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Suppose we have the pair of entangled qubits:

$$ |\psi\rangle = \frac{1}{ \sqrt{2} } ( |00\rangle + |11\rangle ),$$

where $|\psi\rangle$ is one of the $4$ Bell states.

Suppose now that we give one qubit to Alice and the other to Bob. Alice measures her qubit so that to know the spin in the z-direction and, simultaneously, Bob measures the spin in the x-direction. In other words, they simultaneously measure their qubits using different bases.

Alice and Bob both know that these qubits are entangled. Whatever is the value of spin for the first qubit (Alice's qubit) in the z-direction, the same will be the value of the spin in the z-direction for the second qubit (Bob's qubit). Similarly, the spin value in the x-direction, which Bob measures, will be the same for both qubits. Hence, we have known, simultaneously, the spin in both x- and z-directions. But this would violate Heisenberg's uncertainty principle.

Why can (or not) we measure the entangled qubits, simultaneously, in both "bit" (i.e. the basis $\{ |0\rangle, |1\rangle \}$) and "sign" (i.e. the basis $\{ |+\rangle, |-\rangle \}$) bases?

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  • $\begingroup$ We can measure the entangled qubits simultaneously in both bit and sign basis. But that doesn't tell us what state they were originally in. $\endgroup$ – Peter Shor Jun 13 '13 at 13:30
  • $\begingroup$ @PeterShor What do you mean exactly by "state they were originally in"? Are you referring to the entangled state? Anyway, it's not clear, at least to me, how your statement answers the original question or clarifies the original doubt, which I also have. People often present this problem as Alice measuring and then Bob measuring, and not a simultaneously measurements using different bases. Only the fact that they simultaneously measure, for me, it's problematic and not explained anywhere, unless I am missing something fundamental (which is also very likely). $\endgroup$ – nbro Jun 6 '18 at 0:43
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    $\begingroup$ @nbro: comments aren't for answers. My comment was not intended to answer the whole original question. $\endgroup$ – Peter Shor Jun 6 '18 at 3:03
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[EDIT]

0) In the EPR paradox (in fact the CHSH version), based on the hypothesis of local realism , the apparent impossibility of measuring $x$,$z$ polarizations for an entangled state, means that the choice would be between :

a)some interaction exists between the particles, even though they were separated

b) (realism) the information about the outcome of all possible measurements was already present in both particles (hidden parameters).

The first possibility seems incompatible with locality, so Einstein chooses the second possibility (realism), and says that quantum mechanic was incomplete (because we have to add these hidden parameters to the description of each particle).

But experience has showed that Quantum Mechanics violate Bell's inequalities (local realism). So we have to choose between get rid of locality, or get rid of realism. The correct choice is get rid of realism, that is, you cannot consider the 2 particles individually, you have to consider the entangled particles as a whole, you cannot consider the 2 particles separately . This does not mean that Quantum Mechanics violates locality, Quantum Mechanics respects locality. For instance, with entangled particles, it is not possible to send information instantly. Simply, the quantum correlations are stronger than classical correlations.

1) So one first fundamental idea is that the entangled 2-quit state is a whole, and cannot be divided into more little units. You cannot separate the 2 qbits and the operators acting on them.

2) Alice and Bob can freely choose the orientation of their measurement apparatus, and they always obtain an outcome (your video is wrong about this) which maybe 0 or 1.

3) The fact that quantum mechanics does not respect realism can be seen in the mathematical formalism. For instance, measurements are operators like 2*2 Pauli Matrices. A state is a 2-dimensional complex vector. So applying a measurement to a state, is the same thing that looking at the result of a matrix applying to a vector, for instance :

$$\sigma_x |0\rangle = |1\rangle$$

You see that the measurement change the state, the state after the measurement is not the same that the state before the measurement, so there is no more realism

4) In the case of 2 entangled qbits quantum mechanics, you have to use the formalism of tensorial products. If Alice choose a z-axis measurement, and Bob a x-axis measurement, that means that the measurement operator is :

$$\sigma_z \otimes \sigma_x$$ where $\sigma_z$, $\sigma_x$, are

The above expression is a tensorial product of operators, here it is the tensorial product of the matrices $\sigma_z$ and $\sigma_x$. It works like this, suppose a separable state $|a\rangle \otimes ~|b\rangle$, then :

$$(\sigma_z \otimes \sigma_x) (|a\rangle \otimes ~|b\rangle) = (\sigma_z|a\rangle) \otimes ~ (\sigma_x|b\rangle)$$ For instance, with your entangled state $|\Psi\rangle= \frac{1}{\sqrt{2}} (|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle )$, it gives :

$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}(\sigma_z \otimes \sigma_x) (|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle ))$$

$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}((\sigma_z|0\rangle \otimes ~ (\sigma_x|0\rangle\rangle) + (\sigma_z|1\rangle) \otimes ~ (\sigma_x|1\rangle))$$

$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}((|0\rangle \otimes ~ (|1\rangle) - (|1\rangle) \otimes ~ (|0\rangle))$$

You see again that the measurement changes the state, so there is no more realism here too.

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  • $\begingroup$ can u please elaborate ur answer and explain it in simple english, also please have a look at first 4 min of this vedio $\endgroup$ – kansi Jun 12 '13 at 21:45
  • $\begingroup$ I re-write the answer. Hopes it helps $\endgroup$ – Trimok Jun 13 '13 at 11:26
  • $\begingroup$ Thanks for the answer, was able to understand it, but let me get this straight, what ur trying to say is that measurement changes the state. So when alice measures the spin in z direction it causes the state of the entangled qubits to change as a whole and hence when bob measures the spin in x direction, the state of the entangled qubits is not same as it was when alice measured it, so he would get an answer, but the answer would not correspond to the state of qubits we started of with. If this is true, then what if they measure the entangled qubits at the same instant $\endgroup$ – kansi Jun 13 '13 at 17:02
  • $\begingroup$ The measurement is "global", it is $\sigma_z \otimes \sigma_x$, you cannot separate in a $\sigma_z$ measurement and a $\sigma_x$ measurement. It is a "whole" measurement which applyes on a "whole" state. This is the formalism of quantum mechanics. $\endgroup$ – Trimok Jun 13 '13 at 17:08
  • $\begingroup$ thank for ur answer, really cleared some concepts for me. :) $\endgroup$ – kansi Jun 13 '13 at 18:00

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