1. How did Dirac or whoever came up with it know that the momentum operator in quantum mechanics is $-i\hbar\frac{d}{dx}$?

  2. How did he know the $\hbar$ was in there?

  3. How did all these physicists know they had to put $\hbar$?

  4. Does it go back to Planck's paper on blackbody radiation?

  5. I know it is to make the units correct but why couldn't they use another constant with the same units as $\hbar$, like a constant which was a little bigger than $\hbar$?

  6. Also I know that $\hbar = \frac{h}{2\pi}$ but basically I am asking why $h$ and not some other constant?

  • $\begingroup$ Planck fit some now-famous curve to a blackbody and required a value of $4.135\,\times10^{-15}\rm{ eV\cdot s}$ for the curve to work. $\endgroup$
    – Kyle Kanos
    Jul 25, 2015 at 18:32
  • $\begingroup$ Yes I understand that. but how did physicists know that this constant was the one they should use in their formulas. How did they know that this constant related to blackbodies would be the one to use? Does h come up in a lagrangian or hamiltonian in quantum mechanics or physics at the atomic level? $\endgroup$
    – mihirb
    Jul 25, 2015 at 18:34
  • $\begingroup$ Have you opened up an introductory textbook on quantum mechanics? Because this should be Chapter 1 of any worthwhile book. $\endgroup$
    – Kyle Kanos
    Jul 25, 2015 at 18:37
  • 2
    $\begingroup$ See "Derivation of quantum mechanical momentum operator in position representation" $\endgroup$
    – Floris
    Jul 25, 2015 at 18:40
  • 1
    $\begingroup$ It's a constant "of action" because it has units of action ($\text{energy}\cdot\text{time}$). $\endgroup$
    – ACuriousMind
    Jul 25, 2015 at 18:51

1 Answer 1


Note: This is a brief summary.

Wikipedia is helpful, if you can't look anywhere else at the moment. It notes that $$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$ because of the de Broglie relation, $$p=\hbar k$$ where $p$ is momentum and $k$ is the wave vector.

de Broglie's equations, in turn, relate to the de Broglie wavelength, $$\lambda = \frac{h}{p}$$ which follows from a variant of the Planck relation.

That's the very, very, very, very basic way to look at it. A proper treatment would, of course, be found in a proper textbook on the subject.


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