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Scaling of an eigenvalue with the coupling constant

  1. Feb 10, 2015 #1
    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? :(
     
  2. jcsd
  3. Feb 11, 2015 #2

    DrDu

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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##.
     
  4. Feb 11, 2015 #3

    Avodyne

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    I believe you meant ##y=g^\alpha x##.
     
  5. Feb 11, 2015 #4

    DrDu

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    Of course, thank you!
     
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