2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
    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
    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
    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
    Oct 13, 2016 at 23:05

1 Answer 1

1
$\begingroup$

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 $$ \frac{d^2\Psi}{du^2} + \left[\beta \alpha^{-\frac{2}{2+\gamma}} - u^\gamma \right]\Psi = 0 $$

$\endgroup$
2
  • $\begingroup$ 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
    Oct 14, 2016 at 13:14
  • $\begingroup$ Welcome, and yes, it looks right to me. $\endgroup$
    – udrv
    Oct 14, 2016 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.