Help with variatinal method

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[SOLVED] help with variatinal method

Use a trial wavefunction of the form exp(-alpha*x^2/2) to compute an approximate ground state energy for the quartic oscillator with potential V(x) = c*x^4 using the variational method.


Here is what I have so far. We need to use the equation
E = integral((psi*)*Hamiltonian*psi)/integral((psi*)psi)
First we have to use the right hamiltonian. I have H = -h^2/2m(d^2)/(dx^2) + cx^4

No my question is, do I just have to take the integral of top and botton and thats it or am I missing something?

(sorry for the bad equation, but i suck at LaTeX)

thanks
nertil
 

Answers and Replies

  • #4
malawi_glenn
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what?? I cant find anything with "variable a"..

With the varinational method you can with help of a test function get an upper limit of the ground state energy for any hamiltonian.

Then if your test function contains a parameter 'a' (for example), you can get the "lowest upper limit" by minimizing: <psi | H psi > / <psi|psi> , with ordinary calculus.
 
  • #5
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I am still having problems with this. can someone tell me step by step what I am supposed to do? I am having troble integrating this
 
  • #6
malawi_glenn
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I am still having problems with this. can someone tell me step by step what I am supposed to do? I am having troble integrating this

Write down (in some detail) what you have done/tried so far and someone will help you.

Remember that according to the rules of this forum you must supply your attempt before you can get help.

Start with this: 1)First apply your hamiltonian to the trial wave function, then multiply what you got with the complex conjugate of your trial wave funtion. 2)Then you integrate all this, preferly from -infinty to + infinity. Then in the denominator is straightforward.

Now try to do the first part. And the the second. You might want to use standard tables of integrals...
 
  • #7
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Okay I integrated top and bottom and then simplified
here is what i got
E = (hbar^2/4m)*(1-alpha/m) + 3c/4alpha^2

can someone check if thats correct?
 
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  • #8
Avodyne
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You should have got

E = (hbar^2 alpha/4m) + 3c/4alpha^2

Note that your factor of (1-alpha/m) cannot be correct, because alpha has dimensions of length^(-2), and m has dimensions of mass, so alpha/m is not dimensionless and cannot be subtracted from 1. Also, (hbar^2/4m) has dimensions of energy-length^2, not energy.
 
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  • #9
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I found my mistake.
thank you
 
  • #10
malawi_glenn
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Okay I integrated top and bottom and then simplified
here is what i got
E = (hbar^2/4m)*(1-alpha/m) + 3c/4alpha^2

can someone check if thats correct?


Show the procedure instead of just the answer. It is not meant that we should do the whole calculations either. So in the future, post your solution and ask if your procedure was correct.
 

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