# On The Bohr Sommerfeld Model

One of the amazing classical models out there is the Bohr Sommerfeld Model

It's basically a classical relativistic model for a point electron assuming quanta-sized momenta of $$n h$$. Oh well, you may say, what's about it. Well the energy values that are produced are exactly those of Dirac's QED except for a spin contribution. Pretty amazing right. The solution is ellipsoidal of the form

$$1/r = K/w^2 + A\cos(w\phi)$$

And the solution satisfies the equation

$$u'' = -w^2u + K,$$

with $$w,K$$ constants s.t

$$w^2 = 1-\frac{k^2 Z^2 e^4}{c^2 n^2 \hbar^2}$$
$$K = \frac{m k Z e^2}{\hbar^2 n^2}\Big (1 + \frac{W}{mc^2} \Big )$$
$$W = \Big (1+\frac{\alpha^2 Z^2}{(n_r + \sqrt{n^2_{\phi}-\alpha^2Z^2})^2} \Big ) - 1,$$

with $$n=n_{\phi}$$.

The interesting thing now is that we can scale the solution by $$m\to m/N$$ and consider an ensemble of such solutions and we will have exactly the same total energy equation and governing equation as before.

This means that we can consider the sum (or a Stiltjes integral together with a probability measure $$\mu$$,

$$\langle u \rangle = \sum_{k=1} \mu_k u_k(\phi) = \int u(y,\phi)\mu(dy).$$

Now, assuming that this results in a uniform solution that does not depend on $$x$$, and we get that the distribution $$U$$ for all $$u$$ that hits at an angle $$x$$ as,

$$U = K/w^2 + A\cos(2\pi R)$$

with $$R\in Re[0,1]$$ e.g. $$R$$ is the uniform distribution on $$[0,1]$$.

But this means that the the total potential energy is

$$V = k U$$

Let's try to deduce the enrgy values of Bohr Sommerfeld. We start by the solution of the radius as,

$$u(\phi) = 1/r(\phi) = K + A \cos(wt),$$

with,

$$K = (\hbar Z \alpha /p_\phi)\frac{E}{mc}/w^2,$$

and,

$$w^2 = 1+ \Big ( \frac{\hbar}{Z \alpha p_\phi} \Big )^2$$

From the quantization relationship,

$$m v_\phi h = \hbar n_\phi$$

We se that

$$p_\phi = m v_\phi = \frac{\hbar n_\phi}{r} = \hbar n_\phi u.$$

So,

$$p_\phi = \hbar n_\phi(K+A\cos(w\phi)).$$

From the energy relation,

$$p_1^2c^2 + p_2^2c^2 + (m_ec^2)^2 = (E-V)^2$$

we note that r.h.s. has an angular dependency of,

$$B_1+\cos(w\phi)B_2 +\cos(w\phi)^2B_3.$$

From

$$p_\phi = p_2 r = \hbar n_{\phi},$$

we note that the angular part of the momentum operator is,

$$p_2 = p_\phi / r = p_\phi u$$

$$p_1 = C \sin(w t).$$

So.

$$p_1^2 = C^2\sin(w\phi)^2 = C^2(1-\cos(w\phi)^2).$$

Now, note that we have,

$$D K = \frac{1-w^2}{w^2}E,$$

with,

$$D = kZe^2$$

so we can reparametize $$A$$ so that,

$$u = \frac{1-w^2}{Dw^2}E(1+A\cos(w\phi)).$$

And also we have (the electric potential is negative),

$$E-V = E(1/w^2 + \frac{1-w^2}{w^2}A\cos(w\phi))$$

combined term for $$p_2$$ becomes,

$$c\frac{(1-w^2)^2}{w^4}.$$

But,

$$(p_\phi c/D)^2(1-w^2) = 1$$

Hence the setup is consistent for the combined term without actuallt specifying any of the unknowns. Let,

$$C = \frac{1-w^2}{w^2}EAC'.$$

Then another reparametrisation yields Comparing all $$\cos^2$$ terms,

$$\hbar^2 C''^2 = p^2 - D^2/c^2$$

With,

$$C'' = \sqrt{n_\phi^2 - (Z\alpha)^2},$$

and,

$$C' = C'' \hbar c / D.$$

Finally for the constant term we have,

$$C^2c^2 + p^2c^2 \frac{(1-w^2)^2}{w^4D^2} + (mc^2)^2= \frac{E^2}{w^4}$$

or

$$A^2C''^2E^2 + n_p^2 + (mc^2)^2 = E^2 \frac{D^2}{\hbar^2c^2(1-w^2)^2}$$

But this is the same as,

$$A^2C''^2E^2 + n_p^2 + (mc^2)^2n_p^2w^4 = E^2n_p^2$$

Now consider what happens when we do the scaling and consider the meassure version above, then we keep the parameters as is then $$A$$ has to tend to $$0$$ as $$N$$ tend to infinity e.g. the ensemble approaches a circle. with the quantization,

$$mvr = \hbar n_\phi.$$

But we will get another term for the radius term e.g.

$$\frac{1}{2\pi}A \int r^2\sin(w \phi)^2 = Ar^2/2 = n$$

This gives in the limit,

$$\lim_{N \to \infty} (2N-1)A = 2 \hbar n_r / r^2$$

And this can be used for further analysis.