# Commutators, etc.

1. May 5, 2012

### noamriemer

Hi you guys!
I am having a hard time understanding some stuff in this context.
I was not able to find any guidance in books or anything.
say I want to calculate:

$[W^2,p^\alpha]=0, W=\frac{m}{2}\varepsilon^{\mu\nu\lambda\varepsilon}M_{\mu\lambda}p_\delta$

How do I do that? I can't manage to understand the whole process...
If anyone has some wisdom to spare here... :)

2. May 5, 2012

### vanhees71

This is all about the unitary irreducible representations of the proper orthochronous Poinacre group $\mathrm{ISO}(1,3)^{\uparrow}$. These representations operate on single-particle states of free particles.

This subgroup of the Poincare group describes tranlations in space and time as well as boosts and rotations, building the semidirect product of the translation group and the homogeneous proper orthochronous Lorentz group.

This is a connected (but not simply connected) 10-dimensional Lie group with the generators for time and space translation (physically meaning the energy and momentum), spatial rotations (physical meaning the angular momentum) and boosts (physical meaning the center of momentum position of the particle). These generators are thus represented by the 10 selfadjoint operators $\hat{P}^{\mu}$ and $\hat{M}^{\mu \nu}=-\hat{M}^{\nu \mu}$. From the group product (composition of Poincare transformations) the commutation relations follow.

When analyzing the possible, physically meaningful, irreps. it turns out that each irrep. is characterized first by the Casimir operators which are $\hat{M}^2:=\hat{P}_{\mu} \hat{P}^{\mu}$ and $W_{\mu} W^{\mu}$ with the Pauli-Lubanski (pseudo) vector

$$W^{\mu}=\frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}.$$

Since we look at irreducible representations, the Hilbert space of states must be one of definite eigenvalues for $\hat{M}^2$ and $\hat{W}^2$. It turns out that meaningful representations with physical interpretability in terms of local QFTs with stable ground state can be built from the representations with $$M^2 \geq 0$$. From now on let's stick to the massive case, i.e., $M^2>0$ since for massless particles there are some complications, one should avoid in the first studies of these issues.

It turns further out that for $M>0$ one can build a generalized orthogonal basis for the vector space of the representation as energy-momentum eigenstates. Then one first considers the representation in the subspace of vanishing three-momentum, i.e., the basis for single-particle states for particles at rest. The subgroup of the Poincare group, which leave this subspace invariant (the socalled "little group" of this representation), are obviously the spatial rotations, and a further characterization is the representation of this rotation group. In order to have an irrep. this must be one representation you know from the quantum theory of angular momentum (including spin), i.e., it is characterized by the eigenvalue of the of the spin operator squared, which is $s(s+1)$ with $s$ taking an integer or half-integer value since this is the one and only Casimir operator of the rotation group. It turns out that the Pauli-Lubanski vector squared then takes the value $-m^2 s(s+1)$, i.e., $\hat{W}^2$ is the covariant description of which irrep. representation of the rotation group for particles at rest is considered. However, $\hat{W}^{\mu}$ do not commute with boosts and rotations as in the non-relativistic case, which means that the spin has only its usual physical meaning for a particle observed in its rest frame. For given mass $m^2$ and spin $s$, the single-particle basis for particles at rest is given by $|\vec{p}=0,\sigma \rangle$ where $\sigma \in \{-s,-s+1,\ldots,s-1,s \}$.

Then one can define single-particle basis states of any non-zero three momentum by applying an appropriate boost, whose representation is uniquely defined by the representation of the little group. This procedure you should look up, e.g., in Weinberg, Quantum Theory of Fields Vol. I. You also find a detailed description in my qft manuscript on my homepage:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

There's only one more point I have to make on the irreps. for massive particles: The irrep. is not yet uniquely defined! We have so far only characterized the Casimir operators, i.e., the mass squared of the particle, and its spin (in the meaning of the representation of the little group in the subspace for particles at rest), its three momentum eigenvalues (which run over all $\mathbb{R}^3$), and the spin-3 component as measured in the particles rest frame.

Since $\hat{M}^2=\hat{P}^2$, also $\hat{P}^0$ should have a specified value, but that's not completely true, since the eigenvalue can be both, $\omega=\pm \sqrt{\vec{p}^2+m^2}$. I.e., each irrep. of the $\mathrm{ISO}(1,3)^{\uparrow}$ with given $m^2$ and $s$ comes in two versions, one where the eigenvalue of $\hat{P}^0$ is $+\sqrt{\vec{p}^2+m^2}$ and one where it's the negative of this value. Since we consider only orthochronous Lorentz transformations here, the sign of the time component of the four-momentum is conserved, and thus one has two possible irreps. with the specified values of the Casimir operators.

In order to make physical sense of these latter states in the sense of local quantum field theories, one has to take both posibilities but reinterpret the states with negative eigenvalues for $\hat{P}^0$ as anti-particle states, i.e., one builds the relativistic quantum fields out of annihilation operators for momentum states with positive value $p^0$ and creation operators for negative $p^0$ values. This leads to the reinterpretation of "particles with negative energy running backward in time" as "anti-particles with positive energy running forward in time". In this way one has reached two goals at once: One has saved causality, i.e., any particle or anti-particle is always runing forward in time and there are only single-particle states with positive energy, i.e., there exists a state with minimal energy, that is called the vacuum and by definition is the state, where no (real) particles are present.

At the same time, in this way one has constructed a space-time dependent field operator, on which the proper orthochronous Poincare group acts locally, i.e., analogous to the way it acts in classical field theory. E.g., in electrodynamics the four-vector potential transforms as a four-vector field under Lorentz transformations, i.e., with $x'=\Lambda x$, one has

$${A'}^{\mu}(x')={\Lambda^{\mu}}_{\nu} A_{\nu}(x)={\Lambda^{\mu}}_{\nu} A_{\nu}(\Lambda^{-1} x').$$

3. May 6, 2012

### noamriemer

Wow... thank you so much for this wonderful reply...I am still trying to understand the whole thing... :)

4. May 6, 2012

### noamriemer

Is it correct to write:
$M_{\beta\alpha}=g_{\alpha\alpha}g_{\beta\beta}M^{\alpha\beta}$

5. May 6, 2012

### dextercioby

No, of course not, you need to respect the Einstein summation convention fully. No repeated index in one term of an equality unless summed over and free index should be different than a summed index.

$$M_{\alpha\beta} = g_{\alpha\sigma}g_{\beta\rho} M^{\sigma\rho} =g_{\alpha\sigma} M^{\sigma\rho}g_{\rho\beta}$$

6. May 6, 2012

### noamriemer

Oh... then there's a problem here...
what do I do when I am given $M_{\alpha\beta}$
and what I need is $M^{\alpha\beta}$?
I have an exercise in which I have to prove that : $[M^{\alpha\beta},P^{\gamma}]=0$ and I need to use $M_{\alpha\beta}$

7. May 6, 2012

### Fredrik

Staff Emeritus
No, this equality doesn't really make sense, since each index appears three times on the right. I would have to interpret it either without the summation convention, or by using $g_{\alpha\alpha}=g_{\beta\beta}=\operatorname{Tr} g=\pm 2$, where the sign depends on what convention you're using for the metric. Either way, the equality says something that's not true.

You can however write $M_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}M^{\gamma\delta}$.

Note that the right-hand side is the $\alpha\beta$ component of $gMg^T$ (where M is the matrix with $\alpha\beta$ component $M^{\alpha\beta}$ and g is the matrix with $\alpha\beta$ component $g_{\alpha\beta}$).

8. May 6, 2012

### dextercioby

Well, I spelled it out for you in the post just above yours, no. 5 of the thread. You need to use the metric tensor in the way I showed you.

9. May 6, 2012

### noamriemer

I will never get used to this type of writing.... and to think it was invented to make things simpler... I'd rather just write the whole thing.
I am sorry but there is something I fail to understand. What are the relations between
$M_{\alpha\beta}$ and $M^{\alpha\beta}$?
Why can't I switch the two of them without using other letters?
Sorry, I am really having a hard time here....

10. May 6, 2012

### Fredrik

Staff Emeritus
This was answered in post #5 and then again in post #7. (I wrote post #7 before I saw post #5).

Because one of them is defined from the other and the metric, by the formula posted in #5 and #7.

$M_{\alpha\beta}$ is a component of a (0,2) tensor, and $M^{\alpha\beta}$ is a component of a (2,0) tensor. If you define the latter tensor first, then you can define the former as "the (0,2) tensor with components $g_{\alpha\gamma}g_{\beta\delta}M ^{\gamma\delta}$". The convention is to denote the components of this specific (0,2) tensor by $M_{\alpha\beta}$.

Last edited: May 6, 2012