7,088 reputation
21339
bio website
location
age
visits member for 3 years, 10 months
seen 20 hours ago

I once got busy in a Burger King bathroom.


Nov
4
comment Are there QFTs in which a field cannot produce a real particle?
Also, conformal field theories don't have particle excitations.
Nov
1
comment What is the meaning of a state in QFT?
You can't in general identify states with fields. Any operator -- including a field operator -- creates a state. But it's not the case that two different operators must create different states.
Nov
1
comment What is the meaning of a state in QFT?
Sure, but those VEVs are amplitudes between certain states. It's really not any different from QM.
Nov
1
comment What is the meaning of a state in QFT?
QFT is a quantum theory. It has states just as any quantum theory has states. Why do you say that states don't make sense in QFT?
Oct
15
accepted Classic Literature in Quantum Gravity?
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
9
answered Moduli spaces in string theory vs. soliton theory
Jul
27
awarded  Pundit
Jul
10
comment A rigorous treatment of distributions in quantum mechanics
See also: physics.stackexchange.com/q/43515
Jul
4
awarded  Nice Answer
Jun
29
comment QFT's that have no action
Probably easier to get the definition right just by saying how the field transforms under special conformal transformations.
Jun
29
comment QFT's that have no action
The nontrivial half of the state-operator correspondence is gotten by shrinking a sphere down to a point where the operator is defined. Apply state-operator correspondence to eigenvector and you get a field.
Jun
29
comment QFT's that have no action
I'm afraid your edits have made your answer slightly confusing. The Virasaro algebra, on the other hand, is peculiar to 2-dimensional CFT. In general CFTs, one defines primary fields by looking for eigenvectors of the dilation operator.
Jun
28
comment QFT's that have no action
The state-operator correspondence is a feature of CFT in general, not just two-dimensional CFT. Likewise, the use of primary fields to generate the algebra of observables via OPE works in any d-dimensional CFT.
Jun
28
comment Perturbative vs. non-perturbative approaches to a well-defined Yang-Mills theory in 4 dimensions
Yes, of course @TobiasDiez meant $e^{-1/g^2}$.
Jun
25
comment Rigorous QFT on a Torus
You can't avoid dealing with color confinement once the spacetime volume is large enough. This is the big obstacle, the one the Clay prize is aimed at.
Jun
25
comment Rigorous QFT on a Torus
Yes. The problem is that the infinite volume limit leads to divergences not present in finite volume. These divergences reflect real physics; they tell you that the gluons are confined on long distance scales.
Jun
25
comment Rigorous QFT on a Torus
I'm sorry: Are you asking if its harder to construct YM on a torus than on $\mathbb{R}^4$?
Jun
25
answered Rigorous QFT on a Torus