# Do global phases matter when a gate is converted into a controlled gate?

Let's say that we have a unitary matrix M such that: $$M = e^{i\pi/8}\begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/12} \\ \end{pmatrix}$$ If we were to apply this unitary matrix to the state $$|1\rangle$$, we would get: $$M|1\rangle\ =\ e^{i\pi/8+i\pi/12}\begin{pmatrix} 0\\ 1 \end{pmatrix}$$

Where the global phase is $$e^{i\pi/8+i\pi/12}$$.

However, when we want to convert this global phase into a controlled gate, we would use the following implementation: $$CM = |0\rangle\langle0| \otimes I + |1\rangle\langle1| \otimes M$$ But would this mean that the global phase does matter in such cases?

The way I see it, there are two possibilities:

1. We do take the global phase into account in the resulting unitary matrix, as such: $$CM = \begin{pmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & e^{i\pi/8} & 0 \\0 & 0 & 0 & e^{i\pi/8+i\pi/12} \\\end{pmatrix}$$ This option would mean that multiplying with the state $$|{+}{+}\rangle$$, gives us the state: $$\frac{|00\rangle + |01\rangle+e^{i\pi/8}|10\rangle+e^{i\pi/8+i\pi/12}|11\rangle}{2}$$ In this case, M's global phase has changed into a relative phase when applied as a control on the state $$|{+}{+}\rangle$$.
2. There is a rule stating that we should not take the global phase into account when converting a gate into a controlled gate.

• Global phase acts on the whole state. The phase acting on the controlled qubit only is not global. Apr 7, 2021 at 16:13

You are definitely right, for a controlled gate the global phase (of the gate) does matter. That makes your first possibility the valid one. A rule that we should not take it into account would make things very ambiguous.

There is, however, something extra you can say. Let's say we have your gate $$M$$, and a gate $$K = e^{i\phi}M$$, e.g. it acts the same up to a global phase. As you pointed out, we have for the controlled versions:

$$\begin{equation} \begin{split} CM &= |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes M,\\ CK &= |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes K = |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes e^{i\phi}M \not = CM,\\ \end{split} \end{equation}$$ so they are not the same, as there is an extra 'phase' part on only the 'controlled-part'.

However, the 'controlled part' is, of course, the part for which the control qubit is $$|1\rangle$$ - and we can effectively change this (relative!) phase by applying a single-qubit operation.

Thus, if we first apply $$CK$$ and afterwards apply the gate $$R_{z}(-\phi) = \begin{bmatrix}1 & 0 \\ 0 & e^{-i\phi}\end{bmatrix}^{1}$$ on the control qubit, we delete the relative phase. That is:

$$\begin{equation} (R_{z}(-\phi) \otimes I) \bullet CK = (R_{z}(-\phi) \otimes I)\bullet (|0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes e^{i\phi}M) \hat{=} CM. \end{equation}$$ (here, $$\bullet$$ means a 'composition' of two maps/gates)

$$^{1}$$up to your definition, I actually may have omitted a global phase here - but this really is a global phase, so we can forget about it.