Understanding the Use of Controlled Not Gates in Quantum Computing

[ShayanJ]

In an article I'm reading, the author defines an operator as below:
<br /> \hat{U}_{CNOT}(\theta)=\exp{(-i \theta \hat{U}_{CNOT})}=\hat{1} \cos{\theta}-i \hat{U}_{CNOT} \sin{\theta}<br />
Where \hat{U}_{CNOT} is the controlled not gate(http://en.wikipedia.org/wiki/Controlled_NOT_gate).
Then the operator is applied to a state of the form (\alpha |0\rangle+\beta|1\rangle) \otimes |\psi\rangle and the resulting state is:
<br /> (\alpha e^{-i \theta} |0\rangle+\beta \cos{\theta} |1\rangle)\otimes |\psi\rangle-i \beta \sin{\theta} |1\rangle \otimes (\hat{\sigma}_x |\psi\rangle)<br />
where \hat{\sigma}_x=\begin{pmatrix}<br /> 0&amp;1\\<br /> 1&amp;0<br /> \end{pmatrix}
Then the author says:

The relative phase e^{i \theta} between |0\rangle and |1\rangle is a complication in the calculations that will follow.To avoid this problem, instead of \hat{U}_{CNOT}(\theta),we will use interactions of the form \hat{Z}_A(\theta) \hat{U}_{CNOT}(\theta),where\hat{Z}_A(\theta)=(\exp{(i \hat{\sigma}_z \theta/2)}\otimes \hat{1})...The single-bit rotation exactly undoes the extra relative phase produced by \hat{U}_{CNOT},while changing nothing else.

Now my problem is that I can't understand how that happens. I don't know how to work with \hat{Z}_A(\theta)\hat{U}_{CNOT}(\theta).I'll appreciate any suggestion.
Thanks

 

[Simon Bridge]

In an article I'm reading

Meaningless without citations.
Which article - where - when - who is the author?

But I think this is just a handy notation to remove a nasty bit of math.
Z does what it is defined to do. The author is only using it by defining it.
See what happens next.

 

[ShayanJ]

Yeah,I wasn't clear enough,sorry.
This is the link to the article:http://arxiv.org/abs/quant-ph/0108132
But the part that I mentioned,doesn't need much knowledge about the other parts of the article.
My problem is that when I apply \hat{Z}_A(\theta) to \hat{U}_{CNOT}(\theta)(\alpha |0 \rangle +\beta|1\rangle)\otimes |\psi \rangle,there is still a relative phase between |0\rangle and |1\rangle which leads me to think I'm missing sth.
Also I can't get help from the article itself,because it doesn't contain calculations that I want.