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Help with quantum mechanics!

  1. Dec 9, 2004 #1
    I'm currently enrolled in a course about physical chemistry. Being a chemist in ground, I do not have much previous mathemathical education and this have now become a problem. People have told me that math isn't really that necessary for this course, but I can see now that we have come to quantum mechanics that I really should have read some more math before. Anyway, not much I can do now and I ask for your help. :)

    I have 2 questions I need help with, and I would be really glad if you could help me out. It's mostly mathemathical at this point. And maybe I should point out that my mother language is not english, so I may get some things weird/wrong.

    1) Apply the quantum mechanical operators of x, p, p^2 on each of the following functions: x^2, exp(ikx), exp(-ax^2).

    2) Look at a particle in one dimension, x, in the interval 0<x<1. The wave function that describes the particle is psi(x) = sin(pi*x) in the interval and zero outside.

    a: Normalize the wave function
    b: Calculate the uncertainty (Heisenberg's) - deltaXdeltaP

    Any hints, tips, help would be the best. Many thanks.
  2. jcsd
  3. Dec 9, 2004 #2


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    What you should know for this question is that the operator x simply multiplies the function by x and that the operator p differentiates the function with respect to x and multiplies it by [itex]\frac{\hbar}{i}[/itex].


    For 2). A wavefunction [itex]\psi(x)[/itex] is normalised if
  4. Dec 9, 2004 #3

    I've thinkered a little about number 2, and my conclusion follows -
    The integral is: F(x) = x/2 - (sin(2*pi*x)/4*pi) = 1/N^2
    And the normalization constant becomes: (1/(x/2 - (sin(2*pi*x)/4*pi))^1/2)
    In other words, the normalized wave function should be:
    (1/(x/2 - (sin(2*pi*x)/4*pi))^1/2)*sin(pi*x)

    Is this correct?
    The integral was looked up in a table, is this integral solved easily by partial integration otherwise?
  5. Dec 9, 2004 #4


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    There's something wrong with the constant od normalization.As Galileo said,apply that formula knowing that your function (sin) is real (so that should make an idea about that square) and that the domain of integrating (-infinty+plus infinity) is devisedin 3 parts:the (-inf,0)(0,1) and (1,+inf).For the integration u need,u won't be needing a table of integrals,unless u're really slow (sorry,no offense).

    Good luck,anyway!!
  6. Dec 9, 2004 #5
    Yes, that was quite messed up. :/

    Got more help from my intelligent friend who helped me out to understand and guided me trough the problem. So the final word is that the nomralization constant is 2^(1/2) which give s the normalized function as 2^(1/2)sin(pi*x)

    Does this sound more correct?
    And I am really slow. :D The problem is that I have no mathemathical background, but my friend who isn't really slow when it comes to mathemathics sais that he would need ****loads of time to do the integration of |sin(pi*x)|^2 and he seems right, or?

    Thanks for all the help, I really appreciate it!
    Last edited: Dec 9, 2004
  7. Dec 10, 2004 #6


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    Integrals with product of sines or cosines can be a a big pain. Lots of work. Always try to look for symmetry or things that can make you see the answer beforehand. You don't want to integrate for 15 minutes only to find out the answer is zero and saying: 'Gee, the function is odd, I should've seen it coming'.

    You usually have to use some trig. identity for these integrals. The following is handy for sine squared::

    In this case you can also do it by noticing

    [tex]\int_0^1\sin^2(\pi x)dx=\int_0^1\cos^2(\pi x)dx[/tex]
    Since the integration covers a whole period.
    Then combine this with [itex]\sin^2x+\cos^2x=1[/itex]

    [tex]1=\int_0^1(\sin^2(\pi x)+\cos^2(\pi x))dx=2\int_0^1 \sin^2(\pi x)dx[/tex]
    To get [itex]\int_0^1 \sin^2(\pi x)dx=1/2[/itex].

    Anyway, you got the right answer. :wink:
    Last edited: Dec 10, 2004
  8. Dec 12, 2004 #7
    Thanks! :)
    I think we will get a small list of useful integrations for the final test, since the mathematics is a bit beyond scope of this course.

    Anyway, I need some more help with the math when I should calculate the uncertainty product. :grumpy:

    I have understood that this is done by calculating deltaX*deltaP, and deltaX^2/P^2 = <x^2>/<p^2>. Put this in the wave function:
    <x^2> = int dx x^2|f(x)|^2 = 2^1/2int dx x^2sin(pi*x)sin(pi*x)

    But here I get stuck, and I can't get hold of my friend. :D
  9. Dec 12, 2004 #8


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    I think u mean Galileo,but lemme be a substitute for him,his knowledge and his ability to give (useful) advice. :wink:

    U have determined the eigenfunction [itex] \psi (x) [/itex],together with its normalization factor,which became the key issue for this thread.
    Now,u wish to calculate the quantities [itex] \Delta x,\Delta p_{x} [/itex],multiply them to check whether Heisenberg's famous uncertainty relation is satisfied.But u're facing some rather nasty integrals.Lemme explain u why.Textbook definitons:
    [tex] (\Delta x)_{\psi(x)} =:+\sqrt{< \hat{x}^{2}>_{\psi(x)} -<\hat{x}>_{\psi(x)} ^{2}} [/tex]
    [tex] (\Delta p_{x})_{\psi(x)} =:+\sqrt{<\hat{p}^{2}_{x}>_{\psi(x)}-<\hat{p}_{x}>_{\psi(x)}^{2}} [/tex]

    In the coordinate representation:
    [tex] \hat{x}=x\hat{1} [/tex]
    [tex] \hat{p}_{x}=-i\hbar\frac{d}{dx}\hat{1}[/tex]

    So,u'll have to find those averages (4 of them),where the abstract [itex] \psi(x) [/itex] is nothing than your normalized wave function.
    I know these QM calculations seem horrible,but u cannot state a sound answer to your problem without making the calculations.

    Last edited: Dec 12, 2004
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