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- Thread starter alphaone
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strangerep

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Th[...] is exactly my concern: The relativistic E^2=p.p+m^2 only comes from the field satisfying Klein Gordon which is the case when we have a scalar field or a spinor field but for an arbitrary theory I do not see why the e-values associated with the Casimir operator p.p (4-vectors) should be the mass^2.

The short answer is that relativistic quantum fields correspond to

irreducible representations of the Poincare group (else they are not

relativistically covariant). A "representation" of a group just means

that we have some vector space or other (1-dimensional, 2-dimensional,.....,

or even infinite-dimensional) and the group elements can be represented as

matrix operators in that vector space. One says that the space "carries" a

representation of the group. A representation is called irreducible if there

are no non-trivial invariant subgroups, i.e: if any two subspaces of the vector

will mix together, for some group element(s).

One classifies all irreducible representations of a given (abstract) group

by the eigenvalues of the group's Casimir operators (which commute

with all other group operators) and one other operator. For the Poincare

group the Casimirs are mass^2 (i.e: P^2), W^2 (square of Pauli-Lubansky

operator - the covariant version of total spin). The 3rd operator is usually

taken to be a spin component orthogonal to the 4-momentum.

Back in 1939, Wigner found that we can classify all irreducible

representations of the Poincare group in this way, and that the eigenvalues

of the total spin operator (W^2) are 0,1/2,1,..... The spin-1/2 fields can be

represented in a space of 2 complex dimensions - this corresponds to a

neutrino field. One can also complex-conjugate this space and get an

inequivalent representation of spin-1/2 (corresponding to anti-neutrinos).

The direct sum of these two spaces corresponds to the (massive) Dirac

field.

So to answer your question: for a massive spin-1/2 field, the Poincare

group is represented on the direct sum of the two 2-complex-dimensional

spaces mentioned above. The Casimir operator P^2, represented in

that same space, still corresponds to a mass^2 operator when acting

on elements of that vector space, by construction.

And... I think I should stop here until I see whether what I've written

above is making any sense.

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Meir Achuz

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Most physicists define the mass of an object as the invariant length of its four-momentum. This turns out to be a useful definition. It has nothing to do with a wave equation.

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Yes, it is. What is the purpose of builing a lagrangian for an arbitrary classical field ? To use it in the quantization formalism. Therefore, we need a mass term, since the mass of the quanta of the field is essential in the quantization...

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Could somebody please remind me why the e-values of the Casimir operator p.p (4-vector product) are the mass^2.

That's how we define the mass of the quanta of the field.

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strangerep

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[...] rephrase my question: Is the coefficient of the quadratic term of a fundamental field in an arbitrary Lagrangian density always related to the eigenvalue of the Casimir operator p.p (4-momentum squared)? If this is the case, could you pleae tell me how and why it should be related?

A (relativistic) Lagrangian represents a particular physical theory

constructed within the representation carrier space we call "Minkowski" space.

In Minkowski space, the translation generators are represented by simple

differential operators, acting on functions of the Minkowski space coordinates.

Such functions are spanned by a plane wave basis.

Therefore, a representation of P^2 in this carrier space is a quadratic

derivative term.

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sorry but thats not always essential right? for gauge fields you dont write a mass term and its as impotant as any...

Last edited:

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yes, it's true. For classical fields, a mass term destroys gauge invariance.

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on any representation.

The language is based on "group actions" which is still a little vague. As Definitions go is this

For example: If I take an Algebra [tex]\mbox{so(3)}[/tex], identify its infinitesimal generators, [tex]\left{ \tau^i \right}[/tex] and take an operator, call it, [tex]L^2[/tex] and find that

[tex]\left[ L^2, \tau^i \right]=0[/tex]

for each [tex]i=1,2,3[/tex].

Then this is

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samalkhaiat

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the operator commuting with all the infinitesimal generators of the Group Algebra?

This is the one.

For example: If I take an Algebra [tex]\mbox{so(3)}[/tex], identify its infinitesimal generators, [tex]\left{ \tau^i \right}[/tex] and take an operator, call it, [tex]L^2[/tex] and find that

[tex]\left[ L^2, \tau^i \right]=0[/tex]

for each [tex]i=1,2,3[/tex].

Then this issufficientto claim [tex]L^2[/tex] is a Casimir of [tex]\mbox{SO(3)}[/tex]?

Provided that

1) [itex]L^{2}[/itex] is invariant under SO(3) transformations.

2) it is constructed out of the group generators.

regards

sam

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Thank you Sam.

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