0
$\begingroup$

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'm not really sure about this method.

I also found this solution:

and I cannot understand what they compute.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think when your question asks for the "average value" it means the expectation value.

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Labo
    Commented Aug 20, 2018 at 12:15
  • $\begingroup$ No worries. It's a bit annoying when homework questions are imprecise in their language, they should have asked for the expectation value. $\endgroup$
    – TanyaR
    Commented Aug 21, 2018 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.