What are the possible results of measuring $X$ and $Z$ on the state $|01rangle+|10rangle$?

Starting with the state $|psi rangle = dfrac{|01rangle + |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 $M = |+ranglelangle +| otimes |0ranglelangle 0 |$ since $|+rangle = dfrac{1}{sqrt{2}} begin{pmatrix} 1 1 end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $X$ and $|0rangle = begin{pmatrix} 1 0 end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $Z$. So explicitly

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

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

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You can extend this to other cases as well.

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Update:

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

For instance, if we again start with $|psi rangle = dfrac{|01rangle + |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 & 1end{pmatrix} &= dfrac{1}{2} begin{pmatrix} 1 & 1 1 & 1 end{pmatrix} otimes begin{pmatrix} 1 & 0 0 & 1end{pmatrix} &= dfrac{1}{2}begin{pmatrix} 1 & 0 & 1 & 0 0 & 1 & 0 & 1 1 & 0 & 1 & 0 0 & 1 & 0 & 1end{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 & 1end{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 $M = |+ranglelangle +| otimes |0ranglelangle 0 |$ (as indicated why on the top of this answer). Hence this probability is

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

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

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

I would start by rewriting the same state in different bases:

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

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).