Is it true that in Quantum Mechanics action is quantized?
Quantum Mechanics can be mostly derived by sort of coarse-graining phase space, it's called deformation quantization. I'd recommend the article http://arxiv.org/abs/quant-ph/0208163v1 if you're at the level of an advanced undergrad or so.
Although I think that article on deformation quantization is important and fascinating (thank you), it is also pretty technical. I believe the OP question is looking for an answer more along the lines of "yes", or perhaps a little better, that quantum mechanics asserts that physical states must be superpositions of states of discrete action, quantized in multiples of h-bar. That is an important statement-- it means that continuous action must be relegated to the level of a statistical expectation over many trials or particles, rather than an actual determined truth about each one. There is something about our methods for determining what is true about the reductionist building blocks of our theories that require we obtain discrete values for the action. What that "something" is may be a complete mystery, at least it is a complete mystery to me.
I have seen action quatization be used to derive the stationary states of the particle in a box problem, but the method seemed a little sketchy, requiring that the action of the trajectory in travelling from one wall to the opposite one (in one dimention) must be a multiple of h-bar. However, if we go to 3 dimensions, where we can conveive of motion in any arbitrary direction, it is not clear (to me) why only the action along principal diagonals should be quantized, or if this is even sufficient to derive the states.
Also, how can the harmonic oscillator states be derived like this? Along what trajectory do we quantize the action?
I think it would be instructive to consider the 1-D oscillator first. As they show in most textbooks, we would draw an ellipse in the phase space, p^2/2mE + x^2/(2E/k) = 1, and then Wilson-Sommerfeld quantization Int pdx = nh will immediately lead to Plank's quantization law E=nhf. But why do they say that Int pdx is equivalent to action over one oscillation? Action in Classical Mechanics is dS=pdq - Hdt, so if we quantize Int pdx this means we have quantized the quantity S + Int Hdt, but we have not quantized S. So where does action quantization come from?
Action is not quantized in QM. It was part of semi-classical quantum physics like the Bohr-Sommerfeld quantization rule, and has counterparts in modern semi-classical stuff like the Gutzwiller trace formula.
Silly me for thinking your question was introductory! You clearly want a very firm concept of what quantization of action means, and there are so many different types of "action" that get used I'm pretty vague on the concept of just what quantization of action really means anyway. So I'll just wait and see what answers can be given to this! (ETA: I see atyy has given an answer as to why the whole issue seems so vague. I wonder if more can be said about it though, perhaps around issues like the connection between commutators and Poisson brackets?)
I'm confused, as usual. I thought that wrt QM there is a fundamental quantum of action, expressed in joule.seconds, ie., energy.time. And that, therefore, action is quantized.
What am I missing here?
Planck's constant has units of action.
In Bohr-Sommerfeld quantization, the action is quantized in multiples of Planck's constant. However, that should be seen as part of "old quantum physics" like the Bohr atom. It's great for intuition, but after Heisenberg, Schroedinger, Dirac, quantization is specified by making canonically conjugate variables not commute. The old quantum physics is an approximation to the results of the proper quantum formalism.
The "action" in this usage is taken from "action-angle" variables. It only applies to integrable systems. Chaotic systems are examples of non-integrable systems.
The terminology is pretty confusing! A useful reference is Cvitanovic's http://chaosbook.org/. Chapter 31 gives proper quantum mechanics, and the following chapters talk about old quantum physics ("semi-classical") and modern semi-classical developments to classically chaotic systems.
I don't think it's that simple. True, h has units of action (J*s), but that fact by itself does not mean that h IS action. Now, action itself is a classical concept. For example, in classical mechanics it is S=Int L dt where L is Lagrangian. In quantum mechanics, the duration of a process is indeterminate due to energy-time uncertainty principle. So if we know L exactly, then how can we know the limits of integration? This issue did not exist in the old (pre-Heisenberg) quantum theory. But even in the old theory, Wilson-Sommerfeld only assumed quantization of the phase integral Int p dq which does not imply quantization of S (unless you make an additional assumption that Int H dt is also quantized). To complicate matters further, Plank spoke of quantized action well before Sommerfeld, i.e. without access to the formula Int p dq = nh. What was his reasoning then? Was it just the fact that h and S had the same units?
Ok, thanks. So energy.time is not conventionally quantized or quantizable. Is that correct?
Thanks for the feedback.
Yes, in the strict sense that there is no time operator, and no commutation relation between energy and time. This is explained in Demystifier's http://arxiv.org/abs/quant-ph/0609163 (see his section 3).
However, it is useful to have an energy-time uncertainty relation as a heuristic.
Thanks, I will read Demystifier's section on this. Meanwhile, I remain fond of useful heuristics.
Oh yes, I forgot. Actually my main point was that in old quantum theory the action ∫pdq around a periodic orbit was quantized, ie. ∫pdq~nh. However, in modern quantum mechanics, quantization is specified by the commutator [p,q]~ih, which in general is not equivalent to the action quantization rule in old quantum theory.
∫pdq around a periodic orbit is called the "action" by the usage in "action-angle" variables, which are only indirectly related to the action S=∫L(q,dq/dt)dt. Wikipedia calls ∫pdq the "abbreviated action" to distinguish it from the action S=∫L(q,dq/dt)dt. So the OP question is really one of terminology in old quantum theory.
Still, although it is of value to distinguish old and new formulations and the improved rigor achieved, for those basically wanting to know "what is the "quantum" in quantum mechanics", the answer still has to be something along the lines of discrete action. This must be viewed as one of the key differences between quantum mechanics and classical particle physics. The main pillars of quantum mechanics would seem to be quantization of action, superposition of possible states, wave/particle duality, and entanglement of interacting particles. These are all wrapped up in QM's treatment of the wave function and its stationary states. More formal versions of these key advances can and should be investigated, but to most people those are never going to be what really makes quantum mechanics tick at the conceptual level. Uniting the conceptual and the formal levels would be of the greatest value of all, if possible, but if one must choose one or the other, and one is not a mathematical physicist, the conceptual advances have the greater value to understand. (I'm not critiquing the more formal answers, the OP asked if something as "true" and that sounds like a formal statement, I'm just pointing out the role of quantization of action in the more informal conceptual understanding of what quantum mechanics does.)
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