Global phase and single qubit gate: does it actually matter for two (or more) qubit gates?

Question

Consider the $X$ gate. Mathematically, we have $X=i e^{-i\frac{\pi}{2} X}$

But as global phase of unitaries don't matter because they will simply act a global phase to the wavefunction, we can consider implementing $X$ by $e^{-i\frac{\pi}{2} X}$, allright.

Now, consider a CNOT. Formally, we have:

$$CNOT=|0\rangle \langle 0| \otimes \mathbb{I} + |1\rangle \langle 1| \otimes X $$

If at this point I say "well, a $X$ gate or a $\pi$ rotation around $x$ is the same, up to global phase", I could say:

$$CNOT=|0\rangle \langle 0| \otimes \mathbb{I} + |1\rangle \langle 1| \otimes e^{-i\frac{\pi}{2} X} $$

But the two expressions of the CNOT do not differ from a global phase.

My question is the following.

Let's assume we want to implement an algorithm. Is it that we have at the beginning to define once for all how we implement an $X$ gate, and be consistant all along.

For example, if as soon as there is an $X$ in the algorithm and that I replace it by $e^{-i\frac{\pi}{2} X}$, then I will be fine.

But, if sometime I replace it by $ie^{-i\frac{\pi}{2} X}$ and sometime by $e^{-i\frac{\pi}{2} X}$ then I will have problems.

So here, indeed my two definitions of CNOT do not implement the same unitary, but if they were inside of an algorithm and that I had chosen a fixed convention for $X$, then I will be safe ?

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Other question (more important for my purpose).

Let's assume I can only do single qubit rotations on which I might have a quantum control on it (I can do controlled rotation in the end).

How is it possible from this to implement a CNOT operation ? Indeed this example shows that a CNOT is not a controlled $\pi$-pulse around $x$. How could I add the $i$ that is missing in practice then ? Because from respect to the target qubit this $i$ is a global phase. This confuses me.

Answer 1

What is this business of global phase?

Unfortunately, most people encounter this when seeing their first calculation and the teacher says, "we can just get rid of that because overall phases don't matter." But where does this come from? First, the math answer:

States of quantum systems are properly modelled as complex projective spaces. (More loosely: 1 dimensional subspaces of the Hilbert space.)

In direct notation, the statement is saying the state $|\psi\rangle$ is really the projector $|\psi\rangle\!\langle\psi|$. Now consider the state defined as $$ |\psi_\theta\rangle = e^{i \theta}|\psi\rangle. $$ Its corresponding projector is $$ |\psi_\theta\rangle\!\langle\psi_\theta|= e^{i \theta}|\psi\rangle\!\langle\psi| e^{-i\theta} = |\psi\rangle\!\langle\psi|. $$ The exact same projector!

You could just take this as the first axiom of quantum theory and be done with it. But that is not all that useful unless you are interested in mathematical physics. (Which you should be because its cool.)

Another answer is that overall phases always drop out at the end of any calculation. Consider again the state defined as $$ |\psi_\theta\rangle = e^{i \theta}|\psi\rangle. $$ You could carry that phase through the whole calculation. But quantum calculations always end with a modulus inner product: $$ |\langle\phi|\psi_\theta\rangle|^2=|e^{i\theta}\langle\phi|\psi\rangle|^2 = |e^{i\theta}|^2|\langle\phi|\psi\rangle|^2. $$

The important part here is that such a calculation is only ever done on the entire system. Phases can be removed not be because they are unphysical or don't matter, but because they are inconvenient in some calculations.

Another way to think about it is: two states can be considered the "same" if there is no experiment you can do which will distinguish them. The same would be true of two measurements and two operations. In your case, you identified that the two different unitaries could be distinguished with an experiment, so they are not the same. Now, if the entire system were two qubits, then $CNOT$ and $e^{i\theta} CNOT$ could be considered equivalent, but not if they were part of a larger system.