4
$\begingroup$

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 ?

$\endgroup$
3
  • $\begingroup$ But a quantum computer should have more than one qubit, so why is the phase global? $\endgroup$ – M. Stern Aug 20 '20 at 18:52
  • $\begingroup$ @M. Stern yes but even if it acts on a single qubit the phase will multiply the full ket so it will be global $\endgroup$ – StarBucK Aug 20 '20 at 18:58
  • $\begingroup$ If it's not used in a controlled unitary gate. Ok I see your point. $\endgroup$ – M. Stern Aug 20 '20 at 19:25
4
$\begingroup$

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:

$$ CU1\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &1 &0 \\ 0 & 0 &0 &i \end{pmatrix} \qquad CR_z\big(\frac{\pi}{2}\big) = \begin{pmatrix} 1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 0 & 0 &e^{-i \frac{\pi}{4}} &0 \\ 0 & 0 &0 &e^{i \frac{\pi}{4}} \end{pmatrix}$$

So if we are trying to construct a circuit by applying a control version of some unitary, the global phase of the unitary shouldn't be neglected. This scenario is not seldom. For example, in QPE (and thus in HHL) algorithm, we should be careful with the global phase in the unitary whose controlled versions are used in the algorithm.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.