Analyzing Bell's inequality for different measurement angles

  • #1
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Main Question or Discussion Point

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.

However, if we choose ##\bf{a}=e_x##, ##\bf{b}=e_y##, ##\bf{c}=e_z##, then we get
$$\left|0-0\right|\leq 1-0$$
so the inequality is valid. I now wanted to find out exactly when the inequality breaks down and interpolated by choosing ##\bf{a}=e_x##, ##\bf{b}=\cos{\phi}\cdot e_y+\sin{\phi}\cdot \left(e_x+e_z\right)/\sqrt{2}##, ##\bf{c}=e_z## with ##0\leq\phi\leq\pi/2##. Plugging this in we get
$$\left|-\frac{\sin{\phi}}{\sqrt{2}}-0\right|=\frac{\sin{\phi}}{\sqrt{2}}\leq 1-\frac{\sin{\phi}}{\sqrt{2}}$$
and solving for equality yields ##\phi=\pi/4##.

So far so good, since the Bell inequality is valid for ##0\leq\phi\leq\pi/4##, there should be a local-realistic description of the system for these values of ##\phi##. Hence there must be a separable, most probably mixed density operator that yields the same expectation values as ##\left|\psi\right\rangle## for these choices of measurement angles. How can I find this density operator? I'm interested in its structure, and how it breaks down when crossing the magic angle ##\phi=\pi/4## from below.
 

Answers and Replies

  • #2
jfizzix
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  • #3
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That's not exactly what I want, I'm trying to construct an explicitly non-entangled (separable, can be written as a convex combination of product states) density operator for the measurement directions I mentioned.
 
  • #4
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Actually, quantum systems that don't violate any kind of Bell inequality can still be entangled.

In 1989, a prof. R. F. Werner proved there were entangled states which admit a local hidden variable model
http://pra.aps.org/abstract/PRA/v40/i8/p4277_1
Isn't Professor Werner using a product state in the article above?
 
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  • #6
meBigGuy
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I may be connecting two incompatible concepts, but are you simply noticing the pi/4 intersection points of the correlation curves for entangled vs classical systems? Look at the graph on this page.

http://en.wikipedia.org/wiki/Bell's_theorem
 
  • #7
jfizzix
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Isn't Professor Werner using a product state in the article above?
Prof. Werner is using a state which is a mixture of the of the maximally mixed state (the identity matrix) and the matrix representing a swap between systems.

This swap operator on [itex]\psi^{A}\otimes\psi^{B}[/itex] would give you [itex]\psi^{B}\otimes\psi^{A}[/itex]
 

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