Measurement of the spin of the EPR pair in two orthogonal directions and how did Einstein tackle this? Let us consider two types of measurement in the EPR experiment. In Bohm's description of this experiment, the state of the electron-positron (called the EPR pair) is given by
$$
|{\rm EPR} \rangle =\frac{1}{\sqrt{2}}\left(|e+\rangle|p-\rangle -  |e-\rangle|p+\rangle\right)
$$
CASE-I First, suppose that Alice measures the electron-spin in the $z$-direction and obtains $S_z=+1/2$. Then, if Bob measures the positron-spin in the $z$-direction, quantum mechanics predicts that he obtains $S_z=-1/2$ with probability $1$. Einstein would argue that this is because the spins-projections of the EPR pair were fixed from the very beginning the pair was created. If the spin projection of the electron was $+1/2$ in some direction, then that of the positron was $-1/2$ in the same direction or vice-versa (because the pair does not have any net spin). So Einstein's point was that the quantum state is an incomplete description of the state.
CASE-II Now, consider a second situation. Suppose that Alice again measures the spin in the $z$-direction and obtains $S_z=+1/2$. Then, if Bob decides to measure the spin not in the $z$-direction but in the $x$-direction, quantum mechanics predicts that he will obtain $S_x=+1/2$ with probability $0.5$ and $S_x=-1/2$ also with probability $0.5$. How would Einstein explain this?
If this simple prediction is borne out in the experiment, will it not demolish Einstein's argument i.e. the spin-projections were fixed from the beginning? I mean, Einstein would have said since $S_z=+1/2$ for the electron, it must have been that $S_z=-1/2$ for the position from the very beginning. But if this were really so, the probability of getting $S_x=\pm 1/2$ should both be zero. Am I wrong in this last argument?
 A: I'll explain the modern understanding of this situation first, and discuss Einstein briefly at the end. It does not seem to be crucial that the pair involves an electron and positron. Instead, let's assume that two qubits are prepared in the Bell state you have written,
$$\left| \text{Bell} \right\rangle = \left( \left| 01 \right\rangle - \left| 10 \right\rangle \right)/\sqrt{2} \, ,$$
in QI notation where $Z \left| 0 \right\rangle =  \left| 0 \right\rangle$ and $Z  \left| 1 \right\rangle = -  \left| 1 \right\rangle$. We prepare this state and give one qubit to Alice and the other to Bob, who are spatially separated.
The way I represent quantum measurements is formally explained here in the case of qubits. The representation follows from the Stinespring Dilation Theorem (see also Figure 1 of this work for a nice picture). Intuitively, it's the many-worlds picture of measurement. It reduces to Copenhagen when you project onto a particular outcome.
When a quantum system is measured, the system becomes entangled with the measurement apparatus in the measurement basis. To explain the formalism, imagine measuring an operator $A$ on a qubit, where $A$ has two eigenvalues, $a^{\,}_{\pm}$. We can write this observable as
$$ A \, = \, \sum\limits_{\pm} \, a^{\,}_{\pm} \, \mathbb{P}^{(\pm)}_{\,} \, , $$
where $\mathbb{P}^{(\pm)}_{\,}$ projects onto the eigenstates of $A$ with eigenvalues $a^{\,}_{\pm}$ (i.e., $A \, \mathbb{P}^{(\pm)}_{\,} = a^{\,}_{\pm}\, \mathbb{P}^{(\pm)}_{\,}$). For a Pauli matrix $A$ (or any operator with $A^2=\mathbb{1}$), $\mathbb{P}^{(\pm)}_{\,} = \left( \mathbb{1} \pm  A \right)/2$.
If the qubit is initially in the state $\left| \psi \right\rangle$, after measuring $A$, the state becomes
$$ \left| \psi \right\rangle \, \to \, \sum\limits_{\pm} \, \left( \mathbb{P}^{(\pm)}_{\,} \,\left| \psi \right\rangle \right)^{\,}_{\rm ph} \otimes \left| \pm \right\rangle^{\,}_{\rm out} \, , $$
where the state $\left| \pm \right\rangle^{\,}_{\rm out}$ is the post-measurement state of the measurement apparatus, which encodes the measurement outcome. The $\pm$ states are orthonormal. Also note that $\mathbb{P}^{(\pm)}_{\,} \,\left| \psi \right\rangle$ is simply the unnormalized collapsed wavefunction for the physical qubit given that outcome $\pm$ was observed (i.e., in the Copenhagen picture). However, in this representation, the expression above is already normalized, and no collapse is needed! All outcomes occur. Measurement entangles the apparatus / observer with the system along a particular outcome.
Case I: Suppose Alice measures $Z$ on her qubit. The Bell state is updated according to
$$\left| \text{Bell} \right\rangle \to \left( \left| 01 \right\rangle \otimes \left| 0 \right\rangle^{\,}_A - \left| 10 \right\rangle \otimes \left| 1 \right\rangle^{\,}_A\right)/\sqrt{2} \, ,$$
where the $A$ subscript denotes the qubit that encodes the outcome of Alice's measurement. If Bob then measures, the state becomes
$$\left| \text{Bell} \right\rangle \to \left( \left| 01 \right\rangle \otimes \left| 0 \right\rangle^{\,}_A \otimes \left| 1 \right\rangle^{\,}_B - \left| 10 \right\rangle \otimes \left| 1 \right\rangle^{\,}_A \otimes \left| 0 \right\rangle^{\,}_B \right)/\sqrt{2} \, ,$$
where the $B$ state labels Bob's outcome.
Importantly, nothing drastic happened to the physical state of the two qubits. This is what Einstein didn't realize at the time. And performing these operations in either order (or simultaneously) gives the same result. This is most clear from noticing that everything is invariant under relabelling $A \leftrightarrow B$ and then flipping both qubits. Contrary to Einstein's concerns: (1) there is no collapse or update required upon measurement of this state, (2) no information is sent upon measurement, and (3) nothing nonlocal happens (the measurement locally couples the measured qubit with the nearby measurement apparatus, no information is sent, etc.).
The way we know no information is sent upon measurement is that even if Bob knows he shares this Bell state with Alice, there is no operation he can perform on his qubit to tell if she measured, what she measured, or her outcome. If he also knows she intends to measure $Z$, there's still nothing he can do to tell whether she measured yet. The reason is physical: The order of measurements does not matter, and no information is transferred by a quantum measurement alone.
Case II: Now, suppose that Bob intends to measure $X$ instead. If we write the same Bell state in the $Z$ basis for Alice and the $X$ basis for Bob, it is instead
$$\left| \text{Bell} \right\rangle = \frac{1}{2}  \left( \left| 00 \right\rangle - \left| 01 \right\rangle  -\left| 10 \right\rangle -  \left| 11 \right\rangle \right) \, ,$$
where the two numbers refer to different operator bases (this is fine as long as we remember that). However, because the state is written in the measurement basis, basically the same thing happens as in the case of the $Z$ measurements above.
After both measurements are made (in either order), the state becomes
$$\left| \text{Bell} \right\rangle \to \frac{1}{2} \left( \left| 00,00 \right\rangle -  \left| 01,01 \right\rangle - \left| 10,10 \right\rangle -  \left| 11,11 \right\rangle \right) \, ,$$
where the digits before the commas in each ket give the physical states of both qubits (in the $Z$ and $X$ basis for Alice and Bob, respectively), and the digits after the commas are the recorded outcomes (in the same bases).
Again, the structure of the physical part of the state (in the measurement basis) does not change upon measurement. There is also nothing nonlocal and no need for action at a distance. The measurement devices are simply entangled with their respective qubits in the measurement basis. Nothing drastic happens, no information can be extracted, and the order of measurements is unimportant. These are merely properties of the Bell state.
Another important remark is that, if we don't tell either Alice or Bob that they share a Bell state, then if they measure $X$, $Y$, or $Z$ multiple times to construct expectation values and extract statistics, then on average, they get both outcomes with equal probability, no matter what the other party does. If they measure all operators, they conclude that their qubit is in the maximally mixed state $\rho = \mathbb{1}/2$, which is a random classical bit that contains no quantum information. It is only if Alice measures her qubit(s), communicates the outcome(s) to Bob, and Bob correctly performs an operation on his qubit that information can be transferred from Alice to Bob (e.g., quantum teleportation). The information is in the classical signal, which obeys relativistic causality / locality.
Einstein: I personally think Einstein would have been happy with this formulation of measurement, knowing what we know now. It is completely local, the state of the universe is completely deterministic under measurement (as opposed to probabilistic), everything is unitary, and nothing shocking or concerning happens. The only randomness is in the particular experience of classical objects, which see only one outcome. This is due to decoherence, which is needed to explain the emergence of classical physics anyway. Moreover, this formulation of measurements was used in this recent work to establish locality measurements (i.e., a finite speed of quantum information even in the presence of quantum measurements of an entangled state and instantaneous classical communication).
By comparison, Bohmian mechanics intrinsically violates locality, though it is also deterministic. As I understand, the main advantage (on paper) is "realism", meaning that our particular experience upon measurement could be predicted. However, this would require knowledge of the "pilot wave". But it seems to me this knowledge is experimentally inaccessible. So, for all experimental purposes, Bohmian mechanics is not practically a "real" theory. Which means it does no better than what I've written above, but also sacrifices locality. Copenhagen is neither local nor deterministic.
Einstein really liked locality, so that's why I think he would like this. I can't say how he would have reacted in 1935 if I showed him this. But I don't think that's an instructive thought experiment. This formulation obeys the principles he cared most about.
