# Casimir operator

#### alphaone

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?

Related High Energy, Nuclear, Particle Physics News on Phys.org

#### humanino

I'm not too sure this will satisfy you, but this is basically Einstein's $$E^2-\vec{p}^{.2}=m^2$$ (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...

#### alphaone

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.

#### strangerep

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.

#### Meir Achuz

Homework Helper
Gold Member
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.

#### alphaone

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

#### cygnus2

i can think of something though im 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

#### dextercioby

Homework Helper
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...

#### dextercioby

Homework Helper
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.

#### strangerep

[...] 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.

#### cygnus2

"Therefore, we need a mass term, since the mass of the quanta of the field is essential in the quantization..."

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

Last edited:

#### dextercioby

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

#### cathalcummins

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 $$\mbox{so(3)}$$, identify its infinitesimal generators, $$\left{ \tau^i \right}$$ and take an operator, call it, $$L^2$$ and find that

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

for each $$i=1,2,3$$.

Then this is sufficient to claim $$L^2$$ is a Casimir of $$\mbox{SO(3)}$$?

#### samalkhaiat

the operator commuting with all the infinitesimal generators of the Group Algebra?
This is the one.

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

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

for each $$i=1,2,3$$.

Then this is sufficient to claim $$L^2$$ is a Casimir of $$\mbox{SO(3)}$$?
Provided that

1) $L^{2}$ is invariant under SO(3) transformations.
2) it is constructed out of the group generators.

regards

sam

Thank you Sam.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving