# Is this a Bell test?

Inspired by this article which uses a $$|+\rangle$$ state as a control for a $$CSWAP$$, I realised you can conditionally measure a qubit by (maybe) swapping it with an empty ancilla, measuring it and (maybe) swapping it back.

As such, you could prepare a Bell pair, (maybe) measure half of it and then check if the other half was affected as follows:

If measurement changes the top qubit then the second $$H$$ gate should put it back into the |+> state and we should measure random results. If the measurement doesn't change the top qubit, then it should turn the $$|+\rangle$$ back to $$|0\rangle$$, deterministically measuring 0. Right?

If so, then I'm hoping there's some extension of the uncertainty principle that would imply that because there is no knowledge about whether the second qubit was measured, we could learn something about the first qubit. I hope.

Of course I've tried to test this but I don't believe the qiskit simulator and none of the IBM devices allow operations to take place after measurement ("Qubit measurement not the final instruction. [7006]"). So can anyone think of a way I could test it? Does this count as a Bell Test? Has this been tried before? Am I missing something obvious?

Not sure if this is the right place to post this but just interested to hear people's thoughts/predictions.

To run the circuit in the OP, just move the second $$H$$-gate on $$q_0$$ in front of the measurement on $$q_2$$. This will have no impact on your outcomes distribution (i.e. these two operations commute).
• @JohnBot Sorry, that's on me. I was hoping to think of a more satisfying response than pointing you to the tensor product structure, but in the end nothing more straightforward is coming to mind. Consider the state of the circuit after the CSWAP as $\vert \psi \rangle$, and call $\tilde H = H \otimes \mathbf{1} \otimes \mathbf{1} \otimes \mathbf{1}$ and a measurement operator $\tilde M = \mathbf{1} \otimes \mathbf{1} \otimes M \otimes \mathbf{1}$. By the basic properties of operator tensor products $[ \tilde H, \tilde M] = 0$, which is not impacted by entanglement in $\vert \psi \rangle$. Sep 1 '20 at 3:29