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I read this on Wikipedia:

[...] That most tangible way of expressing the essence of quantum mechanics is that we live in a universe of quantized angular momentum and the Planck constant is the quantum. [...]

And I'm wondering if anyone can explain this a little more. Is it saying that ALL action is made up of angular momentum which is quantised in units of Planck's constant? Or is it just saying that all matter is composed of particles which have quantised angular momentum?

If the answer is the prior, this would mean that a photon's movement through space is as a result of angular momentum, which seems strange.

Anyway, if you could help me out with a little explanation, that would be great.

Qmechanic
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JoeRocc
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  • Angular momentum is the most basic observable where QM differs from classical mechanics. In QM angular momentum takes only discrete values whereas (unless the particle is trapped in a finite region of space) momentum and position take continuous values in QM just as in classical mechanics. – user10001 Mar 05 '13 at 12:30
  • However real essence of QM is not in the fact that angular momentum takes discrete values but in the fact that observables generally don't commute (whereas in classical mechanics they always commute). – user10001 Mar 05 '13 at 12:39
  • @user10001 Can you please explain what do you mean by "observables don't commute"? What is "commute"? Thanks in advance! – Cheeku Mar 05 '13 at 13:48
  • @Cheeku Observables in QM are matrices rather than functions so they do not commute in general(ie AB=!BA). In more simple terms suppose S is a physical system (say a particle) and A and B be two (quantum) observables associated with S. Say A= position and B=momentum. Now to know the value of A and B you have two choices - i) first measure A and then measure B ii) first measure B and then measure A. Classically both the ways are equivalent and will give the same result while for quantum systems these two ways of measurement are in general not equivalent and give different values of A and B. – user10001 Mar 05 '13 at 14:20
  • @user10001: surely angular momentum is not quantised for an unbound state any more than position and momentum are. – John Rennie Mar 05 '13 at 16:26
  • @JohnRennie: do you know of a case where spin is unbound? I could think of some for orbital angular momentum, but not spin. – Zo the Relativist Mar 05 '13 at 16:32
  • Ah good point, I was only thinking about orbital angular momentum. – John Rennie Mar 05 '13 at 16:42
  • @JohnRennie Sorry I didn't know that. Do you mean angular momentum observables can have states with continuous eigenvalues? – user10001 Mar 05 '13 at 17:02
  • I think that in order to have angular momentum a potential/force is needed since otherwise the particle would go in a straight line. In that sense there cannot be a continuous angular momentum at the micro level since it will be a solution of a potential QM problem that will be an eigenfunction of the angular momentum operator . – anna v Mar 05 '13 at 17:34
  • @user10001 see my answer – anna v Mar 05 '13 at 17:57
  • @annav: You could have orbital angular momentum in a scattering state of the hydrogen atom, right? ${\vec r} \times {\vec p}$ is perfectly well defined, since both $\vec r$ and $\vec p$ are observables of the theory, so you have neither straight line motion an upper bound to anything. – Zo the Relativist Mar 05 '13 at 18:01
  • I think you would find that the impulse that created the angular momentum would be quantized, just because of the form of the angular momentum QM operator. – anna v Mar 05 '13 at 18:18
  • Relevant: is-all-angular-momentum-quantized? – user10001 Mar 05 '13 at 18:47
  • @annav: I don't see how that would happen at all. And if you want a more trivial example of the same thing, take the scattering states of a $\alpha\delta(\beta r)$ potential, with $\alpha$, $\beta$ constant. You're definitely going to have a continua of momenta and position, which will give you a continuum of allowed angular momentum states. – Zo the Relativist Mar 06 '13 at 16:23
  • @JerrySchirmer did you check Lubos' answer to a similar question linked by user10001 above? also http://www2.ph.ed.ac.uk/~gja/qp/partialwave.pdf – anna v Mar 06 '13 at 16:57

3 Answers3

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This is a comment turned into an answer.

Classically one can define an angular momentum of a straight track with respect to any axis as

classical angular momentum

The real meaning though comes from rotational states about a central axis. In this case a potential exists which constrains the particle to revolve about the axis.

Quantum mechanically one defines an angular momentum in potential problems where there exists a solution as a wave function and the corresponding angular momentum operator acts on these functions:

operator of angular momentum

The result of this action will display the quantization of angular momentum, the value being a multiple of h_bar as shown in the table of the link.

The not very clear argument you are quoting from the wiki article is based on the assumption that quantum mechanically there will be an angular momentum only if a potential exists binding a particle, which will then inevitably have a quantized angular momentum. In the microcosm there cannot be a rotation about a center without a potential problem so it is true, but not, in my opinion the easiest way to understand quantum mechancis !.

anna v
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The Wikipedia page has been modified since the question was posed and the cited text disappeared, which is a good thing as it does not make much sense. Now it can be interesting to note that the preceding sentence was

Action is a general physical concept related to dynamics and is most easily recognized in the form of angular momentum.

So I take it that the original wiki editor had in mind the path integral formulation where the action (a quantity with the same units as angular momentum, and the Planck constant) is responsible for the phase relationship between alternate paths and thus governs their interference.

From this perspective the action could maybe be argued to represent the essence of quantum mechanics. But it would be very misleading to formulate this idea by confusing action and angular momentum; see this post for a related discussion.

Community
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Stéphane Rollandin
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Over the decades generations of teachers have attempted to come up with an intuitive way of introducing quantum mechanics, and all have failed since it's fundamentally non-intuitive. These days students are usually just presented with the Schrödinger equation and told to get on with it. By solving the Schrödinger equation you are deriving energy eigenfunctions, so your starting point would be that the basis of QM is that energy is quantised (well, for most bound systems anyway).

The Wikipedia article is attempting to introduce QM as quantised angular momentum. I don't know if this is technically possible, though I would have guessed not since I can't see how you'd derive the radial part of the hydrogen wavefunction without considering energy. Whether it's technically possible or not, it seems to me to be unhelpful and confusing for students new to QM. I would ignore that article and find some other better introduction.

A formal introduction would probably start from the axioms of quantum mechanics, though these are far from intuitive. The Schrödinger equation follows from the axiom that time evolution acts linearly on states.

John Rennie
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