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In this article on matrix mechanics in quantum theory you can read, in the subsection of the harmonic osscilator, that

$$E=\frac{nh}{2\pi},$$

Where $E$ stands for the possible energies of the harmonic oscillator, $n$ is a positive integer (or zero), and h is Planck's constant. It is written that

$$E=\frac{1}{2}(P^2+X^2),$$

where $P$ and $X$ classical expressions for momentum and position. One can read that after setting the mass and the frequency (extensive quantities) equal to one and substituting sinusoidal functions for $X(t)=\sqrt{2E}\sin{t}$ and $P(t)=-\sqrt{2E}\cos{t}$, it follows that, according to the old quantum condition, that the integral of $PdX$ over an orbit, which is the area of the circle in phase space, must be an integer multiple of Planck's constant. The area of the circle of radius $\sqrt{2E}$ is $2πE$. So $E=\frac{nh}{2\pi}$. In Natural units you can make $\frac{h}{2\pi}=\hbar$ equal to one so the energy is an integer.

My question is: how can $E=\frac{nh}{2\pi}$ if E has the dimension of energy ($(Nm)$) and $h$ has the dimension of an action ($(Nmsec)$)? Aren't the frequency $f(\frac{1}{sec})$ and mass $m$ set equal to one to fit the the old quantum condition, in which no $f$ ($=\frac{\omega}{2\pi}$) or $m$ appear?

See also this article about nondimensionalization.

EDIT: Thanks to a comment below, I see now that in the formula above for the energy,

$$E=\frac{1}{2}(P^2+X^2)$$

can't have the dimension $(Nm)$ or $(J)$.

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    $\begingroup$ The first sentence of the passage of the Wiki article you link literally answers your question, even linking to the article on non-dimensionalization. $\endgroup$ – ACuriousMind Jan 21 '18 at 11:07
  • $\begingroup$ Allright. I think it's clear to me right now. It was just that I thought when they set the frequency to one, they removed it's dimension too. $\endgroup$ – descheleschilder Jan 22 '18 at 12:31
  • $\begingroup$ I find this question to be extremely disingenuous. The initial expression for the energy, $E=\frac{1}{2}(P^2+X^2),$ only makes sense if at least two variables have been de-dimensionalized. Complaining that subsequent expressions have dimensions that are inconsistent with SI units seems just short of being hard-headed on purpose. $\endgroup$ – Emilio Pisanty Jan 22 '18 at 23:23
  • $\begingroup$ I don't know what you mean by hard-headed on purpose, but I can see now that the "problem" is already in the equation $E=\frac{1}{2}(P^2+X^2)$. What dimension has this $E$? $\endgroup$ – descheleschilder Jan 23 '18 at 15:13
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In the old quantum theory is $E=n\hbar \omega$. You missed the factor 1/[seconds] by using $\omega=1$, that's why the dimensionality seems not correct.

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