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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: $$ CM = |0\rangle\langle0| \bigotimes I + |1\rangle\langle1| \bigotimes M $$ 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.

Could somebody please help me with getting this clear?

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  • $\begingroup$ Global phase acts on the whole state. The phase acting on the controlled qubit only is not global. $\endgroup$ – kludg Apr 7 at 16:13
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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.

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