I'm reading Griffiths QM 2nd ed and he talks about how all fermions are antisymmetric in nature. Later he talks about how exchange forces with identical particles. He states that because fermions are antisymmetric (not including spin), they can't perform covalent bonding. However, he then states that by including spin, when the electron is in an antisymmetric spin state, somehow that makes the bonding work. This doesn't seem to follow to me. I've been looking around, and many people mention how somehow the radial equation is symmetric by the spin state being antisymmetric, but this just seems arbitrary to me.

  • $\begingroup$ the total wavefunction must be antisymmetric for fermions! $\endgroup$ – drvrm Mar 3 '16 at 7:08

All half integer particles are antisymmetric by the virtue of the spin-statistics theorem, which states that half integer particles must follow Fermi-Dirac statistics. As for your doubt concerning the spatial part of the wave function is symmetric, the net wave function comprising of the spin part and the spatial wave part must be antisymmetric, and this allows the choice of the position part being symmetric with the spin part being antisymmetric. For our case we take the spatial part to be symmetric and the spin part to be antisymmetric.

Exchange interaction

Exchange interaction arises in identical particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing its sign (antisymmetric) when two particles are exchanged. For identical bosons, this results in attraction, while it results in repulsion for identical fermions (that is, when they have symmetric total wave-functions). But when the wavefunction is antisymmetric, the fermions attract each other.

Why this arises

If you take the Hamiltonian of, lets say a two-electron atom (which can be taken as a simple model for the covalent bond, although not exact, but from which basic features about exchange interaction can be understood from it), then in addition to the Coulumb interaction between nucleus and the electron, there will be a self interaction term also. Treating this self interaction term by Time independent perturbation theory, we can find out that there are two energy eigenvalues for the system, corresponding to the symmetric and the antisymmetric states.

Now in order to include effects due to spin, we take into account the Slater determinant, and find out the eigenvalues for the potential energy, which is now represented by a matrix. The state which is favored more (parallel spins or antiparallel spins) depends on the exchange constant, which is dependent on the difference between the energy eigenvalues of the symmetric and the antisymmetric ones.

Now if the exchange constant is positive, then the symmetric spin state is more preferred, while in the case of a negative exchange constant, the antisymmetric spin state is preferred. In our case of a two electron atom, the antisymmetric spin state is preferred, since it results in a lower energy than what would have been there if there was no spin/parallel spin. That indicates the reason to a good extent behind why covalent bonds are formed better if fermions are involved in the picture with antisymmetric spin wavefunctions, as can be inferred from the analysis of this two electron atom problem.

  • $\begingroup$ in order to circumvent this problem, Heisenberg introduced a model in which he took into account the spin spin coupling by introducing a corresponding term into the Hamiltonian directly...details are given on en.wikipedia.org/wiki/Classical_Heisenberg_model $\endgroup$ – Bruce Lee Mar 3 '16 at 13:47

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