Casimir Operator: Mass^2 e-Values Explained

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In summary, the mass of a field is defined as the invariant length of its four-momentum and is related to the eigenvalue of the Casimir operator p.p (4-momentum squared). This is because a Lagrangian, which is used in the quantization formalism, requires a mass term in order to properly represent the physical theory. The mass term is essential in quantization and is represented by a quadratic derivative term in the carrier space. However, for gauge fields, a mass term is not always necessary and can actually destroy gauge symmetry.
  • #1
alphaone
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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. This relation is clear to me when the field in question fulfils the Klein-Gordon equation but what about the general case?
 
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  • #2
I'm not too sure this will satisfy you, but this is basically Einstein's [tex]E^2-\vec{p}^{.2}=m^2[/tex] (so Klein-Gordon is the general case indeed). For a spin-1/2 field, it si clear that if it satisfies Dirac then it satisfies KG...
 
  • #3
Thanks for the reply, but what you are pointing out 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.
 
  • #4
alphaone said:
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.
 
  • #5
alphaone said:
Thanks for the reply, but what you are pointing out 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.
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.
 
  • #6
I agree with the earlier posts and should 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? Thanks in advance
 
  • #7
i can think of something though I am not sure whether it'll satisfy you. you see a quadratic in the fundamental fields is lorentz invariant and so are the eigenvalues of a casimir.it seems to be necessary, even if not sufficient
 
  • #8
alphaone said:
I agree with the earlier posts and should 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? Thanks in advance

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...
 
  • #9
alphaone said:
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.
 
  • #10
alphaone said:
[...] 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.
 
  • #11
"Therefore, we need a mass term, since the mass of the quanta of the field is essential in the quantization..."

sorry but that's not always essential right? for gauge fields you don't write a mass term and its as impotant as any...
 
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  • #12
yes, it's true. For classical fields, a mass term destroys gauge invariance.
 
  • #13
I am trying to understand Casimirs. The Definition I was given was:

A Casimir operator is an operator which commutes with all group actions
on any representation.


The language is based on "group actions" which is still a little vague. As Definitions go is this sufficiently equivalent to the operator commuting with all the infinitesimal generators of the Group Algebra?

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 sufficient to claim [tex]L^2[/tex] is a Casimir of [tex]\mbox{SO(3)}[/tex]?
 
  • #14
cathalcummins said:
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 is sufficient to 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
 
  • #15
Thank you Sam.
 

1. What is a Casimir Operator?

A Casimir Operator is an operator in quantum mechanics that is used to calculate the mass and energy values of particles in a quantum system.

2. How is the Casimir Operator used to calculate mass squared (M^2)?

The Casimir Operator is used in the Klein-Gordon equation, which relates the mass squared value to the energy and momentum of a particle. By solving this equation, the mass squared value can be obtained.

3. Why is the Casimir Operator important in quantum mechanics?

The Casimir Operator is important because it provides a mathematical framework for understanding the mass and energy values of particles in a quantum system. It allows scientists to make predictions and calculations about the behavior of particles at the quantum level.

4. How is the Casimir Operator related to e-values?

The e-values, or eigenvalues, of the Casimir Operator represent the possible values of mass squared for a given quantum system. These values are obtained by solving the Klein-Gordon equation and can be used to determine the mass and energy of particles within the system.

5. Are there any limitations to the Casimir Operator in calculating mass squared?

Yes, there are limitations to the Casimir Operator in calculating mass squared. It assumes that the particles in the system are point-like and have no internal structure. It also does not take into account the effects of relativity, which can be important for particles with high energies.

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