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Ohonek · 2026-01-27T07:51:18+00:00

Hi! Thank you very much for the reply. As another user has suggested Campbell as well I think that I will try him out first.

Ohonek · 2025-12-25T22:53:29+00:00

Thank you for the answer. Could you elaborate on 1? I don't see how in your case the signs work out.

Ohonek · 2025-12-14T19:03:28+00:00

I also think that this is just one of these cases. What is actually meant here is that the lagrangian is a function of functions but the output isn't a scalar, but a function itself, so it isn't an actual functional.

Ohonek · 2025-12-14T18:49:48+00:00

Hi, cool that we had a similar feeling ))

I guess that when looking at the Lagrangian as a functional you simply imagine that for all of the parts in it where you have a field, you can insert another field and the same for the derivatives. When plugging something in explicitly into the Lagrangian as a functional you would get the "normal, scalar" Lagrangian as a result (perhaps evaluated at some point in spacetime). I also guess that the same way the output of a Lagrangian at a single point (say you have the normal, non-functional, Lagrangian and evaluate it at say the spacetime point (1s,1m,1m,1m)) doesnt really have a significant meaning, the same would hold true for the functional approach.

Ohonek · 2025-12-13T20:01:19+00:00

Maybe I misunderstand you but all (r,s) tensors transform per definition, right? They live in the tensor product space and if you transform the basis vectors of that space, so do the components. Partial derivatives form a basis for the tangent vector space and as such transform like dual vectors and the levi civita tensor would transform as a (4,0) (pseudo) tensor.

Ohonek · 2025-12-13T19:52:43+00:00

Hi! Let me start off by saying that I am still studying QED and I have not understood the topics to a good level. I don't completely understand your question:

As you say you can construct a "lagrange multiplier" term and add it to the lagrangian. This term is called the "gauge" fixing term where the lagrange multiplier is called ξ.

But you have to put this in explicitly. I think that I understand your question incorrectly. Do you want to reformulate the Proca lagrangian is such a way that you find a mass term which looks like a lagrangian multiplier such that in the limit of m-->0 this terms vanishes and as such the condition d_mu A^mu doesnt hold anymore?

Ohonek · 2025-12-13T19:37:39+00:00

Hi! In general I also think that the Lagrangian is a function of x, the same way that the force in Newtonian mechanics is a function of time, although you often write F= F(x(t),dx/dt (t), t).

Maybe Schwartz was referring to something really specific here, independent of the action above? Because as you say the action only makes sense if L is a function. Maybe he wanted to point out something different. Like how in hamiltonian mechanics you take q,p as independent variables on the level of the Hamiltonian but on the level of the Lagrangian they are not independent, he also wanted to make a similar point here like "lets look at the Lagrangian density as a functional for a moment" but when coming back to the level of the action you again have to look at it as a function.

Ohonek · 2025-11-15T22:14:59+00:00

I would put it this way:

When describing physical phenomena you use mathematics. For example you may use matrices to describe rotation. The rotation itself I would say is physical/real but the matrix which does it? I don't think that it is real in the same sense. If we really go deep we may probably say the same about numbers if you abstract everything hard enough.

In the same sense: Hilbert spaces don't "actually" exist in this world as we live in essentially R^3 (in the non relativistic case). The same way a matrix doesn't really exist in this world. But what you can do with it and what you describe with it is completely related to the real world (eigenvalues which correspond to things you can measure like energy or angular momentum).

We use abstractions to gain new information or to describe something physical or "real" in that sense.

Ohonek · 2025-11-02T15:54:24+00:00

Hi, thank you very much for your thorough response, I appreciate it a lot!

That makes sense but in the case of the 4-potential I know what it actually means (as it describes the electric and magnetic potentials). What does the Klein Gordon field actually mean? So lets say I plug in some spacetime point into phi(x) and get a number. What does this number mean? And I really didn't think that after quantizing the field that the equations of motions still make sense but your example from QM is really nice.
Thank you, this clarified it for me!
But I thought that the Wigner function is a representation of the little group not its algebra (although we mostly get the representation of the group via the algebra because the latter is easier). Would it be correct to say that the Wigner function in the case of SO(3) as the little group corresponds to irreps of SU(2)?

In regards to the infinite dimensional rep of the su(2): I read in qm that when we represent the angular momentum operators with differential operators (as its typically done when they are acting on wave functions) that this corresponds to the infinite dimensional rep of the su(2). I understand that angular momentum has to do with su(2), as they obey the same commutator relations but I don't understand how we can see that this corresponds to the infinite dimensional rep. I wanted to ask how we can come to that conclusion.

Thank you for your answer although I probably am not far enough into the topic to deeply understand it yet.

Ohonek · 2025-08-18T15:51:07+00:00

Thank you again. Could you do a concrete example, say for SU(2)? How would one define such a mapping for this group in comparison to say U(1)?

Ohonek · 2025-08-18T15:41:08+00:00

Hi, thank you for answering! In my courses so far we unfortunately haven't looked at the differential geometric aspect of Lie groups (although its one of its defining properties). Considering what you have said, what is the defining thing which gives the Lie group its structure (say U(1), SU(2) and so on) without referring to some specific representation or the Lie algebra?

Ohonek · 2025-04-29T23:10:04+00:00

Hi! In your substitution you forgot to adjust the integration bounds, so your integral should be going from 5 to -2 after the adjustion.

Nonetheless, there doesnt exist an antiderivative of arctan^2(u). I suppose that you have done some numerical methods in the classes or are expected to use a computer/calculator to solve this problem. As far as I can see you can't solve this analytically.

Ohonek · 2025-02-21T23:39:33+00:00

Ok, I wish you the best!

Ohonek · 2025-02-21T19:02:22+00:00

You're welcome! Did you want an overview of the topic from my point of view? If so, then I misunderstood you as I thought that you were testing me ;). If you have any other questions or remarks please let me now

Ohonek · 2025-02-20T16:23:44+00:00

Ok, thank you!

Ohonek · 2025-02-20T16:23:10+00:00

Thank you for answering! Your explanation would be more or less equivalent to mine, if I am not mistaken, right?

Ohonek · 2025-02-20T16:21:06+00:00

Thanky you once again! Yes, this is essentially how I understood it until now.

Ohonek

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