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The momentum representation of ##x## and ##[x,p]##

  1. Oct 27, 2015 #1
    To deduce the momentum representation of ##[x,p]##, we can see one paradom
    ##<p|[x,p]|p>=i\hbar##
    ##<p|[x,p]|p>=<p|xp|p>-<p|px|p>=p<p|x|p>-p<p|x|p>=0##
    Why? If we deduce the momentum representation of ##x##, we obtain
    ##<p|x|p>=i\hbar \frac{\partial \delta (p'-p)}{\partial p'}|_{p'=p}##. This value is not definite. So, why two uncertain values can obtained a certain value ##i\hbar##? In addition, the ##x## should be replace by ##i\hbar \frac{\partial }{\partial p}##. Then the eigenvalue ##p## can't extract. However, if we consider ##i\hbar \frac{\partial }{\partial p}## to act on the bra, not the ket, then the eigenvalue ##p## can be extracted. Is anything wrong here?
     
  2. jcsd
  3. Oct 27, 2015 #2

    bhobba

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    What does <p|p> equal?

    Remember formal manipulations involving infinity are rather dubious.

    The real solution to this sort of stuff requires Rigged Hilbert Spaces which needs a very good background in analysis to understand. As a build up to it I suggest the following book which should be in armoury of any physicist nor even applied mathematician:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    It will not rigorously resolve what's going on, that requires considerable advanced analysis, but will allow an intuitive accommodation.

    Thanks
    Bill
     
    Last edited: Oct 27, 2015
  4. Oct 27, 2015 #3
    I just regard this term as normalization 1.
     
  5. Oct 27, 2015 #4

    bhobba

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    <p|p'> = Dirac delta (p - p'). Substitute p = p' and you get Dirac delta (0) - which strictly speaking is undefined, but intuitively is taken as infinity.

    Thanks
    Bill
     
  6. Oct 27, 2015 #5
    Aha, you are right, thank you!
     
  7. Oct 27, 2015 #6

    bhobba

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    No problem.

    This stuff can be tricky because its not really valid with the math you likely know at the beginner level.

    That's why I STRONGLY suggest you get the book I mentioned. It will not resolve things completely, but things will be a LOT clearer.

    Thanks
    Bill
     
  8. Oct 27, 2015 #7
    Yeah, nowadays books about quantum physics are of many mathematical details. Thank you for your recommend.
     
  9. Oct 28, 2015 #8

    stevendaryl

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    As Bill says, the problem is that [itex]|p\rangle[/itex] is not a normalizable state; there are no normalizable momentum eigenstates. However, you can get an approximate momentum eigenstate [itex]\phi_p(x) = (\frac{2 \lambda}{\pi})^{\frac{1}{4}} e^{i p x - \lambda x^2}[/itex]. Then:

    [itex]\hat{p} \phi_p = (p + 2 i \lambda x) \phi_p[/itex]

    which for small values of [itex]x[/itex] is approximately [itex]p \phi_p[/itex]. So [itex]\phi_p[/itex] is an approximate momentum eigenstate. If you compute [itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle[/itex] using the fact that [itex]\hat{p}[/itex] is Hermitian, you find:

    [itex]\langle \phi_p | \hat{p} x | \phi_p \rangle = \langle \hat{p} \phi_p | x | \phi_p \rangle = \langle (p + 2i \lambda x) \phi_p|x|\phi_p\rangle
    = \langle \phi_p|(p - 2i \lambda x)x|\phi_p\rangle[/itex]
    [itex]= p \langle \phi_p|x|\phi_p \rangle - 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

    [itex]\langle \phi_p | x \hat{p} | \phi_p \rangle = \langle \phi_p|(p + 2i \lambda x)x|\phi_p\rangle[/itex]
    [itex]= p \langle \phi_p|x|\phi_p \rangle + 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

    So [itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex], not zero.

    You might think that by choosing the convergence parameter [itex]\lambda[/itex] to be very small, you could ignore this term, but in fact, [itex]\langle \phi_p |x^2|\phi_p\rangle = \frac{1}{4 \lambda}[/itex], so regardless of the choice of [itex]\lambda[/itex],

    [itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \frac{1}{4 \lambda} = -i[/itex].
     
  10. Oct 28, 2015 #9
    Great! This is a specific handle,and you solve it in the coordinate representation. You miss the quantity ##\hbar##, but I think it is just a unit assumed. I have one question. We know the momentum operater is a observable physical quantity. Why its eigenvalue is complex? It means that ##<\phi_p|\hat p\hat x|\phi_p>-<\phi_p|\hat x\hat p|\phi_p>=<\phi_p|f^*(p,x)x|\phi_p>-<\phi_p|xf(p,x)|\phi_p>##?
     
  11. Oct 28, 2015 #10

    stevendaryl

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    The state [itex]|\phi_p\rangle[/itex] isn't an eigenstate, and the quantity [itex]p + i 2 \lambda x[/itex] isn't an eigenvalue. Notice it depends on [itex]x[/itex], and eigenvalues must be constants.

    But you're right, it's kind of weird that the expectation value of [itex]\hat{p}[/itex] is a complex number.

    I'm going to think about that.
     
  12. Oct 29, 2015 #11

    Demystifier

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    This is really great! Did you devised it by yourself, or is it published somewhere else?
     
  13. Oct 29, 2015 #12

    Demystifier

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    Actually, it is a real number. We have
    ##\int dx \, \phi_p^* ix \phi_p =0##
    so the imaginary part vanishes.
     
  14. Oct 29, 2015 #13

    Demystifier

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  15. Oct 29, 2015 #14
    How to understand this complex number based on physical reasons? Thank you for your recommended paper.
     
  16. Oct 29, 2015 #15

    Demystifier

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  17. Oct 29, 2015 #16
  18. Oct 31, 2015 #17

    stevendaryl

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    Well, I did it off the top of my head, but I'm sure it's been done before. I always try to make sense of dubious mathematics by inserting parameters to make everything converge and then see if the results make sense when I let the parameter go to zero (or infinity, or whatever).

    So, for example, the dubious result that [itex]\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx} dx = \delta(k)[/itex], I always think of as:
    [itex]\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx - \lambda x^2} dx = \frac{1}{2 \sqrt{\pi \lambda}} e^{-\frac{k^2}{4\lambda}}[/itex]

    Or alternatively, I think of it as:
    [itex]\frac{1}{2\pi} \int_{-L}^{+L} e^{ikx} dx = \dfrac{sin(kL)}{\pi k}[/itex]
     
  19. Nov 2, 2015 #18

    Demystifier

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    Perhaps, but it is certainly not well known. I believe you could publish it in a journal for pedagogic papers such as American Journal of Physics or European Journal of Physics.
     
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