# $\hbar$, the angular momentum and the action

Is there anything interesting to say about the fact that $\hbar$, the angular momentum and the action have the same units or is it a pure coincidence?

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Torque and energy also have the same units. For a discussion of this fact see en.wikipedia.org/wiki/Torque#Units –  asmaier May 25 '12 at 10:06
IMO this is a question that is based on confusion as far as necessary and sufficient conditions go. Dimensional analysis ( en.wikipedia.org/wiki/Dimensional_analysis ) is a solid tool when checking consistency in derivations of various expressions describing physical systems. It is necessary that the formulas give the correct dimensions. The consistency that is found is not sufficient to invert the reasoning and go back to the initiating formulas. In a similar way that one cannot go in a one to one way from a two dimensional projection to a higher dimensional one. –  anna v May 25 '12 at 13:07

The dimensions of

1. the Planck constant $\hbar$,
2. the action $S$, and
3. the angular momentum,

are constrained by the following important facts:

1. A conjugated pair of two observables is quantum mechanically related to the Planck constant $\hbar$ via a Heisenberg uncertainty relation.

2. A conjugated pair of two variables is classically related to the action $S$ via Noether's Theorem, cf. e.g. this Phys.SE post. Listen e.g. to Richard Feynman approximately 50 minutes into this Youtube video.

3. The conjugated variable to an angular momentum is an angle (angular position), which is usually treated as dimensionless.

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Correction to the answer(v2): The word conjugated should be conjugate. –  Qmechanic Dec 4 '12 at 17:31

Let me try to answer using different words but with the same spirit as Qmechanic.

It is surely not a coincidence that $\hbar, S, \vec J$ have the same units. First of all, $\hbar$ is the quantum of the angular momentum or the quantum of the action, a universal constant that determines the strength of the quantum effects. So if you adopt one of these two definitions, you explain why $\hbar$ has the same units as either $S$ or $\vec J$ (only one of them) and reduce the question to the question why the angular momentum and the action have the same units.

It's not hard to see why the angular momentum and the action have the same units. Both may be written as $p\cdot x$, dimensionally speaking. The (orbital) angular momentum is defined as $\vec r \times P$; the commutator of $x,p$ is $xp-px=i\hbar$, which you may have included as well, has the units of position times momentum; and the action has the same units because the action has the same units as the Lagrangian times time $Lt$ which is the same as the units of the Hamiltonian times time $Ht$ and because $p\dot x$ appears in the difference/sum between $L$ and $-H$, in $L+H$, it's clear that $Lt$ has to have units of $px$, too.

Because the strength of quantum effects is determined by $\hbar$ that has the same units as the action $S$ or the angular momentum $\vec J$, it follows that both $S/\hbar$ and $\vec J/\hbar$ are dimensionless: they have no units.

Both of these facts have a robust and important explanation in the foundations of quantum mechanics. The action divided by the reduced Planck's constant is what appears in the exponent in Feynman's path integral, $${\mathcal A}_{i\to f} = \int {\mathcal D}\phi\cdot \exp(iS[\phi]/\hbar)$$ and the exponents have to be dimensionless, of course. From this Feynman approach, you could determine that the constant measuring the strength of quantum effects has the same units as the action.

Analogously, you may say a similar thing about the angular momentum. The reason is that the operators $J_x/\hbar$ and $J_y/\hbar$ have a commutator $$[\frac{J_x}{\hbar}, \frac{J_y}{\hbar}] = i \frac{J_z}{\hbar}$$ equal simply to the last component of $\vec J/\hbar$, without any extra coefficients. So these three operators generate a flawless $SU(2)$ or $SO(3)$ "Lie algebra" in the unitless mathematical normalization. (Well, mathematicians would also include the $i$ into each generator so that there would be even no prefactor of $i$ on the right hand side.) For this reason, the eigenvalues of $J_z/\hbar$ are quantized: they are inevitably multiples of $\hbar/2$. We may say that $\hbar/2$ is the elementary quantum of the angular momentum. (The orbital angular momentum is a multiple of $\hbar$ without the factor of 1/2.)

Just with some knowledge of Noether's theorem that links conservation laws and symmetries, one could have been able to guess – before he learned the full quantum mechanics – that the angular momentum should be related to generators of rotations. Because angles of rotations are dimensionless, the generators have to be dimensionless as well which means that quantum mechanics must contain a constant whose units are the same as those of the angular momentum so that it is possible to construct a dimensionless $\vec J/\hbar$ out of them.

It is somewhat difficult to find a more "direct" relationship between the angular momentum and the action, despite their having the same units. In particular, the angular momentum is quantized, a multiple of $\hbar/2$ as I mentioned. On the contrary, the action $S$ is continuous. As the Feynman path integral shows, the action $S$ is actually only meaningful in quantum mechanics up to shifts by multiples of $2\pi\hbar$. Such shifts don't change the exponential. So the angular momentum only allows the integer (or half-integer) values; on the other hand, the action only cares about the fractional parts! So the action and the angular momentum are never really "the same thing" in any sense, despite their identical units. After all, the angular momentum is a pseudovector (a particular set of conserved quantities in rotationally symmetric theories) while the action is the ultimate spacetime scalar defining a theory and invariant under everything.

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Although the answers so far to this questions are very interesting and informative, I think from an analytical point of view, your question is not quite sensible.

In a mathematical structure, one could argue that there are no "coincidences", everything is related through the fundamental basis. Now in practice, the answers expain why "$\hbar$", "angular momentum" and "the action $S$" are related. But if "mass $m$", "position $x$" and "momentum $p$" would have the same units, then there would also be an explaination for that, because these are parts of a physical theory, put into mathematical terms.

So if you ask "Is there anything interesting to say about the fact that ℏ, the angular momentum and the action have the same units or is it a pure coincidence?" (and you do), then the answer is "Yes.", optionally followed by an elaboration of the mathematical structure of the theory, a search for a common denominator.

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Alternatively, you could interprete my question as: "Is QM built so that $\hbar$, $S$ and $J$ have the same units and is it necessary?" –  Isaac May 25 '12 at 13:16
@Isaac: I don't understand. The first question "so that...have the same units" is already implied in your formulation and only a matter of checking any source online and in the second question I don't know what "necessary" is supposted to mean. If you already consider QM, then it's obviously necessary and if you don't, then nothing is really necessary. –  NikolajK May 25 '12 at 13:45
I obviously know that my question is not surely sensible; my question implicitely begins by "suppose there could be a reason". You are only discussing the sense of my question, which is clearly not the point (though I am thankful that you have answered). Why would we not speak about the sense of life, then? By "necessary", I could mean: "Has it been tried to build a kind of QM without that characteritic?" or something in that kind. I'm always trying to be synthetic when I ask a question in physics. –  Isaac May 25 '12 at 19:26
@Isaac: I don't know the context of the question "Why would we not speak about the sense of life, then?" here. Also, I don't know what it means to synthetically ask a question. –  NikolajK May 25 '12 at 19:45
English is not my first language, so I beg you pardon if I do not always choose the right words; by synhtetic, I mean that I try to ask my questions with the fewest words possible. Then, the sentence about life is an answer to yours, "nothing is necessary", which I think is very naive, as I have just explained. I suggest us to leave it at that. –  Isaac May 25 '12 at 19:59