# Hermiticity of a projection operator

I have managed to get myself quite confused about something. Consider a two level system consisting of states $$|0\rangle$$ and $$|1\rangle$$. the projection operator onto the state $$|0\rangle$$ is given by $$\hat{\sigma}_{00}=|0\rangle\langle0|$$, and is clearly Hermitian. Being hermitian, you would expect the operator $$\hat{U}=e^{i\omega\hat{\sigma}_{00}}$$ to be unitary. Expanding the taylor series of this operator we see that $$\hat{U}=\sum_{k=0}^{\infty}\frac{(i\omega)^{k}}{k!}\hat{\sigma}^{k}_{00}=e^{i\omega}\hat{\sigma}_{00}$$ since $$\hat{\sigma}^{k}_{00}=\hat{\sigma}_{00}$$. Applying the Hermitian conjugate to this operator we see that $$\hat{U}^{\dagger}=e^{-i\omega}\hat{\sigma}_{00}.$$ However if this is the case then $$\hat{U}^{\dagger}\hat{U}=e^{-i\omega}\hat{\sigma}_{00}e^{i\omega}\hat{\sigma}_{00}=e^{0}\hat{\sigma}_{00}\neq\hat{I}$$ so it isn't unitary. Have I made a mistake in my logic or calculations at any point?

• You've made a mistake: What is $\sigma^0$? If you've found the answer, consider to answer the question yourself. Feb 27 at 12:28
• The statement that $\sigma^k_{00}= \sigma_{00}$ is only true for positive integer values of k. However, your domain also includes $k=0$ and $k=\infty$, for which the original statement does not necessarily hold. Feb 27 at 13:09

Indeed, you missed that $$\sigma_{00}^0=I$$.
$$$$\hat{U}=I+\sigma_{00}(e^{i\omega}-1), \hspace{6 mm} \hat{U}^{\dagger}=I+\sigma_{00}(e^{-i\omega}-1),$$$$
and it is now easy to see that $$\hat{U}^{\dagger}\hat{U}=I$$.