Planck’s constant relates the frequency of a photon to its energy but how does that relate to other quantum particles? For example in the Schrödinger equation it is used according to my book “to correct the units”. Is that true? And why Planck’s constant and not 1?
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9$\begingroup$ you need to get another book... $\endgroup$– ZeroTheHeroCommented May 30 at 15:39
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1$\begingroup$ In addition to Níckolas Alves's answer see What is the real interpretation of Planck's constant and what are its origins? $\endgroup$– John RennieCommented May 30 at 15:51
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1$\begingroup$ I do not think this answer should be closed. It's meaning seems to clear to me, and obviously also to Níckolas Alves since he posted a great answer. It is just possibly a duplicate of the question I linked above but it is not unclear. $\endgroup$– John RennieCommented May 30 at 15:54
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$\begingroup$ Related: physics.stackexchange.com/q/375755/2451 , physics.stackexchange.com/q/315776/2451 and links therein. $\endgroup$– Qmechanic ♦Commented May 30 at 16:05
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$\begingroup$ "why Planck’s constant and not 1?" Why not both? $\endgroup$– hftCommented May 30 at 16:07
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1 Answer
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It is true that Planck's constant was originally understood as the ratio between the photon's energy and its frequency, but later it was found that it is much deeper.
Planck's constant is actually a measure of the granularity of quantum mechanics. For example, you will see (or have seen) that the position and momentum operators respect $$[\hat{x},\hat{p}] = i \hbar,$$ where $\hbar = \frac{h}{2\pi}$. In this sense, $\hbar$ tells us how position and momentum fail to behave classically (in which case they would commute), and hence it leads to the quantum behavior. $\hbar$ sets the scale in which quantum effects start to be meaningful.
Another way of understanding this is that $\hbar$ is necessary not only to fix the units, but also to have agreement with experiment in the Schrödinger equation. We can use the results of the Schrödinger equation to compute, for example, the energies of the hydrogen atom and use those results to fix what is the constant occurring in the Schrödinger equation. In this sense, the properties of photons end up being a corollary, rather than a definition.
Finally, notice that if we used 1, that would depend on the system of units. In natural units, we indeed take $\hbar = 1$, but this value is simply not true in the SI.
answered May 30 at 15:42
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2$\begingroup$ @JakeLevi It is called the commutator of the two operators (position and momentum). If you apply the position operator and then the momentum operator to a wavefunction, the result is different from when you apply first the momentum operator and then the position operator. The commutator tells you how these two things are different. Notice that in classical mechanics the order doesn't matter $\endgroup$ Commented Jun 1 at 16:18