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

In summary, the need to write psi-bar multiplied by psi instead of h.c. psi multiplied by psi in order for it to transform as a Lorentz scalar is due to the transformation properties of psi and the Clifford algebra of gamma matrices. It is also possible to prove that psi-bar multiplied by psi is a scalar under full Lorentz transformations. Infinitesimal transformations are used because they are easier to work with, and they contain all the necessary information for finite transformations. In more advanced applications, Lie Algebras are preferred over Lie groups.
  • #1
captain
164
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
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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:
 
  • #5
Because physicists find it easier to work with Lie algebras rather than Lie groups.
 
  • #6
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?
 
  • #7
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
 
  • #8
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.
 
  • #9
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.
 

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

1. What is a Lorentz Scalar?

A Lorentz Scalar is a mathematical quantity that remains unchanged under Lorentz transformations, which are transformations that preserve the laws of physics in special relativity. This means that the magnitude of the scalar remains the same regardless of the observer's frame of reference.

2. What does it mean to transform a Lorentz Scalar?

Transforming a Lorentz Scalar refers to applying a Lorentz transformation to the scalar quantity. This can involve changing the coordinates or frame of reference in which the scalar is being measured, but the magnitude of the scalar will remain unchanged.

3. What is psi-bar x psi x y_0 in the context of transforming a Lorentz Scalar?

Psi-bar x psi x y_0 is a mathematical expression that represents the product of three quantities: psi-bar, psi, and y_0. Psi-bar and psi are complex numbers, while y_0 is a real number. This expression is often used in the context of transforming a Lorentz Scalar in quantum field theory.

4. How is transforming a Lorentz Scalar relevant in science?

Transforming a Lorentz Scalar is relevant in science, specifically in the fields of relativity and quantum mechanics. It allows scientists to understand how physical quantities, such as energy and momentum, behave under different frames of reference and how they are related to each other.

5. What are the applications of transforming a Lorentz Scalar?

The applications of transforming a Lorentz Scalar are vast and varied. It is used in many areas of physics, including special relativity, quantum field theory, and particle physics. It is also essential in practical applications, such as GPS technology, which relies on precise measurements of time and space that are affected by Lorentz transformations.

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