# Identical particles and antisymmetric wavefunction

I am just confused about why the overall wavefunction of two identical particles should be antisymmetric.

I was looking at this lecture note from University of Edinburgh's QM course:

So the way they are dealing with this idea is that for the spin triplet state, the spin wavefunction is symmetric and for the spin singlet state, the spin wavefunction is anti-symmetric, which I think I understand. Now, they are requiring the overall symmetry of the wavefunction to be antisymmetric and hence, the spatial wavefunction will take corresponding symmetries.

But why does the overall symmetry of the wavefunction need to be anti-symmetric?

• The short answer: Your total wave function must be fully antisymetric under permutation because you are building states of identical fermions. This is as the symmetrization postulate demands, although I think is fair to say that quantum field theory makes the connection between spin and permutation symmetry explicit. This topic is covered in basically all QM literature but are you asking why do we follow the postulate? Or...? – secavara Feb 6 '18 at 2:08

Consider two identical particles numbered as 1 and 2. Assuming that the particles interact weakly (i.e. no interaction), one can write the wave function of the system as the product of the respective wave functions: $$\Psi_{12}(r_1,r_2) = \Psi_1(r_1)\Psi_2(r_2)$$ where $\Psi$ represents the complete (spatial + spin) wave function.
In words of David J Griffiths, if I were to interchange the two particles, not even God could tell the difference. The system remains unchanged under the interchange operation. Therefore, $\Psi_{12}$ can also be written as: $$\Psi_{12}(r_1,r_2) = \Psi_1(r_2)\Psi_2(r_1)$$ A more generalized representation for the wave function $\Psi_{12}$ would be the linear combination of the above two expression: $$\Psi_{12}(r_1,r_2) = \alpha ~ [~ \Psi_1(r_1)\Psi_2(r_2) \pm \Psi_1(r_2)\Psi_2(r_1) ~] \label{wavef}$$ Since both terms are equally likely, the normalizing constant $\alpha$ remains same, however the signs may vary. Note that the wave function $\Psi_{12}$ can either be symmetric (+) or anti-symmetric (-).
Now, the exclusion principle demands that no two fermions can have the same position and momentum (or be in the same state). For the momentum to be identical, the functional form of $\Psi_1$ and $\Psi_2$ must be same, and for position, $r_1=r_2$. For the exclusion principle to hold true, the total wave function for the two fermions to be in the same state should be ZERO.
This is only possible if you choose $\Psi_{12}$ to be anti-symmetric since you'll get \begin{align} \Psi_{12}(r,r)&=\alpha [ \Psi(r1=r)\Psi(r2=r) - \Psi(r2=r)\Psi(r1=r) ]\\ &=0 \end{align}
• Small point, the parity operator is not the same as the exchange operator. Parity mirrors the system in all the coordinate planes so $r_1\rightarrow-r_1,\;r_2\rightarrow-r_2$. Exchange swaps the places of the 2 particles so $r_1\rightarrow r_2,\;r_2\rightarrow r_1$ – By Symmetry Feb 6 '18 at 10:19