Improve knowledge base

1 files changed, 29 insertions, 237 deletions

diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md
index a01bf9e..1589a7a 100644
--- a/source/know/concept/bells-theorem/index.md
+++ b/source/know/concept/bells-theorem/index.md
@@ -17,13 +17,13 @@ Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$,
 in an entangled [Bell state](/know/concept/bell-state/):
 
 $$\begin{aligned}
-    \Ket{\Psi^{-}}
-    = \frac{1}{\sqrt{2}} \Big( \Ket{\uparrow \downarrow} - \Ket{\downarrow \uparrow} \Big)
+    \ket{\Psi^{-}}
+    = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big)
 \end{aligned}$$
 
 Since they are entangled,
-if we measure the $$z$$-spin of particle $$A$$, and find e.g. $$\Ket{\uparrow}$$,
-then particle $$B$$ immediately takes the opposite state $$\Ket{\downarrow}$$.
+if we measure the $$z$$-spin of particle $$A$$, and find e.g. $$\ket{\uparrow}$$,
+then particle $$B$$ immediately takes the opposite state $$\ket{\downarrow}$$.
 The point is that this collapse is instant,
 regardless of the distance between $$A$$ and $$B$$.
 
@@ -69,21 +69,29 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The product of the outcomes of $$A$$ and $$B$$ then has the following expectation value.
-Note that we only multiply $$A$$ and $$B$$ for shared $$\lambda$$-values:
-this is what makes it a **local** hidden variable:
+Note that we multiply $$A$$ and $$B$$ at the same $$\lambda$$-value,
+hence it is a *local* hidden variable:
 
 $$\begin{aligned}
-    \Expval{A_a B_b}
-    = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda}
+    \expval{A_a B_b}
+    \equiv \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda}
 \end{aligned}$$
 
-From this, two inequalities can be derived,
-which both prove Bell's theorem.
+From this, we can make several predictions about LHV theories,
+which turn out to disagree with various theoretical
+and experimental results in quantum mechanics.
+The two most famous LHV predictions are
+the **Bell inequality** and
+the [CHSH inequality](/know/concept/chsh-inequality/).
+
 
 
 ## Bell inequality
 
-If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins:
+We present Bell's original proof of his theorem.
+If $$\vec{a} = \vec{b}$$, then we know that
+measuring $$A$$ and $$B$$ gives them opposite spins,
+because they start in the entangled state $$\ket{\Psi^{-}}$$:
 
 $$\begin{aligned}
     A(\vec{a}, \lambda)
@@ -94,7 +102,7 @@ $$\begin{aligned}
 The expectation value of the product can therefore be rewritten as follows:
 
 $$\begin{aligned}
-    \Expval{A_a B_b}
+    \expval{A_a B_b}
     = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
 \end{aligned}$$
 
@@ -102,7 +110,7 @@ Next, we introduce an arbitrary third direction $$\vec{c}$$,
 and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$:
 
 $$\begin{aligned}
-    \Expval{A_a B_b} - \Expval{A_a B_c}
+    \expval{A_a B_b} - \expval{A_a B_c}
     &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda}
     \\
     &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
@@ -114,7 +122,7 @@ Taking the absolute value of the whole left,
 and of the integrand on the right, we thus get:
 
 $$\begin{aligned}
-    \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big|
+    \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big|
     &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big)
     \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda}
     \\
@@ -122,24 +130,24 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Since $$\rho(\lambda)$$ is a normalized probability density function,
-we arrive at the **Bell inequality**:
+we arrive at the Bell inequality:
 
 $$\begin{aligned}
     \boxed{
-        \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big|
-        \le 1 + \Expval{A_b B_c}
+        \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big|
+        \le 1 + \expval{A_b B_c}
     }
 \end{aligned}$$
 
 Any theory involving an LHV $$\lambda$$ must obey this inequality.
-The problem, however, is that quantum mechanics dictates the expectation values
-for the state $$\Ket{\Psi^{-}}$$:
+The problem, however, is that quantum mechanics dictates
+the expectation values for the state $$\ket{\Psi^{-}}$$:
 
 $$\begin{aligned}
-    \Expval{A_a B_b} = - \vec{a} \cdot \vec{b}
+    \expval{A_a B_b} = - \vec{a} \cdot \vec{b}
 \end{aligned}$$
 
-Finding directions which violate the Bell inequality is easy:
+Finding directions that violate the Bell inequality is easy:
 for example, if $$\vec{a}$$ and $$\vec{b}$$ are orthogonal,
 and $$\vec{c}$$ is at a $$\pi/4$$ angle to both of them,
 then the left becomes $$0.707$$ and the right $$0.293$$,
@@ -147,222 +155,6 @@ which clearly disagrees with the inequality,
 meaning that LHVs are impossible.
 
 
-## CHSH inequality
-
-The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality**
-takes a slightly different approach, and is more useful in practice.
-
-Consider four spin directions, two for $$A$$ called $$\vec{a}_1$$ and $$\vec{a}_2$$,
-and two for $$B$$ called $$\vec{b}_1$$ and $$\vec{b}_2$$.
-Let us introduce the following abbreviations:
-
-$$\begin{aligned}
-    A_1 &= A(\vec{a}_1, \lambda)
-    \qquad \quad
-    A_2 = A(\vec{a}_2, \lambda)
-    \\
-    B_1 &= B(\vec{b}_1, \lambda)
-    \qquad \quad
-    B_2 = B(\vec{b}_2, \lambda)
-\end{aligned}$$
-
-From the definition of the expectation value,
-we know that the difference is given by:
-
-$$\begin{aligned}
-    \Expval{A_1 B_1} - \Expval{A_1 B_2}
-    = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda}
-\end{aligned}$$
-
-We introduce some new terms and rearrange the resulting expression:
-
-$$\begin{aligned}
-    \Expval{A_1 B_1} - \Expval{A_1 B_2}
-    &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda}
-    \\
-    &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
-    - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
-\end{aligned}$$
-
-Taking the absolute value of both sides
-and invoking the triangle inequality then yields:
-
-$$\begin{aligned}
-    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
-    &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
-    - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
-    \\
-    &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg|
-    + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
-\end{aligned}$$
-
-Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$,
-we can reduce this to:
-
-$$\begin{aligned}
-    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
-    &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
-    + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
-    \\
-    &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
-    + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
-\end{aligned}$$
-
-Evaluating these integrals gives us the following inequality,
-which holds for both choices of $$\pm$$:
-
-$$\begin{aligned}
-    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
-    &\le 2 \pm \Expval{A_2 B_2} \pm \Expval{A_2 B_1}
-\end{aligned}$$
-
-We should choose the signs such that the right-hand side is as small as possible, that is:
-
-$$\begin{aligned}
-    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
-    &\le 2 \pm \Big( \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big)
-    \\
-    &\le 2 - \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
-\end{aligned}$$
-
-Rearranging this and once again using the triangle inequality,
-we get the CHSH inequality:
-
-$$\begin{aligned}
-    2
-    &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
-    \\
-    &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
-\end{aligned}$$
-
-The quantity on the right-hand side is sometimes called the **CHSH quantity** $$S$$,
-and measures the correlation between the spins of $$A$$ and $$B$$:
-
-$$\begin{aligned}
-    \boxed{
-        S \equiv \Expval{A_2 B_1} + \Expval{A_2 B_2} + \Expval{A_1 B_1} - \Expval{A_1 B_2}
-    }
-\end{aligned}$$
-
-The CHSH inequality places an upper bound on the magnitude of $$S$$
-for LHV-based theories:
-
-$$\begin{aligned}
-    \boxed{
-        |S| \le 2
-    }
-\end{aligned}$$
-
-
-## Tsirelson's bound
-
-Quantum physics can violate the CHSH inequality, but by how much?
-Consider the following two-particle operator,
-whose expectation value is the CHSH quantity, i.e. $$S = \expval{\hat{S}}$$:
-
-$$\begin{aligned}
-    \hat{S}
-    = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
-\end{aligned}$$
-
-Where $$\otimes$$ is the tensor product,
-and e.g. $$\hat{A}_1$$ is the Pauli matrix for the $$\vec{a}_1$$-direction.
-The square of this operator is then given by:
-
-$$\begin{aligned}
-    \hat{S}^2
-    = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_1 \hat{B}_2
-    + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1^2 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1 \hat{B}_2
-    \\
-    + &\hat{A}_2^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_2^2 \otimes \hat{B}_2^2
-    + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2^2
-    \\
-    + &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_1^2 + \hat{A}_1 \hat{A}_2 \otimes \hat{B}_1 \hat{B}_2
-    + \hat{A}_1^2 \otimes \hat{B}_1^2 - \hat{A}_1^2 \otimes \hat{B}_1 \hat{B}_2
-    \\
-    - &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_1 \hat{A}_2 \otimes \hat{B}_2^2
-    - \hat{A}_1^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_1^2 \otimes \hat{B}_2^2
-    \\
-    = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_1^2 \otimes \hat{B}_1^2 + \hat{A}_1^2 \otimes \hat{B}_2^2
-    \\
-    + &\hat{A}_2^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2} - \hat{A}_1^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2}
-    + \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_1^2 - \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_2^2
-    \\
-    + &\hat{A}_1 \hat{A}_2 \otimes \comm{\hat{B}_1}{\hat{B}_2} - \hat{A}_2 \hat{A}_1 \otimes \comm{\hat{B}_1}{\hat{B}_2}
-\end{aligned}$$
-
-Spin operators are unitary, so their square is the identity,
-e.g. $$\hat{A}_1^2 = \hat{I}$$. Therefore $$\hat{S}^2$$ reduces to:
-
-$$\begin{aligned}
-    \hat{S}^2
-    &= 4 \: (\hat{I} \otimes \hat{I}) + \comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}
-\end{aligned}$$
-
-The *norm* $$\norm{\hat{S}^2}$$ of this operator
-is the largest possible expectation value $$\expval{\hat{S}^2}$$,
-which is the same as its largest eigenvalue.
-It is given by:
-
-$$\begin{aligned}
-    \Norm{\hat{S}^2}
-    &= 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}}
-    \\
-    &\le 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2}} \Norm{\comm{\hat{B}_1}{\hat{B}_2}}
-\end{aligned}$$
-
-We find a bound for the norm of the commutators by using the triangle inequality, such that:
-
-$$\begin{aligned}
-    \Norm{\comm{\hat{A}_1}{\hat{A}_2}}
-    = \Norm{\hat{A}_1 \hat{A}_2 - \hat{A}_2 \hat{A}_1}
-    \le \Norm{\hat{A}_1 \hat{A}_2} + \Norm{\hat{A}_2 \hat{A}_1}
-    \le 2 \Norm{\hat{A}_1 \hat{A}_2}
-    \le 2
-\end{aligned}$$
-
-And $$\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason.
-The norm is the largest eigenvalue, therefore:
-
-$$\begin{aligned}
-    \Norm{\hat{S}^2}
-    \le 4 + 2 \cdot 2
-    = 8
-    \quad \implies \quad
-    \Norm{\hat{S}}
-    \le \sqrt{8}
-    = 2 \sqrt{2}
-\end{aligned}$$
-
-We thus arrive at **Tsirelson's bound**,
-which states that quantum mechanics can violate
-the CHSH inequality by a factor of $$\sqrt{2}$$:
-
-$$\begin{aligned}
-    \boxed{
-        |S|
-        \le 2 \sqrt{2}
-    }
-\end{aligned}$$
-
-Importantly, this is a *tight* bound,
-meaning that there exist certain spin measurement directions
-for which Tsirelson's bound becomes an equality, for example:
-
-$$\begin{aligned}
-    \hat{A}_1 = \hat{\sigma}_z
-    \qquad
-    \hat{A}_2 = \hat{\sigma}_x
-    \qquad
-    \hat{B}_1 = \frac{\hat{\sigma}_z + \hat{\sigma}_x}{\sqrt{2}}
-    \qquad
-    \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
-\end{aligned}$$
-
-Using the fact that $$\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$,
-it can then be shown that $$S = 2 \sqrt{2}$$ in this case.
-
-
 
 ## References
 1.  D.J. Griffiths, D.F. Schroeter,