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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?

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Jekowl
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    The units of $\hbar$ are in fact J.s/rad. – AV23 May 03 '15 at 11:52
  • @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? – Jekowl May 03 '15 at 12:24
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    @Jekowl Yes. But $\hbar = \frac{h}{2\pi}$, $h$ is an angular momentum and $\pi$ has the unit rad. – Noiralef May 03 '15 at 12:26
  • @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. – Jekowl May 03 '15 at 12:33
  • That is the right way to do it, by the way: leave your question as is and post an answer that explains it. – David Z May 03 '15 at 12:44
  • Possible duplicates: http://physics.stackexchange.com/q/11500/2451 and links therein. – Qmechanic May 03 '15 at 12:45
  • @Qmechanic That doesn't mention the units of $\hbar$ which was the part of this question that I got wrong. – Jekowl May 03 '15 at 13:24

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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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