# Systematic derivation of representations of the polarization tensors for particles with different spin (from Clebsch-Gordan coefficients)?

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For a spin-1 particle with 4-momentum $(p = (E,0,0,p))$, the polarization tensor can be represented as$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0)$

$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0,\frac{E}{m})$

$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0)$

and for a spin-2 particle the polarization tensor can be represented as

$\varepsilon_{\mu\nu}(J_z = +2) = \varepsilon_{\mu}(+1)\varepsilon_{\nu}(+1)$

$\varepsilon_{\mu\nu}(J_z = +1) = \frac{1}{\sqrt{2}}\varepsilon_{\mu}(+1)\varepsilon_{\nu}(0) + \frac{1}{\sqrt{2}}\varepsilon_{\mu}(0)\varepsilon_{\nu}(+1)$

$\varepsilon_{\mu\nu}(J = 0) = \frac{1}{\sqrt{6}}[\varepsilon_{\mu}(+1)\varepsilon_{\nu}(-1) + \varepsilon_{\mu}(-1)\varepsilon_{\nu}(+1)] + \sqrt{\frac{2}{3}}\varepsilon_{\mu}(0)\varepsilon_{\nu}(0)$

$\varepsilon_{\mu\nu}(J_z = -1) = \frac{1}{\sqrt{2}}\varepsilon_{\mu}(-1)\varepsilon_{\nu}(0) + \frac{1}{\sqrt{2}}\varepsilon_{\mu}(0)\varepsilon_{\nu}(-1)$

$\varepsilon_{\mu\nu}(J_z = -2) = \varepsilon_{\mu}(-1)\varepsilon_{\nu}(-1)$

Is there a systematic approach or procedure to build or find such representations of the polarization tensors for particles with different spins (in the script I am reading they fall down from heaven)? The script also says that for higher-spin particles, the polarization tensors for integer spin particles can be constracted from the Clebsch-Gordan coefficients. How and why does this work?

And what about polarization tensor representations for fermions (half-integer spin) particles?

Hasn't Wigner Precisely done this , "On Unitary Representations of the Inhomogeneous Lorentz Group"

I'll write an answer if I get some free time. Roughly speaking, it's the same as in quantum mechanics where you use ladder operators to find spin eigenstates, though there are some small subtleties in the polarization tensor case. The Fermion case is more or less the same, except it's not called tensors but spinors. A comprehensive explanation is embedded in Weinberg chap 5.

Section IV of Weinberg's paper on massless fields constructs the free massless fields for general spin $j$. For integral $j$, he gives formulas for the $\epsilon$ in Section II of this paper. By taking an additional spinor factor, one can get in a similar way the half-integral case, too.

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Let's first focus on the spin-1 case. In the context of QFT, these tensors should be understood as the coefficient functions in front of the creation and annihilation operators, that is

$v^\mu(x)=\sum\limits_\sigma\int\frac{d^3p}{\sqrt{2p^0}}[\varepsilon^\mu(\mathbf{p},\sigma)a(\mathbf{p},\sigma)e^{ip\cdot x}+\varepsilon^{\mu*}(\mathbf{p},\sigma)a^\dagger(\mathbf{p},\sigma)e^{-ip\cdot x}]\cdots(1),$

where I've written a charge-neutral field for simplicity(and without loss of generality when we are only concerned with transformation properties).  The question is what the form of $\varepsilon^\mu(\mathbf{p},\sigma)$  is, so that $v^\mu (x)$ really transforms like a vector field. Let me explain what it means.

Following Wigner's classification scheme, $a^\dagger(\mathbf{p},\sigma)$'s create one-particle states that transform unitarily and irreducibly under Poincare transformations, and in particular the subgroup $SO(3)$. Our $\sigma$ labels the degrees of freedoms of the representation restricted to the subgroup $SO(3)$, and we can choose it to label the angular momenta in the $z$ direction, that is, the values of $J_z$. Let's denote $U(\Lambda, a)$ a representation of the Poincare group , where $(\Lambda, a)$ is any element of the Poincare group. Once we fix a specific representation,  the transformation rules on the creation(annihilation) operators are fixed, i.e.,  the exact form of $U(\Lambda, a)a^\dagger (\mathbf{p},\sigma)U^{-1}(\Lambda, a)$ is fixed, but I won't write it down explicitly since it'll only distract us from the logical development of this answer.

Now, $v^\mu(x)$ must transform like a vector field, that is,

$U(\Lambda, a)v^\mu(x)U(\Lambda, a)=\Lambda^\mu_{\ \ \nu} v^\nu(\Lambda x+a).$

The above equation, together with the transformation rules of creation/annihilation operators, put a severe constraint on the form of $\varepsilon^\mu$. In fact, $\varepsilon^\mu$ is uniquely determined up to an equivalence. I'll refer you to Weinberg Chap 5 for mathematical details, but here is the result:

$\varepsilon^\mu(\mathbf{p},\sigma)=L^\mu_{\ \ \nu}(\mathbf{p})\varepsilon^\nu(0,\sigma)\cdots(2),$

$\sum\limits_\bar{\sigma}\varepsilon^\mu(0,\bar{\sigma})\mathbf{J}_{\bar{\sigma}\sigma}=\mathbf{\mathcal{J}}^\mu_{\ \ \ \nu}\varepsilon^\nu(0,\sigma)\cdots(3)$

where $L^\mu_{\ \ \nu}(\mathbf{p})$ is the boost that boosts a particle at rest to the same particle with momentum $\mathbf{p}$; $\mathbf{J}$'s are the three rotational generators(angular momenta operators) of the representation $U(\Lambda, a)$(technically it also has to be restricted to the space that carries only the discrete index $\sigma$), which are the irreducible spin-1 matrices by assumption(hence $J$'s are $3\times 3$ matrices); $\mathbf{\mathcal{J}}$'s are the three rotational generators of $(\Lambda, a)$(hence $\mathcal{J}$'s are $4 \times 4$). It's clear from equation (2) that if we know the form of $\varepsilon^\nu(0,\sigma)$ we will know all of $\varepsilon^\nu(\mathbf{p},\sigma)$. Now I claim (3) uniquely determines the form of $\varepsilon^\mu(0,\sigma)$(up to equivalence). You can solve it by brute force, but here's a perhaps smarter argument:

The form of equation (3) makes Schur's lemma for Lie algebra readily applicable(think of $\varepsilon^\mu(0,\sigma)$ as a $4 \times 3$ matrix carrying indices $\mu$ and $\sigma$), which claims $\mathbf{\mathcal{J}}$ must have an irreducible block that is equivalent to $\mathbf{J}$, and the blocks of $\mathbf{\mathcal{J}}$  that are not equivalent to $\mathbf{J}$ can only be connected to $\mathbf{J}$ by a zero matrix; in other words, $\varepsilon^\mu(0,\sigma)$ must be zero at the entries that correspond to the blocks that are not equivalent to $\mathbf{J}$, and can only have non-zero entries otherwise.

We know very well that $\mathbf{\mathcal{J}}$ has two irreducible blocks: the 1-dimensional one that acts on the 0th component(rotations leave energy invariant), and the 3-dimensional one that acts on the 3-vector components(rotations transform 3-vectors to 3-vectors). Only the latter is equivalent to the spin-1 representation of $\mathbf{J}$, hence by the previous reasoning, $\varepsilon^\mu(0,\sigma)$  can only have nonzero entries in the 3-vector part. (In fact, the arguments we have developed by now can be used to show that a vector field can only describe spin-1 or spin-0 particles.)

We adopt the most commonly used matrix representation of $\mathbf{J}$ and $\mathbf{\mathcal{J}}$:

$J_z=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&-1\end{bmatrix},$

$J_+=J_x+iJ_y=\begin{bmatrix} 0&\sqrt{2}&0\\0&0&\sqrt{2}\\0&0&0\end{bmatrix},$

$J_-=J_x-iJ_y=\begin{bmatrix} 0&0&0\\ \sqrt{2}&0&0\\0&\sqrt{2}&0\end{bmatrix},$

and $(\mathcal{J}_k)^i_j=-i\epsilon_{ijk}$, and we only write the 3-vector indices for $\mathcal{J}$ since we just argued 3-vector indices are the only relevant part.  These are all we need: to get $\varepsilon^\mu(0,0)=(0, 0, 0, 1)$ just specialize equation (3) to the case

$\sum\limits_\bar{\sigma}\varepsilon^\mu(0,\bar{\sigma})(J_z)_{\bar{\sigma}0}=(\mathcal{J}_z)^\mu_{\ \ \ \nu}\varepsilon^\nu(0,0),$

and to get $\varepsilon^\mu(0,\pm 1)=\frac{1}{\sqrt{2}}(0, \mp 1, -i, 0)$, just specialize to the cases

$\sum\limits_\bar{\sigma}\varepsilon^\mu(0,\bar{\sigma})(J_\pm)_{\bar{\sigma}\ \pm1}=(\mathcal{J}_\pm)^\mu_{\ \ \ \nu}\varepsilon^\nu(0,\pm 1).$

You said you wanted the form of the polarization tensors for a particle traveling along the z-axis with a momentum $p$, and now it's clear you just need to apply our equation (2), with a boost matrix $L^{\mu}_{\ \ \nu}$ boosting in the z direction, and you easily get

$\varepsilon^\mu(p,0)=(\frac{p}{m}, 0, 0, -\frac{E}{m}),$

$\varepsilon^\mu(p,\pm 1)=\frac{1}{\sqrt{2}}(0, \mp 1, -i, 0),$

which is exactly the result you wanted with indices raised.

Spin-2 story is the same, but now that we have the spin-1 results at hand, we don't have to start from scratch but just need to construct the spin-2 irreducible part of the tensor product of spin-1 tensors, and the math is covered in any QM textbook under the title of "Addition of angular momenta", and this is where Clebsch-Gordan coefficients come in. Fermion story is again more or less the same, but you need to start the procedure all over again since it's a spinor not a tensor.

1. For a comprehensive explanation see Weinberg Chap 5, but note Weinberg used 1,2,3,0 ordering of the 4-vector indices, where here OP and I used 0,1,2,3.

2. The first half of this answer is really showing how the polarization tensors, which are finite-dimensional, are related to the spin of particles that live in an infinite-dimensional Hilbert space.

answered Apr 12, 2015 by (2,640 points)
edited Apr 13, 2015

You use fat J and curly J - but they seem to be the same!?

@ArnoldNeumaier, $J$ comes from $U(\Lambda, a)$ restricted to the little group SO(3)(technically it also has to be restricted to the space that carries only the discrete index $\sigma$), which is a bona fide 3 by 3 matrix for spin-1, whereas $\mathcal{J}$ are just the rotational generators of the Poincare group, which are really 4 by 4 matrices, but I've only written it's 3 by 3 block since this block is the only relevant part. And in fact, although $J$ is isomorphic to the 3 by 3 block of $\mathcal{J}$, by convention we've chosen different matrices to represent them, as you can easily see from the definition of matrix entries I gave to these two J's.
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