How Do You Apply the Variational Method to a Quartic Oscillator?

  • Thread starter Thread starter gazepdapi1
  • Start date Start date
  • Tags Tags
    Method
Click For Summary
SUMMARY

The forum discussion focuses on applying the variational method to calculate the approximate ground state energy of a quartic oscillator with potential V(x) = c*x^4. A trial wavefunction of the form exp(-alpha*x^2/2) is utilized, and the Hamiltonian is defined as H = -h^2/2m(d^2)/(dx^2) + cx^4. Participants emphasize the importance of minimizing the energy expression E = integral((psi*)*Hamiltonian*psi)/integral((psi*)psi) with respect to the parameter 'alpha' to obtain the lowest upper limit for the ground state energy.

PREREQUISITES
  • Understanding of the variational method in quantum mechanics
  • Familiarity with Hamiltonian mechanics
  • Knowledge of trial wavefunctions and their applications
  • Proficiency in performing integrals, particularly in quantum mechanics contexts
NEXT STEPS
  • Study the derivation of the variational principle in quantum mechanics
  • Learn about the properties and applications of quartic oscillators
  • Explore standard integral tables relevant to quantum mechanics
  • Investigate the implications of dimensional analysis in quantum calculations
USEFUL FOR

Students and researchers in quantum mechanics, particularly those focusing on variational methods and energy calculations for quantum systems.

gazepdapi1
Messages
53
Reaction score
0
[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 that's it or am I missing something?

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

thanks
nertil
 
Physics news on Phys.org
what?? I can't 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.
 
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
 
nertil1 said:
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...
 
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 that's correct?
 
Last edited:
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.
 
Last edited:
I found my mistake.
thank you
 
  • #10
nertil1 said:
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 that's 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.
 

Similar threads

Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K