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Suppose I have a two-electron system, and they are left alone for a long time, such that I do not know where each of them is. I then measure one electron at position $\vec r_1$. Does this collapse the wave function such that the position of both electrons is sharply determined? If so, which method is used to find out where the other is? Or is the wave function of the other still undeterminable, and is instead described by the one body Hamiltonian where there is a static charge at the position that we measured the first electron? Or something else?

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    $\begingroup$ This of course depends entirely on the initial state of the system. $\endgroup$ – WillO Dec 13 '19 at 16:34
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If your system of two electrons is described by the wave function $\psi(\vec r_a,\vec r_b)$, and you measure one of the electrons to be at $\vec r_a=\vec r_1$, then you collapse the state of the system in such a way that, were you to measure the position of the second electron shortly after the first one, probability density to get the second electron's position would be

$$\rho(\vec r_b)=|\psi(\vec r_1,\vec r_b)|^2.$$

I.e. the measurement of position of one of the electrons chooses the slice of the wavefunction with that parameter fixed to the position you've measured.

But after this measurement, since the electrons interact, you can't really describe further evolution of the second electron. Position eigenstate has completely undefined momentum, so first electron can be anywhere not long after the measurement, and this change of state would have affected the state of the second electron.

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