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Normalising Wavefunctions

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Normalise the wavefunction in the diagram which is given by:

    psi(x) = A {x} < a

    {x} is supposed to be mod x.


    2. Relevant equations
    None specifically
    3. The attempt at a solution
    I know that the square integral of the wavefunction needs to be set equal to 1. I am unsure exactly what I square as I haven't seen a wave-function written with a mod x. I know that mod x just means x with positive values. I know I need to square the wavefunction first so would that be:

    psi(x) squared = A^2 a^2?

    The answer for the square integrable of this wavefunction is A^2(2a) according to my textbook but they don't show you how. You would then have to choose A equal to 1/sqrt 2a and plug that in to the original function to normalise it. I don't see how the square integral is A^2(2a) unless you take the area of the graph to be the integral of the function which is A(2a) but then if you square this then you get A^2(2a)^2?

    upload_2016-1-18_19-2-35.png
     
  2. jcsd
  3. Jan 18, 2016 #2

    blue_leaf77

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    The wavefunction should be written like
    $$
    \psi(x) = A \hspace{0.5cm} \textrm{for} \hspace{2mm} |x|<a
    $$
    and zero otherwise. A mathematical expression like ##|x|<a## is equivalent to ##-a<x<a##.

    The normalization reads as
    $$
    \int_{-\infty}^\infty |\psi(x)|^2 = 1
    $$
    What do you get if you calculate the left side using the given wavefunction?
     
  4. Jan 18, 2016 #3
    Yes sorry, what you wrote is exactly what is written in my textbook. This is what I am not sure about. I know I need to square the function first so I will get A^2 but I don't know what else to square is mod x part of the actual function or not? I can't actually get my teeth into this function. I am happy taking the square integral of complex exponentials but this simple one has thrown me!
     
  5. Jan 18, 2016 #4

    blue_leaf77

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    I was aware that that's part of your confusion, that's why I purposely added "for" to the right of the wavefunction.
    The above lines describe the nature of ##\psi(x)##. This just means that ##\psi(x)## has values of zero within ##-a## and ##a##, and zero outside this range. Now you have to incorporate this behavior of ##\psi(x)## into the normalization integral.
     
  6. Jan 18, 2016 #5
    Thanks. Could you give me some guidance on how to do this? Since the integral of any function is the area under the graph can I not say that the integral for ##\psi(x)## is A(2a)?
     
  7. Jan 18, 2016 #6

    blue_leaf77

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    That's right except for that the "graph" you are talking about should be ##|\psi(x)|^2##, instead of ##\psi(x)##.
     
  8. Jan 18, 2016 #7

    PeroK

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    I recall someone else had a similar problem recently when they couldn't accept a constant as a function, because it didn't have x in it. Try the equivalent

    ##\Psi(x) = A\frac{e^x}{e^x}## for ## |x| < a##

    If you need x in your definition of a function.
     
  9. Jan 18, 2016 #8
    Thanks for your help. Ok so I think I get it. In my graph do we call the y-axis Ψ(x). So |ψ(x)|2 would be the square of that graph which means the amplitude A would become A^2. And the integral will be the area underneath the |ψ(x)|2 graph which would be A^2(2a), namely height multiplied by width?
     
  10. Jan 18, 2016 #9

    blue_leaf77

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    Yes.
     
  11. Jan 18, 2016 #10
    Thanks. The solution to the normalisation is A is set equal to 1/sqrt2a. This means that when you take the square integral of the original function using limits of -a to +a using this expression for A you should get 1. Is there a way of showing this mathematically by actually plugging in the limits and doing the integration as you would normally. For example if you square your new wavefunction you get 1/2a. And you can't really integrate this the usual way as you would end upw ith the integral of 1/a which is ln(a). I can see how you get it by using the area of the graph as the integral because then you get 1/2a multiplied by 2a which gives 1.
     
  12. Jan 18, 2016 #11

    DrClaude

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    When you have a function that is defined by parts, the integral looks like
    $$
    \int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{-a} f(x) dx + \int_{-a}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx
    $$
    In your case, on the right hand side, the first and last integrals are 0 because ##f(x)=0## over those ranges.
     
  13. Jan 18, 2016 #12
    Thanks. But with the middle integral are you not taking the integral of 1/2a and then evaluating from -a to +a? How do you end up with one from that integral?
     
  14. Jan 18, 2016 #13

    DrClaude

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    I just wrote a generic function ##f(x)##. In your case, ##f(x) = \left| \psi(x) \right|^2##.
     
  15. Jan 18, 2016 #14

    PeroK

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    You've really got a problem with the constant function. There's not much to say except that the integral of the constant function 1/a is most certainly not ln(a). It's x/a. And that's just about the simplest integral there is.
     
  16. Jan 18, 2016 #15
    Yes it is really strange I can't quite grasp it as I find other complex integrations such as gaussian functions fine. So in this problem we are integrating 1/a with respect to what? x? So are you saying that the integral of a constant with respect to x is x times the constant?
     
  17. Jan 18, 2016 #16

    DrClaude

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    Yes:
    $$
    \int A \, dx = A \int dx = A x
    $$
     
  18. Jan 18, 2016 #17
    Ok, I see so I have done this to show what I think you mean for the problem in question. So is this what you mean I should get:

    upload_2016-1-18_22-15-27.png
     
  19. Jan 18, 2016 #18

    DrClaude

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    Correct!
     
  20. Jan 18, 2016 #19
    Thank you!!
     
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