# Transformation of photons under Lorentz transformation

This question is a continuation of one of my earlier post. In this post,I asked about the transformation of photon fields under rotation. Here I generalize the question to Lorentz transformation, and the motivate the reason behind asking the question.

Under Lorenz transformation, a fermion field $\psi(x)$ transforms as $$\psi^\prime(x^\prime)=\exp[-\frac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu}]\psi(x).$$ From this we can read off the Lorentz generators (both rotation and boost generators) $$J_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu)+\frac{1}{2}\sigma_{\mu\nu}$$ Using this we can compute $W_\mu W^\mu$ to show that $$W_\mu W^\mu=-m^2\frac{1}{2}(\frac{1}{2}+1)$$ $W_\mu$ is the Pauli-Lubansky 4-vector.

How does a photon field (which is a vector field) transform under Lorentz transformation? Actually I want an expression such the above transformation relating $$A_\mu^\prime(x^\prime)=(...)_\mu^\nu A_\nu$$ so that I can read off the generators and compute $W_\mu W^\mu$.

There are other ways of computing $W_\mu$ for massless particles but I think if the transformation of the photon field can be written down, then commuting $W_\mu W^\mu$ will become more illuminating. Can anyone help me with such an expression?

EDIT: Since, $A_\mu$ is a vector field, my guess would be that it transforms like $x_\mu$, i.e., $$A_\mu^\prime=\Lambda_\mu^\nu A_\nu.$$ If this be the case, the generators are same as those for $x_\mu$. My suspiction is that even if the rotation generators for photons are $J_i$, they cannot satisfy SU(2) algebra because if it were then photons must have transformed like representation of SU(2). How does the masslessness of photon will make a change in the commutation relation between different $J_i$ different from SU(2) algebra?

• Perhaps I have missed something, but does not a photon always travel at speed $c$ in all Lorentz frames? Meaning, is not the photon's momentum frame-invariant? Or are you asking about the electromagnetic field tensor $F^{\nu \lambda} = \partial^{\nu} A^{\lambda} - \partial^{\lambda} A^{\nu}$? – honeste_vivere Jan 19 '16 at 13:16
• @honeste_vivere the magnitude of the momentum is invariant ($p^2=0$ in any frame). But the momentum itself is not invariant: its direction changes if you rotate the frame. – AccidentalFourierTransform Jan 19 '16 at 15:15
• @AccidentalFourierTransform - Ah, I see what the OP was looking for... Thanks for the clarification. – honeste_vivere Jan 19 '16 at 15:27

The electromagnetic potential $A^\mu$ is a four-vector, and hence transforms in the fundamental representation of $\mathrm{SO}(1,3)$, i.e. $A^\mu\mapsto \Lambda^\mu_\nu A^\nu$ where $\Lambda$ is the usual 4x4 matrix associated to a Lorentz transformation.

Your question seems fundamentally confused about the difference between the field and the particle. The field transforms as an ordinary four-vector. A particle does not. This is because you have to differentiate between two different kinds of group representations here:

1. Each (classical) field $\phi: \mathbb{R}^4\to V$ has some target space $V$, where $V$ is a finite-dimensional representation of the relevant symmetry groups. In the case of the electromagnetic potential, $V$ is the fundamental representation - the "vector representation" - of the Lorentz group $\mathrm{SO}(1,3)$.

2. Upon the Hilbert space of states of the quantum theory, there is another, unitary representation. The unitary representations of the Poincare group are necessarily infinite-dimensional (the Lorentz group has no finite-dimensional unitary representations), and the possible infinite-dimensional representations correspond to particles by Wigner's classification.

The finite-dimensional representation $\rho_\text{fin}$ and the unitary representation $\rho_\text{U}$ are related by one of the Wightman axioms: $$\rho_\text{fin}(\Lambda)\phi = \rho_\text{U}(\Lambda)\phi\rho_\text{U}(\Lambda)^\dagger\tag{1}$$ where on the l.h.s. the $\phi$ is acted on as the classical vector it was, and on the r.h.s. it is acted on as the operator-valued field upon an infinite-dimensional Hilbert space it is in the quantum theory. This relation essentially is there to ensure that the transformation of the field as an operator upon the space of states of the quantum theory is consistent with its classical transformation behaviour.

The masslessness of the photon does not reflect in any change of the commutation relations of the Lorentz algebra (and I don't know why you would think it would). What the masslessness does is change the little group (see surface of transitivity or e.g. these notes), which is the group under which the four-momentum of a particle is invariant. For massive particles, the little group is $\mathrm{SU}(2)$, but for massless ones, it is the two-dimensional Euclidean group $\mathrm{E}(2)$ (or $\mathrm{ISO}(2)$).

Different little groups classify different infinte-dimensional unitary representations, and indeed, the rather unusual little group $\mathrm{E}(2)$ of the photon leads to the standard generators of $\mathfrak{so}(1,3)$ acting rather unusually upon a photon state. In particular, since $\mathrm{SU}(2)$ is not the little group, you cannot expect that rotations act in the same manner as for massive particles.

If you are interested in a detailed derivation of the structure of the little group and its allowed projective representations, take a look at e.g. the second half of chapter 2 of The quantum theory of fields, Vol. I by Weinberg.

There is a subtle issue with the relation $(1)$ for massless fields. As can be shown (cf. again Weinberg), a vector field constructed out of creation/annihilation operators for a massless particle can only obey a modified version $$\rho_\text{U}(\Lambda)\phi\rho_\text{U}(\Lambda)^\dagger = \rho_\text{fin}(\Lambda)\phi + \mathrm{d}\Omega\tag{2}$$ for a function of spacetime $\Omega$. Therefore, to make the relation $(1)$ hold on the quantum space of states, we must demand that $A\mapsto A + \mathrm{d}\Omega$ is a gauge symmetry of the theory, which must be quotiented out of the naive space of state upon which $(2)$ holds to obtain the actual space of physical states. Taking the quotient by identifying gauge-related states (and operators), $(2)$ becomes $(1)$ on the actual space of states (since $A+\mathrm{d}\Omega = A$ after taking the quotient). This shows that every massless vector field must be a gauge field in quantum field theory.

• I don't have my copy of Weinberg I right now, but I think I remember that he discusses that the field for massless particles don't in general transform like $A\to\Lambda A$, but something like $A\to\Lambda A+\mathrm d\Omega$ for some $\Omega$. If we want $A$ to transform like a vector, we must identify $A$ and $A+\mathrm d\Omega$, which means massless fields must be gauge fields. If $A$ were massive, it would be true that $A\to\Lambda A$; but as the photon is massless, the transformation law is not $A\to\Lambda A$ (I don't remember the details of this, something about the little group...) – AccidentalFourierTransform Jan 19 '16 at 15:13
• @AccidentalFourierTransform: Indeed, Weinberg proves that a vector field constructed from the c/a operators of massless particles must be a gauge field (chapters 5.9 and 8). However, Weinberg's approach to constructing quantum fields is idiosyncratic - he constructs them from the c/a operators of particles. In this answer, $A^\mu$ is just the electromagnetic four-potential, and it is getting quantized. If you try to decompose it into Fourier modes and interpret the modes as c/a operators, you'll run into trouble. The solution is to implement the BRST procedure, not to modify the field trafo. – ACuriousMind Jan 19 '16 at 15:26
• Fair enough [I must admit that I really like Weinberg's approach :) ]. I just wanted to point out that in some cases the transformation for $A$ is not so trivial, but you're right this is specific to W's approach. – AccidentalFourierTransform Jan 19 '16 at 15:34
• @Acuriousmind Do you mean a classical massless field, like the electromagnetic field, being a vector field, transforms the same way as a classical W boson or Z boson fiels? Will there be no complication due to the masslessness of the photon field at the classical level? – SRS Jan 19 '16 at 17:15
• @SRS: I told you the "photon field" transforms in the usual four-vector representation. So instead of $\sigma_{\mu\nu}$, you just use the usual generators of $\mathfrak{so}(1,3)$ itself, which are $M_{\mu\nu}$ with entries $(M_{\mu\nu})_{\sigma\rho} = \mathrm{i}(\eta_{\mu\sigma}\eta_{\nu\rho} - \eta_{\mu\rho}\eta_{\nu\sigma})$. – ACuriousMind Jan 19 '16 at 19:46