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When calculating the probability of getting +1 on X-basis on the first qubit of Bell's state $|01\rangle+|10\rangle$, the result is 1/2 with the state after measurement |++⟩ while the probability of measuring the second qubit with the collapse state is 1 and the state after measurement also $|++\rangle$.

When calculating the probability of getting +1 on Z-basis on the first qubit of Bell's state $|01\rangle+|10\rangle$, the result is 1/2 with the state after measurement |01⟩ while the probability of measuring the second qubit with the collapse state is 0 and the state after measurement also $|01\rangle$.

What is other possible result of measuring X and Z, for example, XX, XZ, ZX, ZZ ?

How to build a table to compile all of the possible results?

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Starting with the state $|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} $.

If you want to find the probability of measuring $+1$ in observable $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $ for the first qubit, and $+1$ in the observable $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $ for the second qubit then you can calculate it as $\langle \psi| M | \psi \rangle = Tr(\rho M)$ where $\rho = |\psi \rangle \langle \psi |$ and here $M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$ since $|+\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $X$ and $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $Z$. So explicitly

\begin{align} M = |+\rangle \langle+| \otimes |0\rangle\langle 0| &= \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \bigg[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \bigg] \\ &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \end{align} Thus, $$\langle \psi| M | \psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 1 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \dfrac{1}{4} $$

Also note that, the above is the same if we have done $Tr(\rho M)$ since $$Tr\bigg( \rho M \bigg) = Tr\bigg( \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \bigg) = \dfrac{1}{4}$$


You can extend this to other cases as well.


Update:

If you want to do sequential measurement, then you can find the the post measurement state $|\psi\rangle_{post}$ then follow the same procedure.

For instance, if we again start with $|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} $

and we want to find the probability of measuring $+1$ in observable $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $ for the first qubit. Then afterward, finding the probability of measuring $+1$ in observable $X $ for the first qubit, and $+1$ in the observable $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $ for the second qubit on this collapsed state then we can do it as follow:

First Step: To find the probability of measuring $+1$ in observable $X $ we can construct $M $ as \begin{align} M = |+\rangle \langle+| \otimes I = \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\ &= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \end{align}

And therefore,

$$\langle \psi| M | \psi \rangle = \dfrac{1}{2} $$

and the state after measurement, $|\psi_{post}\rangle $, is going to be \begin{align} |\psi_{post}\rangle = \dfrac{ M |\psi \rangle }{ \sqrt{prob(+1)}} = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} }{ \sqrt{ 1/\sqrt{2} } } = \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \end{align}

Second Step: Now the probability of measuring $+1$ in observable $X $ for the first qubit, and $+1$ in the observable $Z $ for the second qubit on this collapsed state $|\psi_{post} \rangle$ can be calculated as $\langle \psi_{post} | M | \psi_{post} \rangle $ where again $M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$ (as indicated why on the top of this answer). Hence this probability is

$$ \langle \psi_{post} | M | \psi_{post} \rangle = \dfrac{1}{2} \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = \dfrac{1}{2} $$

Where the post state after this process, $|\psi_{post 2} \rangle$ is now in the state

$$ |\psi_{post 2} \rangle = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} }{ \sqrt{1/2} } = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \dfrac{|00\rangle + |10\rangle }{\sqrt{2}} $$

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  • $\begingroup$ for the second qubit, suppose the measurement needs to use the collapse state of measuring the first qubit. can you show a measurement of observable Z for the first qubit and then measuring the second qubit using the collapse state? $\endgroup$ Nov 22 '20 at 19:42
  • $\begingroup$ how to relate the measurement with the non-locality of quantum entanglement? how they are correlated? @KAJ226 $\endgroup$ Nov 22 '20 at 19:48
  • $\begingroup$ can you explain why the M = |+><+| \otimes |0><0| not I\otimes |0><0| ? $\endgroup$ Nov 23 '20 at 0:02
  • $\begingroup$ @EaraShahirah That is because I wanted to find the probability of measuring $+1$ in observable $X$ for the first qubit, and $+1$ in the observable $Z$ for the second qubit on this collapsed state... But if you want to find the probability of calculating $+1$ in the $Z$ observable for the second qubit then you would make $M = I \otimes |0\rangle \langle 0|$ like you wrote. Hope that helps. $\endgroup$
    – KAJ226
    Nov 23 '20 at 0:51
  • $\begingroup$ Thank you for your answer. I got it now. But, can you give me insight behind all this calculation? what actually this calculation is for? @KAJ226 $\endgroup$ Nov 23 '20 at 15:33
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I would start by rewriting the same state in different bases:

  • $XX$ basis: $\frac{1}{\sqrt{2}}\left(|++\rangle-|--\rangle\right)$
  • $XZ$ basis: $\frac{1}{2}\left(|+1\rangle+|-1\rangle+|+0\rangle-|-0\rangle\right)$
  • $ZX$ basis: $\frac{1}{2}\left(|0+\rangle-|0-\rangle+|1+\rangle+|1-\rangle\right)$
  • $ZZ$ basis: $\frac{1}{\sqrt{2}}\left(|01\rangle+|10\rangle\right)$

Now we can look at all the possibilities where $M_1M_2$ indicates first measuring the first qubit in the $M_1$ basis and then measuring the second qubit in the $M_2$ basis:

$X_1X_2$:
$P(X_1=1) = \frac{1}{2}, P(X_2=1|X_1=1) = 1, P(X_2=-1|X_1=1) = 0$
$P(X_1=-1) = \frac{1}{2}, P(X_2=1|X_1=-1) = 0, P(X_2=-1|X_1=-1) = 1$

$X_1Z_2$:
$P(X_1=1) = \frac{1}{2}, P(Z_2=1|X_1=1) = \frac{1}{2}, P(Z_2=-1|X_1=1) = \frac{1}{2}$
$P(X_1=-1) = \frac{1}{2}, P(Z_2=1|X_1=-1) = \frac{1}{2}, P(Z_2=-1|X_1=-1) = \frac{1}{2}$

$Z_1X_2$:
$P(Z_1=1) = \frac{1}{2}, P(X_2=1|Z_1=1) = \frac{1}{2}, P(X_2=-1|Z_1=1) = \frac{1}{2}$
$P(Z_1=-1) = \frac{1}{2}, P(X_2=1|Z_1=-1) = \frac{1}{2}, P(X_2=-1|Z_1=-1) = \frac{1}{2}$

$Z_1Z_2$:
$P(Z_1=1) = \frac{1}{2}, P(Z_2=1|Z_1=1) = 0, P(Z_2=-1|Z_1=1) = 1$
$P(Z_1=-1) = \frac{1}{2}, P(Z_2=1|Z_1=-1) = 1, P(Z_2=-1|Z_1=-1) = 0$

In summary

  1. if you're measuring in $XX$, it's $\frac{1}{2}$ for the first measurement and the second measurement is fully correlated to the first one,
  2. if you're measuring in $ZZ$, it's also $\frac{1}{2}$ for the first measurement but the second measurement is fully anti-correlated to the first one,
  3. For the other two combinations, it's all $\frac{1}{2}$ for both measurements regardless of the outcome of the first measurement (because the two measurements happen to be completely uncorrelated).
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