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I am trying to understand how the correlation function in John Bell's paper on EPR is derived for a spin singlet state $|{\Psi}\rangle$. This is defined to be

$$ \langle{\Psi}|\left(\mathbf{\sigma}\cdot\mathbf {a}\right)\left(\mathbf {\sigma}\cdot\mathbf {b}\right)|{\Psi}\rangle=-\mathbf a \cdot\mathbf b. $$ I tried to compute it explicitly by using the Pauli matrices, but was unable to derive the scalar product of the two direction vectors.

One attempt to prove this is by using $$ \langle{\Psi}|\left(\mathbf{\sigma}\cdot\mathbf{a}\right)\left(\mathbf {\sigma}\cdot\mathbf {b}\right)|{\Psi}\rangle=\langle{\Psi}|(\mathbf {a} \cdot\mathbf { b}) \mathbb I +i(\mathbf a \times\mathbf b)\cdot\sigma|{\Psi}\rangle $$ However, I am not able to derive the minus sign or get rid of the cross product term either.

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"I tried to compute it explicitly by using the Pauli matrices, but was unable to derive the scalar product of the two direction vectors.". This is a possibility:

Let measurement system A defined by matrix $$ \sigma\mathbf{a}=\begin{pmatrix}a_z & a_x-ia_y \\ a_x+ia_y & -a_z\end{pmatrix}$$

where $|\mathbf{a}|=1$. This matrix has normalized eigenvectors: $$ a_+ = \frac{1}{\sqrt{2\left(a_z+1\right) }}\begin{pmatrix} a_z + 1 \\ a_x + ia_y \end{pmatrix}; \qquad a_- = \frac{1}{\sqrt{2\left(a_z+1\right) }}\begin{pmatrix} -a_x+ia_y \\ a_z + 1 \end{pmatrix}$$

same notation for measurement system B.

When the first particle is measured by system A, it can result in a measurement equal to +1, this particle in state $a_+$ and the anti-symmetric one in state $a_-$; or a measurement equal to -1, first particle in state $a_-$ and the other in state $a_+$. Lets call $P_1$ the probability of first case, $P_2=1-P_1$ the one of the second.

Now, when second particle is measured by system B, the average result of this measurement will be:

case $P_1$: $\left<\mathbf{a}_-|\sigma\mathbf{b}|\mathbf{a}_- \right> = \frac{1}{2 \left( a_z+1 \right)} \,\, \left( -a_x-ia_y, a_z + 1 \right) \begin{pmatrix}b_z & b_x-ib_y \\ b_x+ib_y & -b_z\end{pmatrix}\begin{pmatrix} -a_x+ia_y \\ a_z + 1 \end{pmatrix} = ... = -a_xb_x-a_yb_y-a_zb_z = -\mathbf{a}\mathbf{b}$

case $P_2$: $\left<\mathbf{a}_+|\sigma\mathbf{b}|\mathbf{a}_+ \right> = ... = \mathbf{a}\mathbf{b}$

As Bell experiment counts the average number of matches between measurements by A and B less the number of mismatches: $$ \left<\Psi_-|\mathbf{\sigma}\mathbf{a}\otimes\mathbf{\sigma}\mathbf{b}|\Psi_- \right> = P_1 \left< \mathbf{a}_-|\mathbf{\sigma}\mathbf{b}|\mathbf{a}_- \right> - P_2 \left< \mathbf{a}_+|\mathbf{\sigma}\mathbf{b}|\mathbf{a}_+ \right> = - \mathbf{a}\mathbf{b} $$

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  • $\begingroup$ "When the first particle is measured by system A, it can result in a measurement equal to +1, this particle in state 𝑎+ and the anti-symmetric one in state 𝑎−; or a measurement equal to -1, first particle in state 𝑎− and the other in state 𝑎+. Lets call 𝑃1 the probability of first case, 𝑃2=1−𝑃1 the one of the second." That's a non-trivial part (maybe the most non-trivial part) of the whole derivation, and you state it like it is obvious. $\endgroup$ Aug 13, 2020 at 12:14
  • $\begingroup$ The question is: Why, if you find particle A in a+, will you find particle B in a-? This is a non-trivial calculation (or, rather, once you have done it and written it at $\sigma_A|\psi^-\rangle = - \sigma_B|\psi^-\rangle$, the rest follows immediately). $\endgroup$ Aug 13, 2020 at 14:16
  • $\begingroup$ @NorbertSchuch: When first measurement results in +1, state of second particle, $\psi_B$ must fulfill $<\psi_B|\sigma\mathbf{a}|\psi_B>=-1$, being the only possible solution $\psi_B=\mathbf{a}_-$. $\endgroup$ Aug 13, 2020 at 14:54
  • $\begingroup$ Why must $\psi_B$ fulfill that? $\endgroup$ Aug 13, 2020 at 16:16
  • $\begingroup$ @NorbertSchuch: An immediate repetition of a measurement must give same result than first one, where "measure" must be understood in the most broad way: any way to obtain information about the system under observation. In first step, a measurement is done that provides information about the state of both antisymmetric particles, fixing (until different measurement) also the states of both $<\psi_A|\sigma\mathbf{a}|\psi_A>=1$ and $<\psi_B|\sigma\mathbf{a}|\psi_B>=-1$ (or viceversa). $\endgroup$ Aug 13, 2020 at 17:44
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You can solve this by applying the definition of the Pauli vector, i.e.,

$${\bf{\sigma}\cdot\bf{a} }=\sigma_xa_x+\sigma_ya_y+\sigma_za_z$$ where $a_x,a_y,a_z$ are the components of $\bf a$ and $\sigma_x,\sigma_y,\sigma_z$ are the Pauli matrices. I will assume that we apply ${\bf{\sigma}\cdot\bf{a} }$ on the first subsystem, and ${\bf{\sigma}\cdot\bf{b} }$ on the second subsystem.

If we use the standard notation for Bell states, i.e., $$|\psi^\pm\rangle=\frac{|01\rangle \pm|10\rangle}{\sqrt{2}}\quad;\quad|\phi^\pm\rangle=\frac{|00\rangle \pm|11\rangle}{\sqrt{2}}$$

You can easily verify, that $${\bf{\sigma}\cdot\bf{a} }|\psi^-\rangle=-a_x|\phi^-\rangle+ia_y|\phi^+\rangle+a_z|\psi^+\rangle$$ $${\bf{\sigma}\cdot\bf{b} }|\psi^-\rangle=b_x|\phi^-\rangle-ib_y|\phi^+\rangle+-b_z|\psi^+\rangle$$ Therefore, \begin{align} \langle\psi^-|({\bf{\sigma}\cdot\bf{a} })({\bf{\sigma}\cdot\bf{b} })|\psi^-\rangle&=\left[-a_x\langle\phi^-|-ia_y\langle\phi^+|+a_z\langle\psi^+|\right]\left[|b_x|\phi^-\rangle-ib_y|\phi^+\rangle+-b_z|\psi^+\rangle\right]\\&=-a_xb_x-a_yb_y-a_zb_z=-{\bf a\cdot b} \end{align} where the second equality follows from the orthogonality of Bell states, and the last one follows from the definition of scalar product.

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  • $\begingroup$ @NorbertSchuch: First component of ${\bf{\sigma}\cdot\bf{a} }|\psi^-\rangle$ is $a_x-ia_y$. First component of $-a_x|\phi^-\rangle+ia_y|\phi^+\rangle+a_z|\psi^+\rangle$ is $-a_x+ia_y$. They are not equal. $\endgroup$ Aug 13, 2020 at 17:02
  • $\begingroup$ @pasabaporaqui Then you need to redefine $i$ (or more precisely $\sigma_y$). More seriously, this is a matter of convention (which I didn't mind checking). If you change $i$ with $-i$ everywhere, everything will be correct. $\endgroup$ Aug 13, 2020 at 19:38
  • $\begingroup$ @NorbertSchuch: assume $a_y=a_z=0$. Still $a_x (right) \ne -a_x (left)$. $\endgroup$ Aug 13, 2020 at 19:44
  • $\begingroup$ @pasabaporaqui But if you apply $\sigma_x$ to the first component of |01>-|10>, you obtain |11>-|00>, so the amplitude of |00> is MINUS ONE, not plus one! $\endgroup$ Aug 13, 2020 at 19:45
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There may be a nice geometric or algebraic argument why we get this, but I'll just brute force it. We have \begin{align} \boldsymbol{\sigma} \cdot \mathbf{a} &= a_x \sigma_x + a_y \sigma_y + a_z \sigma_z\\ &= \frac{\alpha}{2} \sigma_+ + \frac{\bar{\alpha}}{2}\sigma_- + a_z \sigma_z. \end{align} where $\alpha = a_x + i a_y$ and the bar denotes complex conjugation.

Applying $(\boldsymbol{\sigma} \cdot \mathbf{a})(\boldsymbol{\sigma} \cdot \mathbf{b})$ to the ket $|A\rangle = (\left|\uparrow\downarrow\right> - \left|\downarrow\uparrow\right>)/\sqrt{2}$, we get \begin{align} C(\mathbf{a}, \mathbf{b}) = \langle A |(\boldsymbol{\sigma} \cdot \mathbf{a})(\boldsymbol{\sigma} \cdot \mathbf{b})|A\rangle = \langle A |\left(\frac{1}{\sqrt{2}}\left[\bar{\alpha}\beta\left|\downarrow\uparrow\right> - \alpha\bar\beta\left|\uparrow\downarrow\right>\right] - a_z b_z |A\rangle \right) \end{align} where $\beta = b_x + i b_y$ and I've dropped terms that vanish when we take the expectation value. Contracting with $\langle A |$ gets us \begin{align} C(\mathbf{a}, \mathbf{b}) &= \frac{1}{2}(\left<\uparrow\downarrow\right| - \left<\downarrow\uparrow\right|)\left[\bar{\alpha}\beta\left|\downarrow\uparrow\right> - \alpha\bar\beta\left|\uparrow\downarrow\right>\right] - a_zb_z\\ &=-\frac{1}{2}(\alpha\bar\beta+ \bar\alpha\beta)-a_z b_z. \end{align} Multiplying out the term in parentheses, $\alpha\bar\beta+ \bar\alpha\beta = 2 (a_x b_x + a_y b_y)$, so $C(\mathbf{a}, \mathbf{b}) = - a_x b_x - a_yb_y - a_z b_z = - \mathbf{a}\cdot\mathbf{b}$.

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  • $\begingroup$ Ah yes, the dangers of copy and paste on exhibit :) $\endgroup$
    – d_b
    Jul 31, 2020 at 18:51
  • $\begingroup$ One question: if $\mathbf{a}=\mathbf{b}=(1,0,0)$ then $C(\mathbf{a},\mathbf{b})=<A|\sigma_x\sigma_x|A>=<A|I|A>=<A|A>=1$ instead of the expected -1 ? $\endgroup$ Aug 4, 2020 at 17:57
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    $\begingroup$ The two Pauli matrices act on different tensor product factors, so they don't multiply to the identity. Where you've written $\sigma_x \sigma_x$, that really means $$\sigma_x^{(1)} \otimes \sigma_x^{(2)}$$. $\endgroup$
    – d_b
    Aug 4, 2020 at 18:10
  • $\begingroup$ $\vec{\sigma}\vec{a}\otimes\vec{\sigma} \vec{b}=\begin{pmatrix} a_zb_z & a_z\bar{\beta} & \bar{\alpha}b_z & \bar{\alpha}\bar{\beta} \\ a_z\beta & -a_zb_z & \bar{\alpha}\beta & -\bar{\alpha}b_z \\ \alpha b_z & \alpha\bar{\beta} & -a_zb_z & -a_z\bar{\beta} \\ \alpha\beta & -\alpha b_z & -a_z\beta & a_zb_z \end{pmatrix}$ $\endgroup$ Aug 10, 2020 at 19:23
  • $\begingroup$ @d_b the first identity you used is wrong. if you would multiply it out, you will get $a_x\sigma_x-a_y\sigma_y$. Using your notation, the correct expansion is $\frac{1}{2}\overline{\alpha}\sigma_{+}+\frac{1}{2}\alpha\sigma_{-}+a_z\sigma_z$. (finally I got the correct latex symbol :) ) $\endgroup$ Sep 18, 2021 at 13:19

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