Let be two qubits |x> and |y>. The controlled-phase shift diagram is:
With |x>=|1> and |y>=|1>, one has:
Why |x> has the sum of all the phase shits and |y> is unchanged?
Thanks for any help.
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$\begingroup$ Hi and welcome to Quantum Computing SE. What type of rotations are $R$ gates? Or, which axis they rotate about? $\endgroup$– Martin VeselyJan 2 at 6:51
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1 Answer
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Phases do not apply to individual qubits. You have to remember that qubits are composed as a tensor product. So, if you have $$ (e^{i\theta_y}|y\rangle)\otimes (e^{i\theta_x}|x\rangle) $$ This is equivalent to $$ e^{i(\theta_y+\theta_x)}|y\rangle\otimes |x\rangle\equiv (e^{i(\theta_y+\theta_x)}|y\rangle)\otimes |x\rangle\equiv |y\rangle\otimes (e^{i(\theta_y+\theta_x)}|x\rangle). $$ We can move global phases around as we want, and it's generally convenient to collect them all together to see what cancels. (Note, however, that one must be more careful with relative phases between terms on a single qubit.)
