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For the set of all fermionic field operators $\Psi(x) | x \in \mathbb{R}^{3 +1}$, we won't find a $|\phi \rangle$ that is an eigenstate to the complete set of field operators, unless we make use of Gr …

In my previous question, I asked about how to handle Grassmann-number variations of operators. I read a book that uses those variations $\delta \Phi = c \mathbb{1}$, with $c$ being a grassmann number …

In the Book "Quantum Field Theory I" by Manoukian, in section 4.3, from what I understood, he derived the quantum-action-principle of Schwinger only by using unitary time-evolution of the field Operat …

As some other questions on this website suggest, I have a really hard time with the fermionic field operator $\psi(x)$. I'd like to come to terms with this blockade. It serves as the smallest building …

As far as I understood, in the path integral formulation of QFT, a field configuration is modelled by a mapping $$ x \rightarrow \Psi(x) $$ Where $\Psi(x)$ are 4 components, each represented by 4 gras …

I'm following the Book of Brian Hatfield (Quantum Field Theory of point particles and Strings), p.192 here: For real Grassmann numbers (and Functionals thereof): If $\Phi[\psi]$ is a functional, and $ …

For Quantum-mechanics on a Hilbert-space over the complex numbers, the usual scalar product of two states $\langle \phi | \psi \rangle$ and gives the transition amplitude between the two states. The a …

Fermionic field operators do obey anticommutation relations, so for a chosen Field operator (and the field momentum), we have: $$ \{\Psi_a, \Psi_b\} = \{\pi_a, \pi_b\}= 0 $$ with the $\Psi_a$ being …

By extending the Hilbert space, using grassmann numbers instead of complex numbers, we can write down eigenstates of the fermionic annihilation operator $a$ without getting into trouble with the antic …