# How can angular momentum be quantized?

1. Aug 24, 2012

### dEdt

Angular momentum is defined with respect to a certain origin, right? Well, what's stopping me from changing my origin ever so slightly, so that the angular momentum changes ever so slightly?

2. Aug 24, 2012

### Simon Bridge

How would you measure it?

3. Aug 24, 2012

### bcrowell

Staff Emeritus
Nothing is stopping you from doing that. For example, the ground state of the hydrogen atom has its electron in a state of zero orbital angular momentum, if you use the proton as the origin. If you choose an infinitesimally different origin, this will be a mixture of all orbital angular momenta 0, 1, 2, ... The average will shift infinitesimally.

When we say that intrinsic spin has a specific value for a specific type of particle, I guess we mean with the particle itself chosen as an origin. This becomes a little subtle in the case of a massless particle, which doesn't have a rest frame.

4. Aug 24, 2012

### Simon Bridge

I thought "quantized angular momentum" meant that when we try to measure angular momentum, we see it in particular sized lumps.

If we measure angular momentum in one orientation, subsequent measurements get the same thing - but if we rotate the axis we measure against, the new statistics will come from a mixture of different angular momentum states ... so the expectation (= average) can change continuously ... but a subsequent measurement still gets the same lumps doesn't it?

5. Aug 24, 2012

### Bill_K

We're not talking about a rotation, the question posed was, what happens to L under a translation? And I agree with what both you and Ben have said: the result will be a superposition of L values, the same discrete integer values, but the expectation will change slightly.

Quantitatively the coefficients in this superposition can be obtained from the commutator [L2, p]. Another way is to use a representation: consider the harmonic functions F(r) = rP(cos θ). These are solutions of Laplace's equation, each having a definite value of L = ℓ. Consider them as the basis in a function space.

An infinitesimal translation along the z-axis corresponds to applying a derivative, ∂/∂z. For example for ℓ = 2, F2 = ½(3z2 - r2), and ∂F2/∂z = 2z = 2F1. Similarly for each value of ℓ, ∂F/∂z can be written as a linear combination of F's with L < ℓ. Thus the result of an infinitesimal translation on a state of angular momentum L = ℓ will be that linear superposition of states with angular momentum L < ℓ.

6. Aug 24, 2012

### Simon Bridge

Oh fair enough ... though if I measure angular momentum, then use identical apparatus in a different position (i.e. translated, but not rotated) on the same particles, doesn't that just give the same measurement? In fact, would we normally expect a mere translation of axis to change the projections of a vector? Wouldn't a rotation, instead, be more illustrative for OPs question? Or maybe I have the wrong question?

I thought the question was: "How can angular momentum be quantized?" ... which I happily interpreted as "What do we mean by 'angular momentum is quantized'?" ... which, admittedly, may not have been the interpretation intended. OP provided an example of how he may expect to change angular momentum... particularly in terms of how L is defined in relation to axes.

The z-axis projection of a "regular" vector depends on how we've drawn the axes after all.

I understood that the different axis were defined (in a meaningful way) by the measuring apparatus.

eg. The Stern-Gerlach apparatus has an important axis, which gets called "z" a lot. The angular momentum is quantized with respect to this axis, no matter how the apparatus is oriented in space. You can put random x-y position particles (the beam) in with random orientation of spins and still get quantized results out.

I was guessing that this was the sort of thing OP was asking about.

So it seems reasonable to say that the angular momentum is said to be quantized because that is what we get when we measure it. It gets that way (the "how" part), regardless of orientation, because of the interaction with the measuring apparatus. We can (very accurately) account for this in the mathematics by considering superpositions of orthogonal angular momenta.

In a way, what OP seems to be wrestling with is a less obvious manifestation of "wave particle duality" ... the math is continuous but the resulting measurement is discrete. OR I have been overthinking this and looked for deep meaning where there is none ;)

7. Aug 27, 2012

### Zarqon

I like to think about quantization as arising from boundry conditions. In this case, it's a spherical boundry condition that says that the allowed orbits must be such that the wavefunction of the electron "overlaps with itself" when traced around the nucleus. If it's not, then it will destructively interfere with itself and will not form a stable orbit.

I'm also not sure what you gain from "changing origin", you would just be changing the theoretical description of the states, not changing the states themselves. It would be harder to see the quantization because you're using a non-intuitive basis, but it wouldn't remove the actual quantization in any way.

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