# Bohr Quantization with linear potential

1. Jan 27, 2015

### andre220

1. The problem statement, all variables and given/known data
Using Bohr's quantization rule find the energy levels for a particle in the potential: $$U(x) = \alpha\left|x\right|, \alpha > 0.$$

2. Relevant equations
$\oint p\, dx = 2\pi\hbar (n + \frac{1}{2})$

3. The attempt at a solution
Okay so:
$\begin{eqnarray} \oint p\, dx &= \int \sqrt{2m(E-U(x))}\,dx\\ &= \int\limits_{-\infty}^{+\infty} \sqrt{2m(E-\alpha\left|x\right|)}\,dx\\ &= 2\pi\hbar (n+\frac{1}{2} \end{eqnarray}$
So far, I believe this is correct, but the integral doesn't converge so either im missing something or I've done something wrong. I can't seem to see what it is. Any help is greatly appreciated.

2. Jan 27, 2015

### Goddar

There's a turning point where the square-root becomes imaginary, that's your upper limit...

3. Jan 27, 2015

### andre220

Right I see, so I get the two turning points at $$\pm \sqrt{\frac{E^2}{\alpha^2}}$$ and now its just a matter of evalutating the integral.

4. Jan 27, 2015

### Goddar

Yes, and the integral is easy to carry out.
Also, note that it is symmetric about 0 so you can dispose of the absolute value and take twice the integral from 0 to the turning point.

5. Jan 27, 2015

### andre220

Thank you for your help. Got an answer of $$E_n = \alpha\left(\frac{3}{2}\frac{\pi\hbar}{\sqrt{2m}}(n+\frac{1}{2})\right)^{2/3}.$$ which seems okay.

6. Jan 27, 2015

### Goddar

No problem... But i think you want to put this α inside the parenthesis ;)