# Scaling of an eigenvalue with the coupling constant

1. Feb 10, 2015

### spaghetti3451

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. Feb 11, 2015

### DrDu

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. Feb 11, 2015

### Avodyne

I believe you meant $y=g^\alpha x$.

4. Feb 11, 2015

### DrDu

Of course, thank you!