# Unidimensional potential and Schroedinger

1. Nov 21, 2009

### Matthew888

1. The problem statement, all variables and given/known data

Consider a particle with charge $e$ which moves in $x$ axys (in particular in the positive region) with the potential $V(x)=-\alpha/x$ ($\alpha>0$). Find the ground state energy.

2. Relevant equations

I think that in some way I have to solve this ODE:

\$-\bar{h}^2/(2*m)*(d^2\psi)/(dx^2)+(\alpha/x+E)\psi=0

3. The attempt at a solution
I don't know how to solve this awful ODE. May you give me an hint?

2. Nov 21, 2009

### jdwood983

Welcome to Physics Forums Matt!

$$-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2}=-\left(\alpha x+E\right)\psi$$

is an awful ODE? I think you might be confusing yourself. Is this an awful ODE:

$$\frac{d^2y}{dx^2}=\kappa^2 y$$

EDIT: Actually, your ODE is wrong, it should be

$$-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2}+\alpha x\psi=E\psi\Longrightarrow-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2}+\left(\alpha x-E\right)\psi=0$$

Last edited: Nov 21, 2009
3. Nov 21, 2009

### Matthew888

I am sorry for my mistakes.
The potential is in the form $$V(x)=\frac{-\alpha}{x}$$, where $$\alpha$$ is positive. The ODE should be:

$$\frac{-\bar{h}^2}{2m}\frac{d^2\psi}{dx^2}-\frac{\alpha}{x}\psi=E\psi$$

I know that ODEs like $$y''(t)+ky(t)=0$$ are very simple, but this happens if $$k$$ is a real number.

I think that I can solve this ODE in the same way as it is usually done for the hydrogen atom. Is this correct?

4. Nov 21, 2009

### jdwood983

Ah, that does certainly change the problem, but not by much. You are looking at a central potential (in 1D) and usually these potentials are solved as functions of $V(x)$. So, from all the textbooks I've worked through on central potentials, you solve it as

$$-\frac{\hbar^2}{2m}\,\frac{d^2\psi}{d x^2}+V(x)\psi(x)=E\psi(x)\longrightarrow\frac{d^2\psi}{dx^2}=\kappa^2\psi$$

where $$\kappa=\sqrt{2m\hbar^{-2}\left(V(x)-E\right)}$$.

5. Nov 22, 2009

### Matthew888

I am pretty sure that I have to find a set of solutions in function of $$x$$. So the question still remains

6. Nov 22, 2009

### jdwood983

Well to find the energy eigenvalues (and thus the ground state energy), we'd need some boundary conditions. Where there any given in the problem? If there weren't any BC's, then you'll probably need the WKB approximation

7. Nov 22, 2009

### Matthew888

I don't think that this problem is so difficult (I don't even know what WKB approximation is ).
I finally managed to solve this ODE: I found the ground state autofunction

$$\psi_1=\frac{2\sqrt{a}x}{a}\exp{-\frac{x}{a}}$$, where $$a=\frac{4\bar{h}^2}{me^2}$$ (I put $$\alpha=\frac{e^2}{4}$$, where $$e$$ is the charge of electron). Finding the fundamental energy and <x> is straightward.

8. Nov 22, 2009

### jdwood983

Actually, now that I've had a nights sleep on this problem, you probably could find (estimate) the ground state energy a lot more easily by using the Heisenberg uncertainty principle:

Since your Hamiltonian takes the form

$$H=\frac{p^2}{2m}-\frac{\alpha}{x}$$

then we can say that the ground state energy can be estimated via

$$E_0=\langle\frac{p^2}{2m}\rangle-\langle\frac{\alpha}{x}\rangle$$

but through the uncertainty in $p$, we can say $\langle p^2\rangle=(\Delta p)^2\geq\hbar^2/4a^2$ where $a$ is the characteristic length of the atom, such that $\langle \alpha/x\rangle=\alpha/a$. The ground state energy can then be written as

$$E_0\geq\frac{\hbar^2}{8ma^2}-\frac{\alpha}{a}$$

to find $a$, we take the derivative of $E_0$ with respect to $a$ and set it equal to 0. You'll take this value of $a$, place it into the above equation and reduce it as much as possible and this would be your estimated ground state energy (usually close within a factor or two of $\pi^2$). This type of ground state energy estimation can be done in this manner for any central potential.

9. Nov 22, 2009

### Matthew888

The way you solved the problem is pretty interesting.
Thank you for the attention!