What seems like a rather simple question is causing me a lot of difficulty as my base in mathematics is weak.
I want to know how I would scale the Schrodinger equation to find dependence on mass, $m$, given a potential $V(x) = Bx^\gamma$, where $\gamma$ is any even integer (2,4,6,8.....). It is not necessary to solve the entire equation for allowed energies.
I know how to do this for the harmonic oscillator,$V(x) = \frac{1}{2}kx^2$ where we have
$\frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2}+\frac{1}{2}kx^2\Psi = E\Psi$
$\frac{d^2\Psi}{dx^2} + (\frac{2mE}{\hbar^2}-\frac{mkx^2}{\hbar^2})\Psi = 0$
Let $\beta = \frac{2mE}{\hbar^2}$ and $\alpha^2 = \frac{mk}{\hbar^2}$
If we let $x = \frac{u}{\sqrt{\alpha}}$, then, $\alpha\frac{d^2}{du^2} = \frac{d^2}{dx^2}$
And the Schrodinger equation we have in terms of $u$
$\alpha\frac{d^2\Psi}{du^2}+(\beta - \alpha u^2)\Psi=0$
dividing by $\alpha$ gives us
$\frac{d^2\Psi}{du^2}+(\frac{\beta}{\alpha} - u^2)\Psi=0$
from which we can derive the wavefunctions and allowed energies. We end up with
$\frac{\beta}{\alpha} = \frac{2mE}{\hbar\sqrt{mk}} = \frac{2E}{\hbar}\sqrt{\frac{m}{k}}$ if $m,k>0$
where we can discover that energy is proportional to $\frac{1}{\sqrt{m}}$
Here it is easy to put things into better perspective as we know the solution for the harmonic oscillator.
Setting $\frac{2E}{\hbar}\sqrt{\frac{m}{k}} = (2n+1)$, found from the Hermite polynomials, we solve for $E$ as
$E_n = (n+\frac{1}{2})\hbar\sqrt{\frac{k}{m}}$, where it is clear that $E_n \varpropto \frac{1}{\sqrt{m}}$
Now, for the case where $V(x) = Bx^\gamma$, I get the TISE as
$\frac{d^2\Psi}{dx^2} + (\frac{2mE}{\hbar^2}-\frac{2mBx^\gamma}{\hbar^2})\Psi = 0$
Now, letting $\beta = \frac{2mE}{\hbar^2}$ and $\alpha = \frac{2mB}{\hbar^2}$
$\frac{d^2\Psi}{dx^2}+(\beta - \alpha x^\gamma)\Psi=0$
Here is where I am stuck. I think I need to rewrite the TISE in terms of $u$ in order to isolate it and have the differential equation depend on only one constant: some ratio of $\beta$ and $\alpha$.
I'm not sure how to create a meaningful substitution to simplify this case. For the harmonic oscillator, it was $\sqrt{\alpha}x = u$, which clearly succeeds in isolating $u$,but I'm not sure how I could find this expression easily and for the potential given.
When I look at solutions for the eigen energies of this potential function, (http://www.physicspages.com/2014/07/21/wkb-approximation-and-the-power-law-potential/), it shows
$E_n = B[(n-\frac{1}{2})\hbar\frac{\Gamma(\frac{1}{\gamma}+\frac{3}{2})}{\Gamma(\frac{1}{\gamma}+1)}\frac{\sqrt{\pi}}{\sqrt{2mB}}]^\frac{2\gamma}{\gamma+2}$
Does this indicate that $E_n \varpropto \frac{1}{\sqrt{2mB}}$ or $(\frac{1}{\sqrt{2mB}})^\frac{2\gamma}{\gamma+2}$?
Again, I don't need to solve for the eigen energies, simply scale the time independent schrodinger equation in a similar fashion to the harmonic oscillator in order to determine just the mass dependence of the energy eigenvalues.