# Noether vs Ward Takanashi

1. Oct 14, 2014

### the_pulp

This is what wikipedia says. Nevertheless, I dont think that it is true. I mean, the conservation of electric charge can follow from noether theorem field generalization applied to Electrodynamic Lagrangian, am I right? Is it wikipedia wrong by saying that conservation of electric charge can not follow from noether theorem? Or at least, not exactly right?

Thanks!

2. Oct 15, 2014

### dextercioby

Noether's theorem is valid only in classical field theory (which of course doesn't really exist for a Dirac field, for example, or even for a complex scalar field). <Conservation of electric charge> in the wiki context is meant in the sense of QFT, where you work with operators (operator valued distributions) and expectation values.

3. Oct 15, 2014

### the_pulp

Thanks for your answer. Nevertheless, in Ryder's "Quantum Field Theory", in page 90, where he talks about Complex Scalar Lagrangian, he derives a sort of electric current from the symmetry condition. This is a pure Classical Field Theory derivation that could be inferred by Noether theorem.
I know that life does not behave as Complex Scalar Lagrangian in Classical Field Theory says. What I try to be sure (or to rectify) is that in classical field theories, conservation of a sort of electric current can arise and that it is not a property of QFT through TW Identity (which seems to be in conflict with what wiki says). Am I right?

Thanks

4. Oct 15, 2014

### dextercioby

That's completely correct.

Yes, in the realm of classical dynamics (particles or fields), you only have the Noether theorem which provides you with the theory's conserved observables. They are then used in the so-called canonical quantization, so that one may be tempted to say that always classical -> quantum, when actually it's the other way around. But this is how the textbooks are written. They always start with the classical Lagrangians (which for the fields other than the e-m one or the gravitational one don't really exist), then bring in symmetries, Noether theorem, conserved charges then bam, canonical quantization. Makes you think that there's no quantum theory that didn't come from a classical counterpart through the 'Dirac trick'. But quantum theories can be built on their own, that's what axiomatical QFT for example does.

5. Oct 15, 2014

### samalkhaiat

What is wrong with Grassmann numbers? This is beauty of the Lagrangian formalism: the fields can take values in a completely arbitrary algebra.
Is there :) ? This is why he was a great man.
There has been a deep mathematical theory behind each one of “Dirac’s tricks”. His delta function paved the way for the Distribution Theory, and the “Poisson to commutator” trick led to the theory of Quantum Groups and quantization became a Lie algebra Deformation.
The program of the “Axiomatic QFT” is more than 60 years old yet, unfortunately, it failed to achieve its goal. The program produced countless theorems each with “weak and strong” version, but it can not construct interaction Hamiltonian for the simplest non-abelian multiplets. Ironically, the fact that Axiomatic QFT cannot account for Noether theorem is the main (if not the only) reason for its failure.

6. Oct 16, 2014

### dextercioby

Grassmann variables are a necessity stemming from quantum theories. They are mathematical artifacts for 'classical Lagrangians'.
Yes, by failing. :)
I offered this example just to show that one can think of QFT ouside its connection with the classical field theories.

7. Oct 16, 2014

### DarMM

There are quantum theories which are not the quantisation of a classical theory. They aren't too useful in relativistic field theory, although again, there are relativistic field theories which are not quantisations of classical ones, but just to say they do exist.

Axiomatic Field Theory as such wasn't concerned with constructing interacting field theories (that is Constructive Field Theory) rather it was concerned with seeing what mathematical system quantum field theories belong to (i.e. what sort of mathematical objects are fields) and then seeing what are the minimum set of assumptions on those objects in order for the work of physicists to be justified. It was then concerned with seeing what general theorems could be proved from those minimal assumptions.
Also interacting Hamiltonians have been constructed for non-Ableian gauge theories, but only in 1+1 and 2+1 dimensions.

Concerning Noether's theorem, I am not aware of an example of constructing a field theory that failed due to Noether's theorem, or how Noether's theorem prevented the construction of a field theory. Could you expand on this?

My understanding is that due to anomalies we wouldn't want Noether's theorem to be a fundamental part of QFT. Anomalies being examples of where application of Noether's theorem produces conserved currents, which are in fact not conserved in the Quantum Theory.
What prevents them from being conserved is in fact the fact that fields are distributions so that their products are ill defined, for example the chiral current in QCD, under Noether's theorem we have:

$\partial_{\mu}j_{5}^{\mu} = 0$

However since $j_{5}^{\mu}$ is a product of distributions, the fields themselves:

$j_{5}^{\mu} = \bar{\psi}^{a}\gamma^{\mu}\gamma^{5}\psi_{a}$

it is not well defined. We can define it in a mathematically rigorous way (extension of distributions, Epstein Glaser) or a physicists way (momentum space renormalisation), either way we get a new well-define version of $j_{5}^{\mu}$, let's call it $j_{5, R}^{\mu}$ however it then turns out that:

$\partial_{\mu}j_{5, R}^{\mu} \neq 0$

in fact:

$\partial_{\mu}j_{5, R}^{\mu} = \frac{N}{8\pi^2} F \wedge F$

which is responsible for the decay rate of pions and the heavy mass of the eta prime for example, very physical effects coming from the fact that Noether's theorem did not work.

EDIT: Personally I think anomalies are a good example to show that renormalisation is not some silly trick. That is, the fields are distributions, you can't just write down products without thinking and renormalisation is the physicists way of correcting these ill-defined products, i.e. working out the correct product and behold the correct product does not have the properties you'd expect, but its properties do show up in experiment.

Last edited: Oct 16, 2014
8. Oct 16, 2014

### DarMM

By the way, if anybody is wondering the real difficulty in constructing field theories in 3+1 dimensions is coupling constant renormalisation. Even in lower dimensions, theories with coupling constant renormalisation have never been constructed.

Originally, in the 1970s, there were problems with constructing theories that were only renormalisable as opposed to super-renormalisable. However this was successfully solved in the 1980s. Having the coupling constant be renormalised (by which I mean that for the physical coupling to be finite, the bare one must go to infinity or zero in the continuum limit) introduces a host of complications that has so far prevented analytic control over the theory.

Again I do not think that the Ward-Takahashi identities (quantum version of Noether's theorem) have posed any problem.

9. Oct 17, 2014

### samalkhaiat

Actually, I meant to criticize both AQFT and CQFT(after all, CQFT is nothing but synthesis of formal perturbative and axiomatic approaches).
In the Lagrangian formalism, thanks to Noether theorem, the generators of symmetry (including the Hamiltonian) are expressed in terms of fields. This luxury does not exist in the AQFT and CQFT.

See CH6.2 of J. Lopuszanski. “An Introduction to Symmetry and supersymmetry in QFT”, World Scientific.1991.
The departure of $\partial_{ a } J^{ a }_{ 5 }$ from its canonical value follows from: i) the intrinsic singularities of QFT, and ii) preserving gauge invariance. It is not Noether-theorem’s job to regulate those singularities for you.
In massless QED, the anomaly is often described as a “clash between two symmetries”: gauge invariance and axial symmetry. It is not possible to maintain both symmetries in perturbation theory, however, either one can be satisfied.
Due to the presence of a dimension 4 operator $\bar{F}^{ a b } F_{ a b }$ in the divergence, the renormalized axial current fails to generate chiral transformation. The corresponding axial charge is not conserved; it does not have the correct equal-time commutation relations with the renormalized fermions field.
In spite of the fact that conserved gauge-invariant axial current does not exist, axial symmetry can still be implemented in the theory. Indeed, it is possible to construct the following conserved bare current
$$\hat{ J }_{ 5 }^{ a } = J_{ 5 }^{ a } - \frac{ e^{ 2 } }{ 4 \pi^{ 2 } } \bar{ F }^{ a c } A_{ c } .$$
It can be shown that this operator is finite and does not need renormalization; the associated charge
$$Q_{ 5 } = \int d^{ 3 } x \hat{ J }_{ 5 }^{ 0 } ( x ) ,$$
is time-independent and generates the correct symmetry transformations for fermions and photons:
$$[ i Q_{ 5 } ( t ) , \psi ( t , \vec{ x } ) ] = \delta_{ 5 } \psi ( x ) ,$$
$$[ i Q_{ 5 } ( t ) , A ( t , \vec{ x } ) ] = 0 = \delta_{ 5 } A ( x ) .$$
Under the gauge transformations, $\delta \psi = i \Lambda \psi$, $\delta A_{ c } = \partial_{ c } \Lambda$, we find that $Q_{ 5 }$ is guage invariant:
$$\delta Q_{ 5 } = \frac{ - e^{ 2 } }{ 4 \pi^{ 2 } } \int d^{ 3 } x \ \partial_{ j } ( \Lambda \bar{ F }^{ 0 j } ) = 0 .$$
Therefore, chiral symmetry remains a good symmetry in spite of the anomaly in the divergence. This has the consequence that any property, based on axial symmetry, will be maintained in perturbation theory (the anomaly cannot be used to generate a mass term for the fermion).

10. Oct 17, 2014

### DarMM

It is? Most of constructive field theory uses very few ideas from perturbation theory. Rather it uses the scaled cluster (propogator) expansion and even then not as a perturbative expansion. 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?

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. 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?

I don't think it is Noether's theorem's "job" to regulate singularities, I'm actually not sure what that would mean, but it is still an example where the theorem fails. So it cannot be fundamental to quantum field theory. Instead the Ward-Takahashi identities are.
Renormalisation invalidates the conclusions of Noether's theorem in this case regardless of whether regularization is its job.

11. Oct 17, 2014

### atyy

Two questions regarding AQFT and CQFT.

1. Do any of the successful constructions by CQFT also fulfill AQFT axioms, so that the sucess of CQFT also means AQFT is succesful?

2. The Bisogano-Wichmann theorem does show up in modern non-rigourous QFT research, eg. http://arxiv.org/abs/1102.0440. Is this tangential to AQFT, or does it indicate that AQFT ideas are in fact used in non-rigourous research also?

12. Oct 17, 2014

### DarMM

Regarding (1.) all constructed quantum field theories satisfy the axioms of axiomatic quantum field theory. Even Yang-Mills in 3+1 has been shown by Balaban to obey some of the Wightman axioms, although we don't know yet if it obeys them all, we don't have enough control over the continuum limit.

By the way, for gauge field theories it is the observable fields (I'm not sure what people prefer to call them, I mean the gauge invariant fields) that obey the Wightman axioms, not the $A_{\mu}$ gauge potential.

13. Oct 17, 2014

### DarMM

Regarding your second point, I would say that AQFT ideas appear a bit in non-rigorous research (such as the Källén–Lehmann
representation of the propagator). However overall I would say it is tangential to mainstream research.

Personal comment: However I will say, based on papers I have read, previous posts on this forum and the stackexchange site, that people have odd ideas about what AQFT or CQFT are concerned with, as if the were competing with or an alternative to normal quantum field theory. They're really nothing more than areas concerned with showing that what physicists do everyday is mathematically valid, that is all.
Think of it like how in Reed and Simons they actually prove that the Hydrogen atom is self-adjoint, where as a normal textbook will just show it is symmetric. Nothing wrong with either way of doing things. Getting too bogged down in mathematics can slow down physical thinking, but usually somebody will come along and mathematically tidy things up.
A perfect example is the existence of QCD. Now I have no doubt QCD exists mathematically, but I think mathematics should advanced to the point where this can be proven. The fact that we cannot prove that QCD and other 3+1-d theories actually exist due to being unable to handle coupling constant renormalisation indicates that our control of C*-algebra nets in the Algebraic/Hamiltonian view or more abstract stochastic processes (Yang-Mills theory in the path-integral approach is essentially a theory of random fiber bundles) in the path integral view, is not yet mature.

14. Oct 17, 2014

### atyy

I think there are things in classical statistical mechanics that even non-rigourous people find bizarre. The $\epsilon$-expansion and the multiplication of distributions in the KPZ equation. I think replica symmetry breaking used to be another one, but I hear that rigourous folks can handle that nowadays. Any progress on the other two? I hear Hairer's got KPZ, which I guess leaves $\epsilon$-expansion?

BTW, I remember hearing the story that Balaban stopped working on Yang-Mills because he moved house, and the moving company lost a box of his Yang-Mills work. Not sure where I heard that now though.

Last edited: Oct 17, 2014
15. Oct 17, 2014

### dextercioby

I don't find the exact reference in R&S's 4 volumes, but I believe a rigorous proof can be constructed via Kato's results in the 1950's. I could find a proof by theorem 10.2 (p. 219) in G. Teschl's <Mathematical Methods in Quantum Mechanics. With Applications to Schrodinger Operators> and his comment following on page 220.
Having futher checked, one can find the full details in Triebel's <Höhere Analysis> (1972, German) or <Higher Analysis> (1992, English), Chapter 7.

Last edited: Oct 17, 2014
16. Oct 17, 2014

### zoki85

17. Oct 17, 2014

### atyy

The story about Balaban's notes is told by Jaffe just after 1:15:00 in this talk:

Jaffe's talk was part of a workshop on the Mathematical Foundations of Quantum Field Theory at the Simons Center at Stony Brook, highlighted by Peter Woit in http://www.math.columbia.edu/~woit/wordpress/?p=4396.

18. Oct 18, 2014

### DarMM

It's in volume II, theorem X.15. It's part of their presentation of Kato's material, as you mentioned. I find it a nice proof of the self-adjointess. Thank you for your references, it's always nice to see how others explain things.

19. Oct 18, 2014

### DarMM

Interesting talk! I'm actually shocked that that is what happened!

20. Oct 18, 2014

### dextercioby

Just read this: http://www.arthurjaffe.com/Assets/pdf/CQFT.pdf. I found it useful.

21. Oct 19, 2014

### samalkhaiat

How many of them “very few ideas” are there? Do you realize that you have re-stated my exact statement?
This is PERTURBATION SERIES. And the whole business of CQFT is about showing that Schwinger functions satisfy the so-called Osterwalder-Schrader axioms.
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.

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.
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.
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.
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.
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

22. Oct 19, 2014

### atyy

How can CQFT be dead when it has already successfully constructed relativistic quantum field theories?

How can CQFT be opposed to Lagrangians when it justifies the use of Lagrangians to construct relativistic quantum field theories?

Also, if the Lagrangian is the only way to construct a relativistic quantum field theory, does that mean that the conjecture of non-Lagrangian QFTs in https://www.ictp.it/media/101047/schwarzictp.pdf has been disproved?

23. Oct 20, 2014

### DarMM

No it is not, there are several differences between the cluster expansions in constructive field theory and usual perturbation theory.

Firstly the cluster expansion is not an expansion in any parameter of the theory, it is an expansion along a path in the space of propagators (that is an expansion along a path in $\mathcal{D}\left(\mathbb{R}^{2d}\right)$. For anybody reading, the cluster expansion uses an artificial parameter $s$ with the Laplacian operator in the Lagrangian is made dependent on:
$\left(\Delta_{s} + m^2 \right)$
Essentially when $s = 0$ it is the Laplacian with zero Dirichlet boundary conditions along the lines on a lattice grid superimposed on space. So for instance in $\mathbb{R}^{2}$ this would be the lines connecting the points of $\mathbb{Z}^{2}$. At $s = 1$ it is the full Laplacian with this conditions removed. $s = 0$ models a scenario where the different regions do not communicate with each other and are decoupled. However note that inside those regions we have the full Lagrangian so the theory is not free.

Secondly the cluster expansion is only done to finite order in $s$, it's not a Taylor expansion.

For these two reasons, I do not think that this is a "perturbation series".

Okay, that's fine, it's a complicated subject which one could argue has a small return on time invested (I'd disagree, but then again I like the questions it answers) but it means you were just making your points up.

The $\phi^{4}_{2}$ theory does not involve a $\phi^{3}$ power, but I'll assume you mean a general $\mathcal{P}(\phi)$ theory.
(i) Yes, the value of $k$ is simple to compute. It simply removes the singularities present in the field's third power. It can be done in curved spcaetimes as well. The answer does depend on the metric of the spacetime.
(ii) I'm not sure what this means honestly. Nobody ordered it, it's just that $\phi^{3}$ is not well-defined.
(iii) No, it does not have a relation to perturbation theory as you do not need perturbation theory to compute it.

Sorry I am confused now. What exactly are you saying, is it that constructive field doesn't use Lagrangians and this is a weakness of it? That is what I took from your previous post. If so, it does use Lagrangians, it has to, since it is attempting to show that standard QFT, which is built on Lagrangians, can be made rigorous.

Well yes of course. There are proofs without references to Noether's theorem in several introductory QFT texts like Srednicki.

I mean I thought the failure of Noether's theorem due to anomolies is just a known fact, it is not some claim unique to me. For example here is a grad course in anomolies at Buffalo, New York that mentions it in the first lecture:
http://www.physics.buffalo.edu/gonsalves/phy522/

So, Noether's theorem does not give the correct expression, one must renormalise the expression given by Noether's theorem to produce the correct conserved current. Hence Noether's theorem fails.

Yes we can renormalise expressions to get conserved charges, but the point is that these are not the ones given by Noether's theorem.

24. Oct 22, 2014

### samalkhaiat

It is dead in the following sense:
(Small number of researchers) is the reason for (or/and caused by) unwillingness of scientific institutions to fund that type of research.

Some programs, like the “Form Factor Program” of Smirnov and Zamolodchikov (bootstrap, Yang-Baxter equation, ISP) have no Lagrangians.
If you look around you, you see biology, chemistry, nuclear physics, particle physics, etc. This simply means that nature decouples at different length scales, i.e., the interaction strength varies with energy scale. Nature seems to continue to work like this in particle physics: there are regions where the coupling becomes weak and an effective (Lagrangian) description become possible. So, people thought that (strongly coupled) non-Lagrangian QFT must be scale (therefore conformal) invariant field theory. So far, it has been like cat and mouse game between people: the “non-Lagrangian” theory of today becomes Lagrangian tomorrow.
Is physics going to be Lagrangian forever? No, certainly not. However, for now, it looks Lagrangian from GR down to M-theory.

25. Oct 22, 2014

### samalkhaiat

No, it is not as complicated as, say, QCD calculations. I certainly do not want to be lonely. I believe that conventional QFT is justified by its extraordinary empirical success and I see no reason to believe that solving the H atom means that you can solve the He atom.
Yes, I was making my point and that was “axiomatic qft failed to reproduce the results of conventional qft.”
Why is that? Has the notation changed? I thought $( \phi^{ 4 } )_{ 2 }$ is the short hand notation for the 2D non-linear field equation
$$\partial^{ 2 } \phi + \phi + \lambda \phi^{ 3 } = 0 .$$
Is k finite? Can you give me formal expression for k? I don’t know why should k depend on the metric, but you can choose the simplest flat metric if you want.
Constractive algebraic QFT, such as "Form Factor Program" based on factorizable S-matrix has no Lagrangians.
No, at some stage, they all use the canonical relation
$$\delta \phi ( x ) \delta ( x - y ) = [ i J^{ 0 } ( x ) , \phi ( y ) ]$$
with $J^{ a }$ is given by the Noether expression for the current.
Yes, I know that. Anomalies are tricky to describe by ordinary language. It is very common to describe anomalies by the following saying “There is no warranty that a given classical symmetry may be elevated to quantum symmetry” which is equivalent to saying “symmetry of the classical action needs not be invariance of the quantum effective action” Do you find anything wrong with these definitions? Well, the wording is certainly wrong if you want to derive the anomalies by Fujikawa’s method. Indeed, the invariance of the effective action under chiral rotation leads to the axial anomaly!!
Who cares? Noether current is NOT unique. It can be modified in two different ways:
1) adding superpotentials to the current
$$\mathcal{ J }^{ a } = J^{ a } + \partial_{ c } F^{ a c } + \partial_{ c } \partial_{ b } F^{ a c b } + \cdots .$$
We use this freedom to make the energy-momentum tensor symmetric in Poincare invariant theories and traceless in CFT’s.
2) adding a total divergence to Noether identity
$$\frac{ \delta S }{ \delta \phi } \delta \phi + \partial_{ a } J^{ a } = \partial_{ c } \Lambda^{ c }$$
So, when the equation of motion is satisfied, we find that the CANONICAL Noether current is not conserved,
$$\partial_{ a } J^{ a } = \partial_{ c } \Lambda^{ c } .$$
Does this look familiar? This does not mean that we do not have symmetry current, We simply modify the non conserved canonical current to obtain a conserved current
$$\mathcal{ J }^{ a } = J^{ a } - \Lambda^{ a } .$$
This is exactly what we do with the anomalous canonical current.

Last edited: Oct 22, 2014