Calculate Probability of Magnitude 1 Spin with Two Particles

[QuarksAbove]

Homework Statement

Two particles with spin 1/2 and are in the spin state:

|\psi> = |\uparrow_{1}>|\downarrow_{2}>

where |\uparrow_{1}> is a state where particle 1 has spin up along the z-axis and
|\downarrow_{2}> is a state where particle 2 is spin down along the z-axis.

If we measure the magnitude of the total spin of the two particles, what is the probability that the magnitude will be 1?

Homework Equations

Probability = |<n|\psi>|^{2}

The Attempt at a Solution

I immediately thought that this problem was like a simple coin toss. 50% to get heads 50% to get tails. Since each particle has equal chance to be spin up or spin down, then the total probability of both being spin up after a measurement would be (0.5)(0.5) = 0.25 ?

This doesn't feel right to me. I feel like it should be more complicated. =/

 

[TSny]

Hello.

Can you expand the state $|\psi \rangle$ in terms of basis states that have definite values of the total spin? These basis states are the "singlet" and "triplet" states.

See here and here.

 

[QuarksAbove]

Hello =)

so the singlet state is
|\psi_{singlet}>= 1/(√2)(|\uparrow\downarrow> - |\downarrow\uparrow>)

and the triplet state is

|\psi_{triplet}>= 1/(√2)(|\uparrow\downarrow> + |\downarrow\uparrow>)

so,
|\psi> = 1/2(|\psi_{singlet}> + |\psi_{triplet}>)

in order to get S=1, the spin state would need to be triplet. The probability would then be the square of the coefficient of the triplet state?

|1/(2\sqrt{2})|^{2} = 0.125

this doesn't make sense though, because the total probabilities don't add up to 1...

 

[TSny]

|\psi> = 1/2(|\psi_{singlet}> + |\psi_{triplet}>)

Make sure $|\psi \rangle$ is properly normalized.

 

[QuarksAbove]

Oh right, I forgot.

After normalization, |\psi> = 1/\sqrt{2}(|\uparrow\downarrow> - |\downarrow\uparrow>) +1/\sqrt{2}(|\uparrow\downarrow> + |\downarrow\uparrow>)

leaving the probability to be in the triplet state (S=1) to be 1/2.

Thanks for your help! I think I got it!

 

[TSny]

That looks correct.