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Vibrating Nuclei_potential energy

  1. Oct 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate <K>, the expectation value of the kinetic energy for the ground state of a pair of vibrating nuclei (assume internuclear distance--hence -infinite to +infinite)

    2. Relevant equations
    K = -h2/2(mu) d2/dx2

    where (mu) = reduced mass; m1m2/(m1+m2)
    and
    wave(x) = (a/pi)1/4 e(-ax2/2)
    where a = (mu) (omega)/h

    (Assume h = planck's constant with dash)

    3. The attempt at a solution
    <K> = integral (x = -infinite to infinite) wave2(x) K(x) dx
    = integral (a/pi)1/2 e(-ax2) -h2/2(mu) d2/dx2 dx
    = -ah2/(2pi(mu)) integral e(-ax2) -h2/2(mu) d2/dx2 dx
    = -a5/4h2/(2pi5/4(mu)) integral {-ae-3ax2/2 + a2x2e(-3ax2/2)}dx
    ...beyond, I don't know how to simplify the integral--have I done something wrong in the midway? I don't know how to integrate anymore. Any suggestions would be welcome!
     
  2. jcsd
  3. Oct 8, 2014 #2

    Simon Bridge

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    You'll notice that the operator K does not make sense without some function in front of it. $$\langle K \rangle = \int_{-\infty}^\infty \psi^\star(x)\hat K \psi(x)\; dx$$... would be more usual.
     
  4. Oct 8, 2014 #3
    That's what I did--wave(x) is a psi (x)....
    So, that's why I did psi2(x) which is wave2(x), and I still have trouble getting to the solution....

    Plus, what's with asterisk on one psi? Could you pls explain? Thanks!
     
    Last edited: Oct 8, 2014
  5. Oct 8, 2014 #4

    BvU

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    Asterisk = complex conjugate

    And ##{d\over dx}e^{-ax^2}## is not something with ##e^{-2ax^2}## !
     
    Last edited: Oct 8, 2014
  6. Oct 8, 2014 #5
    Is there a way to solve integral {-ae-3ax2/2 + a2x2e(-3ax2/2)}dx?
    (where integral is from x = -infinite to + infinite)​
     
  7. Oct 8, 2014 #6

    BvU

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    It looks suspiciously identical to the Schroedinger equation for the harmonic oscillator, only (if you do it right) a factor 1/2 different....
     
  8. Oct 8, 2014 #7
    Could you please explain what you mean by:

     
  9. Oct 8, 2014 #8

    BvU

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    work out ##\left ({a\over \pi}\right )^{1\over 2}\;\int_{-\infty}^{+\infty}\;\left (e^{-ax^2/2}\right )\; \left (-{h^2\over 2\mu}{d^2\over dx^2}\right ) e^{-ax^2/2}\ dx \ ## first and see what you get.
     
  10. Oct 8, 2014 #9

    Simon Bridge

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    I got that, it's the next bit I wanted to direct you attention to.
    BvU is doing a decent job so I'll just reinforce and sit back for a bit:

    Which is not correct.
    The ##\hat K## operator has a second derivative in it - that means the ##\psi## to the right of it gets altered.
    You should have some course notes about how to apply operators to wavefunctions.
    $$\hat K \psi(x) = -\frac{\hbar^2}{2\mu}\left(\frac{d^2}{dx^2}\psi\right)$$

    The star refers to the complex conjugate.
    If ##\psi = a+ib## then ##\psi^\star = a-ib## (a,b both real and i2=-1)
    In your case, ##\psi## is real - but what I wrote is the formal definition and it's a step you missed out and it is important to your thinking.

    Aside: you are also reaching the stage where you will need to learn to use LaTeX ;)
     
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