How can you tell the spin of a particle by looking at the Lagrangian?

In summary, people generally omit some indices when writing out spinor indices, which makes the spinor transform as a vector.
  • #1
BiGyElLoWhAt
Gold Member
1,630
134
TL;DR Summary
I'm just starting to work with QFT's and am trying to understand the construction of the Lagrangian's as well as how to read/work with them.
I'm just starting to get into QFT as some self study. I've watched some lectures and videos, read some notes, and am trying to piece some things together.

Take ##U(1)_{EM}: L = \bar{\psi}[i\gamma^{\mu}(\partial_{\mu} - ieA_{\mu}) - m]\psi - 1/4 F_{\mu\nu}F^{\mu\nu}##
This allegedly governs spin 1 bosons, where as something like the Klein-Gordon Lagrangian deals with spin 0 and Dirac deals with spin 1/2 (fermions).

It is extremely mysterious to me how to tell that one Lagrangian should govern one spin as opposed to another. The only thing that sticks out to me at all here are the gamma matrices, which act on spinors. However, a similar term is also present in the Dirac equation, so this can't be the full story. How does the "spin content" drop out?

Not sure whether to mark this as Intermediate or Advanced. I've studied both GR and QM so am familiar with the notation and the concepts in both, but am not all that familiar with quantum field theory, only a little bit with classical field theory.

Thanks
 
  • Like
Likes atyy
Physics news on Phys.org
  • #2
You look how the fields transforms under Lorentz Transformations.

Scalars - spin 0, transforms as scalars.
Fermions - spin ½, transforms as spinors (½,0) or (0,½) representation (depending if you have a "left-handed" or a "right-handed" spinor).
Vectors - spin 1, transforms as vectors (½,½) representation.

http://sharif.edu/~sadooghi/QFT-I-96-97-2/LorentzPoincareMaciejko.pdf

Note, this is applicable to classical fields as well.

Try to get one book and read it from start to finish instead of scattered notes etc. I can recommend the book by Mark Srednicki, you can get a free draft version on his homepage. The book is divided into several sections depending on spin and discusses at some lenght what is "special" about the various cases in terms of their Lorentz Transformations http://web.physics.ucsb.edu/~mark/qft.html
 
Last edited:
  • Like
Likes atyy, vanhees71, PeroK and 3 others
  • #3
Ahh ok. So Klein-Gordon is basically ##E^2 - p^2 = m^2## which is Lorentz invariant, which is why this is spin 0. If that's correct, this makes complete sense.
So then is it the presence of the ##\bar{\psi}## in the EM Lagrangian that is the main difference between Dirac and ##U(1)_{EM}##? That doesn't seem to change how ##\psi## should actually change though, unless it's something that just drops out eventually.
Thanks for the intuitive reply, that's effectively the type of answer I was looking for, I just wasn't sure if there was a clear cut and simply answer like that available.
 
  • #4
BiGyElLoWhAt said:
So then is it the presence of the ##\bar{\psi}## in the EM Lagrangian that is the main difference between Dirac and ##U(1)_{EM}##? That doesn't seem to change how ##\psi## should actually change though, unless it's something that just drops out eventually.
##\psi## in the QED Lagrangian is still a spin-1/2 field. It is the gauge field ##A_\mu## that is the spin-1 field.
 
  • Like
Likes dsaun777, topsquark and BiGyElLoWhAt
  • #5
And the QED Lagrangian is invariant under a local U(1) transformation.
 
  • Like
Likes BiGyElLoWhAt and topsquark
  • #6
So ##A_{\mu}## is the EM vector potential, which obviously transforms as a (co)vector. The naive observer (myself) would argue that ##\partial_{\mu}\psi## should also transform as a (co)vector. I don't completely understand spinors yet, but from what I understand, when ##\gamma^{\mu}## acts on a vector (I would assume that means covectors as well) they become spinors. Does ##\gamma^{\mu}## not act on ##A_{\mu}## in this Lagrangian? If so, would that not make it a spinor? And thus this should transform as a spinor, no? At least the first bit. Gamma doesn't seem to act on the scalar quantity F^2.

I'm very clearly missing something here.
 
  • #7
No. The ##\gamma##s have two sets of indices. The one that is written out explicitly and the two matrix indices belonging to the representation of ##\psi##. It holds that ##\bar\psi \gamma^\mu \psi## transforms as a vector.
 
  • Like
Likes vanhees71, BiGyElLoWhAt and malawi_glenn
  • #8
It will be much clearer if you wrote down the spinor indices. See Srednicki ch 36.
 
  • Like
Likes vanhees71 and BiGyElLoWhAt
  • #9
It's extremely convenient that Mark Srednicki offers a free pre-publication of his book. I also enjoy the quotes at the top of the page =D.
https://web.physics.ucsb.edu/~mark/qft.html
Thanks for the help, I will look into it more and post back if I have any further questions.
 
  • Like
Likes vanhees71
  • #10
BiGyElLoWhAt said:
It's extremely convenient that Mark Srednicki offers a free pre-publication of his book. I also enjoy the quotes at the top of the page =D.
https://web.physics.ucsb.edu/~mark/qft.html
Thanks for the help, I will look into it more and post back if I have any further questions.
Note that there can be typos, in particular in the draft version.
 
  • Like
Likes BiGyElLoWhAt
  • #11
I did read the disclaimer. I plan to work through it, so hopefully I can catch any that I run into.
 
  • Like
Likes vanhees71
  • #12
Orodruin said:
No. The ##\gamma##s have two sets of indices. The one that is written out explicitly and the two matrix indices belonging to the representation of ##\psi##. It holds that ##\bar\psi \gamma^\mu \psi## transforms as a vector.
On a related note, people will generally not write out all the indices. Imagine doing so in the quark sector of the Standard Model Lagrangian… it would simply look horrific.
 
  • Like
Likes atyy, vanhees71, malawi_glenn and 1 other person
  • #13
Quite frankly, we've all seen the T-Shirt.
 
  • Haha
Likes atyy, topsquark, vanhees71 and 1 other person
  • #14
OK, so now my curiosity has peaked. With regards to string theory and the graviton, how does its field transform? Is it vector ##\otimes## vector, or is it something else entirely?
 
  • #15
The graviton — if it exists — would be a spin-2 particle and therefore the corresponding field (the metric) indeed would transform as a rank 2 tensor.
 
  • Like
Likes vanhees71 and BiGyElLoWhAt
  • #16
malawi_glenn said:
Try to get one book and read it from start to finish instead of scattered notes etc. I can recommend the book by Mark Srednicki, you can get a free draft version on his homepage. The book is divided into several sections depending on spin and discusses at some lenght what is "special" about the various cases in terms of their Lorentz Transformations http://web.physics.ucsb.edu/~mark/qft.html
I didn't notice that you had edited in this addition. I actually have "Quantum Field Theory - for the Gifted Amateur" by Lancaster & Blundell. Actually, I believe I got this particular one based off of a book recommendation thread here on PF. However, I bought this book well in advance of my ability to use it (~5-7 years ago), and have only recently been able to understand and follow what were on the pages.

I definitely understand the value in following one train all the way through. I do, as well, have a bad habit of jumping around between sources. However, the way people explain things differently, I find valuable. Take the way Leonard Susskind (Stanford) teaches GR vs. the way Alex Flournoy (Colorado Mines) teaches GR. Susskind places a significantly larger emphasis on the conceptual bits, and Flournoy places a significantly larger emphasis on the mathematical elements. Susskind ##\cup## Flournoy would, in my opinion, be one of the best educators on this subject matter.
 
  • #17
1669013499090.png


and there are still some suppressed indices …
 
  • Like
  • Wow
  • Haha
Likes atyy, DennisN, topsquark and 3 others
  • #18
^^^
Ladies and Gentlemen, I give you, the most beautiful equation in the world!

Really though, that's absolutely insane.

Things like this really make me feel like we're brute forcing things too hard.
 
  • #19
BiGyElLoWhAt said:
Things like this really make me feel like we're brute forcing things too hard.
It is, however, remarkably accurate.
 
  • Like
Likes topsquark and BiGyElLoWhAt
  • #20
Yes. It is... :-(
 
  • #21
BiGyElLoWhAt said:
I didn't notice that you had edited in this addition. I actually have "Quantum Field Theory - for the Gifted Amateur" by Lancaster & Blundell. Actually, I believe I got this particular one based off of a book recommendation thread here on PF. However, I bought this book well in advance of my ability to use it (~5-7 years ago), and have only recently been able to understand and follow what were on the pages.

I definitely understand the value in following one train all the way through. I do, as well, have a bad habit of jumping around between sources. However, the way people explain things differently, I find valuable. Take the way Leonard Susskind (Stanford) teaches GR vs. the way Alex Flournoy (Colorado Mines) teaches GR. Susskind places a significantly larger emphasis on the conceptual bits, and Flournoy places a significantly larger emphasis on the mathematical elements. Susskind ##\cup## Flournoy would, in my opinion, be one of the best educators on this subject matter.
I think that book is ok, but is not as comprehensive in terms of particle physics as Srednicki.

Try to stay on one path, and write down questions. Maybe next book you read on that subject will cover it, or base your next read by your questions
 
  • Like
Likes vanhees71 and BiGyElLoWhAt
  • #22
Well I guess then, forgive me if this should be a different topic, the question is, why are 0-spin rank 0 tensor, 1-spin rank 1 tensor, 2-spin (if it exists) rank 2 tensor, why do we have to basically invent something to account for this 1/2 thing?
 
  • #23
I mean the gifted amateur book is probably better in introducing the "big picture" of qft in many fields whereas Srednicki is more detailed but at the expence of being 100% particle physics
 
  • Like
Likes vanhees71 and BiGyElLoWhAt
  • #24
malawi_glenn said:
I mean the gifted amateur book is probably better in introducing the "big picture" of qft in many fields whereas Srednicki is more detailed but at the expence of being 100% particle physics
TBH, what I'm gathering from this, is that I really should be looking at both. I am very much so "just coming into" QFT.

To be frank, what I'm really interested in understanding effectively boils down to why we're doing what we're doing.
It seems like "it's too complicated".
There are such beautiful and eloquent solutions to many things. The Dirac equation predicting anti-matter, for instance. Taking Einstein's E^2-p^2=m^2 and plugging in quantum operators to get the klein-gordon equation. I'm not sure what to say beyond that.
 
  • #25
BiGyElLoWhAt said:
Taking Einstein's E^2-p^2=m^2 and plugging in quantum operators to get the klein-gordon equation. I'm not sure what to say beyond that.
That is not how qft works.
Klein-Gordon equation is still classical (non-quantum) field theory. Perhaps you should study some non-quantum field theory first? I can give you some references
 
  • Like
Likes BiGyElLoWhAt
  • #26
I'm not going to say no.
 
  • #27
But really, is box^2=m^2 any different from E^-p^2=m^2?
 
  • #28
BiGyElLoWhAt said:
But really, is box^2=m^2 any different from E^-p^2=m^2?
That’s still a classical equation.
 
  • Like
Likes malawi_glenn and BiGyElLoWhAt
  • #29
Ok, so there's the piece that I'm missing then. Why does replacing operators in E total = KE + PE work, but replacing operators in E^2 - p^2 =m^2 not work.
 
  • #30
BiGyElLoWhAt said:
Ok, so there's the piece that I'm missing then. Why does replacing operators in E total = KE + PE work, but replacing operators in E^2 - p^2 =m^2 not work.
? replacing operators?

I think we are on a side-track now, you are essentially asking about how quantization of a non-quantum field theory is done, which has nothing to do with the OP regarding spin of fields.

BiGyElLoWhAt said:
I'm not going to say no.
I think most students actually encouner KG eq, Dirac eq and Procca eq for the first time as they enter a QFT course. This is not ideal, since then the classical vs quantum gets lost somewhat.
 
Last edited:
  • Like
Likes Orodruin
  • #31
malawi_glenn said:
You look how the fields transforms under Lorentz Transformations.

Scalars - spin 0, transforms as scalars.
Fermions - spin ½, transforms as spinors (½,0) or (0,½) representation (depending if you have a "left-handed" or a "right-handed" spinor).
Vectors - spin 1, transforms as vectors (½,½) representation.

http://sharif.edu/~sadooghi/QFT-I-96-97-2/LorentzPoincareMaciejko.pdf

Note, this is applicable to classical fields as well.

Try to get one book and read it from start to finish instead of scattered notes etc. I can recommend the book by Mark Srednicki, you can get a free draft version on his homepage. The book is divided into several sections depending on spin and discusses at some lenght what is "special" about the various cases in terms of their Lorentz Transformations http://web.physics.ucsb.edu/~mark/qft.html
Concerning the representation theory of the Poincare group, I think you should refer to Weinberg, QT of fields vol. 1, which is very detailed and complete.

A very good somewhat simpler treatment, restricted to the special cases needed for the standard model, i.e., spin 0, 1/2, and 1, can be found in Quantum Field Theory Lectures by Sidney Coleman. This book also nicely works out in detail, why the first-quantization approach doesn't work in relativistic QT.
 
  • Like
Likes BiGyElLoWhAt and malawi_glenn
  • #32
vanhees71 said:
Quantum Field Theory Lectures by Sidney Coleman
I have still failed to receive my copy :(
 
  • Sad
Likes vanhees71
  • #33
Don't order it from World Scientific directly. It's awful. I never got my copy from them and just had to write several times for getting my payment back. Then I ordered it from Amazon, and also for them it took pretty long, but it finally arrived, and they take the money from the credit card only after they really shipped it!
 
  • Wow
  • Love
Likes atyy and malawi_glenn
  • #34
malawi_glenn said:
? replacing operators?
Should read "replacing with operators". So ##p \to \nabla## and ##E \to \frac{\partial}{\partial t}## or ##E^2 - p^2 \to \Box^2##
##\frac{p^2}{2m} + V = E \to \frac{1}{2m}\nabla^2 + V = \partial_t##
##E^2 - p^2 = m^2 \to \Box^2 = m^2##

I do see though that those are just wave equations.

Edit
I think I have (since last night) answered my own question.
 
Last edited:
  • #35
malawi_glenn said:
It will be much clearer if you wrote down the spinor indices. See Srednicki ch 36.
And for a student, restoration of hbars is also a helpful practice.
 
  • Like
Likes malawi_glenn, BiGyElLoWhAt and topsquark

Similar threads

Replies
1
Views
2K
Replies
2
Views
3K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
7
Views
2K
Replies
5
Views
2K
Back
Top