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.

  • $\begingroup$ Hi again. There's a straight bracket missing in your $E_n$ equation. So it's definitely $(\frac{1}{\sqrt{2mB}})^\frac{2\gamma}{\gamma+2}$. Nice find, that page! Gotta love these ingenuous substitutions... :-) $\endgroup$
    – Gert
    Commented Oct 13, 2016 at 22:51
  • $\begingroup$ Heya. So judging from how the energies scale with mass according to $\gamma$, how could I work backwards to find the properly scaled TISE? $\endgroup$
    – bleuofblue
    Commented Oct 13, 2016 at 22:58
  • $\begingroup$ Knowing $E_n = (\frac{1}{2mB})^\frac{2\gamma}{\gamma+2}$, what would this say about my constants $\beta, \alpha$? In other words, is there a way to discover how to determine the result for how $E_n$ scales with $m$? $\endgroup$
    – bleuofblue
    Commented Oct 13, 2016 at 23:01
  • $\begingroup$ Sorry, don't really know enough about that technique to give a qualified answer. Look forward to reading some actual answers though. +1 for good question. $\endgroup$
    – Gert
    Commented Oct 13, 2016 at 23:05

1 Answer 1


Just take the initial scaling constant as arbitrary at first.

Setting $u = \chi x$ gives $\frac{d^2}{dx^2} = \chi^2 \frac{d^2}{du^2}$ and the TISE as $$ \chi^2 \frac{d^2\Psi}{du^2} + \left[ \beta - \frac{\alpha}{\chi^\gamma} u^\gamma \right]\Psi = 0 $$ Isolating the term in $u^\gamma$ obtains $$ \frac{\chi^{2+\gamma}}{\alpha}\frac{d^2\Psi}{du^2} + \left[ \frac{\beta\chi^\gamma}{\alpha} - u^\gamma \right]\Psi = 0 $$ But since $\chi$ is arbitrary, we can always set the factor of the second derivative to unit, $$ \frac{\chi^{2+\gamma}}{\alpha} = 1 \;\;\;\;\; \Rightarrow \;\;\;\;\; \frac{\beta\chi^\gamma}{\alpha} = \beta \alpha^{-\frac{2}{2+\gamma}} $$ which gives the scaled TISE in the form $$ \frac{d^2\Psi}{du^2} + \left[\beta \alpha^{-\frac{2}{2+\gamma}} - u^\gamma \right]\Psi = 0 $$

  • $\begingroup$ Ah very awesome. The arbitrary constant threw me off at first, but I understand this now. Thank you. From the scaled TISE, it appears that $E \varpropto \frac{(2mB)^\frac{2}{2+\gamma}}{m}$, which checks out for the harmonic oscillator case. Does this seem correct? $\endgroup$
    – bleuofblue
    Commented Oct 14, 2016 at 13:14
  • $\begingroup$ Welcome, and yes, it looks right to me. $\endgroup$
    – udrv
    Commented Oct 14, 2016 at 16:52

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