# Solving Schrödinger's equation for a specific potential

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!

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Your problem is this: you're assuming a form of the answer that is incorrect. Plot the original potential and notice that it's still parabolic like the regular harmonic oscillator, but the minimum is no longer at $x=0$. But eventually, you make the Ansatz $\chi(\epsilon) = f(\epsilon)\exp[-\epsilon^2/2]$ and then try to show that $f(\epsilon)$ is a power series that terminates, i.e. a polynomial. This isn't correct. You essentially bake into the cake the assumption that you have (Hermite) polynomials times a Gaussian centered at $\epsilon=0$ when you factor $\chi$ that way you do. Try instead $\chi(\epsilon) = f(\epsilon-\epsilon_0)\exp[-(\epsilon-\epsilon_0)^2/2]$ for some judiciously chosen $\epsilon_0$. Or equivalently, when you rescale your $x$ coordinate to $\epsilon$, also make a shift, $\epsilon = Ax+B$ to put the minimum of your potential at $\epsilon = 0$.