# Probabilities of the States of a Spin 1 Particle

1. Jul 8, 2015

### tomdodd4598

I have been following a series of Leonard Susskind's lectures called 'Quantum Entanglements' (Part 1). In general, he explains how to find the probabilities of measurements of spin ½ particles' states, both single particles and pairs of them. I have learned the following: how to use the 2x2 spin matrices for single particles, and the 4x4 spin matrices for pairs of particles, how to construct projection operators to help with the calculations and how to find the probabilities of the results of measurements of different states, using the projection operators (I may have missed a few things).

Now that I understood, to an extent that I was happy with, how to work with these spin ½ particles, I wanted to move on to looking at spin 1 particles. I have found the 3x3 spin matrices and their eigenvectors, but then I have no clue whether I can construct the projection operators in the same way that I did from the spin ½ matrices. Is this possible or is it more difficult than that?

(I can't see how it is possible myself, as the original equation for the projection operators was ½(I ± Sn), where the '±' is a '+' for a measurement of the spin being up and '-' for a measurement of the spin being down, but since there are three possible outcomes for the measurement of the spin of a spin 1 particle, this doesn't seem to work)

It is difficult for me to stick to the post template as this is more of an adventure into a new area of physics rather than a set assignment, but if it helps, the work I have done so far, for spin half particles, is here (effectively the relevant equations): LINK

Thanks for any help in advance.

2. Jul 8, 2015

### BiGyElLoWhAt

Isn't the (1/2)(I +/- Sn) just (eigenvector) (I +/- Sn) where the possible states are 0,-1,+1 ?
I've only worked through this a little bit, but the paper I read used a general formula, and then talked about spin 1/2 particles. The +/- wouldn't matter for spin 0. I'll see if I can't dig it up.

3. Jul 8, 2015

### BiGyElLoWhAt

See 1.3
This paper is basically for people who don't know anything, like me. =]
So maybe I have misinterpretted it, but I think that's how it works. I could be wrong.

4. Jul 8, 2015

5. Jul 9, 2015

### tomdodd4598

I tried that, but then I began getting all of my probabilities of different outcomes adding up to numbers bigger than 1.
I now have two different ways to find the probabilities of the outcomes of measurements, but they sometimes give different answers when I do calculations. The crucial difference is that one always gives me a 0 probability that the spin can be measured as 0, or, in other words, the only possible results are spins of 1 and -1, while the other shows that 1, 0 and -1 are all possible measured spins. I guess my question is whether a spin of 0 can actually be measured. If it can, then I have a problem as my projection operators don't work, but if it can't, then that's great as I can still form the projection operators as usual.

6. Jul 10, 2015

### tomdodd4598

If the particle is in the state (0, 1, 0), so has spin 0 in the z direction, is the probability of measuring the spin along the z direction to be 0 equal to 1, or is there a probability of 1/2 for getting 1 and a probability of 1/2 for getting -1 as an answer to the measurement?

7. Jul 13, 2015

### BiGyElLoWhAt

Well if you know the state, then you can't measure something other than the state.

8. Jul 13, 2015

### tomdodd4598

Yeh, in hindsight I should have known that - there is another question that I don't know the answer to:

If the particle has spin 0 in the z direction, what are the probabilities of measuring the spin along the x direction to be 1, 0 and -1?

I get the answer of probabilities of 1/2, 0 and 1/2 for measuring a spin of 1, 0 and -1 respectively, but am not sure if this is correct.

9. Jul 13, 2015

### BiGyElLoWhAt

It is for a free particle. If you set up the Hamiltonian and there's bias for spin in one direction, it'll come out in the probabilities.

10. Jul 13, 2015

### tomdodd4598

Ok, thanks - my next question is this - are the spin matrices the following:

σ(z) = \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}

σ(x) = \begin{array}{ccc} 0 & 1/√2 & 0 \\ 1/√2 & 0 & 1/√2 \\ 0 & 1/√2 & 0 \end{array}

σ(y) = \begin{array}{ccc} 0 & -i/√2 & 0 \\ i/√2 & 0 & -i/√2 \\ 0 & i/√2 & 0 \end{array}

11. Jul 13, 2015

### BiGyElLoWhAt

Yea, that looks right.

12. Jul 13, 2015

### tomdodd4598

Ok, so at least I got that right :)

I'm starting to look at what happens when you have more than one of these spin 1 particles together, or 'n' spin 1 particles. I learned that for 2 spin 1/2 particles, there are, for each direction, 2 spin matrices, one for each particle, and that these are constructed in the following way:

σ(w)₁ = σ(w) ⊗ I₂
σ(w)₂ = I₂ ⊗ σ(w)

Where w is the direction and the subscript is the matrix/particle number.

From that, I can only assume that the following is the case:

Where w is the direction, a is the matrix/particle number, n is the number of particles and dim(V) = (2s + 1)^n, where s is the spin of each particle (1/2 for spin 1/2, 1 for spin 1 etc.).

Is this correct, or are the spin matrices formed in a different way?