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We are told that $$Z|0 \rangle = | 0\rangle \\ Z | 1\rangle =-|1\rangle \\ X|0\rangle =|1\rangle \\ X|1\rangle =|0\rangle$$

and we have the state $$|\psi\rangle =|0\rangle |1\rangle +|1\rangle |0\rangle$$

The question is: Without doing an explicit calculation explain why expectation of Z*X and X*Z in the given state are the same?

My answer: I kind of done the calculation in my head and said that "For the right hand side of both inner products Z will only change the coefficient of a ket while X will switch the ket, resulting in tensor products of similar kets i.e either $|0\rangle|0\rangle$ and $|1\rangle |1\rangle$. This won't match up to the left hand side $\langle\psi|$ which means both expectations will be zero."

I don't think this is the non-calculation answer they were looking for. X and Z do not commute, my only other guess is because the state is entangled. Are there any properties linear operators can have that make their expectations the same either way round?

Martin
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Traceback
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I assume $Z*X$ means the operator that acts as $Z$ on the first tensor factor and as $X$ on the second, and analogously for $X*Z$. If so, then $Z*X=T\cdot(X*Z)\cdot T$, where $T$ is the linear operator that interchanges the two tensor factors. Using this and the fact that your particular state $|\psi\rangle$ satisfies $T|\psi\rangle=|\psi\rangle$ and $\langle\psi|T=\langle\psi|$, you get $$ \langle\psi|Z*X|\psi\rangle=\langle\psi|T(X*Z)T|\psi\rangle=\langle\psi|X*Z|\psi\rangle. $$ I'm not sure whether this counts as an explicit computation and is therefore prohibited. If so, I recommend meditating on the equations (or on symmetry) instead of writing them.

Andreas Blass
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