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I have a question on the article

J. S. Bell, On the Einstein Podolsky Rosen paradox, Physics 1, 195, 1964. (link)

My question concerns the expression (3) of the article, at page 196. I don't understand what is the reasoning that leads to this expression of the expectation value... I think I miss something but I don't know what.

This is what I understood from now on :

$\vec{\sigma_1}$ and $\vec{\sigma_2}$ are the spins of the two particles that move apart and must be exactly opposite according to quantum mechanics when measured in a direction of the component $\vec{a}$.

First, did I understand well and do we really have

$$A(\vec{a},\lambda) = \vec{\sigma_1}.\vec{a} = \pm 1 \\ B(\vec{b},\lambda) = \vec{\sigma_2}.\vec{b} = \pm 1$$

then ? If not, what does $A(\vec{a},\lambda)$ and $B(\vec{b},\lambda)$ correspond to ? A sort of $sign$ function or something like in the next section?

Secondly, why

$$ <\vec{\sigma_1}.\vec{a}\; \vec{\sigma_2}.\vec{b}> = -\vec{a}.\vec{b}$$

Is it because $\vec{\sigma_1}$ and $\vec{\sigma_2}$ are opposite ?

Thanks !

mwoua
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1 Answers1

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No, all your $\vec \sigma_1, \vec \sigma_2$ are simply the Pauli matrices, but applying to the first or the second particle, so $\vec \sigma_1. \vec a, \vec \sigma_2. \vec b$ are the measurement operators applying respectively to particles $1$ and $2$. The outcome of a measurement can be $1$ or $-1$, but that does not mean that $\vec \sigma_1. \vec a = \pm 1$ or $\vec \sigma_2. \vec b = \pm 1$. This is false, this is not a operator equality.

In fact, a better notation for the measurement operator of the 2-particles system, is $\vec \sigma. \vec a \otimes \vec \sigma. \vec b$.

Here $\vec \sigma$ are also the Pauli matrices

This notation means that the operator $\vec \sigma. \vec a$ is applyed to the first particle, and that the operator $\vec \sigma. \vec b$ is applyed to the second particle.

The mean value is referred to the singlet state :

$$\psi = \frac{1}{\sqrt{2}} (|+ \rangle |- \rangle - |- \rangle |+ \rangle) \tag{1} $$

So, you have :

$$M=<\vec \sigma. \vec a \otimes \vec \sigma. \vec b>_{Singlet} = \langle\psi|\vec \sigma. \vec a \otimes \vec \sigma. \vec b|\psi\rangle \tag{2}$$

That is : $$M = \frac{1}{2}(\langle +| \langle -| - \langle -| \langle +| )|\vec \sigma. \vec a \otimes \vec \sigma. \vec b(|+\rangle|-\rangle - |-\rangle|+\rangle) \tag{3}$$

So, you have four terms for $M$ ($M= M_1 + M_2 +M_3 + M_4$):

$$M_1 = \frac{1}{2} \langle +|\vec \sigma.\vec a|+\rangle \langle -|\vec \sigma.\vec b|-\rangle\tag{4}$$

$$M_2 = -\frac{1}{2} \langle +|\vec \sigma.\vec a|-\rangle \langle -|\vec \sigma.\vec b|+\rangle\tag{5}$$

$$M_3 = -\frac{1}{2} \langle -|\vec \sigma.\vec a|+\rangle \langle +|\vec \sigma.\vec b|-\rangle\tag{6}$$

$$M_4 = \frac{1}{2} \langle -|\vec \sigma.\vec a|-\rangle \langle +|\vec \sigma.\vec b|+\rangle\tag{7}$$

With the help of the expressions :

$$\langle +|\vec \sigma.\vec c|+\rangle = c_3, \langle +|\vec \sigma.\vec c|-\rangle = c_1 - ic_2,\langle -|\vec \sigma.\vec c|+\rangle = c_1 + ic_2, \\ \langle -|\vec \sigma.\vec c|-\rangle = -c_3\tag{8}$$ you will find easily that :

$$M = -(a_1b_1 + a_2b_2 + a_3b_3) = - \vec a.\vec b \tag{9}$$

Trimok
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  • I'm quite new to all this, sorry about it. But I still don't understand everything... > > $\vec{a}$ is a direction of measure, in the form of $(a_1,a_2,a_3)$? So if we want only the measure of $x$-component, we take $\vec{a}=(1,0,0)$ ? And in the same way, $\vec{\sigma_1}=(\sigma^x_1,\sigma^y_1,\sigma^z_1) (=\sigma^x_1\sigma_x+\sigma^y_1\sigma_y+\sigma^z_1\sigma_z)$, with $\sigma_i$ being the Pauli matrices? > > How is that the measurement of $\vec{\sigma_1}.\vec{a};\psi$ can only be 1 or -1? – mwoua Aug 21 '13 at 14:14
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    Yes, $\vec a$ is the direction of axis corresponding to the spin measurement. So, you are right, if you want only the measure of x-component, you have to take $\vec a=(1,0,0) $. $\vec \sigma$ is a notation for $(\sigma^x,\sigma^y,\sigma^z)$, or $(\sigma^1,\sigma^2,\sigma^3)$ - it is the same thing $(\sigma^x=\sigma^1,\sigma^y=\sigma^2,\sigma^z=\sigma^3)$. So $\vec\sigma. \vec a$ is a matrix which is $\vec\sigma. \vec a = \sigma^1a_1 + \sigma^2a_2 + \sigma^3a_3$. – Trimok Aug 21 '13 at 14:40
  • The measurement can be always be $1$ or $-1$, because the final state, after the measurement, must be an eigenvector of the measurement operator $\vec\sigma. \vec a$, and the measurement outcome is the corresponding eigenvalue. The only possible eigenvalues for the the matrix $\vec\sigma. \vec a$ (with $\vec a$ normed, that is $\vec a^2=1$) are $1$ and $-1$. You can check this, by taking the square of the matrix : $(\vec\sigma. \vec a)^2=\vec a^2 \mathbb{Id} =\mathbb{Id}$. So, the square of the eigenvalues is obviously $1$, as expected. – Trimok Aug 21 '13 at 14:40
  • Or are $\vec{\sigma_1}.\vec{a}$ simply another way to write a linear combination of $S_x$, $S_y$ and $S_z$ ? – mwoua Aug 21 '13 at 14:41
  • $\vec \sigma.\vec a$ is the measurement operator of the spin in the direction defined by $\vec a$ – Trimok Aug 21 '13 at 14:47
  • I finally understood the whole thing. Thank you for everything ! :) – mwoua Aug 21 '13 at 15:10
  • @Trimok Whence come the expressions $\langle +|\vec \sigma.\vec c|+\rangle = c_3, \langle +|\vec \sigma.\vec c|-\rangle = c_1 - ic_2,\langle -|\vec \sigma.\vec c|+\rangle = c_1 + ic_2, \langle -|\vec \sigma.\vec c|-\rangle = -c_3$? – The Ledge Jun 27 '18 at 18:09