# Homework Help: Spin probability

1. Dec 11, 2013

### QuarksAbove

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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. =/

2. Dec 11, 2013

### 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.

Last edited: Dec 12, 2013
3. Dec 12, 2013

### 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...

4. Dec 12, 2013

### TSny

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

5. Dec 12, 2013

### 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!

6. Dec 12, 2013

### TSny

That looks correct.