Bohr-Sommerfeld Quantization Rule for 3D Systems

[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:
<br /> \int \sqrt{2E - \frac{l^2}{r^2}+\frac{\alpha c \hbar}{r}} = n h<br />
But I don't know which the contour I should chose.

 

[ytuab]

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:
<br /> \int \sqrt{2E - \frac{l^2}{r^2}+\frac{\alpha c \hbar}{r}} = n h<br />
But I don't know which the contour I should chose.

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.

 

[ytuab]

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)