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Basis definition: Begin with the 2 qubit basis corresponding to circular polarisation of photons

$\{\lvert++\rangle, \lvert+-\rangle, \lvert-+\rangle, \lvert--\rangle\}$

, where $\lvert +\rangle$ is clockwise polarisation and $\lvert -\rangle$ is anticlockwise, and e.g. $\lvert +-\rangle =\lvert +\rangle\otimes \lvert -\rangle$, thus the first ket is Alice and second ket is Bob:

So today in trying to understand more about the CHSH inequality and how a combination of $C_{NOT}$ and Hadamard gate will form the triplet bell state $\frac{1}{\sqrt{2}}(\lvert ++\rangle + \lvert --\rangle)$ with input state $\lvert ++ \rangle$, I decided to pick a rotation operator $R(\theta)$ in place of $C_{NOT}$ to slowly rotate the separable state into the bell state and track how the quantum correlations changes

For the given input state, $\lvert ++\rangle$, Alice's state after transformed by the Hadamard gate will become the linearly polarised state

$\lvert \updownarrow\rangle=\frac{1}{\sqrt{2}}(\lvert +\rangle+\lvert -\rangle)$ while Bob's state remains unchanged. Therefore the composite state at this stage is the product state

$\lvert \updownarrow -\rangle=\lvert \updownarrow\rangle\otimes \lvert - \rangle$

The rotation gate under this basis is given as follows

$$R (\theta)=\begin{pmatrix}1 & 0& 0& 0 \\ 0 & 1& 0& 0 \\ 0 & 0& \cos \theta& -\sin \theta\\ 0 & 0& \sin \theta& \cos \theta\end{pmatrix}$$

or more explicitly, the matrix elements (where $i,j \in \{++, +-, -+, --\}$) are

$$\langle i\rvert R (\theta)\lvert j\rangle=\begin{pmatrix}\langle ++\rvert R (\theta)\lvert ++\rangle & 0& 0& 0 \\ 0 & \langle +-\rvert R (\theta)\lvert +-\rangle& 0& 0 \\ 0 & 0& \langle -+\rvert R (\theta)\lvert -+\rangle & \langle -+\rvert R (\theta)\lvert --\rangle\\ 0 & 0& \langle --\rvert R (\theta)\lvert -+\rangle & \langle --\rvert R (\theta)\lvert --\rangle\end{pmatrix}$$

Now this product state is fed into the rotation gate, which gives:

$$\lvert m\rangle=R(\theta)\lvert \updownarrow -\rangle=\frac{1}{\sqrt{2}}(\lvert +-\rangle-\sin \theta \lvert -+\rangle+\cos \theta \lvert --\rangle)$$

To compute the CHSH correlation function, we need to pick two obsevables. The observables we will be using are $J,I$ where $I=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ is the identity and $J=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$

Define:

$J_A=J \otimes I_B$

$J_B=I_A \otimes J$

Letting the two eigenvalues of each observable be $1,-1$. Therefore, the quantum correlations $E(,)$ are computed as follows:

$E(1,1)=\langle J_AI_B\rangle = \langle m\rvert J_AI_B\lvert m\rangle = -\sqrt{2}\sin \theta$

$E(1,-1)=\langle J_AI_B\rangle = \sqrt{2}\cos \theta$

$E(-1,1)=\langle J_BI_A\rangle = \sqrt{2}\cos \theta$

$E(-1,-1)=\langle I_BI_A\rangle = 1$

And the correlation function gives $E(1,1) - E(1,-1) + E(-1,1) +E(-1,-1) = 1 -\sqrt{2}\sin \theta$

However, this is incorrect because if we pick $\theta = \frac{3\pi}{2}$, then $\lvert m\rangle$ becomes the aforementioned bell state, which is maximally entangled, thus the CHSH correlation function should give the Tsirelson bound $2\sqrt{2}$, but instead the above calculation gives $1+\sqrt{2}< 2 \sqrt{2}$

I also noticed that $I$ has one eigenvalue which is $1$ and the whole hilbert space is an eigenspace, whereas for $J$, it has two distinct eigenvalues $1,-1$ with eigenvectors $\begin{pmatrix}1 \\ 1\end{pmatrix}$ and $\begin{pmatrix}1 \\ -1\end{pmatrix}$ respectively. Therefore I suspect the reason I don't get the expected $2 \sqrt{2}$ might have something to do with picking an observable that does not have 2 distinct eigenvalues.

Is the computation of the CHSH correlation function works only for observables with exactly two distinct eigenvalues, and fail otherwise?

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  • $\begingroup$ am not totally following but the bell nonlocality measurement varies depending on the rotation of Bobs/ Alices detectors and is maximal only for some rotations (apparently $\theta$ in your formulation). as for CHSH inequality, after many years of study think that it may have some extremely subtle sampling/ statistical assumptions/ loopholes and suspect experimenters have not considered photon (anti)bunching effects wrt different polarizations. invite further discussion here chat.stackexchange.com/rooms/9446/theory-salon $\endgroup$ – vzn Apr 11 '17 at 15:23
  • $\begingroup$ I think the formulation used in the OP is to have two spin ups prepared, and then rotate the overall 2 qubit state slowly into the triplet bell state. Tsirelson's bound should be achieved for any pair of observables that has $\pm 1$ as eigenvalues if the state is maximally entangled such as the bell state, yet the above calculation give a result less than that even when the bloch sphere angle is set such that you get the triplet state, and this forms the question in the OP $\endgroup$ – Secret Apr 12 '17 at 6:48

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