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Diffstat (limited to 'source/know/concept/bells-theorem')
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1 files changed, 42 insertions, 42 deletions
diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md index 3b71dbf..a01bf9e 100644 --- a/source/know/concept/bells-theorem/index.md +++ b/source/know/concept/bells-theorem/index.md @@ -13,7 +13,7 @@ layout: "concept" cannot be explained by theories built on so-called **local hidden variables** (LHVs). -Suppose that we have two spin-1/2 particles, called $A$ and $B$, +Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$, in an entangled [Bell state](/know/concept/bell-state/): $$\begin{aligned} @@ -22,23 +22,23 @@ $$\begin{aligned} \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}$$. The point is that this collapse is instant, -regardless of the distance between $A$ and $B$. +regardless of the distance between $$A$$ and $$B$$. Einstein called this effect "action-at-a-distance", and used it as evidence that quantum mechanics is an incomplete theory. -He said that there must be some **hidden variable** $\lambda$ -that determines the outcome of measurements of $A$ and $B$ +He said that there must be some **hidden variable** $$\lambda$$ +that determines the outcome of measurements of $$A$$ and $$B$$ from the moment the entangled pair is created. However, according to Bell's theorem, he was wrong. -To prove this, let us assume that Einstein was right, and some $\lambda$, +To prove this, let us assume that Einstein was right, and some $$\lambda$$, which we cannot understand, let alone calculate or measure, controls the results. We want to know the spins of the entangled pair -along arbitrary directions $\vec{a}$ and $\vec{b}$, -so the outcomes for particles $A$ and $B$ are: +along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$, +so the outcomes for particles $$A$$ and $$B$$ are: $$\begin{aligned} A(\vec{a}, \lambda) = \pm 1 @@ -46,8 +46,8 @@ $$\begin{aligned} B(\vec{b}, \lambda) = \pm 1 \end{aligned}$$ -Where $\pm 1$ are the eigenvalues of the Pauli matrices -in the chosen directions $\vec{a}$ and $\vec{b}$: +Where $$\pm 1$$ are the eigenvalues of the Pauli matrices +in the chosen directions $$\vec{a}$$ and $$\vec{b}$$: $$\begin{aligned} \hat{\sigma}_a @@ -59,8 +59,8 @@ $$\begin{aligned} = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z \end{aligned}$$ -Whether $\lambda$ is a scalar or a vector does not matter; -we simply demand that it follows an unknown probability distribution $\rho(\lambda)$: +Whether $$\lambda$$ is a scalar or a vector does not matter; +we simply demand that it follows an unknown probability distribution $$\rho(\lambda)$$: $$\begin{aligned} \int \rho(\lambda) \dd{\lambda} = 1 @@ -68,8 +68,8 @@ $$\begin{aligned} \rho(\lambda) \ge 0 \end{aligned}$$ -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: +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: $$\begin{aligned} @@ -83,7 +83,7 @@ which both prove Bell's theorem. ## Bell inequality -If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins: +If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins: $$\begin{aligned} A(\vec{a}, \lambda) @@ -98,8 +98,8 @@ $$\begin{aligned} = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}$$ -Next, we introduce an arbitrary third direction $\vec{c}$, -and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$: +Next, we introduce an arbitrary third direction $$\vec{c}$$, +and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$: $$\begin{aligned} \Expval{A_a B_b} - \Expval{A_a B_c} @@ -109,7 +109,7 @@ $$\begin{aligned} \end{aligned}$$ Inside the integral, the only factors that can be negative -are the last two, and their product is $\pm 1$. +are the last two, and their product is $$\pm 1$$. Taking the absolute value of the whole left, and of the integrand on the right, we thus get: @@ -121,7 +121,7 @@ $$\begin{aligned} &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda} \end{aligned}$$ -Since $\rho(\lambda)$ is a normalized probability density function, +Since $$\rho(\lambda)$$ is a normalized probability density function, we arrive at the **Bell inequality**: $$\begin{aligned} @@ -131,18 +131,18 @@ $$\begin{aligned} } \end{aligned}$$ -Any theory involving an LHV $\lambda$ must obey this inequality. +Any theory involving an LHV $$\lambda$$ must obey this inequality. The problem, however, is that quantum mechanics dictates the expectation values -for the state $\Ket{\Psi^{-}}$: +for the state $$\Ket{\Psi^{-}}$$: $$\begin{aligned} \Expval{A_a B_b} = - \vec{a} \cdot \vec{b} \end{aligned}$$ Finding directions which 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 $0.293$, +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 $$0.293$$, which clearly disagrees with the inequality, meaning that LHVs are impossible. @@ -152,8 +152,8 @@ meaning that LHVs are impossible. 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 $\vec{b}_2$. +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: $$\begin{aligned} @@ -196,7 +196,7 @@ $$\begin{aligned} + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| \end{aligned}$$ -Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$, +Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$, we can reduce this to: $$\begin{aligned} @@ -209,7 +209,7 @@ $$\begin{aligned} \end{aligned}$$ Evaluating these integrals gives us the following inequality, -which holds for both choices of $\pm$: +which holds for both choices of $$\pm$$: $$\begin{aligned} \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| @@ -235,8 +235,8 @@ $$\begin{aligned} &\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}$$ -The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$, -and measures the correlation between the spins of $A$ and $B$: +The quantity on the right-hand side is sometimes called the **CHSH quantity** $$S$$, +and measures the correlation between the spins of $$A$$ and $$B$$: $$\begin{aligned} \boxed{ @@ -244,7 +244,7 @@ $$\begin{aligned} } \end{aligned}$$ -The CHSH inequality places an upper bound on the magnitude of $S$ +The CHSH inequality places an upper bound on the magnitude of $$S$$ for LHV-based theories: $$\begin{aligned} @@ -258,15 +258,15 @@ $$\begin{aligned} 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. $S = \expval{\hat{S}}$: +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. +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: $$\begin{aligned} @@ -292,15 +292,15 @@ $$\begin{aligned} \end{aligned}$$ Spin operators are unitary, so their square is the identity, -e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to: +e.g. $$\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}$$ -The *norm* $\norm{\hat{S}^2}$ of this operator -is the largest possible expectation value $\expval{\hat{S}^2}$, +The *norm* $$\norm{\hat{S}^2}$$ of this operator +is the largest possible expectation value $$\expval{\hat{S}^2}$$, which is the same as its largest eigenvalue. It is given by: @@ -321,7 +321,7 @@ $$\begin{aligned} \le 2 \end{aligned}$$ -And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason. +And $$\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason. The norm is the largest eigenvalue, therefore: $$\begin{aligned} @@ -336,7 +336,7 @@ $$\begin{aligned} We thus arrive at **Tsirelson's bound**, which states that quantum mechanics can violate -the CHSH inequality by a factor of $\sqrt{2}$: +the CHSH inequality by a factor of $$\sqrt{2}$$: $$\begin{aligned} \boxed{ @@ -359,8 +359,8 @@ $$\begin{aligned} \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}} \end{aligned}$$ -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. +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. |