# Solving $-\chi''(\epsilon)+\Big[\epsilon^2+\frac{2F}{hw}\sqrt{\frac{h}{hw}}\epsilon \Big]\chi(\epsilon)=\mu\chi(\epsilon)$

I am trying to solve this differential equation:

$$-\chi''(\epsilon)+\Big[\epsilon^2+\frac{2F}{hw}\sqrt{\frac{h}{hw}}\epsilon \Big]\chi(\epsilon)=\mu\chi(\epsilon) \tag1$$

This was found using Schrödinger's equation, modelling a particle in potential $V(x)=\frac{1}{2}mw^2x^2+Fx$. I have changed variables and got to this point.

Here I then look for solutions in the form

$$\chi(\epsilon)=f(\epsilon)e^{-\epsilon^2/2}$$

I then substiution into $(1)$ giving $$f''-2\epsilon f'+f(\mu-1-\frac{2F\epsilon}{w\sqrt{hwm}})=0$$

Here I seek a power series solution for $f$. This gives a recurrence relation for the terms in the power series. This power series must terminate for the solution to make physical sense. If it terminates there exists a $n$ such that $a_n=0$. We can use this to get information about $\mu$, as this term describes the energy levels of the particles, this is the goal.

However the recurrence relation I get is:

$$a_{k+3}=\frac{a_{k+1}(2k+3-\mu)+\frac{2Fa_k}{w\sqrt{hwm}}}{(k+3)(k+2)}$$

If we set this equal to zero the idea is to eliminate the remaining terms in the series, as they are non-zero, to leave useful information. I can't see how to do this! any help would be greatly appreciated!

There is a simpler way to manage this. Consider the potential you start from

$$V(x)=\frac{1}{2}m\omega^2x^2+Fx=\frac{1}{2}m\omega^2\left(x^2+\frac{2F}{m\omega^2}x\right).$$

This is just

$$V(x)=\frac{1}{2}m\omega^2\left(x+\frac{F}{m\omega^2}\right)^2-\frac{1}{2}\frac{F^2}{m\omega^2}$$

and this is just the potential of the harmonic oscillator properly shifted by $x_0=\frac{F}{m\omega^2}$. Then , the solution of the Schroedinger equation is just

$$\psi_n(x,t)=C H_n\left(\sqrt{\frac{m\omega}{\hbar}}(x+x_0)\right)e^{-\frac{m\omega}{2\hbar}(x+x_0)^2}$$

with $H_n$ the Hermite polynomials and the eigenvalues are

$$E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{F^2}{2m\omega^2}.$$

This is the way you have to manage your equation (1).

Further clarifications as asked by OP:

So, let us consider the Schroedinger equation

$$-\frac{\hbar^2}{2m}\frac{d^2\psi_n}{dx^2}+\left(\frac{1}{2}m\omega^2x^2+Fx\right)\psi_n=E_n\psi_n.$$

Rewriting of the potential in the way given above one has

$$-\frac{\hbar^2}{2m}\frac{d^2\psi_n}{dx^2}+\left[\frac{1}{2}m\omega^2\left(x+\frac{F}{m\omega^2}\right)^2-\frac{1}{2}\frac{F^2}{m\omega^2}\right]\psi_n=E_n\psi_n.$$

We move the constant term on the rhs and put $E'_n=E_n+\frac{1}{2}\frac{F^2}{m\omega^2}$ and then introduce the new coordinate $\xi=x+x_0$. You will get

$$-\frac{\hbar^2}{2m}\frac{d^2\psi_n}{d\xi^2}+\frac{1}{2}m\omega^2\xi^2\psi_n=E'_n\psi_n.$$

This is the standard form for the equation of a harmonic oscillator. Now you can change the variable to $\epsilon=\sqrt{\frac{m\omega}{\hbar}}\xi$ and $\mu_n=\frac{2E'_n}{\hbar\omega}$ and you are left with your equation in a more manageable form.

• Thank you for this, I very much appreciate a theoretical physicist's input. Would you think that in principle my method is correct? Feb 12, 2012 at 16:11
• @LHS: Yes, of course. But you are in trouble to uncover the correct solution using a series like yours as the powers of the shifted polynomials will mix up with the powers of the expansion of the exponential and all is messed up. The approach I showed you is indeed a standard one in this case.
– Jon
Feb 12, 2012 at 16:18
• Ah thank you, do you think you could elaborate how you got to your wave function? What effect does adding a constant to a potential have on the wave function, as I can see why adding a constant to the x value would shift it. But the $-x_0/2$ term appears to have not affected your wave function. Many thanks. Feb 12, 2012 at 19:14