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Representations of the Poincare group

Last updated Dec 29, 2024

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 $R^{3+1}$

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

# Representations

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

# Physical representations

Therefore we can define particles as

Definition: 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

Physically this is intuitive, since they form a set of states which only mixes amongst each other under poincare transformations, as one would expect. An electron stays an electron under poincare transformation.

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

This will lead to only fields beeing suitable candidates for particle states

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 independent states for every momentum $p$
    2. For $m>0$ there are $2J+1$ independent states for every $p^2=m^2$
  • 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.

Note:

On Unitatrity:

# Spin $J=0$

# Massive Spin $J=1$

We extend to the most general Lorentz invariant Lagrangian for a vector field $A_\mu$:

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

The equations of motion is given by $$a\Box A_\mu + b\partial_\mu\partial^\nu A_\nu + m^2A_\mu=0$$

$$\partial_\mu A^\mu = 0$$

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 $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$

Equation of motion for a Vector field

The equation of motion for $A_\mu$ from the Proca Lagrangian is given by $$\Box A_\mu - \partial_\mu\partial_\nu A_\nu + m^2A_\mu=0$$ This contains two equations (the equivalence can be seen by taking $\partial_\mu$ of the full equation and substituting the result back in): $$(\Box + m^2)A_\mu = 0$$ $$\partial_\mu A_\mu = 0$$

The most general solution to the first of these equations can be written as $$A_\mu(x) = \sum_i\int\frac{d^3\vec{p}}{(2\pi)^3}a_i(\vec{p})\epsilon^i_\mu(p)e^{ipx},$$ with some basis vectors $\epsilon^i_\mu$

Note:

$$\boxed{p^\mu\epsilon^i_\mu(p)=0}$$

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 $p^\mu = (E,0,0,p_z)$ one can chooose:

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

# Massless Spin $J=1$

We get the Lagrangian of the massless theory by taking $m\to 0$.

Then we find the 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 $$\partial^2_jA_0=0$$$$\Box A_j = 0$$ From the first one it follows together with the Coulomb gauge condition that $$A_0=0$$

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 ($p_\mu = (E,0,0,E)$), these equations have two solutions for the polarization (we choose) $$\epsilon^\mu_1 = (0,1,0,0) \quad \epsilon^\mu_2=(0,0,1,0)$$ 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 be imposed on a local lagrangian to enable massles spin one particles to be described by two polarization vectors

# Higher spin fields

  1. Massive: 5 polarizations ($2J+1$)
  2. Massless: 2 polarizations

Ghosts

A ghost is a state with negative norm, or wrong sign kinetic term (?)

Example: Spin 1

We can split up any vector field $A_\mu(x)$ into $$A_\mu(x) = A^T_\mu(x)+\partial_\mu \pi(x)$$ with $\partial_\mu A^T_\mu = 0$, since we can choose any $A^T_\mu(x)$ and $\partial_\mu \pi(x)$ by shifting $A^T_\mu \to A^T_\mu + \partial\alpha$ and $\pi \to \pi-\alpha$. So using this shift we can choose any $A^T_\mu, \pi$ combination we want. However we already know that there must be an appropriate $\alpha$ so that $\partial_\mu A^T_\mu = 0$ (This is Lorenz gauge). Here $A^T_\mu$ are the transverse and $\pi$ the longitudinal polarizations

We can use the decomposition to see if the non transeverse polarizatoins $\pi(x)$ are physical or not.

The most general lorentz-invariant lagrangian for $A_\mu$ is $$\mathcal{L}=aA_\mu\Box A_\mu + bA_\mu\partial^\mu\partial^\nu A_\nu + m^2A^2_\mu$$ Putting in $A_\mu$ gives a Lagrangian in terms of $A^T$ and $\pi$ (two separate fields) $$\mathcal{L}=aA^T_\mu\Box A^T_\mu + m^2(A^T_\mu)^2 - (a+b)\pi\Box^2\pi - m^2\pi\Box\pi$$ Calculating the poropagator for $\pi$ gives $$\Pi_\pi = \frac{1}{2m^2}\left[\frac{1}{k^2}-\frac{a+b}{(a+b)k^2-m^2}\right]$$ When writing it this way we see that the $\pi$ field has a part with negative norm (??), relating to the negative probability following from the negative propagator (probability to propagator from $x$ to $y$.)

In general a knietik term with more than two derivaties is always related to a theory which is not unitary (and thus unphysical) We can fix the propagator by choosing $a = -b$ (removing the $\Box^2$ term). With appropriate choices for $a=-b$ this leads to the known physical Lagrangian for a massive spin-1 field: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu$$

In addition, a mode (a field like e.g. $\pi$) which does not have a kinetic term, is also not allowed to have an interaction term (the interaction strength would be ininite.)

For a massless Spin 1 that means if $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, under $A_\mu(x) = A^T_\mu(x)+\partial_\mu \pi(x)$ the kinetic $\pi$ terms cancel out, which means interaction terms should not have a $\pi$. This is equivalent to the statement of gauge invariance!

For the case of the massless Spin one this means only interactoin terms with $\mathcal{L}{int}=A\mu J^\mu$ are allowed when $\partial_\mu J^\mu=0$ (since after inserting the transformation $\partial_\mu\pi J^\mu$ vanishes with partial integration).

A similar thing like for the Spin 1 can be done for the Spin 2 case:

  1. Decompose Spin 2 field in terms for polarizations
  2. Write down most general Lagrangian
  3. Fix factors in the lagrangian by requirering that $\Box^2$ terms are zero
  4. Fix interactions by requiering that $\pi_\mu$ does not appear in interactions

The massive Spin 2 field has 5 polarizations: 2 transverse and three longitudinal why?? $$h_{\mu\nu}=h^T_{\mu\nu}+\partial_\mu\pi_\nu+\partial_\nu\pi_\mu$$ with $\partial_\mu h^T_{\mu\nu}=0$ and $$\pi_\mu = \pi^T_\mu + \partial_\mu\pi^L$$ with $\partial_\mu\pi^T_\mu = 0$.

In this way one can derive the kinetic terms (with 3.) and the interactions (with 4.) of the Spin 2 lagrangian to $h_{\mu\nu}T_{\mu\nu}$ with $\partial_\mu T_{\mu\nu}=0$ similar to the Spin 1 case. There are other possibilites for interactions. The requirement 4. will lead to allowed interaction terms and transformation rules for all fields.

Deriving the lagrangian in this way is a lot of work, it is easier to use symmetry arguments. However this is basically the underlying reason why these symmetry arguments work.

# Building Lagrangians

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