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?


1 Answer 1


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, 2021 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, 2021 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, 2021 at 11:17
  • $\begingroup$ okay, I think I got it now. Thanks for the explanation $\endgroup$
    – eli44
    Apr 30, 2021 at 12:41

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