1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    (1/(1-ib)^(1/2)*(1/(1+ib)^(1/2))=1?

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gaussian Wave packet- Griffiths 2.22
  1. Gaussian Wave Packet (Replies: 2)

  2. Gaussian Wave packet (Replies: 8)

Loading...