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In this exercise I need to find the expectation value of the observable $M=X_1 \otimes Z_2$ for two qubit system measured in the state $\dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$.

I know that $E[M]=\langle\psi|M|\psi\rangle$ = $\left(\dfrac{\langle00| + \langle11|}{\sqrt{2}}\right) M \left(\dfrac{|00\rangle + |11\rangle}{\sqrt{2}}\right)$

But I'm having trouble finding the correct result (which is 0). Thank you!

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Taking the last two terms of last expression you gave, we can do the following

$$ \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\rangle \otimes |0\rangle+|0\rangle \otimes -|1\rangle}{\sqrt{2}}\right) = \left(\frac{|10\rangle-|01\rangle}{\sqrt{2}}\right) \end{align} $$

Now, you can plug this in into the equation for the expectation value

$$ \begin{align} E[M]&=\left(\frac{\langle00|+\langle11|}{\sqrt{2}}\right)\left(\frac{|10\rangle-|01\rangle}{\sqrt{2}}\right) \\ &= \frac{1}{2}\left( \langle00|10\rangle-\langle00|01\rangle+\langle11|10\rangle-\langle11|01\rangle \right) = 0 \end{align} $$

As you can see, you end up with four inner products, all between orthogonal states, which means all of them evaluate to $0$.

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    $\begingroup$ Thank you so much!!! A special thanks also for your response speed. $\endgroup$
    – user18257
    Sep 11 at 14:37
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An intuitive way to think about it is that $E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$

If we only think about $E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of $Z_2$ on the second qubit. Consider that our second Qubit in the entangled state $\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$ is measured to be $\frac{+\hbar}{2}$ half the time and $\frac{-\hbar}{2}$ half the time. Therefore by observation $E[\mathbb{1} \otimes Z_2]=0$.

$$E[M]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]=0$$

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