How one might derive the fermionic anticommutation relations? For bosonic particles, there is no ordering issue, and its commutation relation could be easily derived.

However, for fermion, is there any easy way to show that this must be true?


Basically, you need to use the Heisenberg-Lagrange-Hamilton approach...starting by the field Lagrangian that leads to Dirac wave equation; then you have to quantize the field by using a mode expansion in which the fermionic field is expressed in terms of the free solutions of the Dirac equation which as you know are spinor solutions. For instance, the expansion for a spin-${1}/{2}$ fermion field is:

$\hat{\Psi} = \int \frac{d^3k}{(2\pi)^3\sqrt{2\omega}} \sum_{s=1,2} [\hat{c}_s(k)u(k,s)\exp{(+ik.x)}+\hat{d}^{\dagger}_s(k)v(k,s)\exp{(-ik.x)}]$

Now consider a state with two fermions created by $\hat{c}^{\dagger}_{s_1}(k_1)\hat{c}^{\dagger}_{s_2}(k_2)$ from the vacuum. If you consider one of these double states consisted of exactly two same fermions, then that state must be anti-symmetric under the exchanges $k_1 \leftrightarrow k_2$ and $s_1 \leftrightarrow s_2$. Now if the operators $\hat{c}^{\dagger}_{s_1}(k_1)$ and $\hat{c}^{\dagger}_{s_2}(k_2)$ anti-commute, then the resulting doublets would be anti-symmetric. Lots of textbooks do have such proofs, for instance the proof here is from Aitchison and Hey's "Gauge Theories in Particle Physics Vol.1". Only a simple correction has been made here in the expansion.

Regarding the fundamental anti-commutation relation, we need to postulate the same thing as here. A worthful exercise would be to calculate the Hamiltonian of this field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.