# Old quantum mechanics

1. Dec 23, 2009

### paweld

"Old" quantum mechanics

There is famous Bohr Sommerfeld quantization rule which says that $$\oint p dx = n h$$. Is it possible to apply this rule to three dimensonal system e.g. hydrogen atom.
The integral will probably look as follow:
$$\int \sqrt{2E - \frac{l^2}{r^2}+\frac{\alpha c \hbar}{r}} = n h$$
But I don't know which the contour I should chose.

2. Dec 23, 2009

### ytuab

Re: "Old" quantum mechanics

In the hydrogen of the Bohr-Sommerfeld model, the orbital becomes elliptical or circular.
So simply I use the two-dimensonal system.
The orbital length becomes a integer times the de Broglie's wavelength ($$\lambda =h/mv =h/p$$).

For example, when the orbital is elliptical, we divide the electron's movement into the two directions at short time intervals.

The $$\perp$$ is the direction of the angular momentum(tangential), and the $$r$$ is the radial direction.
The two directions are rectangular at each point.

$$\oint p_{\perp} dq_{\perp} = n_{1}h, \qquad \oint p_{r} dq_{r} = n_{2}h$$

At each point, the de Broglie's wavelengths are $$\lambda_{\perp}= h/p_{\perp}, \lambda_{r}= h/p_{r}$$
The wavelengths are changing at each point. So, the above equations are equal to,

$$\oint dq_{\perp}/\lambda_{\perp} = n_{1}, \qquad \oint dq_{r}/\lambda_{r}= n_{2}$$

This means that in the Bohr-Sommerfeld model, the sum of the number of the de Broglie's waves contained in each short segment becomes a interger.

In the case of the three-dimensional system or more complex system, we had better use the computer or something, I think.

Last edited: Dec 23, 2009
3. Dec 25, 2009

### ytuab

Re: "Old" quantum mechanics

Sorry. I should have added one more thing in the above statement(#2).

I divided the electron's motion into the two directions which are rectangular at each point on the elliptical orbital.

But actually the electron is moving in one direction.
And its momentum $$p$$ satisfies $$p^2 = p_{\perp}^2 + p_{r}^2$$ at each point.

So in a short time($$dt$$), the electron moves $$dq=\frac{p}{m} dt$$.
Of course this satisfies,

$$dq^2 = dq_{\perp}^2 + dq_{r}^2$$

The number of de Broglie's waves ($$\lambda$$ in length) contained in the short segment($$dq$$) is,

$$dq/\lambda=dq/(\frac{h}{p})=p dq /h = p^2dt /hm$$

As I said above, due to $$p^2 = p_{\perp}^2 + p_{r}^2$$, we arrive at the following relation,

$$dq /\lambda = dq_{\perp}/\lambda_{\perp} + dq_{r}/\lambda_{r}$$

As a result, the number of de Broglie's waves contained in one-round of the orbital satisfies,

$$n = n_{1} + n_{2}$$

For example, in the energy level (n=2) of the hydrogen(the Bohr model), there are two patterns as follows,
n1=1, n2=1 ----- elliptical (angular momentum=1)
n1=2, n2=0 ----- circular (angular momentum =2)