# What does it mean to 'combine' spin 1/2 particles [duplicate]

I'm trying to understand a phenomenon covered in several resources that I think I'm struggling with due to my lack of understanding of its meaning.

The idea is that 2 spin 1/2 particles can combine to form 4 spin states. These are broken down into the following prescriptions:

1. $$|1,1\rangle = |\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle$$

2. $$|1,0\rangle = \sqrt{\frac{1}{2}}(|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},\frac{-1}{2}\rangle+|\frac{1}{2},\frac{-1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle)$$

3. $$|1,-1\rangle = |\frac{1}{2},\frac{-1}{2}\rangle|\frac{1}{2},\frac{-1}{2}\rangle$$

4. $$|0,0\rangle = \sqrt{\frac{1}{2}}(|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},\frac{-1}{2}\rangle-|\frac{1}{2},\frac{-1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle).$$

My understanding is that each of the $$|s,m\rangle$$ states refers to a particle. In this case, $$s = 1/2$$ and $$m = 1/2$$ or $$-1/2$$ referring to spin up or spin down respectively.

These individual states then form a system which has an overall spin of 1 or 0 and another quantity which I think is the projection of the spin onto the $$z$$-axis but I'm not really sure as to the meaning/importance of this.

By this understanding, I read the above as you can get the first state as a combination of 2 spin up spin 1/2 particles and the third state as a combination of 2 spin down spin 1/2 particles.

I'm a little bit more confused by the others though. Is this 4 particles in the state or just a statement that it could be up/down or down/up with 50/50 odds. If so, what is the difference between these? Also, where does the root 1/2 come from? I assume some normalisation thing but can't see the exact reasoning.

Sorry, I appreciate that this is a big question. I just essentially want a breakdown of what $$s$$ and $$m$$ are, what the 4 statements physically represent and how the total state is derived from the individual states/vice versa.

• Mar 21 at 18:56
• I don't think the proposed duplicate is particularly helpful in answering the OP here. It's certainly related, but it assumes a pretty high level of math background not needed to answer this question. Mar 23 at 19:31

From the four independent tensor-product basis states: $$|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle$$ $$|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle$$ $$|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle$$ $$|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle$$

You can form whatever different linear combination you would like.

However, usually, you would like to form linear combinations that make it easier to solve a problem. And usually the problem you would like to solve involves a Hamiltonian that is invariant under rotations, and thus commutes with the total angular momentum squared, and total z-angular momentum.

The four basis states written above do not transform simply under rotations. This is because they are not eigenfunctions of the total angular momentum squared ($$\vec L^2 = (\vec L^{(1)}+\vec L^{(2)})^2$$). Rather, they are eigenfunctions of the individual angular momenta squared ($${L^{(1)}}^2$$, $${L^{(2)}}^2$$) and z-angular momenta ($$L_z^{(1)}$$, $$L_z^{(2)}$$).

This is why we like to form linear combinations that make up three new independent basis states: $$|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle$$ $$\frac{1}{\sqrt{2}} \left(|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle +|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle \right)$$ $$|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle\;,$$ which transform like a vector. This "vector" representation is also referred to as "triplet" or "spin $$1$$" or sometimes just "$$1$$".

And a linear combination of a fourth independent state: $$\frac{1}{\sqrt{2}} \left(|\frac{1}{2},\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle -|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},\frac{1}{2}\rangle \right)$$ which transforms like a scalar. This scalar combination is also referred to as "singlet" or "spin $$0$$" or sometimes just "$$0$$".

This can be written symbolically as: $$\frac{1}{2}\otimes\frac{1}{2} = 1 \oplus 0$$

Similarly, if we wanted to "add" three spin-$$1/2$$ angular momentum we would find: $$\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2} =(1 \oplus 0)\otimes \frac{1}{2} =\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{1}{2}$$

Similarly, if we wanted to "add" four spin-$$1/2$$ angular momentum we would find: $$\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2} =(1 \oplus 0)\otimes \frac{1}{2}\otimes\frac{1}{2} =(\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{1}{2})\otimes\frac{1}{2} =2\oplus 1\oplus 1\oplus 0 \oplus 1 \oplus 0\;,$$ and so on.

Is this 4 particles in the state or just a statement that it could be up/down or down/up with 50/50 odds.

In the case you are asking about, there are only ever two particles.

If so, what is the difference between these?

The difference is how they transform under rotations.

Also, where does the root 1/2 come from?

It comes from normalizing the state.