Analyzing Bell's inequality for different measurement angles

[greypilgrim]

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.

 

[jfizzix]

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

There are also now necessary and sufficient conditions for violating a Bell inequality for pairs of two-level systems which may help you in your search
http://www.sciencedirect.com/science/article/pii/037596019500214N#

Hope this helps:)

 

[greypilgrim]

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.

 

[StevieTNZ]

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?

 

[StevieTNZ]

Just seen this: http://pra.aps.org/accepted/20076N60Sa11d318e22f12e2030ab6559c9459dd3

I thought all entangled states violate a Bell, or other, inequality?

Now, all states that are nonseparable and pure (i.e., vector states) violate a Bell’s inequality (Popescu & Rohrlich)

 

[meBigGuy]

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

 

[jfizzix]

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 $\psi^{A}\otimes\psi^{B}$ would give you $\psi^{B}\otimes\psi^{A}$