Representations of the Poincare group

Last updated Aug 5, 2023

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In QFT we want physics to be invariant under translations and Lorentz transformations. These transformations together form the Poincare Group

Definition: Poincare group

The Poincare Group is the group of translations and Lorentz transformations. It can be written as the semidirect product of the Lorentz group $O(3,1)$ and the translations group in $\reals^{3+1}$

In general a state $\ket{\psi}$ transforms as $$\ket{\psi}\to \mathcal{P}\ket{\psi}$$ under a poincare transformation $\mathcal{P}$.

Definition: Group representation

Mathematically a representation of a Group $G$ is a homomorphism to the Group of bijective linear transformations on some vector space $V$. So formally a representation $\rho$ is map $$\rho: G \to GL(V),$$ where $GL(V)$ is a linear map on the repesentation space $V$. The representation satisfies $$\rho(g_1g_2) = \rho(g_1)\rho(g_2)$$

If $V$ is finite dimensional, a representation is usually associated with a set of inverteble (because bijective) $n\times n$ matrices.

In QFT a representation is a set of objects ${\psi_i}$ that mix under a group transformation. This is equivalent, since in this case the set of objects ${\psi_i}$ is the representation space $V$ and the relation of the mixing $\psi \to \mathcal{P}\psi$ is the map on the representation space defining the representation.

For the objects in the representation space (lets take them to be states $\ket{\psi})$ there is a basis ${\ket{\psi_i}}$, so that a transformation of the basis states is given by $$\ket{\psi_i} \to \mathcal{P}_{ij}\ket{\psi_j}.$$ From the decomposition of a general state in terms of the basis states follow the general transformation property.

Definition: Irreduceable representation

A representation is said to be irreduceable if no subset of basis states in the representation space only transforms among one another.

In the picture of vectors in the representation space and matrices acting on them as transformations, this would amount to blockdiagonal parts of a matrix.

In addition, unitary representaitons are of interest, since they lead to poincare invariant matrix elements $$\mathcal{M} = \braket{\psi_1|\mathcal{P}^\dagger\mathcal{P}|\psi_2}$$

Definition: Unitary representation

A representation is unitary if the transformations $\mathcal{P}$ acting on the representation space satisfy $$\mathcal{P}^\dagger\mathcal{P} = 1$$

For unitary representations the norm of vectors in the representation space $\braket{\psi|\psi}$ is preserved

Therefore we can define particles as

Particles

Particles are objects that transform under irreducible unitary representations of the Poincare group. Therefore they build the representation space and with every representation there is a particle associated to it

There are no finite-dimensional unitary representations of the Poincare group.

classification of representations of the Poincare group

The classification of representations of the Poincare group was done by Wigner

• They are infite dimensional (since they depend on the momentum $p$)
• Uniquiely classified by two quantum numbers, one positive real, one half integer. Found to be
  1. mass m
  2. spin J
• If $J>0$:
  1. For $m=0$ there are 2 states for every representation
  2. For $m>0$ there are $2J+1$ states for every representation What means states??
• If $J=0$:
  1. For any $m$ there is only one state

Definition: Little Group

The subgroup of the Poincare Group for a fixed $p$ is called little group. It is finite dimensional.

It would be desireable to associate the representations with the fields $\phi$, $A_\mu$ etc…

• Representation $A_\mu$ is not unitary (?)
• Lorentz invariance and unitarity conflict ($\Lambda^\dagger \neq \Lambda^{-1}$), which can change the norm of vectors when boosted ($\braket{\psi^\prime|\psi^\prime}\neq\braket{\psi|\psi}$)

• There is only one d.o.f
• Put one d.o.f into a single scalar field $\phi$:
• General (?) Lagrangian $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi(x)\partial^\mu\phi(x) - \frac{1}{2}m^2\phi^2(x)$$
• Leads to equation of motion $$(\Box + m^2)\phi = 0$$
• Energy desnity is positive definite (classical equivalent to unitarity)

• There are three d.o.f.
• Real vector field $A_\mu$ has $4 = 3 \oplus 1$ d.o.f.
• Analoge Lagrangian like in the $J=0$ case would give negative energy densities for some of the four components

Most general Lorentz invariant Lagrangian is $$\mathcal{L} = \frac{a}{2}A_\mu\Box A^\mu + \frac{b}{2}A_\mu\partial^\mu\partial_\nu A^\nu + \frac{1}{2}m^2A^2$$

With the equations of motion $$a\Box A_\mu + b\partial_\mu\partial^\nu A_\nu + m^2A_\mu=0$$

• Take the derivative of the equation of motion leads to $\partial_\mu A^\mu = 0$ for $a = -b$.
• This condition removes a degree of freedem (Spin zero component)
• For $a = -b = 1$ we get the Proca Lagrangian

Definition: Proca Lagrangian

The Proca Lagrangian is given by $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu,$$ with the field strength tensor $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$

The equation of motion is $$(\Box + m^2)A_\mu = 0$$

From which follows the additional condition $$\partial_\mu A^\mu = 0$$

• One can show the energy for $A_\mu$ is positive
• The most general solution to the equation of motion is $$A_\mu(x) = \sum_i\int\frac{d^3\vec{p}}{(2\pi)^3}a_i(\vec{p})\epsilon^i_\mu(p)e^{ipx},$$ with some basis vectors $\epsilon^i_\mu$.
• Since we also want $\partial_\mu A^\mu=0$ we have $$\boxed{p^\mu\epsilon^i_\mu(p)=0}$$
• For a given $p^2 = m^2$ are now only three linearly independent basis vectors

Definition: Polarization vectors

The Polarization vectors are three independent basis vectors satisfying $$p^\mu\epsilon^i_\mu(p)=0$$ They are normalized to $$\epsilon^*_\mu\epsilon^\mu = -1$$

For $p^\mu = (E,0,0,p_z):$

1. Transverse polarizations: $$\epsilon^\mu_1 = (0,1,0,0),\quad \epsilon^\mu_2(0,0,1,0)$$
2. Longitudinal plarizations: $$\epsilon^\mu_L = (p_z/m,0,0,E/m)$$

• The three polarization vectors generate the irreduceble representation (?)
• Since they depend on $p$ the dimension is infinite dimensional

• There are two d.o.f.

Most general Lagrangian $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$

This Lagrangian is gauge invariant under $$A_\mu(x)\to A_\mu(x)+\partial_\mu \alpha(x).$$ Gauge invariance gives us the freedom to choose a particular gauge $\alpha$

Definition: Coulomb Gauge

Coulomb Gauge defines $\alpha(x)$ so that $$\partial_j A_j(x) = 0,$$ where the sum over $j=1,2,3$ is understood.

The equations of motion are $$\Box A_\mu - \partial_\mu(\partial_\nu A^\nu)=0,$$ in coulomb gauge this is $$\Box A_j = 0$$

From the equations of motion under coulomb gauge and the representation $A_\mu = \int \frac{d^4p}{(2\pi)^4} \epsilon_\mu(p) e^{ipx}$ one finds that

1. $A_0=0$
2. $p^2 = 0$
3. $p_j \epsilon_j =0$
4. $\epsilon_0 = 0$

Chosing WOLOG a frame along the $z$-axis, these equations have two solutions for the polarization $$\epsilon^\mu_1 = (0,1,0,0) \quad \epsilon^\mu_2=(0,0,1,0)$$ Therefore we only have two d.o.f. as suggested for a massless Spin 1. Another common basis is the basis of circular polarization called helicity eigenstates: $$\epsilon^\mu_R = \frac{1}{\sqrt{2}}(0,1,i,0) \quad \epsilon^\mu_L=\frac{1}{\sqrt{2}}(0,1,-i,0).$$

Gauge invariance has to imposed on a local lagrangian to enable massles spin one particles to be described by two polarization vectors