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I'm reading Dirac's Principles of Quantum Mechanics. He defines $\hbar$ to be the real number satisfying the following relation $$ uv - vu = i\hbar[u,v]$$ where $u$ and $v$ are dynamical variables, and $[u,v]$ is the classical Poisson bracket. He later defines the left hand side of this equation (with the variables replaced with the corresponding operators) to be the quantum Poisson bracket.

He then says that from experiments, we must have $$\hbar=\frac{h}{2\pi}$$ where $h$ is the constant that was introduced by Planck. How does one get the $2\pi$? Is it an approximation? How can we be certain that it is exactly $2\pi$ to an arbitrary degree of precision?

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  • $\begingroup$ Doesn't that relationship follow just from the relationship between (ordinary) frequency $\nu$ and angular frequency $\omega = 2\pi\nu$ (or wave number $k$ and wavelength $\lambda = 2\pi/k$). For example, the energy of a photon of frequency $\nu$ is $E_\gamma = h\nu = h\omega / 2\pi = \hbar\omega$. Or are you looking for something else? $\endgroup$
    – Alfred Centauri
    Jun 23, 2019 at 15:01
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    $\begingroup$ I personally was taught that it was a convention to simplify the notation. $\hbar/2\pi$ was often seen in the equations so they said. Heck it, let's say it's a reduced constant. $\endgroup$
    – Dominik Car
    Jun 23, 2019 at 15:03
  • $\begingroup$ @AlfredCentauri How does one know it's the same constant appearing here? $\endgroup$
    – Anonymous_original
    Jun 23, 2019 at 15:05
  • $\begingroup$ @DominikCar Sometimes it is introduced that way. But Dirac doesn't introduce it that way. $\endgroup$
    – Anonymous_original
    Jun 23, 2019 at 15:06

2 Answers 2

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These are connected, though perhaps in a roundabout way. For example:

  • Starting from $[x, p]$, one gets the canonical commutation relation in terms of $\hbar$.
  • One can write the Hamiltonian of a quantum harmonic oscillator $p^2/2m + kx^2/2$ in terms of raising and lowering operators. Using the canonical commutation relations, one finds $H = \hbar \omega (a^\dagger a + 1/2)$.
  • One then concludes the spacing between energy levels is $\hbar \omega$.
  • This is the energy of photons released in QHO transitions, which is $E = hf$ according to Planck. This gives $\hbar = h / 2 \pi$.

A rather different question is whether this constant $\hbar$ has to be the same for all pairs of classical variables. I asked a question about this here.

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  • $\begingroup$ It seems ok. But isn't very satisfying and is a little "roundabout" as you said. Is there a direct connection? $\endgroup$
    – Anonymous_original
    Jun 23, 2019 at 15:19
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This is a question about the history of Quantum Mechanics. The reduced Planck's constant $\hbar$ has been introduced by Dirac in the 3rd Ed. of his texbook in 1947. The $2\pi$ is in Quantum Mechanics essentially since 1925, from Heisenberg's article "Über quantentheoretische Umdeutung kinematischer und mechanischer Beziehungen" on page 7 of the article (page 885 of the Journal "Zeit. f. Physik", Vol. 33) and then formula 23 on page 11 of the article, the energy spectrum of a quantum harmonic oscillator:

$$ W = \frac{\left(n+\frac{1}{2}\right) h\omega_0}{2\pi}$$

The $2\pi$ is of course older, it can be tracked down in Sommerfeld and Wilson's equations used by Heisenberg as a starting point and it comes really from Bohr's 1913 angular momentum treatment of the Hydrogen atom using circular orbits for the electron.

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