Skip to main content

All Questions

1 vote
1 answer
66 views

Anticommutator Relation of Quantized Fermionic Field and Fermi–Dirac statistics: How are these related?

I'm reading the Wikipedia article about Fermionic field and have some troubles to understand the meaning following phrase: We impose an anticommutator relation (as opposed to a commutation relation ...
user267839's user avatar
  • 1,395
0 votes
0 answers
67 views

Rewriting two-body operator in second-quantized form

I would like to understand the following identity for fermion field operators: $$\psi^\dagger(x) \psi^\dagger(y) \psi(y) \psi(x) = \psi^\dagger(x) \psi(x) \psi^\dagger(y) \psi(y) - \delta(x - y) \psi^\...
DJMeister's user avatar
2 votes
2 answers
1k views

Using Grassmann variables on fermionic theories

I think the best way to put my question is the following: what (fermionic) theories make use of Grassmann variables? Let me clarify my question a little further. I remember some discussions in quantum ...
JustWannaKnow's user avatar
3 votes
1 answer
722 views

Minus Sign in Fermionic Creation and Annihilation Operators

I have the same question as the person here: Action of Fermionic Creation and Annihilation Operators The question actually wasn't anwered, because using anticommutation relations between creation $c_\...
Olexot's user avatar
  • 182
1 vote
1 answer
201 views

Second quantisation for fermions

I am trying to build a model for reactions on a lattice in the Doi-Peliti formalism. Suppose there exists a lattice of $N$ sites indexed by $i$. Each site can be either occupied or unoccupied. ...
jcp's user avatar
  • 141
2 votes
1 answer
1k views

Bogoliubov transformation for fermionic Hamiltonian

I have the Hamiltonian $H=\sum\limits_k [Ab^{\dagger}_{k}b_{k} + B(b^{\dagger}_kb^{\dagger}_{-k}+b_{k}b_{-k})]$, where $b^{\dagger}_k$ and $b_k$ are fermionic creation and annihilation operators. ...
C-Roux's user avatar
  • 179
1 vote
1 answer
288 views

Can a general many-body Hamiltonian with quadratic and biquadratic terms be diagonalized?

Can an arbitrary many-body hamiltonian in second quantization form with quadratic and biquadratic terms $$H=\sum_{v_1,v_2} \alpha_{v_1 v_2}\ c_{v_1}^{\dagger}c_{v_2}+ \sum_{v_1,v_2,v_3,v_4}\beta_{v_1 ...
norman's user avatar
  • 13
1 vote
2 answers
2k views

Deriving anti-commutation relation between creation/annihilation operators for Dirac fermions

Starting from Dirac fields: $$\Psi(x) = \dfrac{1}{(2\pi)^{3/2}} \int \dfrac{d^3k}{\sqrt{2\omega_k}}\sum_r\left[ c_r(k)u_r(k)e^{-ikx}+d^\dagger_r(k)v_r(k)e^{-ikx} \right]_{k_0=\omega_k}$$ $$\Psi^\...
Mors's user avatar
  • 23
2 votes
1 answer
194 views

Spinor quantization: contradiction between covariant anticommutator and canonical rules?

Starting from the free lagrangian $$\mathscr L = \bar\Psi(i\displaystyle{\not}\partial - m)\Psi$$ I compute the canonical momenta $$\Pi =\frac{\partial \mathscr L}{\partial\dot{\Psi}}=i\Psi^\dagger ...
M. M. R.'s user avatar
  • 523
1 vote
0 answers
83 views

Some subtleties in quantizing a fermi field

Consider the quantization conditions for a complex Fermi field $\Psi=\Phi_1+i\Phi_2$: $$\{\Psi(x),\Psi(y)\}=\{\Psi^\dagger(x)\Psi^\dagger(y)\}=0,~~~~ \{\Psi^\dagger(x),\Psi(y)\}=\delta(x-y)$$ Compare ...
David Roberts's user avatar
1 vote
0 answers
146 views

Definition of partity in quantized Dirac Theory.

I'm studying from the book "An Introduction to Quantum Field Theory" from Michael E. Peskin and Daniel V. Schroeder, and I read the following: "The operator P should reverse momentum of a particle ...
OnTheHighlands's user avatar
1 vote
1 answer
295 views

Matrix elements of a one-fermion operator (first and second quantizations)

I'm currently struggling with the expression of operators in second quantization. I did an exercise in which I had to consider a fermion in a central potential $V(\vec{r})$ and show that the matrix ...
mackenshin's user avatar