# 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? – M. Stern 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 – StarBucK Aug 20 '20 at 18:58
• If it's not used in a controlled unitary gate. Ok I see your point. – M. Stern 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 [1] and in this this answer [2] the control version of this gates are not equivalent to each other:
$$CU1\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &1 &0 \\ 0 & 0 &0 &i \end{pmatrix} \qquad CR_z\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &e^{-i \frac{\pi}{4}} &0 \\ 0 & 0 &0 &e^{i \frac{\pi}{4}} \end{pmatrix}$$