Universality of Planck's Constant

As I understand it, the value of Planck's constant $h$ was originally introduced, in some sense, as a fitting parameter to understand blackbody radiation and the quantization of photons. In that sense, one can think of it as simply the proportionality constant between the energy and frequency of a photon.

However, we also have other relations such as the de-Broglie wave relation which use $h$, and some at a deeper theoretical level like in Schrodinger's wave equation, the commutation relation $[x,p] = i\hbar$, and the quantum path integral in the path integral formulation of quantum mechanics or quantum field theory.

Why is $\hbar$, which is essentially a proportionality constant related to light, so deeply tied to and seemingly universal in quantum theory? To be concrete, why does the value of $\hbar$ for the photon have to be the same as that of the de-Broglie wave relation, and why is $\hbar$ the same for all massive particles? Why does it have to be the same as that in Schrodinger's wave equation or the weight in the path integral or in the commutation relation? Why does classical theory generically stop being relevant on action scales close to $\hbar$? Qualitatively speaking, it doesn't feel like quantum theory intrinsically restricts $\hbar$ to be the same in all these cases, in the same way that general relativity identifies inertial mass with gravitational mass. Is it simply black magic that all of these are related, or is there something deeper? Or does it have something to do purely with the observables we can measure being deeply tied to the photon in some way?

One way to think of Planck's constant is as the factor that relates momenta to spatial frequencies, ie reciprocal lengths. This includes temporal momenta (aka energies), which will be related to temporal frequencies, ie reciprocal durations.

Whenever you see a pairing $p\cdot x$ or $E\cdot t$ (or by implication $p\cdot\dot x$), it might be appropriate to stick a factor $1/\hbar$ in there. An economic way to do so is to just scale all momenta by $1/\hbar$, which is implicitly achieved by using natural units.

• Right, but $\hbar$ can also be given a physical interpretation as, in some moral sense, the scale at which quantum mechanics is important. However, ab initio, I don't understand why the value of this scale is universal. For example, let me imagine a physical oscillator, say a rope or a membrane, and cool it to its quantum ground state. Why must its minimum quantum of energy be $\hbar\omega /2$? Why couldn't proportional to some other $\hbar' = 100*\hbar$, for example? – Aaron Oct 22 '16 at 21:36
• @Aaron You can derive that $\hbar \omega/2$ value starting from $[x, p] = i\hbar$, or any of the other starting points listed in my answer, so I don't think that's an independent question. – knzhou Oct 22 '16 at 22:22

All of these appearances of $$\hbar$$ aren't really independent, they're closely linked.

• First off, the equations $$E = \hbar \omega$$ and $$p = \hbar k$$ aren't independent; they're linked by relativity. Space and time are related in the same way that $$E$$ and $$p$$, and $$\omega$$ and $$k$$ are.
• The equation $$[x, p] = i \hbar$$ is more fundamental, and you can use it to derive the other results. Specifically, the Stone von-Neumann theorem states that $$[x, p] = i\hbar$$ implies that states of definite $$p$$ are plane waves of wavenumber $$k$$, with $$p = \hbar k$$.
• Starting from either of the previous results, you can heuristically "derive" the Schrodinger equation. In this derivation, the $$\hbar$$ just 'comes along for the ride' the whole time, so it's not really a new appearance.
• The result $$p = \hbar k$$ can also be used to derived the Heisenberg uncertainty principle $$\Delta x \Delta p \sim \hbar$$. Again, $$\hbar$$ just comes along for the ride.
• Starting from these results, you can derive various semiclassical results, such as the Born quantization rule $$\int p\, dx = n \hbar$$. The $$\hbar$$ is still just tagging along here, since it generally appears with every combination of $$p$$ and $$x$$.
• By relativity, $$\hbar$$ should also appear with every combination of $$E$$ and $$t$$. This means it should be the unit of action $$S$$, which is why the path integral weights by $$e^{iS/\hbar}$$.
• Regarding your first bullet, that's true but the actual value of $\hbar$ can be determined via experiment. Why must that value be the same across all particles? I also don't quite understand what you mean by the last bullet. Isn't that just a dimensional analysis statement? Hence, one could imagine that the value of $\hbar$ depends on what specific action you're looking at. Regarding the relation $[x,p] = i\hbar$, this by itself doesn't set the value of $\hbar$. You must match it to experiment. But then I don't see why these operators must be quantized universally in the same way. – Aaron Oct 22 '16 at 22:12
• @Aaron It looks like you're thinking of quantum mechanics as a bunch of independent almost-classical particles whizzing around, each blurred by some value $\hbar$. I guess in that picture, it does look weird that all the $\hbar$'s are the same. – knzhou Oct 22 '16 at 22:17
• Maybe it's more natural in quantum field theory with path integrals. In this case, there's just a single field with a single action $S[\phi]$, which clearly has a "single" $\hbar$ value. Then it follows that every excitation of this field has "the same" $\hbar$. This also makes sense because all particle excitations of a field are completely identical. There's nothing that could distinguish them with different $\hbar$s. – knzhou Oct 22 '16 at 22:19
• Right, I'm kind of thinking in that way. So if we take the minimal set of particles, say the standard model, you still have multiple fermion families and gauge fields. If you turn off interactions, then one could imagine the corresponding free action for each particle to be weighted with different $\hbar$ values. It could be that because of interactions, somehow this magically constrains $\hbar$ to be universal, but I don't see why this has to be the case. In essence, the strange thing to me is that $\hbar$ is primarily measured using electrons/photons, but yet we "assume" its universal. – Aaron Oct 22 '16 at 22:41
• Perhaps the question can be formulated as one regarding the Hilbert space in which $x$ and $p$ live and its interpretation. These are operators that "measure" the position and momentum of quantum states. But one could consider, say, quantum states of electrons living in a Hilbert space disjoint to that of photons. Then, we could define operators $x$ and $p$ in each Hilbert space separately such that they are quantized with different $\hbar$ values. To me, this seems legal and doesn't seem to change anything qualitatively about QM physics. But somehow nature wants $\hbar$ to be universal. – Aaron Oct 23 '16 at 0:18

As I understand it, the value of Planck's constant h was initially introduced, in some sense, as a fitting parameter to understand blackbody radiation and the quantization of photons.

Pretty much. Max Planck introduced it to get past the ultraviolet catastrophe.

In that sense, one can think of it as simply the proportionality constant between the energy and frequency of a photon.

Don't. It's more than that. It's telling you something important about the nature of light.

However, we also have other relations such as the de-Broglie wave relation which use h, and some at a deeper theoretical level like in Schrodinger's wave equation, the commutation relation [x,p]=iħ, and the quantum path integral in the path integral formulation of quantum mechanics or quantum field theory.

Yes we do. Note that it's the de Broglie wave relation and the Schrodinger wave equation. Also note that the path integral formulation is to do with wave function.

Why is ħ which is essentially a proportionality constant related to light, so deeply tied to and seemingly universal in quantum theory?

Because quantum theory is the strange theory of light and matter.

To be concrete, why does the value of ħ for the photon have to be the same as that of the de-Broglie wave relation, and why is ħ the same for all massive particles?

Because of the wave nature of matter.

Why does it have to be the same as that in Schrodinger's wave equation or the weight in the path integral or in the commutation relation?

Because of the wave nature of matter. We make matter out of light in pair production. Then we can diffract electrons and other particles. Because in atomic orbitals electrons exist as standing waves. Then we can annihilate particles with their antiparticles and get the light back. See Newton's Opticks query 30: "Are not gross bodies and light convertible into one another?" Yes they are.

Why does classical theory generically stop being relevant on action scales close to ħ.

It doesn't. We have classical electromagnetism and optics.

Qualitatively speaking, it doesn't feel like quantum theory intrinsically restricts ħ to be the same in all these cases, in the same way that general relativity identifies inertial mass with gravitational mass. Is it simply black magic that all of these are related, or is there something deeper?

There's something deeper. See Leonard Susskind in this video. At 2 minutes 50 he rolls his marker round and round saying angular momentum is quantized. You doubtless know that a sine wave is associated with rotation. The quantum nature of light is to do with that: roll your marker round fast or slow, but roll it round the same circumference, because Planck's constant of action h is common to all photons regardless of wavelength. Photon momentum is p=h/λ. The momentum can vary, the wavelength can vary, but h doesn't. The dimensionality of action can be given as momentum x distance. And it's the same distance for all wavelengths. Ever played Spanish guitar? You vary the wavelength with your left hand and you pluck with your right. In most situations the amplitude of your pluck doesn't change. Like Maxwell said, light consists of transverse undulations. Take a look at some pictures of the electromagnetic spectrum. Notice how the depicted wave height is the same regardless of wavelength? Note how all those transverse undulations have the same amplitude? You are looking at the quantum nature of light, at that something deeper, and it is hidden in plain view:

• You use the word "wave" a lot in this answer, but it's being used like "energy makes it go" in the old Feynman story. Just saying "because of the wave nature of matter" is so vague that all you're doing is transferring the confusion to a different word. – knzhou Oct 22 '16 at 21:07
• Sorry, but I don't quite understand why the value of $\hbar$ is fixed across all particles. Ostensibly I can imagine doing a diffraction experiment of various different particles and calculate an experimental $\hbar$ for each particle by plotting wavelength against momentum. In my understanding, it does not seem fundamentally necessary to quantum theory that $\hbar$ takes a particular value universally in the same way gravitational and inertial mass must be the same in general relativity. Is this, in some sense, a postulate of quantum mechanics? – Aaron Oct 22 '16 at 21:49
• @Aaron : it isn't a postulate of quantum mechanics, it's a constant of action. It's a real thing. And see the CSIRO image above - it applies to various particles because you can render them all down to light. – John Duffield Oct 23 '16 at 11:16