# Adding a phase to qubit: why is it necessary for arbitrary single qubit gate

An arbitrary single qubit gate can be decomposed as:

$$U=e^{i \alpha} R_z(\beta) R_y(\gamma) R_z(\delta)$$

We notice that in addition to the three rotations, there is a coefficient $$e^{i \alpha}$$. What disturbs me is that this extra phase $$e^{i \alpha}$$ shouldn't really matter as it will only add a global phase in the computation. Thus, why is it usually written ? It it because we want to "mathematically" identify the expression of the unitary but in term of physics this phase will never be added in practice on a quantum computer ?

• But a quantum computer should have more than one qubit, so why is the phase global? Aug 20 '20 at 18:52
• @M. Stern yes but even if it acts on a single qubit the phase will multiply the full ket so it will be global Aug 20 '20 at 18:58
• If it's not used in a controlled unitary gate. Ok I see your point. Aug 20 '20 at 19:25

A reason why we need that $$e^{i \alpha}$$ term:
It is right that the global phase $$e^{i \alpha}$$ will not change the action of the gate, but let's consider these two gates:
$$U1\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} \qquad R_z\big(\frac{\pi}{2}\big) = \begin{pmatrix} e^{-i \frac{\pi}{4}} & 0 \\ 0 & e^{i \frac{\pi}{4}} \end{pmatrix}$$
It can be easily seen that $$R_z\big(\frac{\pi}{2}\big) = e^{-i \frac{\pi}{4}} U1\big(\frac{\pi}{2}\big)$$. So both gates are differ by a global phase $$e^{-i \frac{\pi}{4}}$$ which means that they are equavalent when we apply them in the circuits. Nevertheless, as was discussed in this question  and in this this answer  the control version of this gates are not equivalent to each other:
$$CU1\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &1 &0 \\ 0 & 0 &0 &i \end{pmatrix} \qquad CR_z\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &e^{-i \frac{\pi}{4}} &0 \\ 0 & 0 &0 &e^{i \frac{\pi}{4}} \end{pmatrix}$$