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Consider two electrons in the spin singlet state. If a measure of $S_z$ from one of the electrons show that it is in a state with $m = 1/2$, what is the probability of a measure of $S_z$ does the other electron result in $m = 1/2$?

I am not sure how to answer this problem: The possible states are

$$\uparrow\uparrow-\downarrow\downarrow-\uparrow\downarrow-\downarrow\uparrow$$

We have three options having at least one spin up. So, considering these three, one has the other spin up and other two have spin down.

So, the probability that we have two $m=1/2$ is 1/3? I mean, of course I am considering that the electrons are indistinguishable, because otherwise I could say "the electron 1 has spin up, so the first spin need to be up arrow, and it leads 1/2 probability to the other electron have spin up too"

Is my answer/interpretation right?

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2 Answers 2

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You might be confused about what a spin singlet state means. As far as I know it is a system in which all electrons are paired, whose net angular momentum is zero i.e whose overall spin quantum number s=0. So, if in a singlet state one electron is measured to be +1/2 other should be -1/2. So both electrons in a singlet cannot have the same spin, similar to the Pauli's exclusion Principle.

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  • $\begingroup$ Did you believe that i didn't read the singlet part? Jesus christ ... You are right $\endgroup$
    – LSS
    Commented Jul 18, 2021 at 19:58
  • $\begingroup$ Pauli's exclusion principle has nothing to do with this. $\uparrow\uparrow$ is not an excluded state from the Hilbert space of two spin-$1/2$ particles. See this post for a nice discussion of the subject. $\endgroup$
    – user87745
    Commented Jul 18, 2021 at 21:04
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Conceptually, you are wrong that there are four possibilities for the configuration of the electrons. There are infinitely many. A general state could be any linear combination of the four states you wrote. Of course, a singlet state means a particular linear combination, namely, $\frac{1}{\sqrt{2}}\big(\uparrow\downarrow-\downarrow\uparrow\big)$.

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