In the context of quantum teleportation, my lecturer writes the following (note that I assume the reader is familiar with the circuit):
If the measurement of the first qubit is 0 and the measurement of the second qubit is 0 , then we have the state $\left|\phi_4\right>=c_0\left|000\right>+c_1\left|001\right>=\left|00\right>\otimes \left(c_0\left|0\right>+c_1\left|1\right>\right)=\left|00\right> \otimes \left|\psi '\right>$.
Now to get the final teleported state we have to go through the final two gates $Z, X$.
My lecturer writes this as;
$\left|\gamma\right>=Z^0X^0\left|\psi '\right>= \left|\psi'\right>= c_0\left|0\right>+c_1\left|1\right>$
Here are my questions:
Why is it that the we do not have $\left|\gamma\right>=Z^0X^0\left(\left|00\right>\otimes \left|\psi '\right>\right)$? I don't understand why we cut the state $\left|\phi_4\right>$ "in half ", to use bad terminology, at this step.
What does the superscript 0 on the operators refer to?
In the final state again to use bad terminology we only use half of the state $\left|\phi_4\right>$ can we always assume that the final state will be the $\left|\psi'\right>$ part of $\left|\phi_4\right>=\left|xy\right>\otimes\left|\psi'\right>$ state, and if so what is the significance of the final step.
If my question is unanswerable due to a deep misunderstanding of the math processes here, I'd still really appreciate some clarification on whatever points can be answered and I can edit it to make more sense as I learn.