Transforming a Lorentz Scalar: psi-bar x psi x y_0

  • Context: Graduate 
  • Thread starter Thread starter captain
  • Start date Start date
  • Tags Tags
    Lorentz Psi Scalar
Click For Summary

Discussion Overview

The discussion revolves around the transformation properties of the expression involving the Dirac spinor, specifically the product of the Hermitian conjugate of the spinor and the spinor itself, in the context of Lorentz invariance. Participants explore the necessity of including the gamma matrix in this expression and its implications for transforming as a Lorentz scalar.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the requirement to use the product of the Hermitian conjugate of the spinor and the spinor itself, questioning why the standard Hermitian conjugate form does not suffice for Lorentz scalar transformation.
  • Others argue that the transformation properties of the spinor dictate that the expression must include the gamma matrix to ensure Lorentz invariance.
  • It is noted that the product \(\psi^{\dagger} \gamma^0 \psi\) is a scalar under full Lorentz transformations, and this can be shown similarly to infinitesimal transformations.
  • One participant mentions the preference of physicists for working with Lie algebras over Lie groups, suggesting it simplifies the analysis of transformations.
  • There are challenges regarding the correctness of certain mathematical expressions related to infinitesimal transformations, with participants questioning the validity of specific operator manipulations.
  • Some participants emphasize that infinitesimal transformations contain all necessary information for finite transformations, countering the notion that they are merely approximations.
  • Concerns are raised about potential topological complications in the limit of infinitesimal transformations, although it is suggested that these issues are generally not significant.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of including the gamma matrix in the transformation expression. There is no consensus on the correctness of certain mathematical manipulations related to infinitesimal transformations, indicating ongoing debate and uncertainty.

Contextual Notes

Some limitations in the discussion include unresolved assumptions about the definitions of the terms used and the mathematical steps involved in proving the transformation properties. The scope of the discussion is primarily focused on theoretical aspects of quantum field theory and Lorentz transformations.

captain
Messages
163
Reaction score
0
i am confused why you need to write psi-bar multiplied by psi (where psi-bar is the hermitian conjugate of psi multiplied by y_0 (dirac matrix or gamma matrix)) instead of the h.c. psi multiplied by psi in order for it to transform as a lorentz scalar.

Note: h.c. means hermitian conjugate
 
Physics news on Phys.org
captain said:
i am confused why you need to write psi-bar multiplied by psi (where psi-bar is the hermitian conjugate of psi multiplied by y_0 (dirac matrix or gamma matrix)) instead of the h.c. psi multiplied by psi in order for it to transform as a lorentz scalar.

Note: h.c. means hermitian conjugate

It simply follows from the transformation of psi. If you look up how psi transforms, you will see that psi^dagger psi is not invariant but "psi bar psi" is. Just check it.
 
You must have that [itex]\gamma_{0}[/itex] there. It follows from the Clifford algebra of the gamma matrices and how the psi transforms under infinitesimal LT-s.
 
dextercioby said:
You must have that [itex]\gamma_{0}[/itex] there. It follows from the Clifford algebra of the gamma matrices and how the psi transforms under infinitesimal LT-s.

It is also possible to prove that [itex]\psi^{\dagger}\gamma^0 \psi[/itex] is a scalar in full Lorentz transformations. It is not much more difficult than in infinitesimal transformations. Either way, you have to merely multiply three matrices (but you have to first solve the matrices of full transformations). I've always been slightly confused about why physicists always want to do stuff in infinitesimal scale :confused:
 
Because physicists find it easier to work with Lie algebras rather than Lie groups.
 
But for example

[tex] (1-iA\alpha) X (1+iA\alpha) = (1+B\alpha) X + O(\alpha^2) \quad\implies\quad e^{-iA\alpha} X e^{iA\alpha} = e^{B\alpha} X[/tex]

isn't really correct, is it?
 
See any text on the Dirac eq. You'll find that (psi adjoint)*psi is charge, the time component of a four-vector, (psi bar )*psi is a scalar. It has to do with Lorentz xforms and normalization.

Regards,
Reilly Atkinson
 
kweh?

jostpuur said:
But for example

[tex] (1-iA\alpha) X (1+iA\alpha) = (1+B\alpha) X + O(\alpha^2) \quad\implies\quad e^{-iA\alpha} X e^{iA\alpha} = e^{B\alpha} X[/tex]

isn't really correct, is it?

This doesn't even make sense as a statement with Lie Algebra... Recall these are operators we're dealing with.
 
jostpuur said:
But for example

[tex] (1-iA\alpha) X (1+iA\alpha) = (1+B\alpha) X + O(\alpha^2) \quad\implies\quad e^{-iA\alpha} X e^{iA\alpha} = e^{B\alpha} X[/tex]

isn't really correct, is it?

This isn't really what's going on. To see how infinitesimal rotations give you finite rotations, think of a finite rotation as nothing but an infinite number of infinitesimal rotations. For example, if you want to rotate by a finite amount x, then break x up into N pieces x/N and rotate N times. For N very large, this is an infinitesimal rotation so your rotation operator is:

[tex] (1+i(x/N)A)^N[/tex]

where A is your rotation generator. Now imagine taking [tex]N\rightarrow\infty[/tex], and you get [tex]e^{ixA}[/tex], the usual finite rotation.

It is very important to understand that when we are doing these infinitesimal transformations, we are NOT making any approximations - all of the information for finite transformations is there in the infinitesimal transformation! It's not that physicists are sloppy and feel content to only consider "leading order" effects. This is all mathematically well defined and justified.

I should say that I'm being a little cavalier in the above paragraph: there could be so called "topological" complications, which is just to say the way you take the above limit could be important. But these issues are not very important in general, and they don't change anything.

Finally, let me just say that in more advanced applications of these topics (gauge theories, QFT, etc.), it is more appropriate to formulate the physics in terms of Lie Algebras (infinitesimal) rather than Lie groups (finite). For a wonderful explanation of why this is the case, I refer you to your favorite textbook. A very nice one is H. Georgi's "Lie Algebras for Particle Physics", but there are others out there as well.
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K