- #1
spaghetti3451
- 1,344
- 33
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? :(
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? :(