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, Pieter Feb 16 '18 at 18:53

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  • $\begingroup$ Why do you believe the given version of the uncertainty relation is tight? $\endgroup$ – Norbert Schuch Feb 11 '18 at 7:48
  • $\begingroup$ @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... $\endgroup$ – m137 Feb 11 '18 at 13:33
  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Feb 11 '18 at 13:52
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    $\begingroup$ 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.) $\endgroup$ – Norbert Schuch Feb 11 '18 at 13:55
  • $\begingroup$ @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. $\endgroup$ – 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

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

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

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

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