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So. We have a singlet sate $$ \dfrac{1}{\sqrt{2}}(\vert\uparrow\downarrow\rangle-\vert\downarrow\uparrow\rangle)$$ And two pauli matrices for z axis - one that acts on 1st spin (lets denote it with $\sigma_{z}$), other for 2nd spin (denoted $\tau_{z}$). When I calculate expectation values, I get $$\langle\sigma_{z}\rangle=0$$ $$\langle\tau_{z}\rangle=0$$

Which I do understand physically (I think, I do). We prepare many pairs of spins in a singlet state and measure first or second spin along z axis. Summing up results gives average spin value 0.

However expectation for two observables together $$\langle\sigma_{z}\tau_{z}\rangle=-1$$ I assumed that two observables in the row means two measurements (we measure 1st spin and then second, or vice versa). So expectation value should be zero, because if one spin is measured, it gives direction to other spin instantly, so average value of one measurement is always zero. What is wrong in my reasoning? Why expectation value is -1?

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If you measure one particle to be in $\left|\uparrow\right\rangle$, then the other will be in $\left|\downarrow\right\rangle$, just as you said. So your two eigenvalues are $1$ and $-1$. Multiplying (not averaging) the results together: $$1\times (-1) = -1.$$ You'll get the same answer if you measure the particles to be down/up rather than up/down.

I think you were overlooking that you don't average the two eigenvalues you get; you average the product. And the product is always $-1$ for any given pair of measurements. You were likely thinking $\langle \sigma_z\rangle\langle\tau_z\rangle$ rather than $\langle\sigma_z\tau_z\rangle$.

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  • $\begingroup$ So stacking together observables like that (no matter what state we act on) has nothing to do with changing a system, but is about multiplying expectation values of each measurement? Does order matter in general case? $\endgroup$ – Rena Mar 28 '15 at 23:50
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    $\begingroup$ I believe in general order matters. The general procedure is to figure out what state you're in after your first measurement, and then use that new state as input for your second. $\endgroup$ – BMS Mar 29 '15 at 20:16

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