Basic Introduction to Higher-Spin Theories

Abstract

This is a collection of my lecture notes on the higher-spin theory course given for students at the Institute for Theoretical and Mathematical Physics, Lomonosov Moscow State University. The goal of these lectures is to give an introduction to higher-spin theories accessible to master level students which would enable them to read the higher-spin literature. I start by introducing basic relevant notions of representation theory and the associated field-theoretic descriptions. Focusing on massless symmetric fields I review different approaches to interactions as well as the no-go results. I end the lectures by reviewing some of the currently available positive results on interactions of massless higher-spin fields, namely, holographic, Chern-Simons and chiral higher-spin theories.

Appendices

A Conventions

We use the mostly plus convention $\eta = \textrm{diag}(-,+,\dots ,+)$.

To deal with symmetric tensors we use the following notation

$$\begin{aligned} \varphi ^{a(s)}=\varphi ^{a_1\dots a_s}. \end{aligned}$$

(A.1)

Moreover, we will use identical indices to indicate that indices have to be symmetrized. Symmetrization is defined with the normalization that makes it a projector. For example,

$$\begin{aligned} \partial ^a\varphi ^{a(s)}=\frac{1}{s+1}\left( \partial ^{a_1}\varphi ^{a_2\dots a_{s+1}}+\dots + \partial ^{a_{s+1}}\varphi ^{a_1\dots a_{s}}\right) , \end{aligned}$$

(A.2)

where the r.h.s. contains $s+1$ non-trivial permutations of indices in the expression on the l.h.s. Alternatively, one can sum over all $(s+1)!$ permutations of indices and then the overall factor in front of the bracket would be $[(s+1)!]^{-1}$.

Levi-Civita Tensor

We will introduce the Levi-Civita tensor with lower indices so that

$$\begin{aligned} \epsilon _{1 \dots d}= 1. \end{aligned}$$

(A.3)

By raising indices we find the Levi-Civita tensor with upper indices

$$\begin{aligned} \epsilon ^{a_1 \dots a_d} = \eta ^{a_1b_1}\dots \eta ^{a_d b_d} \epsilon _{b_1\dots b_d}. \end{aligned}$$

(A.4)

It should be remarked that

$$\begin{aligned} \epsilon ^{1 \dots d} =\sigma , \qquad \sigma \equiv \textrm{det}[\eta ], \end{aligned}$$

(A.5)

so for the Minkowski metric $\sigma =-1$.

There are a couple of useful formulas that involve the Levi-Civita tensor

$$\begin{aligned} A_{a_1}{}^{b_1}\dots A_{a_d}{}^{b_d} \epsilon _{b_1\dots b_d}=\textrm{det}[A]\epsilon _{a_1\dots a_d} \end{aligned}$$

(A.6)

and

$$\begin{aligned} \epsilon ^{k[n]l[d-n]}\epsilon _{k[n]m[d-n]}=\sigma n!(d-n)!\delta ^{[l_1}_{m_1}\dots \delta ^{l_{d-n}]}_{m_{d-n}}, \end{aligned}$$

(A.7)

where, as usual, antisymmetrization is understood as a projection.

B The Coleman-Mandual Theorem

In this appendix we sketch the proof of the Coleman-Mandula theorem stated in Section 10.2. For a more rigorous discussion we refer the reader to the original paper [195] and to [7] for shortcuts and clarifications. Here we mostly follow [7].

B.1 Step 1: Generators that Commute with Translations

The proof goes in several steps. We will start by considering only those symmetry transformations $B_\alpha$ that commute with the translation generator

$$\begin{aligned}{}[B_\alpha ,P_\mu ]=0. \end{aligned}$$

(B.1)

It is convenient to choose the basis for the states in the theory so that they have definite momenta. Then, the action of B on single-particle states is of the form⁵⁰

$$\begin{aligned} B_{\alpha }|p\rangle ^m = (b_\alpha (p))^m{}_{m'} |p\rangle ^{m'}, \end{aligned}$$

(B.2)

where $m, m'$ label the states with given momentum p, which also includes spin labels. Note that according to our assumptions – more precisely, the assumption of having finitely many mass shells with the mass below any fixed value – the space of states with fixed p is finite-dimensional.

The argument then works differently for $B_{\alpha }$ that map into the pure trace $b_\alpha$ – that is those proportional to the unit matrix $\delta ^m{}_{m'}$ – and for those that map into traceless $b_{\alpha }$. We will first consider the pure trace part of b.

By one of our assumptions, B act on two-particle states by acting on each single-particle state separately. Together with (B.2), this gives

$$\begin{aligned} (b_\alpha (p,q))^m{}_{m',}{}^{n}{}_{n'}= (b_\alpha (p))^m{}_{m'} \delta ^n{}_{n'}+(b_\alpha (q))^n{}_{n'} \delta ^m{}_{m'}. \end{aligned}$$

(B.3)

Next, we consider a two-particle scattering process of particles with momenta p and q into $p'$ and $q'$,

$$\begin{aligned} S(p,m;q,n \rightarrow p',m';q',n')\equiv \delta ^d(p'+q'-p-q) S(p',q';p,q)^{m'n'}{}_{mn}, \end{aligned}$$

(B.4)

where $p^2 = (p')^2$ and $q^2 = (q')^2$. Invariance of the S-matrix with respect to B implies

$$\begin{aligned} b_\alpha (p',q')S(p',q';p,q)=S(p',q';p,q)b_{\alpha }(p,q). \end{aligned}$$

(B.5)

Assuming that S is non-vanishing, we multiply both sides of (B.5) with $S^{-1}$ and take the trace. Due to the cyclicity of the trace, we find

$$\begin{aligned} \textrm{tr}\, b_\alpha (p',q') = \textrm{tr}\, b_\alpha (p,q). \end{aligned}$$

(B.6)

The trace of the tensor product of two matrices is the product of their traces. Together with (B.3) and (B.6) this leads to

$$\begin{aligned} N(q^2)\textrm{tr}\, b_{\alpha }(p')+N(p^2)\textrm{tr}\, b_{\alpha }(q')=N(q^2)\textrm{tr}\, b_{\alpha }(p)+N(p^2)\textrm{tr}\, b_{\alpha }(q), \end{aligned}$$

(B.7)

where $N(q^2)$ and $N(p^2)$ result from taking the traces of the Kronecker delta’s and count the numbers of states on each mass shell. Equivalently, we have

$$\begin{aligned} \frac{\textrm{tr}\, b_{\alpha }(p')}{N(p^2)}+\frac{\textrm{tr}\, b_{\alpha }(q')}{N(q^2)}=\frac{\textrm{tr}\, b_{\alpha }(p)}{N(p^2)}+\frac{\textrm{tr}\, b_{\alpha }(q)}{N(q^2)}. \end{aligned}$$

(B.8)

Its general solution reads

$$\begin{aligned} \frac{\textrm{tr}\, b_\alpha (p)}{N(p^2)} = c^\mu _{\alpha } p_\mu +d_\alpha , \end{aligned}$$

(B.9)

where both c and d are p-independent. Indeed, the d-part on both sides of (B.8) cancels out identically, while the c-part cancels out due to momentum conservation. We, therefore, find that the pure trace part of $b_\alpha$ is either proportional to momenta or is an internal symmetry.

We now proceed with $B_{\alpha }$ that map into traceless $b_\alpha$, which we will denote

$$\begin{aligned} (b^{\sharp }_\alpha )^{n'}{}_n\equiv (b_\alpha )^{n'}{}_n - \frac{\textrm{tr}\, b_\alpha (p)}{N(p^2)}\delta ^{n'}{}_n. \end{aligned}$$

(B.10)

From (B.9) it follows that the generators represented by $b^{\sharp }_\alpha$ are

$$\begin{aligned} B^{\sharp }_{\alpha } \equiv B_\alpha - c^\mu _{\alpha } P_\mu -d_\alpha . \end{aligned}$$

(B.11)

On the two-particle states these act by

$$\begin{aligned} (b^{\sharp }_\alpha (p,q))^m{}_{m',}{}^{n}{}_{n'}= (b^{\sharp }_\alpha (p))^m{}_{m'} \delta ^n{}_{n'}+(b^{\sharp }_\alpha (q))^n{}_{n'} \delta ^m{}_{m'}. \end{aligned}$$

(B.12)

Our next goal is to show that $B_\alpha ^{\sharp } \rightarrow b^{\sharp }_\alpha (p,q)$ is an isomorphism of Lie algebras. To this end, we need to show that

$$\begin{aligned} l^\alpha b^{\sharp }_\alpha (p,q)=0 \quad \Rightarrow \quad l^\alpha B_\alpha ^{\sharp }=0, \end{aligned}$$

(B.13)

where $l^\alpha$ are some coefficients. Equation (B.13) implies that $B_\alpha ^{\sharp } \rightarrow b^{\sharp }_\alpha (p,q)$ is invertible, so it is an isomorphism.

Since $B^{\sharp }$ are symmetries of the S-matrix, we have

$$\begin{aligned} b^{\sharp }_\alpha (p',q')S(p',q';p,q)=S(p',q';p,q)b^{\sharp }_{\alpha }(p,q). \end{aligned}$$

(B.14)

Again, we assume that the S-matrix is non-vanishing. Then, (B.14) implies that $b^{\sharp }_\alpha (p',q')$ and $b^{\sharp }_\alpha (p,q)$ are related by the similarity transformation. This, in turn, implies that

$$\begin{aligned} l^\alpha b^{\sharp }_{\alpha }(p,q) = 0 \qquad \Rightarrow \qquad l^\alpha b^{\sharp }_{\alpha }(p',q') = 0. \end{aligned}$$

(B.15)

Considering (B.12) and that $b^{\sharp }_\alpha (p)$ are traceless, one finds

$$\begin{aligned} l^\alpha b^{\sharp }_{\alpha }(p',q') = 0 \qquad \Rightarrow \qquad l^\alpha b^{\sharp }_{\alpha }(p') = 0. \end{aligned}$$

(B.16)

We, thus, have shown that $l^\alpha b^{\sharp }_{\alpha }(p,q) = 0$ implies $l^\alpha b^{\sharp }_{\alpha }(p') = 0$, where $p'$ is constrained to be on the same mass shell as p and, moreover, there should exist $q'$ on the same mass shell with q, so that $p+q=p'+q'$. With some extra work, this limitation on $p'$ can be lifted, that is one can prove that

$$\begin{aligned} l^\alpha b^{\sharp }_{\alpha }(p,q) = 0 \qquad \Rightarrow \qquad l^\alpha b^{\sharp }_{\alpha }(k) = 0, \end{aligned}$$

(B.17)

where k is an arbitrary on-shell momentum. The fact that $l^\alpha b^{\sharp }_{\alpha }(k)$ vanishes for any k means that $l^\alpha B^{\sharp }_{\alpha }=0$. Thus, we managed to show (B.13), which implies that the Lie algebra generated by $B^{\sharp }_\alpha$ is isomorphic to its representation on two-particle states with the trace part removed, $b^{\sharp }_{\alpha }(p,q)$.

Next, we apply the standard theorem, see e.g. [7], which tells us that a Lie algebra of finite-dimensional Hermitian matrices – like $b^{\sharp }_{\alpha }(p,q)$ for fixed p and q – is at most the direct sum of a semi-simple Lie algebra and some number of U(1) Lie algebras. We will now explore the consequence of this theorem focusing on the semi-simple part. The associated symmetry generators will be denoted $B^{\flat }_\alpha$.

The Lorentz group acts on these generators in the standard way⁵¹

$$\begin{aligned} G(\Lambda ,0): \qquad B^{\flat }_{\alpha } \quad \rightarrow \quad B^{\flat }_{\alpha }(\Lambda )\equiv U(G(\Lambda ,0))B^{\flat }_{\alpha }U^{-1}(G(\Lambda ,0)). \end{aligned}$$

(B.18)

Since $B^{\flat }_{\alpha }$ commute with $P_\mu$ it follows that $B^{\flat }_{\alpha }(\Lambda )$ commute with $\Lambda _\mu {}^\nu P_\nu$. Considering that $\Lambda _\mu {}^\nu P_\nu$ is just a linear combination of translations, we conclude that $B^{\flat }_{\alpha }(\Lambda )$ commutes with $P_\mu$. This, in turn, entails that $B^{\flat }_{\alpha }(\Lambda )$ is some linear combination of $B^{\flat }_{\alpha }$

$$\begin{aligned} U(G(\Lambda ,0))B^{\flat }_{\alpha }U^{-1}(G(\Lambda ,0)) = D^{\beta }{}_{\alpha }(\Lambda )B^{\flat }_{\beta }. \end{aligned}$$

(B.19)

Therefore, $B^{\flat }_{\alpha }$ realise a representation of the Lorentz group. We would like to show that it is unitary.

To do that one notices that $B^{\flat }_{\alpha }(\Lambda )$ commute the same way as $B^{\flat }_{\alpha }$ – U and $U^{-1}$ factors cancel out. This means that the structure constants of the algebra generated by $B^{\flat }_{\alpha }$ are invariant under Lorentz transformations, that is

$$\begin{aligned} f^{\gamma }{}_{\alpha \beta } = D^{\alpha '}{}_\alpha (\Lambda )D^{\beta '}{}_\beta (\Lambda )D^{\gamma }{}_{\gamma '}(\Lambda )f^{\gamma '}{}_{\alpha '\beta '}. \end{aligned}$$

(B.20)

As a consequence, the Lie algebra metric

$$\begin{aligned} g_{\beta \delta }\equiv f^{\gamma }{}_{\alpha \beta }f^{\alpha }{}_{\gamma \delta } \end{aligned}$$

(B.21)

is also Lorentz invariant. Moreover, since the Lie algebra generated by $B^{\flat }_{\alpha }$ is semi-simple, metric (B.21) is positive-definite. Altogether, this implies that $B^{\flat }_{\alpha }$ realise a unitary finite-dimensional representation of the Lorentz group. As we mentioned in Section 2, for finite-dimensional representations, this is only possible if the representation carried by $B^{\flat }_{\alpha }$ is trivial. Thus, $B^{\flat }_{\alpha }$ generate internal symmetries.

With some extra arguments, one can show that the U(1) part of $B^{\sharp }_{\alpha }$ also commutes with the Lorentz algebra.

Summarising the results of the first part of the proof, we found that symmetries of the S-matrix that commute with momenta are either momenta – the first term on the right hand side of (B.9) – or internal symmetries – the second term in (B.9) and all generators $B^{\sharp }_{\alpha }$.

B.2 Step 2: Locality in Momentum Space

On the next step, we take up the possibility of symmetry generators that do not commute with translations. In general, the symmetry generator in the momentum basis reads

$$\begin{aligned} A_\alpha |p\rangle ^n =\int d^dp' \mathcal{A}_\alpha (p',p)^n{}_{n'}|p'\rangle ^{n'}. \end{aligned}$$

(B.22)

Since the kernel $\mathcal{A}$ maps physical states to physical states, it should vanish unless both p and $p'$ are on the mass shell. Our goal is to show that $\mathcal{A}$ vanishes for any $p\ne p'$.

To achieve this, one considers a generator

$$\begin{aligned} A_\alpha ^f = \int d^dx e^{iPx}A_\alpha e^{-iPx}f(x), \end{aligned}$$

(B.23)

where f is an arbitrary function. It is a symmetry generator as it is defined via a composition of symmetry generators P and $A_\alpha$. It is straightforward to show that $A^f$ acts on the single-particle states as

$$\begin{aligned} A^f_\alpha |p\rangle ^n =\int d^dp' \tilde{f}(p'-p)\mathcal{A}_\alpha (p',p)^n{}_{n'}|p'\rangle ^{n'}, \end{aligned}$$

(B.24)

where $\tilde{f}$ is the Fourier transform of f

$$\begin{aligned} \tilde{f}(p)\equiv \int d^d xe^{ixp}f(x). \end{aligned}$$

(B.25)

Next, we return to the analysis of the 2-to-2 scattering, $p+q=p'+q'$. Let $\Delta$ be such that $p+\Delta$ is still on-shell, while all $q+\Delta$, $p'+\Delta$ and $q'+\Delta$ are off-shell. Then, picking $\tilde{f}$ in (B.24) with the support in the vicinity of $\Delta$, we find that

$$\begin{aligned} A_\alpha ^f|q\rangle ^m=0 ,\qquad A_\alpha ^f|p'\rangle ^{n'}=0,\qquad A_\alpha ^f|q'\rangle ^{m'}=0. \end{aligned}$$

(B.26)

Indeed, outside the support of $\tilde{f}$, the $\tilde{f}$ factor vanishes in (B.24), while inside the support of $\tilde{f}$, the $\mathcal{A}$ factor vanishes, as $\mathcal{A}$ only relates physical states. The condition of invariance of the S-matrix with respect to $A^f$ reads

$$\begin{aligned} \langle p',q'| S |A^f p,q\rangle + \langle p',q'| S | p,A^f q\rangle = \langle A^f p',q'| S | p,q\rangle + \langle p',A^f q'| S | p, q\rangle , \end{aligned}$$

(B.27)

where we again used our assumption about the action of symmetries on multi-particle states. Due to (B.26) this reduces to

$$\begin{aligned} \langle p',q'| S |A^f p,q\rangle =0. \end{aligned}$$

(B.28)

This can happen for two reasons: either S or $A^f$ is vanishing. By invoking some geometric considerations, it is not hard to see, that one can change on-shell p, q, $p'$ and $q'$, so that momentum conservation is still satisfied, moreover, $p+\Delta$ remains on-shell, while $q+\Delta$, $p'+\Delta$ and $q'+\Delta$ remain off-shell. In other words, the argument presented above holds in a certain continuous range of the Mandelstam variables. Keeping in mind our assumption that the S-matrix can only have isolated zeros, we conclude that (B.28) entails $A^f=0$. This, in turn, implies that

$$\begin{aligned} A(p',p)=0,\qquad \text {for} \qquad p-p'=\Delta . \end{aligned}$$

(B.29)

The same argument can be applied to other $\Delta$ that shift an on-shell momentum p to an on-shell momentum. Typically, one can choose the remaining three momenta so that the momentum conservation is still satisfied, while after a shift by $\Delta$ they all go off-shell. The value of $\Delta$ which is excluded by these arguments is $\Delta =0$ as, clearly, all on-shell states remain on-shell. Accordingly, we find that A is supported only on $p=p'$ or

$$\begin{aligned} A(p',p)=0, \qquad \forall \quad p'\ne p. \end{aligned}$$

(B.30)

B.3 Step 3: Constraining Derivatives in Momenta

With some technical assumptions the fact that the integral kernel $A(p,p')$ is only supported on $p=p'$ implies that it is given by $\delta ^d(p-p')$ and its derivatives of finite order. In other words, the action of A on one particle states is given by

$$\begin{aligned} A_\alpha |p\rangle ^n = \sum _{i=0}^k A_\alpha ^{(i)|\mu _1\dots \mu _i |n}{}_{n'}(p)\frac{\partial }{\partial p^{\mu _1}}\dots \frac{\partial }{\partial p^{\mu _i}}|p\rangle ^{n'}. \end{aligned}$$

(B.31)

The last step of the proof is to reduce the analysis of symmetries of this type to those that commute with momenta, discussed in the first part of the proof.

To achieve this one considers a k-fold commutator of (B.31) with momenta, which is also a symmetry of the S-matrix

$$\begin{aligned}{}[P^{\mu _1},[P^{\mu _2}, \dots [P^{\mu _k},A_\alpha ]\dots ]]= (-1)^k A_\alpha ^{(k)|\mu _1\dots \mu _k}(p), \end{aligned}$$

(B.32)

where A is symmetric in $\mu$ indices. It no longer contains derivatives of momenta, hence, it commutes with the translation generators. Therefore, the results of the first step of the proof can be applied and we have

$$\begin{aligned} A_\alpha ^{(k)|\mu _1\dots \mu _k|n}{}_{n'}(p) = b_\alpha ^{\mu _1\dots \mu _k|n}{}_{n'}+c_{\alpha }^{\mu | \mu _1\dots \mu _k}p_\mu \delta ^n{}_{n'}. \end{aligned}$$

(B.33)

Here c is just the pure trace c term from (B.9), while b combines p-independent internal traceless and pure-trace symmetries. Note that we dropped $N(p^2)$ from (B.9) for the pure trace part. This can be done for the following reason. First, by our assumption, there are finitely many mass shells in the system, so $p^2$ takes discrete eigenvalues. Moreover, as we showed, A acts locally in momentum space. Altogether, this implies that A acts within a single mass shell, so $N(p^2)$ can be replaced with a number $N(-m^2)$.

We will first focus on the case with $m^2\ne 0$ and take into account that A may only act within a single mass shell. In general, invariance of the mass shell $p^2+m^2=0$ with respect to transformation O implies

$$\begin{aligned} (P^2+m^2)O = O'(P^2+m^2), \end{aligned}$$

(B.34)

where $O'$ is an arbitrary operator. Condition (B.34) can be rewritten as

$$\begin{aligned}{}[P^2,O] = (O'-O)(P^2+m^2). \end{aligned}$$

(B.35)

We would like to apply this conclusion to O defined by

$$\begin{aligned} O=[P^{\mu _2},[P^{\mu _3}, \dots [P^{\mu _k},A_\alpha ]\dots ]], \qquad k\ge 1. \end{aligned}$$

(B.36)

By evaluating $[P^2,O]$, we find

$$\begin{aligned}{}[P^2,O]= (-1)^k 2p_{\mu _1}(b_\alpha ^{\mu _1\dots \mu _k}+c_{\alpha }^{\mu | \mu _1\dots \mu _k}p_\mu ). \end{aligned}$$

(B.37)

This should be compared with the admissible form for $[P^2,O]$ on the right-hand side of (B.35). We find that this requires

$$\begin{aligned} (-1)^k 2p_{\mu _1}(b_\alpha ^{\mu _1\dots \mu _k}+c_{\alpha }^{\mu | \mu _1\dots \mu _k}p_\mu )=0. \end{aligned}$$

(B.38)

This constraint is applicable for $k\ge 1$, because otherwise O does not exist, see (B.36).

Equation (B.38) should be satisfied for any p on the mass shell with $m^2>0$. This leads to

$$\begin{aligned} b_\alpha ^{\mu _1\dots \mu _k}=0, \qquad c_{\alpha }^{\mu | \mu _1\dots \mu _k}=-c_{\alpha }^{\mu _1| \mu \dots \mu _k}. \end{aligned}$$

(B.39)

The symmetry condition on c can be solved non-trivially only for $k=1$, see exercise 5 for $k=2$ case. We, thus, find the only non-trivial solution to be

$$\begin{aligned} c^{\mu |\mu _1}=-c^{\mu _1|\mu }, \end{aligned}$$

(B.40)

which generates Lorentz transformations.

In summary, we are left with the following possibilities for symmetries of the S-matrix in massive theories: for $k=1$ these may only contain Lorentz transformations, while for $k=0$ A’s commute with momenta and, as was shown on previous steps of the proof, may be either internal symmetries or momenta themselves. This finishes the proof of the Coleman-Mandula theorem for massive particles.

This argument can be naturally extended to include massless particles. For massless particles the right-hand side of (B.35) does not have the mass term, so we find

$$\begin{aligned} (-1)^k 2p_{\mu _1}(b_\alpha ^{\mu _1\dots \mu _k}+c_{\alpha }^{\mu | \mu _1\dots \mu _k}p_\mu ) = (O'-O)p^2. \end{aligned}$$

(B.41)

In addition to the solutions that we have already discussed in the massive case, (B.41) can be solved as

$$\begin{aligned} c_{\alpha }^{\mu | \mu _1\dots \mu _k} = \eta ^{\mu ( \mu _1} d_\alpha ^{\mu _2 \dots \mu _k)} \end{aligned}$$

(B.42)

for some d. Solutions to (B.42) with $k=1$ correspond to conformal symmetries. Solutions with $k\ge 2$ can be argued away, e.g. by noticing that the associated A under commutator generate an unbounded number of derivatives in p, which contradicts (B.31). Summarising, we find that for massless particles, the symmetry of the S-matrix may consist of a direct product of the conformal algebra and the algebra of internal symmetries.

C Helicity

For the 4d Poincare group it is conventional to introduce the Pauli-Lubanski pseudovector

$$\begin{aligned} W_{\mu }\equiv \frac{1}{2}\varepsilon _{\mu \nu \rho \sigma }J^{\nu \rho }P^\sigma . \end{aligned}$$

(C.1)

It is straightforward to compute that $[P,W]=0$, so one can pick a basis in the space of states so that both P and W take definite values. We will use this basis in the following.

Focusing on massless fields, we take the standard momentum as in (2.23). Then, the only non-vanishing components of P are $p_3=p_0$ and the only component of J, that is non-trivially realised is $J^{12}$. It then straightforward to see that the only non-vanishing components of W are given by

$$\begin{aligned} W_0 = J^{12}p_3, \qquad W_3 = J^{12}p_0. \end{aligned}$$

(C.2)

This implies that for the chosen basis, for the states with the standard momentum, $J^{12}$ also takes a definite value

$$\begin{aligned} J^{12}|p,\lambda \rangle = \lambda |p,\lambda \rangle . \end{aligned}$$

(C.3)

Equation (C.3) holds for the frame, in which momentum takes the standard form. To write it in the Lorentz-covariant form we note that (C.2) and (C.3) together entail

$$\begin{aligned} W_\mu |p,\lambda \rangle =\lambda P_\mu |p,\lambda \rangle . \end{aligned}$$

(C.4)

Since both $W_\mu$ and $P_\mu$ transform as vectors under parity-preserving Lorentz transformations, by keeping $\lambda$ Lorentz invariant, (C.4) takes a manifestly covariant form. Together with the Wigner approach of induced representations, this implies that $\lambda$ defined as the proportionality coefficient between W and P is the same for all the states in the representation and, hence, can be used to label different massless representations in the same way as spin does.

To understand the connection between helicity $\lambda$ and spin, let us return to representations of the Wigner little group. In a given case it is SO(2), which for the standard momentum is generated by $J^{12}$. In the main body of the text irreducible representations of SO(2) were given as traceless symmetric tensors of SO(2) with spin being the rank of a tensor. By counting the number of independent components of such a tensor, it is not hard to see that it is two for $s>0$ and one for $s=0$. This may seem to be in contradiction with (C.3), which suggests that a representation space of the Wigner little group is generated by a single vector $|p,\lambda \rangle$, so it is one-dimensional.

To clarify what actually happens, we consider a simple example of the SO(2) vector representation. In this case $J^{12}$ acts via a matrix

$$\begin{aligned} J_{12}= \left( \begin{array}{cc} 0 &{} i\\ -i &{} 0 \end{array} \right) . \end{aligned}$$

(C.5)

It is straightforward to find that its eigenvalues are $\lambda =\pm 1$ and the associated eigenvectors are

$$\begin{aligned} v_{\pm }=\left( \begin{array}{c} 1\\ \pm i \end{array}\right) . \end{aligned}$$

(C.6)

Obviously, these are not real, so there is no contradiction with irreducibility of the real vector representation of SO(2).

Still, when we are dealing with a real vector representation, it may be convenient to use basis (C.6) in which $J^{12}$ acts diagonally. At the same time, the coordinates of a vector in this basis should satisfy certain reality conditions to ensure that the associated vector is, indeed, real. The same refers to representations of the Poincare algebra obtained from these by the Wigner induced representation technique. A similar result holds for fields of any spin $s>0$: a symmetric traceless tensor of rank s has two eigenvectors with respect to $J^{12}$ with eigenvalues being $\pm s$ and both eigenvectors corresponding to complex tensors.

C.1 Helicity in the Light-Cone Gauge

With the necessary background reviewed, let us demonstrate that the representation given in (13.5)-(13.8), indeed, has helicity $\lambda$. To this end, we evaluate $W_0$ for the standard momentum, see (C.2). With some simple algebra one finds that only the spin part of J contributes, moreover,

$$\begin{aligned} S^{12} = -i S^{x\bar{x}}. \end{aligned}$$

(C.7)

Taking into account, in addition, the relative factors due to the definition of generators (12.40), we find that (13.8) leads to

$$\begin{aligned} W_0 = \lambda p^3 = \lambda p_0. \end{aligned}$$

(C.8)

Therefore, the proportionality coefficient between W and P is, indeed, given by $\lambda$.

Finally, let us mention the intuitive meaning of helicity. Remembering that only the spin part of J contributes to W, we have

$$\begin{aligned} W_0 = \frac{1}{2}\varepsilon _{0\nu \rho \sigma }S^{\nu \rho }P^\sigma = \lambda P_0. \end{aligned}$$

(C.9)

For the expression to be non-vanishing, indices $\nu$, $\rho$ and $\sigma$ may take only spatial values. Moreover, the Levi-Civita tensor reduces to the spatial one

$$\begin{aligned} \frac{1}{2}\varepsilon _{ij k}S^{ij}P^k = \lambda P_0. \end{aligned}$$

(C.10)

Then, helicity becomes

$$\begin{aligned} \lambda = \frac{S_k p^k}{p_0}, \qquad S_{k} \equiv \frac{1}{2}\varepsilon _{ij k}S^{ij}. \end{aligned}$$

(C.11)

Since $p_0=\pm \sqrt{p^k p_k}$, (C.11) implies that helicity, up to a sign, is a projection of spin on the spatial part of momentum.

D Fourier Transform for the Light-Cone Approach

In the light-cone deformation procedure it is convenient to make the Fourier transform with respect to spatial coordinates (13.48), which is then followed by the change of variables $p=iq$. For readers convenience, we present here some of the useful formulas in the Fourier transformed form.

In these terms the canonical commutator reads

$$\begin{aligned}{}[\Phi ^{\lambda _1}(q^\perp _1,x^+),\Phi ^{\lambda _2}(q^\perp _2,x^+)] = \frac{\delta ^{\lambda _1+\lambda _2,0}\delta ^3(q^\perp _1+q^\perp _2)}{\beta _1-\beta _2} \end{aligned}$$

(D.1)

and the Noether charges are

$$\begin{aligned} \nonumber P_2^i&= \sum _{\lambda }\int d^3q^\perp _1 d^3q^\perp _2 \delta ^3(q^\perp _1+q^\perp _2)\beta _1 \Phi ^{-\lambda }(q^\perp _1,x^+) p^i_2(q_2,\partial _2) \Phi ^\lambda (q^\perp _2,x^+),\\ J_2^{ij}&= \sum _{\lambda }\int d^3q^\perp _1 d^3q^\perp _2 \delta ^3(q^\perp _1+q^\perp _2)\beta _1 \Phi ^{-\lambda }(q^\perp _1,x^+) j^{ij}_2(q_2,\partial _2) \Phi ^\lambda (q^\perp _2,x^+), \end{aligned}$$

(D.2)

where

$$\begin{aligned} {\begin{matrix} p_2^+ &{}=q^+, \qquad p_2^- = -\frac{q \bar{q}}{\beta }, \qquad p_2 = q, \qquad \bar{p}_2= \bar{q},\\ j_2^{+-} &{}= \frac{\partial }{\partial \beta }\beta , \qquad j_2^{x\bar{x}} = N_q- N_{\bar{q}} -\lambda ,\\ j_2^{x+}&{} = -\beta \frac{\partial }{\partial \bar{q}}, \qquad j_2^{x-} = \frac{\partial }{\partial \bar{q}}\frac{q \bar{q}}{\beta } + q \frac{\partial }{\partial \beta }+\lambda \frac{q}{\beta },\\ j_2^{\bar{x}+}&{} = -\beta \frac{\partial }{\partial q},\qquad j_2^{\bar{x}-} = \frac{\partial }{\partial q}\frac{q \bar{q}}{\beta } + \bar{q} \frac{\partial }{\partial \beta } -\lambda \frac{\bar{q}}{\beta } \end{matrix}} \end{aligned}$$

(D.3)

and

$$\begin{aligned} N_q \equiv q\frac{\partial }{\partial q}, \qquad N_{\bar{q}} \equiv \bar{q}\frac{\partial }{\partial \bar{q}}. \end{aligned}$$

(D.4)