Bell experiment: Rotate measurement device 180°

  • #1
greypilgrim
443
28
Hi,

I was just writing another thread when I stumbled upon something strange:
Hi,

I'm trying get a better understanding of Bell's inequality in the form
$$\left|E\left(\bf{a},\bf{b}\right) -E\left(\bf{a},\bf{c}\right)\right|\leq 1+E\left(\bf{b},\bf{c}\right)\enspace.$$
I'm considering the Bell state
$$\left|\psi\right\rangle= \frac{1}{\sqrt{2}}\left(\left|+\right\rangle_1\left|-\right\rangle_2- \left|-\right\rangle_1\left|+\right\rangle_2\right)\enspace.$$
and the expectation value ##E^{qt}## of the product of the result of a measurement of spin 1 in direction ##\bf{a}## and spin 2 in direction ##\bf{b}##
$$E^{qt}\left(\bf{a},\bf{b}\right)= \left\langle\psi\right|\vec{\sigma_1} \cdot\bf{a}\otimes \vec{\sigma_2} \cdot\bf{b} \left|\psi\right\rangle=-\bf{a}\cdot\bf{b}$$
which is a straigthforward calculation. I'm using the notation from 1.5.1 in
http://www.uibk.ac.at/exphys/photonik/people/gwdiss.pdf
which is, however, in German.

We can violate the inequality by choosing e.g. ##\bf{a}=e_x##, ##\bf{b}=\left(e_x+e_z\right)/\sqrt{2}##, ##\bf{c}=e_z## which yields
$$\left|-\frac{1}{\sqrt{2}}-0\right|=\frac{1}{\sqrt{2}}\leq 1-\frac{1}{\sqrt{2}}$$
which is obviously wrong. I think this choice of vectors also maximally violates the inequality.

What if I now make a slight change and replace ##\bf{b}\rightarrow-\bf{b}##? The expectation values containing ##\bf{b}## change signs and I get
$$\left|\frac{1}{\sqrt{2}}-0\right|=\frac{1}{\sqrt{2}}\leq 1+\frac{1}{\sqrt{2}}$$
which is correct! But I've only turned my measurement device 180°, i.e. exchanged the (+1) and (-1) results, or more mathematically, I've permuted my measurement basis. How can this lead to such a profound change, i.e. this would allow a local-realistic description for these measurement angles? Surely there has to be something wrong.
 

Answers and Replies

  • #2
jk22
723
24
I don't see something wrong, but Bell's inequality is violated only by some cases of the measurement configuration by QM.
 
  • #3
greypilgrim
443
28
Yes I know that, but in this case I'm not even really changing the configuration but only swapping the up/down-outcomes. The measurement basis is the same as before (only permuted), that confuses me.
 
  • #4
jk22
723
24
no your b vector is rotated by 180 degrees so the angles are not the same any more. Usually Bell inequalities are violated when the angles between them is 45 degrees.

You don't change +/- outcome, those are always given by probability 1/2 1/2 by QM.

But it's not because you can simulate the result by hidden variables that nature will do it. How will it know the configuration permits to do that ?
 
Last edited:
  • #5
DrChinese
Science Advisor
Gold Member
7,664
1,513
Usually Bell inequalities are violated when the angles between them is 45 degrees.

This is not correct. There are combinations of angles, usually 3 or more, that lead to an inequality. Not all combinations do have that feature. 45 degrees can be used with other specific ones.
 
  • #6
greypilgrim
443
28
I'm still confused. ##\vec{\sigma_2} \cdot\bf{b}## and ##-\vec{\sigma_2} \cdot\bf{b}## have exactly the same eigenvectors. This means the measurement statistics (probabilities) are exactly the same, and so are the post-measurement states. Only the results (+1) and (-1) are swapped. I still don't see how this is "significant enough" such that the Bell inequality is suddenly satisfied.
 

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