Anharmonic Oscillator - Energy Shift Calculation Using 1st Order Perturbation

  • Thread starter Hart
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  • #26
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[tex].. = 3\sigma^{2} [/tex]
 
  • #27
vela
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Yup!
 
  • #28
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Yay! :approve:

So just to check, that is the result for the energy shift of the ground state?

i.e. [tex]E_{10}^{'} = 3\sigma[/tex]

If so, where does this factor in (defined at the start):

[tex]E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda[/tex]

I seem to have two values for the same thing :(
 
  • #29
vela
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Yay! :approve:

So just to check, that is the result for the energy shift of the ground state?

i.e. [tex]E_{10}^{'} = 3\sigma[/tex]
No. Go back and look at what you were trying to calculate. The result you got earlier was just part of the calculation. Also, I assume you had a typo here. If not, you're doing something I don't understand again.
If so, where does this factor in (defined at the start):

[tex]E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda[/tex]

I seem to have two values for the same thing :(
What is [itex]\sigma[/itex] defined as?

I think at this point it would help you to go back over this problem and write up what you have so far cleanly. Get rid of the false starts and the dead ends. Whittle it down the solution to its essentials. Think about how you'd explain how to solve this problem to someone else.
 
  • #30
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Just substituted:

[tex]\sigma = \frac{\hbar}{2m\omega}[/tex]

Into:

[tex]E_{10} = 3\sigma[/tex]

Which becomes the previously defined statement:

[tex]E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda[/tex]
 

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