How do antisymmetric wavefunctions behave for 3 or more identical particles? I get that the basic gist of what antisymmetric wavefunctions are is that switching the variables means flipping the sign of the wavefunction, which can be written concisely as $\Psi\left(x_1,x_2\right)=-\Psi\left(x_2,x_1\right)$ in the case of two particles. But what does this mean for wavefunctions with more than two particles? Does switching two of the variables still flip the sign, i.e. $\Psi\left(x_1,x_2,x_3\right)=-\Psi\left(x_2,x_1,x_3\right)$?
 A: Here's two ways to express it.
First, given a totally antisymmetric wavefunction $\Psi(x_1,x_2,x_3)$, we have 5 relationships among the 6 possible permutations of $x_1, x_2, x_3$:
\begin{eqnarray}
\Psi(x_1,x_2,x_3) = - \Psi(x_2,x_1,x_3) = \Psi(x_3,x_1,x_2) = -\Psi(x_3,x_2,x_1) = \Psi(x_2,x_3,x_1) 
 = - \Psi(x_2,x_3,x_1)
\end{eqnarray}
Every even permutation gets a plus sign, and every odd permutation gets a minus sign.
For $N$ particles, you would have $N!-1$ relationships among $N!$ permutations. (Note for $N=2$, you get $1$ relationship among $2$ permutations, which is $\Psi(x_1,x_2)=-\Psi(x_2,x_1)$).
Second, one way to satisfy these relationships, is to start with a general wavefunction $\psi(x_1,x_2,x_3)$, and form a totally antisymmetric wavefunction by adding permutations with the right signs:
\begin{equation}
\Psi(x_1,x_2,x_3) = \psi(x_1,x_2,x_3) - \psi(x_2,x_1,x_3) + \psi(x_3,x_1,x_2) -\psi(x_3,x_2,x_1) + \psi(x_2,x_3,x_1)   - \psi(x_2,x_3,x_1)
\end{equation}
This also generalizes to $N$ particles.
In the case when the wavefunction separates in a product of one-particle states, a concise way to write this expression is the Slater determinant.
