
#1
Feb1112, 07:27 PM

P: 381

In QFT, Lagrangian is often mentioned. While in QM, it's the Hamiltonian, Is the direct answer because in QFT "position" of particle is focused on and Lagrangian is mostly about position and coordinate while in QM, potential is mostly focus on and Hamiltonian is mostly about potential and coordinate?




#2
Feb1112, 09:04 PM

Sci Advisor
P: 8,006

The Lagrangian is a way to get the Hamiltonian. Sometimes it's easier to incorporate symmetries by using the Lagrangian. The Hamiltonian is still required because it generates "unitary" time evolution. Roughly, unitarity is is what makes probabilities sum up correctly all the time.




#3
Feb1112, 09:09 PM

P: 194

No;
The reason people talk about Lagrangians in QFT is because theories are expressed Lorentzinvariantly using the Lagrangian, but not using the Hamiltonian (since the Hamiltonian is an energy which depends on the Lorentz frame). In QM, the Hamiltonian is more useful since many problems involve, in some form or other, diagonalizing the Hamiltonian operator (i.e. solving the Schrodinger eqn). This is much more tractable due to fact that most often, the number of degrees of freedom are much fewer than that of QFT. 



#4
Feb1112, 09:19 PM

P: 381

Lagrangian vs Hamiltonian in QFT vs QMLagrangian = position/velocity Hamiltonian = position/momentum, potential You mentioned the Lagrangian is a way to get the Hamiltonian, meaning the position/velocity is a way to get the position/momentum and potential? How come? 



#5
Feb1112, 09:23 PM

P: 381





#6
Feb1112, 09:28 PM

P: 194

Neither the Hamiltonian nor the Schrodinger equation solves for potential. The potential is entered into the Hamiltonian manually. The Hamiltonian is part of the Schrodinger equation. You solve that equation. 



#7
Feb1112, 09:36 PM

P: 381





#8
Feb1112, 09:45 PM

P: 194

You may then ask, "why are we using the Lagrangian in QFT if we're dealing with a quantum mechanical system?" The answer to that question is because the quantum field theorists are being sneaky: technically, they are using the Hamiltonian which is derived from the Lagrangian they write down, but do not say so. The reason for this is the Hamiltonian is a messy and Lorentznoninvariant object. Ironically, it also turns out Feynman rules can be read off the Lagrangian much more easily than from the Hamiltonian. But keep in mind that the actual derivation of the Feynman rules must come from the Hamiltonian (unless you use the pathintegral approach). 



#9
Feb1212, 04:32 AM

Sci Advisor
P: 5,307

In QFT one learns how to derive the Lagrangian path integral from the Hamiltonian formulation. Then in most cases one forgets about this derivation and uses directly the Lagrangian PI w/o ever mentioning the Hamiltonian. This is reasonable especially when dealing with Lorentzcovariant scattering calculations where Lorentz covariance is manifest in the Lagrangian whereas it would have to be checked terme by term in a Hamiltonian approach
Strictly speaking there may be problems with the Lagrangian PI b/c going from the Hamiltonian PI to the Lagrangian PI involves a Gaussian integral which in not be there in general; one theory were this equivalence between Hamiltonian (canonical) and Lagrangian PI is not well understoof ist LQG/SF b/c  we do not know the correct (physically, regularized) Hamiltonian yet  we do not have a rigorous definiton of the SF model yet  and we do not know a rigorous transformation between Hamiltonian and SF model In lowenergy hadran physics like calculations of properties of bound states like confinement, masses, electromagnetic form factors, ... (even in QCD ) the Hamiltonian is used. Here Lorentz covariance does not play a major role b/c we are looking at one particle (e.g. a proton) at rest and we never want to boost it (we could do that using the Hamiltonian plus other generators of the Lorentz algebra and we would find Lorentzcovariance again, but it would be an awfully difficult calculation). 



#10
Feb1212, 06:32 AM

P: 132

I dont understand why physicists tend to "believe" more in hamiltonian than in Langangians. I dont understand neither why they tend to "believe" in canonical quantization than in PI. First of all, from both we can derive the other, so, there is no mathematical obvious choice. And second, there Lagrangian  PI formulation is shorter, clearer (it gives directly the probability amplitude summing up "everything that can happen") and Lorentz invariant. I really think that the "real physics" (meaning the mathematical model that perhaps describes reality without having problems of renormalization, deals with the 4 interactions and so on) is described by an extrapolation of the Lagrangian approach (that is to say, I think that Lagrangian approach is closer to the truth than hamiltonian).
Why all you people (that studied much more than me) tend to think the other way round? Thanks! 



#11
Feb1212, 06:42 AM

P: 381

(and the kinetic energy is a function of velocity v and potential energy will typically be a function of position), while Hamiltonian is kinetic energy plus potential energy (and the Hamiltonian should be expressed as a function of position x and momentum p (rather than position and velocity, as in the Lagrangian) Now how come lorentz covariance is manifest in the Lagrangian? I know it is not manifest in the Hamiltonian because time is said to be separate whatever that means. Also how come one can derive the Lagrangian from the Hamiltonian and vice versa? For example. Lagrangian = 5  3 = 2 Hamiltonain=5 + 3 = 8 Given 2, how do you figure out its 5 and 3 and not say 10098? Given 8, how do you figure out it's 5 and 3 and not say 6+2? Btw.. what is SF in LQG/SF? (I know LQG is Loop Quantum Gravity). Thanks. Thanks. 



#12
Feb1212, 04:18 PM

Sci Advisor
P: 817

Sam 



#13
Feb1212, 05:48 PM

P: 381

If the Hamiltonian formulation didn't exist before the discovery of the Schroedinger Equation. Would Schroedinger reinvent the Hamiltonian? Or can you formulate the SE without the concept of the Hamiltonian? How?
Maybe we have a situation now where Superstings Theory haven't discovered the right math or the Hamiltonian equivalent in SE? Also it is said the Hamiltonian is an operator. And the solutions of QFT are operators (not observables). So in the context of the Hamiltonian as operators. The solutions of the fields in QFTs are Hamiltonian formulas without any final solutions? I still can't understand what good is to have operators as answer without solving for the values like position or momentum in QFT. Hope someone can address this using the context here. Thanks. 



#14
Feb1312, 01:38 AM

Sci Advisor
P: 5,307

If you look at research topics in QM you will mostly find Hamiltonian methods; this could be an indication (but is perhaps no good reason to believe in the Hamiltonian method in QFT) But as soon as you want to calculate something that's not always true. My impresion is that in the Hamiltonian picture you get a much clearer understanding of the really big problems. In the Lagrangian you can write down many things easily w/o ever thinking about the (mathematical) definiton. With the Hamiltonian things become more complicated  but that's a benefit  at least partially  b/c these problems should alert you that something goes fundamentally wrong (here I mean that you have a fundamental problem, not only a problem with the canonical method). But as I said in the beginning, this may be a personal impression based on my experience. I remember some conversations where the Lagrangian approach seemed to be much simpler, but it was not the case that the problems could be solved or did not show up at all; they were simply swept under the carpet. there ain’t no such thing as a free lunch ... 


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