How to determine whether an eigenstate of total spin is symmetric or antisymmetric? Here we have two identical paticles with spin $I$, integer or half-integer,
and there are $(2I+1)^2$ states.
Each one of them can be uniquely determined by total spin and its orientation, we can use $|J,m\rangle$ to represent this state. And because of its uniqueness, it is either symmetric or antisymmetric. 
How to determine whether $|J,m\rangle$ is symmetric or antisymmetric based on $I$, $J$ and $m$? 
 A: Let's denote the spins of the individual particles by $j_1 = j_2 = I$, the quantum numbers for the $z$-components of their angular momentum by $m_1$ and $m_2$, the spin of their combined state by $J$, and the $z$-component of the angular momentum of the combined state by $M$.  We have two bases for the states of these particles:  the "individual particle" basis, denoted by
$$
|I \, m_1 \, I \, m_2 \rangle
$$
and the "combined particle" basis, denoted by 
$$
|J \, M \rangle.
$$
(Note that $j_1$ and $j_2$ are also still "good" quantum numbers for this latter state;  but including them in both notations is redundant and can confuse things, so I'll omit them.)  Finally, let's denote by $\hat{E}$ the exchange operator between particles 1 & 2, i.e., we define $\hat{E}$ such that
$$
\hat{E} |I \, m_1 \, I \, m_2 \rangle = |I \, m_2 \, I \, m_1 \rangle.
$$
We can perform a basis transformation to express any state $|J \, M \rangle$ in terms of the basis $|I \, m_1 \, I \, m_2 \rangle$:
$$
|J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I  \,m_1  \,I \, m_2 | J \, M \rangle 
$$
The coefficients $\langle I  \,m_1  \,I \, m_2 | J \, M \rangle$ are known as the Clebsch-Gordan coefficients.  We want to know what happens when we apply the exchange operator $\hat{E}$ to this state:
$$
\hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_2 \, I \, m_1 \rangle \langle I  \,m_1  \,I \, m_2 | J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I  \,m_2  \,I \, m_1 | J \, M \rangle
$$
(The second step is just a relabelling of dummy indices $m_1$ and $m_2$ in the sum.)  We can see that $|J \, M \rangle$ will be an eigenstate of $\hat{E}$ if and only if all of the Clebsch-Gordan coefficients $\langle I  \,m_1  \,I \, m_2 |J \, M \rangle$ are multiplied by the same factor when we exchange $m_1 \leftrightarrow m_2$.  Thankfully, the Clebsch-Gordon coefficients satisfy the identity
$$
\langle j_1  \,m_1  \, j_2 \, m_2 | J \, M \rangle = (-1)^{j_1 + j_2 - J} \langle j_2  \,m_2  \, j_1 \, m_1 | J \, M \rangle
$$
and so we have
\begin{multline}
\hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \left[ (-1)^{2I - J} \langle I  \,m_1  \,I \, m_2 | J \, M \rangle  \right] \\ = (-1)^{2I - J} \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle  \langle I  \,m_1  \,I \, m_2 | J \, M \rangle = (-1)^{2I - J} |J \, M \rangle.
\end{multline}
Thus, the combined states are symmetric when $2I$ and $J$ are both even or both odd, and antisymmetric when one quantity is even and the other is odd.
A: For spin-1/2 particles, the entire wavefunction has to be antisymmetric under particle exchange. Also, the spatial component has parity $(-1)^\ell$, so if $\ell$ is even, the spin component must be odd under exchange.
