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Homework Help: Gaussian Wave packet- Griffiths 2.22

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi, The problems asks to calculate multiple things for a Gaussian wave packet. Steady state function: psi(x,0)=A*exp(-ax^2).My problem is that I'm stuck at calculating <p^2>.

    2. Relevant equations
    <p^2>=Int(|psi|^2*(-1*h^2*d^2/dx^2))dx or
    <p^2>=Int(psi*(-1*h^2*d^2/dx^2(psi-conjugate)))dx ?
    I am not sure if I can use the expression for |psi|^2- which is already defined and I think calculating the second expression will be wrong since it will give us an answer with i and t sine the exponent from complex conjugate of psi will come down or am i wrong?

    3. The attempt at a solution
    I know I'm doing something wrong since after i evaluate integrals I get:
    <p^2>=4*(2/m)^(1/2)*w^6h^2*(4w^2*(pi/w^6)^(1/2)-2(1/2(pi/w^2)^(1/2)) where w=(a/(1+(2ihat/m)^2)^1/2 which cancels out to 0, somehow I should get ah^2. Now I'm not sure is I should use <p^2>=Int(|psi|^2*(-1*h^2*d^2/dx^2))dx where its a second derivative of |psi|^2 or <p^2>=Int(psi*(-1*h^2*d^2/dx^2(psi-conjugate)))dx where the second derivative is only of the complex conjugate and not the |psi|^2. Or I should not use the w just use the expression defined above? (I thought since w is independent of x it would not matter). Any help will do, just before Sunday evening as its due on Monday morning.
    Last edited: Oct 17, 2009
  2. jcsd
  3. Oct 17, 2009 #2
    [tex]<p^2>=|A|^2 \int \psi^*(-h^2 \frac{d^2}{dx^2} \psi)dx [/tex]

    Is the correct way to evaluate the expectation value.

    As for the integral, I believe there is a very cleaver u-sub to (easily) solve this problem.
  4. Oct 17, 2009 #3
    Sorry just came to me:
    Can I use:
    <p^2>=m^2*(d^2(<x>^2)/dx^2) or it doesnt work like that?
    nvm...<x>=0 :(
  5. Oct 17, 2009 #4
    Ok, I dont know any U-sub, it will help to know about it anyway- but i understand if you dont want to give me a straight answer to my problem- maybe after monday you can tell me how to do that the "easy way", now I am going through that mess and im left with something like this:
    (1/(1+ib)^(1/2))*(1/(1-ib)^(1/2)) where b=2hat/m and i is imaginary number, do those cancel? I cannot solve this and I tried to use one of those equation solvers on internet and wont do it neither..... *the second expression suppose to be a complex conjugate of the first one.... please help
  6. Oct 18, 2009 #5
    Well [tex]\int_{\infty}^{\infty} e^{-ax ^2} dx = \frac{\sqrt{\pi}}{\sqrt{a}} = f(a)[/tex] right?

    What's f'(a)?

    The function you're trying to integrate looks like:

    [tex]-2 a h^2 \int_{\infty}^{\infty} -e^{-2 a x^2} + 2 a x^2 e^{-2 a x^2} dx[/tex]

    Which looks sort of like b f(a) + c f'(a) if b and c are just constants, doesn't it?
  7. Oct 18, 2009 #6
    well... yes... but i solved it using int(exp(-ax^2)dx=(pi/a)^(1/2) and int(x^2*exp(-ax^2)dx=1/2(pi/a^3)^(1/2) which i believe yields the same output.... I kind of understand the logic here... thanks for the tip as this will be easier then memorizing the common integrals :].... the only problem now is that i have an answer looking like this:

    <p^2>=h^2*a*(1/(1-2ihat/m)^(1/2))*(1/(1b2ihat/m)^(1/2)) and basically my question is:


    I know this is multiplying a complex number by its complex conjugate but does this case works since the complex number is under a square root? or maybe i defined wrong complex conjugate for this? the original expression is:

    (1/(1-(2ihat/m))^1/2))) what is its complex conjugate? what does this reduce to? it seems like it should be 1 since the real part is 1/1 so 1^2=1 but how this works in this case with above expression? I hope you can understand which part im having a problem with.....
    Last edited: Oct 18, 2009
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