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0 votes
0 answers
26 views

How can interacting field operators in $2D$ still satisfy the canonical commutation relation?

Free fields in any dimensions are well-known to be Gaussian, act on the Fock space and satisfy the canonical commutation relations. By definition, interacting field operators are NOT such cases, as ...
Keith's user avatar
  • 1,669
2 votes
0 answers
81 views

Why Fock representation holds only in a free quantum field theory?

With a quantum system with $N$ degrees of freedom, all the representations are unitarily equivalent to Fock representation. However, if the number of degrees of freedom goes to infinity, there are ...
MBlrd's user avatar
  • 159
1 vote
1 answer
332 views

How is the interacting vacuum defined in QFT?

I have seen this in a couple of textbooks (Schwartz and Zee), where the author would use the interacting vacuum $|\Omega \rangle$ in a calculation, but would never mention how the state is defined. ...
Tachyon's user avatar
  • 1,896
1 vote
0 answers
134 views

Clarification on interaction picture in QFT

Say we want to calculate $\langle f(t_2)|O|i(t_1)\rangle$. Where $O$ is an arbitrary operator. We can treat the states as stationary and then evolve the operator $$\langle f(0)|O(t)|i(0)\rangle\\O(t) =...
user avatar
5 votes
2 answers
1k views

How is Hilbert space constructed in interacting theory?

In the free field theory, we an decompose the field with creation and annihilation operators $a^{\dagger}_k$ and $a_k$. $a_k$ acts on some state $|0\rangle$ and outputs $0$. We call that state the ...
user avatar
7 votes
1 answer
321 views

Hilbert space of free theory vs interacting theory

In view of Haag's Theorem, it seems the Hilbert spaces of a free theory and an interacting theory are not the same. Though it seems very believable, I could not find a result that states that this is ...
CBBAM's user avatar
  • 3,350
4 votes
1 answer
184 views

Why is the $S$-matrix calculated using the free vacuum state and not the full interacting vacuum state?

Let $H = H_0 + H_I$ be a Hamiltonian that is the sum of a free Hamiltonian and an interacting Hamiltonian. Denote the free vacuum state by $| 0 \rangle$ and the full vacuums state by $|\Omega \rangle$....
CBBAM's user avatar
  • 3,350
5 votes
1 answer
575 views

Vacuum, creation and annihilation operators in interacting QFT

I am reading the QFT book by M. Schwartz. More specifically, I have issues with the section about LSZ. I am puzzled with the way the creation and annihilation operators from the free theory act there. ...
Dr.Yoma's user avatar
  • 705
3 votes
1 answer
223 views

What makes coming up with a mathematically solid, non-shaky relativistic quantum field theory (RQFT) so hard?

This is something I know of but I'm not quite sure I understand the details. Particularly, when it comes to interacting RQFTs, such as even QED, where some posts here have pointed out that it cannot ...
The_Sympathizer's user avatar
5 votes
1 answer
328 views

What does Haag's theorem say about the Schrodinger picture?

Suppose there are two interacting fields $\phi _1 $ and $\phi_2 $. Let $\psi [\phi_1, \phi_2]$ be a functional with the two fields as the input functions and complex numbers as the output, such that ...
Ryder Rude's user avatar
  • 6,355
4 votes
0 answers
247 views

Commutation relations interacting fields

I am reading Schwartz's "Quantum field theory and the standard model". I have a question on how he derives the Feynman rules for an interacting scalar field from a Lagrangian formalism. In ...
Lukas's user avatar
  • 133
6 votes
2 answers
412 views

How are annihilation/creation operators used to reach an external state of $|0 \rangle$ in an $S$-matrix?

I'm trying to understand how to compute the $S$-matrix element for $\phi \phi \to \phi \phi$. In "Peskin and Schroeder's Ch. 4.6". I'm lead to believe that, in $\phi^4$ theory, $$ S = \...
user7077252's user avatar
-3 votes
1 answer
174 views

General phenomenology of quantum field theory [closed]

Resume: Unfortunately, the moderators of the forum, like some of the participants, were unable to understand that the question was not about "personal theories", but rather about a widely ...
Arman Armenpress's user avatar
2 votes
1 answer
274 views

Vacuum of the QED and the free vacuum

I was reading in the Schwartz, "Quantum Field Theory and the Standard Model". On page 88 he finds an expression for proportionality of the vacuum of the interacting theory (the free theory ...
Joao Vitor's user avatar
3 votes
1 answer
481 views

Why do operator valued distributions in QFT refer to free fields only?

Operator valued distributions in quantum field theory refer to free fields. The creation operators act in Fock space. They create particle states of various energies and momenta. Every state has a ...
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