DarMM said:
Most of constructive field theory uses very few ideas from perturbation theory.
How many of them “very few ideas” are there? Do you realize that you have re-stated my exact statement?
Rather it uses the scaled cluster (propogator) expansion and even then not as a perturbative expansion.
This is PERTURBATION SERIES. And the whole business of CQFT is about showing that Schwinger functions satisfy the so-called Osterwalder-Schrader axioms.
Could you comment on how Constructive field theory uses perturbation theory, with reference to, let us say, the construction of ##\phi^{4}_{3}## as a simple example?
I am sorry to disappoint you, the last time I wasted my time on this subject was in 1995. So, I’m neither an expert nor do I have an up to date knowledge about this (dead) subject. However, since you seem to like the subject and know about it, can you explain the following from \phi^{ 4 }_{ 2 } model?
In the process of “renormalization”, one replaces the “\phi^{ 3 }” with the polynomial \phi^{ 3 } - k \phi.
i) Can you tell people what value does the constant k take?
ii) Who ordered this constant into the construction?
iii) Does it have any relation to perturbation theory? This last question is just to see if your above question to me has an answer. Thanks.
In the examples of the construction of a field theory that I am familiar with, the Langragian formalism along with symmetry generators written in terms of the field do appear.
What do you mean by “appear”?. The examples you are familiar with, start with field equations, i.e. you
already have Lagrangians. When you have a Lagrangian, you do have a Noether current.
Could you explain a little, with reference to a paper involving an actual construction of a field theory in constructive field theory how does luxury does not exist?
Yes, always true:
No Lagrangian \sim No Luxury, or
Give me the Lagrangian, I give you everything. I’ve already mentioned the section, chapter and the book which explain that for you. The book also points you to some more relevant paper on the subject.
"job" to regulate singularities, I'm actually not sure what that would mean,
It means: Noether theorem gives you a CLASSICAL expression for the symmetry current, which may or may not hold as an operator expression; in general, one should allow for “quantum corrections”. Also, because of (anti)commutation relations the current is necessarily singular and it is
our job to regulate the singularity by, say, point-splitting technique. For example, for the axial current of massless fermion in the presence of EM interaction, we write
J_{ 5 }^{ a } ( x ; k , \epsilon ) = i \bar{ \psi } ( x + \epsilon / 2 ) \gamma^{ a } \gamma_{ 5 } \psi ( x - \epsilon / 2 ) e^{ i e k \int_{ x - \epsilon / 2 }^{ x + \epsilon / 2 } d y^{ c } A_{ c } ( y ) } .
The local physical current is obtained by averaging over the directions of \epsilon_{ a } and taking the limit \epsilon^{ 2 } \to 0.
Now considering one fermion coupled to an external em field, we can by some lengthy but non-trivial algebra, involving the use of the field equation and the free fremion propagator, calculate VEV of \partial_{ a } J_{ 5 }^{ a } ( x ; k ,\epsilon ). You should obtain, after 4 pages of calculation, the following
\langle 0 | \partial_{ a } J_{ 5 }^{ a } ( x ; k ,\epsilon ) | 0 \rangle = \frac{ e^{ 2 } ( 1 + k ) }{ 16 \pi^{ 2 } } \bar{ F }_{ a b } F^{ a b } .
This shows that, in perturbation theory, it is impossible to maintain both gauge invariance and chirality, though either one can be satisfied; k = 1 for gauge invariance and k = - 1 corresponds to the equally good global axial symmetry.
but it is still an example where the theorem fails.
No, because (as I explained previously) you can take the
CONSERVED SYMMETRY CURRENT of Noether to be the
bare
\mathcal{ J }^{ a }_{ 5 } = J^{ a }_{ 5 } - \frac{ e^{ 2 } }{ 4 \pi^{ 2 } } \bar{ F }^{ a c } A_{ c } .
This current is
a finite operator and does not need
renormalization. Plus, again as I have pointed out before, the
charge associated with this current satisfies
all properties of a
genuine Noether charge. So, if it looks like a dog, barks like a dog, runs like a dog and bites like a dog, it can not be a fish.
So it cannot be fundamental to quantum field theory. Instead the Ward-Takahashi identities are
I am not sure if that statement makes any sense. I can argue for the opposite, because we derive the W-T identities either from the Noether identity
\frac{ \delta S }{ \delta \psi } \delta \psi + \partial_{ a } J^{ a } = 0 ,
or from the canonical commutator
\delta \psi ( x ) \delta ( x - y ) = [ i J^{ 0 } ( x ) , \psi ( y ) ] .
Unless of course, you can show us that the W-T identities are derivable without any reference to Noether theorem! Can you?
Sam