Creating a Quantum Circuit for Measuring X⊗Z, Z⊗X, & Y⊗Y Simultaneously

[Travis091]

I'm trying to create a quantum circuit which does the following very simple thing:

Alice has two qubits, and she wants to measure the observables: X⊗Z, Z⊗X, and Y⊗Y, each taking values in {-1,1}.

Note that this is a commuting set of observables, so it should be possible to measure all three simultaneously.

Now when I try to build a quantum circuit which does this using only measurements in the computational basis, what I would do is the following:

I start with the observable X⊗Z, i.e. X on the first qubit and Z on the second qubit, then for the first qubit I would have a measurement in the computational basis conjugated by two Hadamard gates (so effectively measuring in the Hadamard basis then converting back) , whereas for the second I would simply measure directly in the computational basis.

Usually we throw away qubits after measuring them, but here we propagate them further to do Z⊗X. I would have a computational basis measurement on the first qubit, and a measurement conjugated by two Hadamard gates for the second qubit.

To do Y⊗Y, I do something analogous.

The problem I run into is that when I do the analysis things don't work out as they should. A simple property of the product (X⊗Z)(Z⊗X)(Y⊗Y)=(XZY⊗ZXY)=Identity, is that the product of the outcomes must always equal 1, but outcomes obtained in the above manner do not necessarily follow this rule. Therefore there must be something wrong in my conceptual understanding of how to measure observables in a quantum circuit. I would appreciate any help.

Thank you.

 

[VantagePoint72]

I start with the observable X⊗Z, i.e. X on the first qubit and Z on the second qubit, then for the first qubit I would have a measurement in the computational basis conjugated by two Hadamard gates (so effectively measuring in the Hadamard basis then converting back) , whereas for the second I would simply measure directly in the computational basis.

The problem is that that is not a measurements of X⊗Z, it's a measurement of X⊗I followed by a measurement of I⊗Z. That the three observables commuting with one another implies measurement commutativity hinges on the assumption that each measurement yields a single bit of information: the product of the individual qubit measurements. The procedure you're describing yields two bits of information and so you're running up against the usual fact that the Pauli matrices, taken individually, do not commute. What you need to do is devise a measurement that reveals the value of $x_1 z_2$ (and similarly with the other two measurements) that doesn't reveal $x_1$ and $z_2$ (etc) individually.