# Total spin angular momentum of a two spin-1/2 particle system

We have a system of two spin-1/2 particles. The states of the system in the "uncoupled representation" are

$$\mid m_1, m_2 \rangle\ =\ \mid \uparrow, \uparrow \rangle,\ \mid \uparrow, \downarrow \rangle,\ \mid \downarrow, \uparrow \rangle,\ \mid \downarrow, \downarrow \rangle$$

where the up arrow represents spin +1/2 and the down arrow represents spin -1/2. The states are eigenstates of the operators $\hat{s}_{1z}$ and $\hat{s}_{2z}$. Uncoupled in this case means the two particles are independent of (uncorrelated to) each other.

We can also have a "coupled representation" of the system, in which case, the states are

$$\mid S, M_S \rangle\ =\ \mid 1, 1 \rangle,\ \mid 1, 0 \rangle,\ \mid 1, -1 \rangle,\ \mid 0, 0 \rangle$$

where the states are eigenstates of the operators $\hat{S} = \hat{s_1} + \hat{s_2}$ and $\hat{S_z}$. Coupled in the sense that the two particles are no longer independent.

This sounds really stupid but here's my question. If $s_1 = s_2 = \frac{1}{2}$, the possible values of $S$ can be $1$ and $0$ but why can't it be $-1$? Why can't we have a state(s) with $S = -1$?

In the $\vert{S, M_S}\rangle$ the $S$ it's related the norm of the added spin and $M_S$ it's the orientation (proyection in a axis). So, the old intuitive $-1$ spin it's the $\vert{1, -1}\rangle$ ket.

Yes they can, it's the $\left| 1, -1 \right>$ state, but $S$ is the magnitude of the spin. You're thinking of eigenvalues of the operator $\hat{S}$. Those eigenvalues are $M_s$, don't let the notation confuse you.

Consider $\mathbf{J}_1$ and $\mathbf{J}_2$ be two angular momentum operators, acting on state space $\mathcal{E}_1$ and $\mathcal{E}_2$. The total angular momentum $\mathbf{J}=\mathbf{J}_1+\mathbf{J}_2$ acts on the total state space $\mathcal{E}_1\otimes \mathcal{E}_2$.

You are just dealing with the particular case $\mathbf{J}_i=\mathbf{S}_i$.

There is a natural basis on the total space, formed by the eigenvectors common to $J_1^2,J_2^2,J_{1z},J_{2z}$ which is

$$|j_1,m_1;j_2,m_2\rangle=|j_1,m_1\rangle\otimes |j_2,m_2\rangle,$$

this is the one you call uncoupled. This basis is not convenient to deal with the total angular momentum.

The whole point is that we can consider the new set of commuting observables $J_1^2,J_2^2,J^2,J_z$. Since $\mathbf{J}$ is also angular momentum, $J^2$ has eigenvalues $j(j+1)\hbar^2$ and $J_z$ has eigenvalues $m\hbar$. The eigenvectors forming the basis corresponding to this commuting set of observables are of the form

$$|j_1,j_2,j,m\rangle$$

It remains to determine the possible values of $j$. Now one can prove that given $j_1$,$j_2$ the only possible values of $j$, and hence the only possible kets $|j_1,j_2,j,m\rangle$, are the following ones

$$j=|j_1+j_2|,|j_1+j_2-1|,\dots,|j_1-j_2|$$

so you pick $j_1,j_2$ start at the sum, and keep subtracting $1$ until you reach the difference $|j_1-j_2|$. These are the possible values for $j$.

In your case $j_1,j_2=1/2$, hence the only pair $j_1,j_2$ is $1/2,1/2$ and for this pair you can check that you have only $j=1,0$.

This is the basic idea. I strongly suggest reading the full development of these ideas. A good text is the chapters on this matter on Cohen's "Quantum Mechanics, Vol. 2" book.

Why can't we have a state(s) with $s=−1$?
1. Well, this follows from definitions in the representation theory for the $su(2)$ Lie algebra with generators $\hat{S}_i$. The operator $\hbar^{-2}\hat{\bf S}^2$ is semi-positive definite and a Casimir. This means its eigenvalues $\lambda\in [0,\infty[$ are non-negative, $$\hat{\bf S}^2|v\rangle~=~ \hbar^2\lambda|v\rangle.$$ Here $\hbar^2$ is included to make $\lambda$ dimensionless.
2. We next define a bijective map $$[0,\infty[~\ni~ s\quad\mapsto \quad s(s+1)~=~\lambda~\in~[0,\infty[.$$ Instead of labelling irreducible representations with $\lambda\in [0,\infty[$, we might as well label irreducible representations with $s\in [0,\infty[$. In practice, physicists use $s$ (and mathematicians use $2s$), since for finite-dimensional irreducible representations, $s$ turns out to be a half-integer, $$s~\in~\frac{1}{2}\mathbb{N}_0.$$
3. In particular, $s$ is by definition never negative, cf. OP's question. In the semiclassical regime $s \gg 1$, the variable $s$ has the physical interpretation of being approximately the magnitude $$\sqrt{\lambda}~=~\sqrt{s(s+1)}~\approx~ s\qquad\text{for}\qquad s ~\gg~ 1$$ of the spin vector $\hbar^{-1}\hat{\bf S}$.