
    "How do you define your inner product?"

    Correct, using the L2 definition. Both f and g lie in Hilbert space, so they are complex. I must show that A is either hermitian or not hermitian more generally. I do not think the professor would allow us to select specific vectors. I think I've got it, though. A is non-hermitian, as <f|Ag>=-<Af|g>.
    If you want to show A is not hermitian, then showing a single example where it does not apply suffices.
  6. 23 Feb '10 08:12 / 1 edit
    Good point.

    That said, our professor has instructed us to consider the wavefunctions localized, which leads certain operators (such as the derivative) to be hermitian in these cases, whereas in general (from a mathematical, and not a physical perspective) they would not be. Such results seem to be most readily apparent by actually performing the calculus on abstract vectors, at least to me. If you have a more efficient method, please do share.

    Thanks,

    Amol

    Edit: Just some musing on my part -- please let me know if it sounds reasonable --

    Localizing the wavefunction must put all the work we're doing in some subspace of the complex function space. Using your point above, if I could find two vectors for which A was not hermitian in this subspace, I would have proved non-hermiticity for A. That said, I'm having a hard time convincing myself that I would be able to accurately pick out vectors which lie in the appropriate subspace.

    I believe I've solved the problem from a more abstract perspective, but what you said never dawned on me for some reason. Does everything above sound reasonable to you?
  7. 23 Feb '10 10:59
    I'm not sure what your professor means by "considering the wavefunctions localized", but if certain wavefunctions f and g do not overlap, <f|g> = 0, and for these wavefunctions, A is probably hermitian (not sure though). It's an important concept in quantum physics because in principle, all e.g. electrons in the universe are correlated with each other due to the symmetrization requirement - but if the individual wavefunctions do not overlap, you can describe the (clusters of) electrons seperately.
  8. 23 Feb '10 15:14
    Originally posted by KazetNagorra
    I'm not sure what your professor means by "considering the wavefunctions localized",
    The wavefunction goes to 0 at infinity.

    Thank you for your help in this thread.
  9. 23 Feb '10 16:59
    Originally posted by amolv06
    The wavefunction goes to 0 at infinity.

    Thank you for your help in this thread.
    No, the wavefunction going to 0 at infinity is a consequence of the wavefunction being normalized.
  10. 23 Feb '10 17:25 / 2 edits
    I am sorry, this is all very new to me, but I don't quite understand. Perhaps this is just semantics, but wouldn't normalizability be a consequence of the convergence of the inner product of the wavefunction with itself, rather than the other way around?

    Furthermore, I understand that when referring to a localized particle, the wave function vanishes outside a certain region of space. Would it be inappropriate, then, to call a wavefunction which vanishes at infinity localized? I believe this is how he has been doing it.

    I'm sorry, I don't mean to challenge you, and I appreciate your help, I am just trying to get a better grasp on the terminology. We are still early in my class, so it is very possible, even likely, that I'm confusing some of the terminology and/or concepts.
  11. 23 Feb '10 17:45
    Originally posted by amolv06
    I am sorry, this is all very new to me, but I don't quite understand. Perhaps this is just semantics, but wouldn't normalizability be a consequence of the convergence of the inner product of the wavefunction with itself, rather than the other way around?

    Furthermore, I understand that when referring to a localized particle, the wave function vanishes o ...[text shortened]... is very possible, even likely, that I'm confusing some of the terminology and/or concepts.
    Oh I get what you mean, yeah I think you can see it that way.