# Find expectation value of the observable $X_1\otimes Z_2$ for a maximally entangled two-qubit system

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!

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 $$0$$.

• Thank you so much!!! A special thanks also for your response speed. Sep 11, 2021 at 14:37

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

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