# Average value of non-projective observables

I am quite confused about how to measure observables (like Pauli spins).

For example, in the exercise 2.66 of Nielsen and Chuang's textbook:

Show that the average value of the observable $X_1Z_2$ for a two qubit system measured in the state $(\vert00\rangle+\vert11\rangle)/\sqrt2$ is zero.

My first take is to decompose the system into eigenvalues of the operators:

$$(\vert00\rangle+\vert11\rangle)/\sqrt2 = \left(\frac{\vert+\rangle+\vert-\rangle}{\sqrt2}\vert0\rangle + \frac{\vert+\rangle-\vert-\rangle}{\sqrt2}\vert1\rangle\right)/\sqrt2$$

So, each one of the four possibilities appears with the same probability 1/4, and the average is $$\frac{1*1\ +\ (-1)*1\ +\ 1*(-1)\ +\ (-1)*(-1)}{4}=\frac{1-1-1+1}{4}=0$$

Now, I observe that just computing $\langle\phi\vert X_1Z_2\vert\phi\rangle$ also works although $X_1Z_2$ is not projective:

\begin{align} 2 * \langle\phi\vert X_1Z_2\vert\phi\rangle &= \langle00\vert X_1Z_2 \vert00\rangle + \langle00\vert X_1Z_2 \vert11\rangle + \langle11\vert X_1Z_2 \vert00\rangle + \langle11\vert X_1Z_2 \vert11\rangle \\ &= \langle00|10\rangle - \langle00|01\rangle + \langle11|10\rangle - \langle11|01\rangle\\ &= 0+0+0+0=0 \end{align}

I also found this solution:  and I cannot understand what they compute.

The expectation value of the operator $A$ in the state $|\psi>$ is given as $<\psi|A|\psi>$. https://en.wikipedia.org/wiki/Expectation_value_(quantum_mechanics)
In your case, $A=X_1Z_2$ and $|\psi> = 1/\sqrt(2) (|00>+|11>)$. Now just plug and chug. If |0> or |1> is not an eigenstate of $X$ or $Z$ then you will need to re-express your state vector in the correct eigenbasis, as you mentioned in your question. I don't know how any of your operators and their eigenvectors/eigenvalues are defined, as I don't have that textbook on hand. But I think your first step is likely correct, and you just need to plug that into the expression for the expectation value.
• Thank you! The operators corresponded to the Pauli matrices. I didn't know this formula and was under the impression that $A^*A$ should be used. arxiv.org/pdf/1701.01409 presents some demonstrations of the formula in the introduction. – Labo Aug 20 '18 at 12:15