# Spin probability

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"

• 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.
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, $$\frac{1}{\sqrt{2}}\big(\uparrow\downarrow-\downarrow\uparrow\big)$$.