In the hello many worlds tensorflow tutorial and in the lockwood paper (2020) I have seen that often in QVC the following combination of gates is used:

$R_z(\theta), R_y(\theta), R_x(\theta)$

I am wondering if not just two of them would suffice for reach every measurable quantum state.

I thought, that certain quantum states are only theoretically different, so for example:

$|\psi> = \frac{1}{\sqrt2}\big(|0> + |1>\big)$ and $|\psi> = \frac{1}{\sqrt2}\big(i|0> + i|1>\big)$

are, when measured collapsing to 50% to $|1>$ and to 50% to $|0>$

So my question is, why would we need to add a third gate, if already 2 gates suffice, to gain every possible probability when measuring?


At the start of the circuit you're right that you only need two parameters. This is actually easy to show if you decompose into a sequence of rotations starting with a Z rotation, because Z rotations have no effect on $|0\rangle$, so clearly that Z rotation angle would be irrelevant.

But in the middle of a circuit, a gate is likely operating on a state that is entangled with other qubits. For these states, all three parameters are relevant. You can see this for yourself by preparing the state $|00\rangle + |11\rangle$ and putting a gate on one of the qubits.

  • $\begingroup$ Okay, I think I don't get it, sorry. Let $\psi$ be $\frac{1}{\sqrt2}(|00> + |11>)$ I've calculated $R_x \psi , R_y \psi , R_z \psi$ and it seems to me, that only $R_z$ induces a measurable difference in probabilities. I calculated all 3 of them with the state $|00>$ and the state $|01>$ only, and again, it seems that $R_x and R_y$ do not change the probabilities.. but just the complex part, which actually doesn't change probabilities. Sorry, that I don't understand it yet. Did I miscalculate? Should the states differ in measurable ways? $\endgroup$
    – eli44
    Apr 30 at 6:24
  • $\begingroup$ ah, to be precise: I have applied the gates to the first qubit. Afterwards I have applied it (in a separate calculation) to the second qubit, but it didn't change the measurable probabilities. Which makes sense, because they are only differ in an -i on the left bottom entry.. And doesn't that mean, that in general I don't get $M_z(R_x \psi) \ne M_z(R_y \psi)$, where $M_z()$ is a measurement on z-basis? $\endgroup$
    – eli44
    Apr 30 at 7:24
  • $\begingroup$ @eli44 Well what affects measurements depends on what comes later. The simplest thing you can do to make them all matter in a tiny circuit is H, CNOT, (your gate), CNOT, H, measure both. $\endgroup$ Apr 30 at 11:17
  • $\begingroup$ okay, I think I got it now. Thanks for the explanation $\endgroup$
    – eli44
    Apr 30 at 12:41

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