# What is the probability of finding one spin of a doublet system as spin up?

How can I calculate the probability of finding one spin of a doublet as spin up? Suppose I have the following system: $$|\psi\rangle=(|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle)\frac{1}{\sqrt 2}.$$

How could I find the probability of a spin going up? I suppose I have to do the same as with singlet states, like: $$( \left \langle \uparrow \right| \otimes 1)\ |\psi\rangle= ( \left \langle \uparrow \right | \otimes 1)(\left|\uparrow \right\rangle \otimes \left|\downarrow \right\rangle + \left |\downarrow \right\rangle \otimes \left |\uparrow \right\rangle)\frac{1}{\sqrt 2}= (1\otimes \left |\downarrow \right\rangle)\frac{1}{\sqrt 2}$$ But then I don't know what else to do. Thanks!

• One spin is always up. May 18, 2022 at 12:49
• I do not understand your second equation at all: there are some typos there. Can you fix them? May 18, 2022 at 15:47
• I tried to improve the style of the second equation and added an additional step. May 18, 2022 at 18:37

I guess you mean the joint probability of finding s1=+ while measuring any other spin for s2, right? In that case, one can prepare a -test- state for a fully polarized s1 and unpolarized s2, for example: $$|\psi\rangle=\frac{1}{\sqrt{2}}\left[|+\rangle \otimes (|+\rangle+|-\rangle)\right]$$. This gives you a what you would intuitively expect, P=1/2. Hope that helps!