# Insights Mathematical Quantum Field Theory - Lagrangians - Comments

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1. Nov 18, 2017

### strangerep

Hmm, ok. Indeed, I had wondered about correct handling of boundary terms at temporal infinity earlier when you mentioned integration by parts when deriving the E-L eqns. If the variations are not fixed to be $0$ at $t=\pm\infty$, then either one has a nontrivial term remaining, as you say, or else one must rely on the term being a function of conserved quantities (so that what comes in at $t=-\infty$ necessarily goes out at $t=+\infty$).

Now you've lost me again. I don't follow at all how $\Omega_{BFV}$ "induces" a Poisson bracket.

2. Nov 19, 2017

### Urs Schreiber

Do you know how a symplectic form induces a Poisson bracket? The $\Omega_{BFV}$ becomes the (pre-)symplectic form of the field theory after integration over a Cauchy surface.

3. Nov 19, 2017

### strangerep

Well,... I'll take the simplest case in classical mechanics where the basic dynamical variables are a single configuration variable $q$ and corresponding canonical momentum $p$. Then, suppose we have 2 functions $f,g$ on the corresponding phase space, i.e., $f\leftrightarrow f(q,p)$ and $g\leftrightarrow g(q,p)$, then their total differentials are phase space vectors given by $$df ~=~ f,_q dq + f,_p dp ~;~~~~~~ (\mbox{and similarly for}~ dg).$$A symplectic form $\omega$ on this space is (by definition) a non-degenerate skewsymmetric bilinear 2-form, so$$\omega(df,dg) ~=~ f,_q g,_p - f,_p g,_q ~=:~ \{f,g\}_{PB} ~,$$
Hmm. I need to work that through some more to see it properly...

If we have a simple action $I$ as follows: $$I ~=~ \int\!dt\; L(q,\dot q)$$ then, proceeding with the usual variation computation...
$$0 ~=~ \delta I ~= \int\!dt\;\Big[L(q+\delta q,\, \dot q + \delta\dot q) - L(q,\dot q)\Big] ~= \int\!dt\;\Big(L_q \delta q + L_{\dot q} \delta\dot q \Big) ~.$$Performing an integration-by-parts on the 2nd term gives: $$\Big[L_{\dot q} \delta q \Big]^{t=+\infty}_{t=-\infty} ~-~ \int\!dt\; L_{\dot q} \delta q ~.$$Normally, we'd assume that the variations $\delta q$ vanish at $t\to\pm\infty$, so the term in square brackets vanishes, and the remaining terms in the integrand yield the usual E-L equations of motion.

For the simple case of a free particle, with $L=m\dot q^2/2$ we have $p := L_{\dot q} = m\dot q$, so the square bracket term would be $$\Big[p \, \delta q \Big]^{t=+\infty}_{t=-\infty} ~~,$$but how do we get from here to a "(pre-)symplectic form" ?

Last edited: Nov 20, 2017
4. Nov 20, 2017

### Urs Schreiber

You did rederive the presymplectic potential $\Theta_{BFV}$, which in the case you considered is $p \delta q$. The presymplectic form is the variational derivative of that $\Omega_{BFV} := \delta \Theta_{BFV}$. In the case you considered this yields $\delta p \wedge \delta q$, which is the standard symplectic form that you expect to see.

5. Nov 20, 2017

### strangerep

If it had been $p\,dq$, where $d$ is the exterior derivative, then $d(p\,dq) = dp \wedge dq$ (since $d^2q=0$).
But we're using variational (a.k.a. functional) derivatives here, aren't we? What am I missing?

6. Nov 20, 2017

### Staff: Mentor

Same here - but its still fascinating reading.

As Asimov said, except for the very greatest of mathematicians there comes a point where it really becomes a slog, whereas at lower levels you cruise.

I hit that when I read Gelfland's opus on Generalized Functions. But persevered and did eventually learn a lot.

I suspect its the same here.

Thanks
Bill

7. Nov 21, 2017

### Urs Schreiber

The variational derivative becomes the de Rham differential after "transgression" to the space of field histories. This is the content of the section Local observables and Transgression of chapter 7. Observables, the very statement is item 2 of prop. 7.32.

The full proof is given there, but this should be intuitively clear:

The variational derivative measures the change of field values at a point of spacetime, or of change of spacetime derivative of the field value at a point of spacetime, etc. and the field values at spacetime points are precisely the "canonical variables" for the field theory, while the spacetime derivatives of the field values are precisely the "canonical momenta". It it is in this way that the variational derivative on field eventually becomes the de Rham differential on the phase space.

8. Nov 21, 2017

### strangerep

The full abstract account of transgression, etc, has reached the point of being 99% gobbledegook in my mind.

Can you illustrate by continuing my simple concrete example from post #28 ?

9. Nov 21, 2017

### Urs Schreiber

You need a concept of differential forms on a space of field histories. The way to go is to have for each $U$-parameterized family of field histories $\Phi_{(-)}$ an ordinary differential form on $U$, thought of as the pullback of the differential form on the whole space of field histories. Now the variational derivative $\delta$ is the full de Rham differential on the space of field values and their derivatives minus the de Rham differential along spacetime. But the latter disappears anyway as we integrate over spacetime, due to Stokes' theorem. In conclusion, the variational derivative measures the remaining change of fields as they vary, notably as they vary in families parameterized by some $U$. This way it becomes the ordinary de Rham differential along $U$ , for each $U$-parameterized family of fields.

Last edited: Nov 21, 2017
10. Nov 21, 2017

### Urs Schreiber

The way to go is to read the development step-by-step. Each single step is trivial. If you let me know which one is the first step that you don't follow, then I can add explanation or improve the wording, or add an example, or whatever it takes. But for that I need coordinates on which step you find opaque.

Also, I might re-emphasize that the idea being developed is really simple, much simpler than what you get elsewhere. At the heart of it is this: A field history is some function $$\Phi(-) : \Sigma \longrightarrow E$$ and a family of field histories that vary with some parameter $u \in U$ is a function $$\Phi_{(-)}(-) \colon U \times \Sigma \longrightarrow E$$ Given this, it is completely obvious what the variational derivative in the direction of variations parameterized by $U$ should be: it's nothing but the de Rham differential $d_U$ along $U$. Finally, to get the full variational derivative, the one that knows about all possible variations, we simply look at this for all possible $U$ and for all families of field histories $\Phi_{(-)}$ parameterized by these, and collect all the ordinary de Rham differential $d_U$ on all these $U$.

That simple idea is the realization of the variational derivative as a differential form in the sense of "diffeological spaces".

Next we want to do some actual computations with this. For this it is exceedingly helpful to make use of the fact that the differential forms which we want to vary are actually "local" in that at each spacetime point they depend only on the values of the fields and their derivatives at that point. Suppose first it depends in fact only on the field values, not on that of their derivatives. This means that we simply have a differential form on $E$, which we then pull back along field histories to $U \times \Sigma$.

(Ah, are you familiar with the concept of pullback of differential forms? If not, we need to pause here and say a word about that.)

Now the de Rham differential generally commutes with pullback of differential forms. In view of the above, this has the key consequence that in order to compute the variation of a differential form on the space of field histories, we may just compute the ordinary de Rham differential on $E$ and then pull this back. If in addition we integrate the result over $\Sigma$, then any "horizontal" or "total spacetime" derivatives drop out, and this is how we decompose the de Rham differential on $E$ into a horizontal piece and the remaining $\delta$.

Finally, to have the local differential form also depend on the values of the derivatives of the fields, we do the same with $E$ replaced by its jet bundle, which is nothing but the collection of all these values.

To amplify, if we consider some finite jet order (which we can do in all examples of interest here) then this gives a way to speak of variational derivatives and how they are the de Rham differential on the space of field histories entirely using elementary finite-dimensional differential geometry. There is just the spacetime $\Sigma$, the manifold $E$ where fields take their values, and the Cartesian space $U$ with which we parameterize field histories, and smooth functions of the form $\Phi_{(-)} : U \times \Sigma \to E$. It doesn't get simpler than that , really.

11. Nov 21, 2017

### vanhees71

Well, for us normal theoreticians it's always good, if you - from time to time - express everything in our normal notation with indices and usual differential operators (partial derivatives or covariant derivatives or functional derivatives). Often you have quite complicated looking formal equations that become understandable at one glance when translated to old-fashioned 19-century Ricci-calculus notation. I know that's nearly heretic to mathematicians, but it's the common notation in the physics community (and imho for the good reason of clarity and "calculational safety").

12. Nov 21, 2017

### Urs Schreiber

I feel there is a lot of standard physics notation in the notes, plenty of examples that unwind the general machine to the standard formulas. I'll be happy to improve further, but please give me more concrete pointers. Which formula do you feel is lacking examples? I'll add them.

By the way, I like to caution against the habit of organizing people into camps, such as "normal theoreticians" here and "mathematicians" there. There is just one subject, QFT, and we need all the tools we can get hold of to understand it. It seems strange to insist that a theoretical physicist is by necessity one who may not be bothered with learning the tools that it takes to make sense of QFT, or that progress in the field should be impossible due to the inertia of "we have always done it differently".

I am a theoretical physicist, and I managed to learn some of what it takes, and here I am trying to explain it. I can try to add more explanation where need be, but for that please let me know which particular bit bothers you. Best to start from the very beginning, since the whole series develops strictly incrementrally: If you start reading the series in chapter 1, page 1 and keep reading, what is the first point where you feel lost?

13. Nov 26, 2017

### ftr

Urs

I don't know what is exactly the purpose of these threads of yours since they seem to be an order of magnitude more complicated than standard textbooks. I think PF readers are vast mix. The majority usually come here for clarification of concepts and simplified setups not mathematical gymnastics. I think you can use your wide expertise to make the physics understandable. Of course, that does not mean it is not useful for some. Please do not take this personal, this is just my opinion and could be wrong.

14. Nov 26, 2017

### dextercioby

It is precisely because physics textbooks are filled with so much mathematical hand-waving, that one needs an authoritary treatment from a mathematician's viewpoint of what is really going on in fundamental physics. From this perspective, I applaud @Greg Bernhardt 's decision to invite Urs here and allow him to share his knowledge with us.

IIRC, it was the famous mathematician's David Hilbert interest in physics that was so magnificiently summed by his (approximately quoted into English) words: "Physics has become too difficult to be left to physicists".

15. Nov 26, 2017

### vanhees71

I must also stress that it's great that Urs posts this series of lecture notes about formal QFT. I know the usual hand-waving approach quite well, and for sure it's a good opportunity to read about it from the more formal mathematical point of view to understand this most important but mathematically least understood methodological framework better. The problem is that the two communities between practitioners and mathematicians are nearly completely disjoint, and I think it's great to have the opportunity in these forums to have somebody you can interact with via the forum.

I only wish, I had the time to start studying the notes seriously :-((.

16. Nov 26, 2017

### Urs Schreiber

If you make this more specific, I can react. Which passage do you find overly complicated?

17. Nov 28, 2017

### Boing3000

It may be then that I am part of some strange minority, but even if standard textbook also eludes me, I still find those articles very interesting, full of lively cross reference, and new terms that are actually very refreshing to read.

18. Nov 28, 2017

### vanhees71

Well, then I'm also a minority, because I think I haven't understood a problem or, even more difficult, the solution of a problem before it is formulated mathematically.