# Bohr quantization hypothesis [closed]

To explain Rydberg formula, Bohr have assumed somewhat general hypothesis which is applicable to various classical system. As far as I know he assumed that for any classical system with periodic motion, $$\oint p_i dq_i = n_i h \qquad (1)$$ where $q_i$s are generalized coordinates and the integral is taken over full period of the coordinate. I have tried to recover the quantization condition $mvr = n \hbar$ for a circular motion with constant velocity. Pick any potential with rotational symmetry and assume that the particle is in a circular motion at radius $r$. Let $v$ be the speed of the particle. Then $$\oint p_1 dx = \oint m \dot{x}^2 dt = \oint mv^2 \sin^2 \omega t dt = \frac{1}{2} mv^2 T$$ where $T = \frac{2 \pi r}{v}$. Substituting yields $\pi r mv = n_1h$ or $$rmv = 2n_1 \hbar$$ This is not only contradicts to the known result $rmv=n \hbar$ but also is an absurdity by itself since, we can also perform the same calculation for $i=2$ where $q_2 = y$ and obtain $rmv = 2n_2 \hbar$. So it is apparent that I have misunderstood the Bohr quantization hypothesis.

As appeared in the answer, if we regard the index on the right hand side are under influence of summation convention, then problem vanishes for this case since $\oint \sum p_i \frac{dq_i}{dt} dt = \oint \sum mv_i^2 dt = mv \cdot 2\pi r$. However, this interpretation does not fit into the equation (1) since then the index on the left hand side is dummy but that in the right hand side is not. So how can I understand equation (1)?

• Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". – Kyle Kanos Jul 24 '18 at 10:13
• @seoneo You are asking for someone to check where there's a mistake in that proof, which is off-topic according to the meta post linked by KyleKanos. The homework-and-exercise tag is for questions involving a specific algebraic or numerical computation, and is hence applicable to this question. To answer your second comment, see this meta post. But relax a bit, it's not too hard to fix questions according to the guidelines set by Physics SE. – user191954 Jul 24 '18 at 13:08
• Now I'm done and satisfied. I don't care if this question is erased. I'll not edit my question because it is in its right form which should be. However, it was hard to recognize the necessity of separable condition. I hope that this question may be useful someone like me as it really was for some of my student. The student was also satisfied with the my explanation below. – seoneo Jul 26 '18 at 16:17

One may regard the Bohr-Sommerfeld quantisation as the quantisation of the action. We can see that the action is

$$S=\int dt L = \int dt \frac{p^2}{2m} = \frac{1}{2m}\int (\mathbf{p}dt)\cdot\mathbf{p} = \frac{1}{2m}\int d\mathbf{q} \cdot\mathbf{p}$$

Therefore, your quantisation condition is not quite correct. When you write $$\oint p_i dq_i$$

The Einstein summation must be applied (Note that we write $\oint$ instead of $\int$ here because we quantise the action of a closed orbit).

Hence, the correct form of the original Bohr-Sommerfeld quantisation is

$$\oint \mathbf{p} \cdot d \mathbf{q} = nh$$

By applying this equation to the circular orbit with a spherical symmetry potential, you should be able to obtain the right answer.

• Thanks for your attention. In his book Quantum Mechanics, page 2, Merzbacher wrote "... the classical phase integrals for a conditionally periodic motion were required to be quantized according to $$\oint p_i dq_i = n_i h$$ where the quantum number $n_i$ are integers, and each contour integral is taken over the full period of the generalized coordinate q_i." So it looks like that the equation is given for each $i=1,2,\ldots$. Am I misunderstanding? – seoneo Jul 23 '18 at 12:35

For the Bohr quantization makes sense, the generalized coordinates should be separable. In the example of the question, the variables $x$, $y$ are not separable. Hence weirdo results are obtained. The right choice for the problem is $r$ and $\theta$. Then $\oint p_\theta d \theta = 2\pi L = n_\theta h$ so $L = n_\theta \hbar$ as desired.