# Physical meaning of symmetric and antisymmetric wavefunction

On describing Bosons and Fermions, the symmetry of wavefunction is introduced first. Here, If two particles a and b, are in two states n and k respectively, we get the wavefunction individually. On changing the position of particles between two states earns another wavefunction. But ,the overall general wavefunction is written in two linear combinations (this complicated description is simplifed in the picture below). Why if particles are interchanged, we get symmetric and anti symmetric wavefunctions. But what is the physical meaning of sign reversal in anti-symmetric case?

Edit: In the picture, the second case should be - particles, a in k state and b in n state.

## 2 Answers

Firstly, the question refers to the symmetry about particle exchange - the title is confusing, since there are many other important symmetries, starting with $$\psi(\pm x)=\pm \psi(x)$$, which has to do with the number of nodes in the eigenstates vs. the ground state.

Regarding the particle exchange, the argument given in most quantum mechanics books is that, since the particles are indistinguishable, the physical properties should be independent when we switch any two particles. Wave function is not a measurable quantity itself, but the particle density is. Thus, in a system of two particles we have $$\rho(x_1,x_2)=|\psi(x_1,x_2)|^2=\rho(x_2,x_1)=|\psi(x_2,x_1)|^2$$ (Note that a system of particles is generally described by a single wave function.) This means that the wave function under particle exchange may acquire a phase factor: $$\psi(x_1,x_2)=e^{i\phi}\psi(x_2,x_1).$$ If we exchange the particles twice, we should return to the initial state (this condition is however relaxed in the case of anyons): $$\psi(x_1,x_2)=e^{i\phi}\psi(x_2,x_1)=e^{2i\phi}\psi(x_1,x_2),$$ hence $$e^{2i\phi}=1\Rightarrow e^{i\phi}=\pm\sqrt{1}=\pm 1.$$

I.e., the wave function is either symmetric or antisymmetric in respect to exchange of any two particles (the discussion above is easily generalized to exchanging any two particles among a larger number of particles.)

In case of non-interacting particles, the eigenstates of the multiparticle system can be represented in terms of one-particle eigenstates, of which one has to form symmetric/antisymmetric combinations to satisfy the condition derived above: $$\psi(x_1,x_2)=\phi(x_1)\varphi(x_2)\pm \varphi(x_1)\phi(x_2).$$ This is easily generalized to more particles with Slater determinants and permanents.

This innocent game with the phase factor turn out to have far reaching consequences in terms of the particle properties (spin and magnetic moment), their quantum statistics (Bose-Einstein vs. Fermi-Dirac), their interactions, etc.

• (+1) What if $\phi(x)$ and $\varphi(x)$ are the same function, for instance in the case of the 2 electron Helium atom? It seems that the anti-symmetric wavefunction $\psi(x_1,x_2)=\phi(x_1)\phi(x_2) - \phi(x_1)\phi(x_2) = 0$ will vanish everywhere? Commented Jul 18 at 19:13
• @James indeed, this is called "Pauli exclusion principle" - two fermions cannot occupy the same state. Commented Jul 18 at 19:56
• thank you! I tried to implement it and got 0 everywhere even for different locations $x_1, x_2$, so I must be understanding something wrong... For example, $\Psi([1,2,3], [4,5,6]) = \phi([1,2,3])\phi([4,5,6]) - \phi([4,5,6])\phi([1,2,3])$ $= \phi([1,2,3])\phi([4,5,6]) - \phi([1,2,3])\phi([4,5,6]) = 0$ even though the locations $x_1, x_2$ are different. The whole antisymmetric wavefunction seems to vanish for any combination of $x_1, x_2$... Is there a mistake in my calculation method? Commented Jul 18 at 20:21
• @James your notation is not clear to me: how many particles do you have? The numbers ranges from 1 to 6, but you have only two factors of $\phi$. For six particles you would need a six-row Slater determinant. I added to the answer a few links to other answers, which may clarify your reasoning for more than 2 particles. Commented Jul 19 at 7:53

First of all, states of indistinguishable particles have to be either symmetric or antisymmetric with respect to a permutation $$P$$ of two particles, which is a parity operation $$P^2 = \mathbb{I}$$. If a state would not be an eigenstate of $$P$$, one could use the transformation behaviour to distinguish the two particles.

Now the sign seems like a technicality, but has a vast impact on how many particles interact with each other. Take for instance two fermions in the non-trivial states $$\psi_1$$ and $$\psi_2$$, respectively. If two fermions (admitting an antisymmetric wavefunction) would occupy the same state, the two-particle wave function would vanish

$$\psi_1 \otimes \psi_2 - \psi_2 \otimes \psi_1 = 0 \qquad \iff \qquad \psi_1 = \psi_2.$$

Bosons (symmetric wavefunction) on the other hand are allowed to occupy same states, in fact there is a phenomenon called Bose-Einstein condensation that takes this further. So the sign really contains information about allowed states. In the presence of Lorentz symmetry, one can show a relation of the statistics with the spin of the particles. Particles admitting a half-integral spin quantum number can be thought of as a physical interpretation of the sign, see Spin-Statistics Theorem for details.

• What is the RHS of parity operator P^2 = ? Is it I. I am new to quantum mech. , So I don't understand how the parity is related to indistinguishable particles? Commented Jul 18 at 9:48
• $\mathbb{I}$ is the identity operator. Every hermitian operator $P$ that satisfies $P^2 = \mathbb{I}$ can only have eigenvalues $\pm 1$ (there is a generalization to anyons, see @Roger V.'s answer). The permutation of two indistinguishable particles is such a parity operation, because it squares to identity (switching two particles twice is equivalent to doing nothing, again anyons are exceptions). Commented Jul 18 at 12:04
• The last statement is incorrect. The Spin-Statistics Theorem relies on special facts about the Lorentz Group. It's not equivalent to multi-particle states being eigenvectors of the permutation operator. The existence of anyons in 2D demonstrates this. Commented Jul 18 at 21:18
• True, let me edit my post accordingly. Commented Jul 19 at 7:29
• So.. if two particles are Fermions, the asymmetric equation does not allow the two particles to occupy the same state, right? Commented Jul 19 at 8:10