# Why are 3, rather than 2 gates used in quantum variational circuits?

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.
• 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? Apr 30 at 6:24
• 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? Apr 30 at 7:24