EPR-Bohm paradox and Robertson uncertainty relation [closed]

Consider Bohm's version of the EPR paradox, where spin-1/2 are used. The bipartite system is in the state $$\vert \Psi\rangle = \dfrac{1}{\sqrt{2}} \left( \vert \uparrow_x \rangle_A \vert \downarrow_x \rangle_B - \vert \downarrow_x \rangle_A \vert \uparrow_x \rangle_B \right) \;.$$ Measuring particle $A$'s spin allows to predict particle $B$'s spin with certainty. This is true for any spin measurement direction.

Following the EPR argument it looks like $\sigma_x^B$, $\sigma_y^B$ and $\sigma_z^B$ could be simultaneously measured, which should be paradoxical.

However, the uncertainty relation for spins $$\Delta^2 \sigma_z^B \Delta^2 \sigma_x^B \geq \dfrac{1}{4} \left| \langle \left[ \sigma_z^B, \sigma_x^B \right] \rangle \right|^2 = \dfrac{1}{4} \left| \langle \sigma_y^B \rangle \right|^2 \;,$$ is not $=0$ for the state considered above? This would mean that $B$ can indeed measure simultaneously its spin components, and that no paradox subsists.

closed as unclear what you're asking by Norbert Schuch, Chris♦, Emilio Pisanty, sammy gerbil, PieterFeb 16 '18 at 18:53

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• Why do you believe the given version of the uncertainty relation is tight? – Norbert Schuch Feb 11 '18 at 7:48
• @NorbertSchuch, I do not believe it is tight, in fact is should be $1/4*1/4 \geq 0$. But the problem is not this, I think... – m137 Feb 11 '18 at 13:33
• Then what is the problem, in your opinion? If the inequality is not tight is see no contradiction. --- In any case, it is not clear what your question is. The only thing terminated by a question mark refers to the Robertson inequality. – Norbert Schuch Feb 11 '18 at 13:52
• You might e.g. consider the introduction of arxiv.org/abs/1512.02383. (Random google hit on Robertson and tight, but there has been quite a body of work on different settings for uncertainty relations recently.) – Norbert Schuch Feb 11 '18 at 13:55
• @NorbertSchuch: I guess the question is : “Where does the paradox comes from ?”. And actually (see my anwer), one can deduce it from the Robertson inequality plus a simple argument, even if the latter is not tight. – Frédéric Grosshans Feb 14 '18 at 16:29

You should not forget that the Pauli obervables $\sigma_x$, $\sigma_y$ and $\sigma_z$ can only take $±1$ values. From this, you can easily deduce that $\langle \sigma_y \rangle =0$ iff $\Delta^2\sigma_y=1$ which means that perfectly knowing $\sigma_x$ or $\sigma_z$ implies that $\sigma_y$ is maximally unknown. By symmetry over permutations of $x$, $y$ and $z$, it implies that only one of them can be known perfectly.

More quantitatively, it is easy to deduce from the $±1$ values of $\sigma_y$ that: \begin{gather} \left|\left<\sigma_y\right>\right|^2=1-\Delta^2\sigma_y\\ \Delta^2\sigma_x \Delta^2\sigma_z + \frac14 \Delta^2\sigma_y ≥ \frac14 \end{gather} which, if not very elegant, clearly shows that all tree variances cannot be simultaneously $0$. And, of course, it should be completed by the two other relations to be complete: \begin{gather} \Delta^2\sigma_x \Delta^2\sigma_y + \frac14 \Delta^2\sigma_z ≥ \frac14\\ \Delta^2\sigma_y \Delta^2\sigma_z + \frac14 \Delta^2\sigma_x ≥ \frac14 \end{gather}

Norbert, this version appears from the mean square uncertainty of a set of operators are:

$\Delta <A>^2 =\ <\psi|(A - <A>)^2|\psi>$

$=\ <\psi|A^2|\psi>$

$\Delta <B>^2 =\ <\psi|(B - <B>)^2|\psi>$

$=\ <\psi|B^2|\psi>$

The scalar product of $A|\psi> + i \lambda B|\psi>$ - as a modulus, the scalar product must be greater or equal to zero. Expanding you get

$<\psi|A^2|\psi> + \lambda^2<\psi|B^2|\psi> + i \lambda <A\psi|B\psi>\ \geq 0$

After reorganizing the inequalities in terms of uncertainties you can find the following identity:

$\Delta <A>^2 \Delta <B>^2\ \geq - \frac{1}{4}<\psi|[A,B]|\psi>$

The arbtitrary operators A and B are given as

$A = (A - <A>)^2$

$B = (B - <B>)^2$

• Don’t worry, @NorbertSchuch knows perfectly well how to derive Robertson’s inequality and he probably has taught it many times to his students. – Frédéric Grosshans Feb 14 '18 at 17:21
• deleted, I misread him. – Gareth Meredith Feb 14 '18 at 18:00