The creation and annihilation operators - also known as ladder operators are; $ \hat{a}^\dagger$ and $\hat{a}$ respectively.

Using the equation $\hat{H} = \hbarω\left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\right)$

and knowing that the units of $\hat{H}$ are J,

the units of $ω$ are Rad/s

and the units of $\hbar$ are J.s

I think that the ladder operators should have units of $\frac{1}{\sqrt{Rad}}$

But I have never seen a square root of an angle in units before. Is this correct?

  • 1
    $\begingroup$ The units of $\hbar$ are in fact J.s/rad. $\endgroup$
    – AV23
    Commented May 3, 2015 at 11:52
  • $\begingroup$ @AV23 Well that would make things much neater but wikipedia (that exceedingly trustworthy source) does not mention that. Nor does hyperphysics. Is this because Rad are not considered 'real' units? $\endgroup$
    – Jekowl
    Commented May 3, 2015 at 12:24
  • 1
    $\begingroup$ @Jekowl Yes. But $\hbar = \frac{h}{2\pi}$, $h$ is an angular momentum and $\pi$ has the unit rad. $\endgroup$
    – Noiralef
    Commented May 3, 2015 at 12:26
  • $\begingroup$ @Noiralef that make sense! thanks. Not really sure what the correct procedure here is but I guess I will leave false assumptions in the question and they can be corrected in the answer. $\endgroup$
    – Jekowl
    Commented May 3, 2015 at 12:33
  • $\begingroup$ That is the right way to do it, by the way: leave your question as is and post an answer that explains it. $\endgroup$
    – David Z
    Commented May 3, 2015 at 12:44

1 Answer 1


The units of $\hbar$ are in fact J.s/rad. (thanks AV23) this is because $\hbar = \frac{h}{2\pi}$ the units of h are J.s and the units of $\pi$ are rad. Thus we have J.s/rad. (thanks Noiralef)

Thus the ladder operators are in fact unitless.

On reflection this is the only logical possibility as they move between different eigenstates - which must all be in the same units.


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