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What is \begin{align} \sum_{\mu=0}^{3} \langle \sigma_{\mu} \rangle^2 = ? \end{align} $\sigma_{\mu}$ are the Pauli matrices. The Bra-Ket notation is used in this question: \begin{align} \langle \sigma_{\mu} \rangle = \langle \Psi \lvert \sigma_{\mu} \lvert \Psi \rangle , \end{align} where $\Psi$ is the Pauli spinor of two complex components.

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  • $\begingroup$ Why don't you just calculate it directly for arbitrary |Ψ= |(α,β)〉 and its Hermitean conjugate, so that it is normalized, αbar α+βbar β=1 ??? What stands between you and 1 for the sum of the squares of the expectation values you computed for the 3 Pauli matrices, + 1 for your euclidean metric, as defined, and -1 if you chose to subtract the identity term, instead? $\endgroup$ – Cosmas Zachos Feb 14 '16 at 15:30
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The answer to a slightly different question—what is the $\langle\sigma_{\mu} \sigma_{\mu}\rangle$—is 4 since the Pauli's independently square to the identity.

EDIT: This is incorrect. The answer should be 2 since the $\mu = 0$ term has a minus sign attached to it, as @JeffDror pointed out in the comments.

Onto your question. Let's start off by asking what is $\langle\sigma_{\mu}\rangle$ in the state $\left|n\right\rangle$ (my notation means that the spin is aligned along the unit vector n).

Well, let's say $\mu = 3$—then if the spin makes an angle $\theta$ with the z-axis the expectation of $\sigma_3$ will be $\cos(\theta)$, just the projection of the spin onto the z direction. Similarly, $\sigma_1$ will have expectation $\sin(\theta)\cos(\phi)$, where $\phi$ is the azimuthal angle, and $\sigma_2$ will have expectation $\sin(\theta)\sin(\phi)$ (this can all be proved explicitly using the fact that $\left|n\right\rangle = (\cos(\theta/2),e^{i\phi}\sin(\theta/2))$).

In any event, it is clear that $\langle\sigma_{\mu}\rangle\langle\sigma_{\mu}\rangle = 1$.

EDIT: As @JeffDror pointed out, I neglected the $\mu = 0$ term. This trivially has expectation $\langle\sigma_0\rangle = \langle\sigma_0\rangle\langle\sigma_0\rangle = 1$. However, there is a minus sign built into the Minkowski metric with which the sum over $\mu$ is being performed, so this term cancels the positive one contribution from the $\mu = 1,2,3$ components, and the total sum is indeed zero.

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  • $\begingroup$ I think you are missing the $<\sigma_0>$ term which gives a $1$ and including the relative minus sign between the spatial and temporal components the answer should be $0$. $\endgroup$ – JeffDror Mar 15 '14 at 16:27
  • $\begingroup$ Haha, I know. I just get lazy in comments. $\endgroup$ – JeffDror Mar 15 '14 at 16:37
  • $\begingroup$ Good start to answering the question. Can you prove it for any spinor orientation and spin orientation? $\endgroup$ – linuxfreebird Mar 15 '14 at 16:43
  • $\begingroup$ The above discussion does prove it for any orientation. The angles, $\phi$ and $\theta$ are arbitrary. $\endgroup$ – JeffDror Mar 15 '14 at 17:11
  • $\begingroup$ @JeffDror you're completely right thanks for catching my error $\endgroup$ – mhodel Mar 15 '14 at 18:27

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