Scaling of an eigenvalue with the coupling constant

  • #1
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Main Question or Discussion Point

Consider the Hamiltonian ##H = - \frac{d^2}{dx^2}+gx^{2N}##.

Scaling out the coupling constant ##g##, the eigenvalues scale as ##\lambda \propto g^{\frac{2}{N+2}}##.

So, we can drop the g dependence and just consider the numerical value of the eigenvalues and the associated spectral functions at ##g=1##.



I understand that if the eigenvalues do scale as ##\lambda \propto g^{\frac{2}{N+2}}##, then the eigenvalues remain on the same order of magnitude for increasing values of N (as a power of g). As a result, the value of g makes little difference to the value of the eigenvalues. That, I understand.

What I don't understand though is why the eigenvalues scale as ##\lambda \propto g^{\frac{2}{N+2}}## in the first place. Could somebody pleas explain? :(
 

Answers and Replies

  • #2
DrDu
Science Advisor
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Try to introduce a new variable ##y=g^\alpha## and find a value of ##\alpha##, so that both the kinetic and the potential energy have the same pre-factor ##g^\beta##.
 
  • #3
Avodyne
Science Advisor
1,396
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I believe you meant ##y=g^\alpha x##.
 
  • #4
DrDu
Science Advisor
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Of course, thank you!
 

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