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Shift in energy due to weak interaction

  • Thread starter samee
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Homework Statement



A particle is in a box of length L on x-axis. 0<x<L. What is the first order shift due to V(x) of the wavefunction?

Homework Equations



H=p2/2m
Sin=sqrt(2/L)cos(npix/L)
En=n2pi2hbar2/2mL2
V(x)=kx2/2

The Attempt at a Solution



I'm super confused because the shift in the energy is usually determined by a sum, not an integral. Unless I calculate En(1) by taking;

En1=<n|H|n>
=<n|V|n>
=V * integral(2/Lcos2(nxpi/L)dx) with limits 0 to L

Is this the correct way to do this one? I didn't include k very well unless I put it in for V at the end...
 
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Answers and Replies

  • #2
vela
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You have to put V(x) inside the integral since it depends on x.
 
  • #3
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So I set up my integral correctly? Thank god... So now I have;

En(1)=(k/2)*(2/L)*integral(x2*cos2(npix/L) dx) with limits 0 and L.

=k/L * integral(x2*cos2(npix/L) dx) with limits 0 and L.

Is this what the question is asking for? Also- How exactly do I solve this integral by hand? My calculator is giving me;

k/L*{180*(360npi2x cos(npix/L)+(n2pi4x2-64800L2)sin(npix/L)}/n3pi6
 
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  • #4
vela
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Your wavefunctions aren't right. If one end of the box is at x=0, you should have sines, not cosines.

I assume you're studying perturbation theory. You should be able to tell me what you're calculating.
 
  • #5
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Yes, this is perturbation theory. I think that I'm finding the 1st order shift in the energy. The way the question asks though, "shift in energy of wavefunction due to k, to 1st order in k" sounds as if I should be shifting V(x) instead of the energy.

The question was written with cos in the wavefunction, but I can email my professor and ask if it's a typo... Or do you mean that instead of cos^2 in the integral I should have cos*sin?
 
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  • #6
vela
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In this problem, you're treating the spring potential as a perturbation. The shift in energy due to the perturbation can be expressed as a series in k. The problem is just asking you to find the first term in that series.

The solutions to the Schrodinger equation are linear combinations of sine and cosine. When you require that the wavefunction vanish at x=0, that forces the cosine term to disappear, so you're left with just sine.

I'd have guessed the intent was to have the well go from x=-L/2 to x=+L/2, so that the spring potential is centered in the well, but then you should have half the wavefunctions with cosine and the other half with sine.
 
  • #7
vela
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Also- How exactly do I solve this integral by hand? My calculator is giving me;

k/L*{180*(360npi2x cos(npix/L)+(n2pi4x2-64800L2)sin(npix/L)}/n3pi6
First, use the trig identity cos^2 x = (1+cos 2x)/2. You can then integrate the first term. The second term will require repeated application of integration by parts.
 
  • #8
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I'm not sure what to do now, that's how the question was written. I've emailed my prof. and hopefully will get a response in the morning explaining it. Sadly, this is due very soon. Is there no way to solve the problem as is? Or maybe I should say, were the wavefunction sine, then am I doing it correctly and will solving the integral be the only hurdle I need to jump?
 
  • #9
vela
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Yes, if the wavefunction was sine instead of cosine, it would be correct.
 
  • #10
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Ha! You were correct! The wavefunction IS sine. Thank you so much! You're a lifesaver.
 

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