# What are the units of the creation and annihilation operators?

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?

• The units of $\hbar$ are in fact J.s/rad. – AV23 May 3 '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 3 '15 at 12:24
• @Jekowl Yes. But $\hbar = \frac{h}{2\pi}$, $h$ is an angular momentum and $\pi$ has the unit rad. – Noiralef May 3 '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 3 '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 3 '15 at 12:44

## 1 Answer

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.