# Sommerfeld quantisation:relation between energy and potential

1. Feb 22, 2010

### krishna mohan

I found the following question in a question paper.

If the potential experienced by a particle in one dimension is given by $$ax^4$$, how does the energy of the particle depend on the parameter a? Use Bohr-Sommerfeld quantisation formula..

I know how to find the dependence of energy on n, the integer that occurs in the quantisation.

But how does it depend on a?

Since
Hamiltonian is $$\ K+V$$ ,
Energy is $$\ K+ax^4$$.

I dont know what else I can say.

2. Feb 22, 2010

### ytuab

For example, in the harmonic oscillator, the potential V is,
$$V = ax^2 = \frac{1}{2}m \omega^{2}x^2$$
So, the total energy is,
$$E= K+V = \frac{p_{x}^2}{2m}+\frac{1}{2}m \omega^{2}x^2(=ax^2)$$

As shown in this thread, The Bohr-Sommerfeld quantization formula is, (in a quarter period)
$$\oint p_{x}dx = nh \to \sqrt{2m}\int_{0}^{\sqrt{\frac{E}{a}}}\sqrt{E-ax^2}dx = \frac{1}{4}nh$$

change the variable to $$\theta$$ as follows,
$$\sqrt{\frac{a}{E}}x=sin\theta, \quad E\sqrt{\frac{2m}{a}}\int_{0}^{\frac{\pi}{2}}cos^2\theta d\theta =\frac{\sqrt{2m}E\pi}{4\sqrt{a}}=\frac{1}{4}nh$$

So if we erase the a, the the total energy E is,
$$E=\frac{nh\sqrt{a}}{\sqrt{2m}\pi}=\frac{nh\omega}{2\pi}=n\hbar\omega$$

This result shows the interval of the energy E is $$\hbar\omega$$

But as shown in the ground state electron of the Bohr hydrogen model, the electron has the orbital angular momentum 1 (x $$\hbar$$) instead of the spin 1/2. So we have to consider the minimum rotation (not 1-D) in the Bohr model.

This magnetic moment which can be measured by the experiment is the same as the QM hydrogen model because the magnetic moments(angular momentum x g-factor) are 1 x 1 (in the Bor model) = 1/2 x 2 (in the QM model).
If the angular momentum is $$\hbar$$, the electron naturally return by one rotation as follows, (also in the result of this experiment)

$$L_{z}(=-i\hbar\frac{d}{d\varphi})\phi_{m}(\varphi) = m\hbar\phi_{m}(\varphi), \to \quad \phi_{m}(\varphi)=e^{im\varphi}$$

Sorry, the case of $$V=ax^4$$ is a little complicated, so I'm glad if you calculate like this, and tell me the result.

3. Feb 27, 2010

### krishna mohan

My question was not about the dependence on the quantum number n obtained from the quantisation condition....

It was about the dependence of the energy on the parameter a in the Hamiltonian..that is the potential energy being given as $$a x^4$$...

As for dependence on n, it can be easily accomplished..see this video...at around the thirty minute mark...

Last edited by a moderator: Sep 25, 2014
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