# What is the group transformation property of photons under rotation?

Both the photon and the W boson are spin-1 particles. Under rotation W boson must transform under the 3-dimensional representation of SU(2). However, the photon has two degrees of freedom (or helicity states), unlike W boson. How does it transform under the rotation of coordinates? What is the underlying group and group representation which describes the transformation of photons under rotation.

• U(1) as a result of the gauge symmetry of electromagnetic field. Jan 18, 2016 at 19:10
• I think you are looking for the E(2)-like little group (E(2) is the two-dimensional Euclidean group), which Wigner (1939) found to be the maximal subgroup of the Lorentz group which leaves the four-momentum of massless particles invariant. Jan 18, 2016 at 19:29

The general theory

For each irreducible representation of the Poincare group with a given helicity $\pm s$, which is realized by relativistic creation-destruction field, $$\hat{\psi}_{l}(x) = \sum_{\sigma = \pm s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2\epsilon_{p}}}\left(u_{l}(\sigma , \mathbf p)\hat{a}(\sigma, \mathbf p)e^{-ipx} + v_{l}(\sigma ,\mathbf p)\hat{b}^{\dagger}(\sigma, \mathbf p)e^{ipx} \right)_{\epsilon_{p} = |\mathbf p|}$$ under Lorentz group transformation you must have $$\tag 1 \begin{cases} u_{l}(\sigma , \mathbf k)e^{i\sigma \theta (\mathbf k, \Lambda)} = \sum_{l {'}}D_{l l{'}}(R(\theta, \alpha , \beta))u_{l{'}}(\sigma , \mathbf k)\\ v_{l}(\sigma , \mathbf k)e^{-i\sigma \theta (\mathbf k, \Lambda)} = \sum_{l {'}}D_{l l{'}}(R(\theta, \alpha , \beta))v_{l{'}}(\sigma , \mathbf k)\end{cases}$$ Here $D_{ll}$ is the the unitary irreducible representation of the little group $R(\theta , \alpha, \beta) = S(\alpha , \beta)\tilde{R}(\theta)$, which leaves invariant the standard vector $k^{\mu} = (k, |\mathbf k|)$. It is Euclide group $E(2)$.

The problem with 4-potential

It can be shown that the relativistic field which represents unitary irreducible representations of Poincare group with zero mass and helicity $\sigma = \frac{1}{2}(A - B)$ obeys equations (here $(...)$ means complete symmetrization) $$\tag 2 \begin{cases} \hat{\psi}_{a_{1}...a_{A}\dot{b}_{1}...\dot{b}_{B}} = \hat{\psi}_{(a_{1}...a_{A})(\dot{b}_{1}...\dot{b}_{B})},\\ \partial^{c \dot{c}}\hat{\psi}_{ca_{2}...a_{A}\dot{b}_{1}...\dot{b}_{B}} = 0, \\ \partial^{c \dot{c}}\hat{\psi}_{a_{1}a_{2}...a_{A}\dot{c}...\dot{b}_{B}} = 0 \end{cases}$$ Here, not strictly speaking:

the first condition defines $\hat{\psi}_{a...\dot{b}...}$ as irreducible representation $\left( \frac{A}{2}, \frac{B}{2}\right)$ of the Lorentz group, $a, \dot{b}$ are spinor indices;

the second and third conditions contains information about masslessness of the representation and the number of independent components (which is equal to one).

If we want to describe state with both helicities $+\sigma , -\sigma$, we need to take the direct sum of representations $\left( \frac{A}{2}, \frac{B}{2}\right) \oplus \left( \frac{B}{2}, \frac{A}{2}\right)$, which obey $(2)$.

What is photon? It is massless irreducible representation of the Poincare group with the helicity $\pm 1$. It can be shown by direct computations and the relation between spinor and vector representations of the Lorentz group, that for $\sigma = \pm 1$ Eqs. $(2)$ are equal to free Maxwell equations on EM strength tensor $F_{\mu \nu} \in \left( 0, 1\right) \oplus \left( 1, 0\right)$: $$\begin{cases} \partial_{\mu}F^{\mu \nu} = 0, \\ F_{\mu \nu} = -F_{\nu \mu} , \\ \epsilon_{\mu \nu \alpha \beta}\partial^{\nu}F^{\alpha \beta} = 0\end{cases}$$ But due to the experimental face of inversed square law of static EM interaction we need to introduce 4-potential $A_{\mu}$, which is defined as $$F_{\mu \nu} \equiv \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$ The full action of EM field and matter now depends not only on $F$ but on $A$ too: $$S = S[A, F]$$ In fact, $A_{\mu} \in \left(\frac{1}{2}, \frac{1}{2} \right)$. Could it be both irreducible representation of the Poincare group with zero mass and helicity $1$, and to be Lorentz invariant? The answer is no. Let's use Eqs. $(1)$: we have that (D_{ll{'}}(R) = R_{\mu \nu} for the vector representations of the Lorentz group) $$\tag 3 D^{\mu}_{\ \nu}(R(\theta , \alpha , \beta))e^{\nu}(\mathbf k , \pm 1) = e^{\mp i\theta}\left( e^{\mu}(\mathbf k , \pm 1) + \frac{(\alpha \pm i\beta)}{2\sqrt{|\mathbf k|}}k^{\mu}\right)$$ Here $e^{\mu}(\mathbf p , \sigma)$ is the polarization vector: $$A_{\mu} = \sum_{\sigma = \pm 1}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2\epsilon_{\mathbf p}}}\left(\hat{a}(\mathbf p , \sigma)e^{-ipx}e_{\mu}(\mathbf p ,\sigma) + \hat{a}^{\dagger}(\mathbf p ,\sigma)e^{ipx}e^{*}_{\mu}(\mathbf p ,\sigma) \right)$$ The second summand of Eq. $(3)$ breaks the Lorenz invariance. In fact, however, we obtain $(3)$ by the requirement that 4-potential $A_{\mu}$ represent massless particles with helicity $1$.

This implies the statement: and object $A_{\mu}$ can't both be Lorentz 4-vector and represent massless particles with helicity $1$.