1
vote
0answers
59 views

Normal ordering

If I understood correctly there are two terms called normal ordering: $:c c^\dagger: = c^\dagger c \hspace{.5cm}$so shifting all creation operators to the left and all annihilation operators to the ...
0
votes
0answers
47 views

Tricks at manipulating creation/annihilation operators

Manipulation of terms in algebras different from the standard one (e.g. boolean algebra) can be a bit unnatural but there are always shortcuts that can help you. I was wondering if there is a list ...
2
votes
1answer
156 views

Reason behind canonical quantization in QFT?

Reason behind canonical quantization in QFT? In the scalar field theory we simply promote the scalar field, $\phi(x)$ to a set of operators: $\hat{\phi}(x)$. What is the reason behind this?
0
votes
0answers
97 views

Time ordering and Fermions

Having time ordering operator for fermions, should it reverse sign if it swaps operators with opposite spin variable? In other words should $T[c_{t_1,\uparrow}c_{t_2,\downarrow}^\dagger]$ return ...
5
votes
1answer
505 views

How to evaluate spin operators in second quantization for spin symmetry-broken Slater determinants?

Suppose we have the following Slater determinant: \begin{equation} | \Psi \rangle = \prod \limits_{i,i'} a^+_{i\alpha} a^+_{i'\beta} | \rangle \end{equation} where $a^+_{i\alpha}$ creates an electron ...
2
votes
1answer
222 views

A missing factor of 2 in the standard Hartree-Fock mean field?

Let's start from a very simple argument: If $A$ and $B$ are some operators, then I can write their product as $$AB = (A-\langle A\rangle)(B - \langle B \rangle) + \langle A \rangle B + A \langle B ...