Can Linear Momentum Be Quantized Like Angular Momentum?

[wakko101]

This question deals with the Wilson-Sommerfelt quantization rule:

(integral) Pdq = nh

The first part of my question asks you to derive Bohr's quantization condition from this, with the hint that angular moment does not depend on the angle. That part I got.

The second part asks, in terms of the situation of a particle in a box moving back and forth with no other forces acting upon it, that we do the same for linear momentum, so that we're integrating with respect to dx. The thing I'm confused about is, although velocity is taken in terms of dt, can we take v out of the integral that's being calculated wrt dx? In which case, the answer would be:

2mva=hn.

Cheers,
Lauren. =)

 

[Jhero]

Dear Lauren,

Thank you for your question regarding the Wilson-Sommerfelt quantization rule and its application to Bohr's quantization condition and the case of a particle in a box. I will address each part of your question separately.

1. Deriving Bohr's quantization condition:

As you correctly noted, the Wilson-Sommerfelt quantization rule states that (integral) Pdq = nh. This means that the integral of momentum (P) with respect to position (q) is equal to an integer (n) multiplied by Planck's constant (h). We can rearrange this equation to solve for the momentum: P = nh/dq.

Now, in the case of Bohr's quantization condition, we are dealing with the angular momentum (L) of an electron orbiting the nucleus in an atom. The hint given in the question is that angular momentum does not depend on the angle. This means that the angular momentum is a constant, and therefore, its derivative with respect to time (dL/dt) is equal to zero. We can use this fact to rewrite the equation for momentum as: P = L/dt.

Substituting this into the Wilson-Sommerfelt quantization rule, we get: L/dt = nh/dq. Rearranging this equation, we get: Ldq = nhdt. This is the same equation as the one given for Bohr's quantization condition: Ldq = nh. Therefore, we have derived Bohr's quantization condition from the Wilson-Sommerfelt quantization rule, taking into account the hint that angular momentum does not depend on the angle.

2. Applying the Wilson-Sommerfelt quantization rule to linear momentum:

In the case of a particle in a box moving back and forth with no other forces acting upon it, we can apply the Wilson-Sommerfelt quantization rule to linear momentum (p) instead of angular momentum. The equation becomes: (integral) pdx = nh. This means that the integral of linear momentum with respect to position (x) is equal to an integer (n) multiplied by Planck's constant (h).

Now, you are asking whether we can take the velocity (v) out of the integral when we are calculating with respect to dx. The answer is yes, we can. This is because we know that velocity is equal to the derivative of position with respect to time (dx/dt