Grassmann numbers and Fermions

[kelly0303]

Hello! I am a bit confused about fermions in QFT when they are considered grassmann numbers. If you have 2 grassmann numbers $\theta_1$ and $\theta_2$, something of the form $\theta_1\theta_2\theta_1\theta_2$ gives zero. However, a term in a QED lagrangian of the form $\bar{\psi(x)}\psi(x)\bar{\psi(x)}\psi(x)$ is not automatically zero. If fermions are Grassmann numbers, why isn't that term also zero?

 

[jambaugh]

Fermions are not Grassmann numbers. Fermions are quantum particles with Fermi-Dirac statistics.
That statistics can be expressed by the fact that their creation operators are elements of a Grassmann algebra, and dually their annihilation operators are also elements of a (dual) Grassmann algebra. But the full algebra of Fermi-Dirac creation and annihilation operators forms a neutral signature Clifford Algebra.

Specifically:
\bar{\psi}_a(x)\psi_b(y) +\psi_b(y)\bar{\psi}_a(x) = \delta_{ab} \delta(y-x)\boldsymbol{1}
(That's the Kronecker and Dirac delta functions on the r.h.s.)

 

[haushofer]

Hello! I am a bit confused about fermions in QFT when they are considered grassmann numbers. If you have 2 grassmann numbers $\theta_1$ and $\theta_2$, something of the form $\theta_1\theta_2\theta_1\theta_2$ gives zero. However, a term in a QED lagrangian of the form $\bar{\psi(x)}\psi(x)\bar{\psi(x)}\psi(x)$ is not automatically zero. If fermions are Grassmann numbers, why isn't that term also zero?

Well, how do you do derive that $\theta_1\theta_2\theta_1\theta_2$ gives zero? I guess you write $\theta_1\theta_2\theta_1\theta_2 = - \theta_1\theta_1\theta_2\theta_2 = -0 \times 0 = 0$.

So the question is: if you write out the term $\bar{\psi}(x)\psi(x)\bar{\psi}(x)\psi(x)$ explicitly in spinorcomponents of $\psi$, what do you get? I leave that explicit writing out to you; that's the essential part of your question, I think ;) So pick a basis and write it out.

edit: your tex-code should read \bar{\psi}(x) instead of \bar{\psi(x)} for readability if you ever want to use that in your own texts.