Massive N = 1 supermultiplets
with arbitrary superspins
Yu. M. Zinoviev ∗
Institute for High Energy Physics
Protvino, Moscow Region, 142280, Russia
Abstract
In this paper we give explicit construction of massive N = 1 supermultiplets in
flat d = 4 Minkowski space-time. We work in a component on-shell formalism based
on gauge invariant description of massive integer and half-integer spin particles where
massive supermultiplets are constructed out of appropriate set of massless ones.
∗E-mail address: yurii.zinoviev@ihep.ru
http://arxiv.org/abs/0704.1535v1
Introduction
In a flat space-time massive spin s particles in a massless limit decompose into massless spin
s, s− 1, . . . ones. This, in particular, leads to the possibility of gauge invariant description
of massive spin s particles, e.g. [1]-[13]. In this, two different approaches could be used.
From one hand, one can start with usual non gauge invariant description of massive particle
and achieve gauge invariance through the introduction of additional fields (thus promoting
second class constraints into the first class ones). From the other hand, one can start with
the appropriate set of massless particles having gauge invariance from the very beginning and
obtain massive particle description as a deformation of massless theory. This last approach
closely mimic situation in spontaneous gauge symmetry breaking where gauge field has to
eat some Goldstone field(s) to become massive.
In the supersymmetric theories all particles must belong to some supermultiplet, massive
or massless. Till now most of investigations in supersymmetric theories where bounded to
massless supermultiplets. Quite a few results on massive supermultiplets mainly devoted
to superspins 1 and 3/2 exist [14]-[21]. The aim of this paper is to extend these results to
include massive N = 1 supermultiplets with arbitrary superspins. Certainly, it would be
nice to have superfield off-shell description of such supermultiplets, but as previous results
clearly show it is a highly non-trivial task. So in this paper we restrict ourselves with
component on-shell formalism in terms of physical fields. The same reasoning on the massless
limit means that massive supermultiplets could (should) be constructed out of the massless
ones in the same way as massive particles out of the massless ones. So our approach will
be supersymmetric generalization of the second approach to massive particle description
mentioned above. Namely, we will start with appropriate set of massless supermultiplets
and obtain massive one as a smooth deformation.
The paper is organized as follows. Though our previous examples on massive superspin
1 [21] and superspin 3/2 [15] supermultiplets already give important hints on how general
case of arbitrary superspin could looks like, due to peculiarities of lower spin fields they are
not enough to achieve such generalization. Thus, in the first two sections we give two more
concrete examples, namely massive supermultiplets with superspin 2 and 5/2, correspond-
ingly. All these and subsequent results heavily depend on the gauge invariant description of
massive particles with integer [2, 3] and half-integer [10] spins as well as on the known form
of massless supermultiplets [22]. For reader convenience and to make paper self-contained,
in the next two sections we give all necessary formulas in compact condensed notations. One
of the lessons from previous investigations is that the structures of massive supermultiplets
with integer and half-integer superspins are different, so in the last two sections we consider
these two cases separately. We will see that, in spite of large number of fields, all calculations
are pretty straightforward and mainly combinatorical.
1 Superspin 2
Massive superspin 2 supermultiplet contains four massive particles with spins 5/2, 2, 2’ and
3/2, correspondingly. In the massless limit massive supermultilets must decompose into the
appropriate set of massless ones in the same way as massive spin s particles — into massless
spin s, s-1, ... ones. Simple counting of physical degrees of freedom immediately gives:
0, 0′
So we will start with five massless supermultiplets (Φµν , hµν), (fµν ,Ψµ), (Ωµ, Aµ), (Bµ, ψ)
and (χ, z). From our previous experience with massive superspin 1 and superspin 3/2 su-
permultiplets we know that it is crucial for the construction of massive supermultiplets to
make dual rotation of vector Aµ and axial-vector Bµ fields mixing massless supermultiplets
containing these fields. But now we have two tensor fields hµν and fµν as well, moreover they
necessarily must be tensor and pseudo-tensor ones. Thus we have to consider the possibility
to mix massless supermultiplets with these fields as well and the real structure of massless
supermultiplets we are going to work with looks like:
hµν fµν
Aµ Bµ
Then, introducing a sum of the massless Lagrangians for bosonic fields:
∂αhµν∂αhµν − (∂h)µ(∂h)µ + (∂h)µ∂µh−
∂µh∂µh+
∂αfµν∂αfµν − (∂f)µ(∂f)µ + (∂f)µ∂µf −
∂µf∂µf −
(∂µϕ)
(∂µπ)
2 (1)
as well as sum of the massless Lagrangians for fermionic fields:
Φ̄µν ∂̂Φµν − 2i(Φ̄γ)µ(∂Φ)µ + i(Φ̄γ)µ∂̂(γΦ)µ + i(Φ̄γ∂)Φ−
Φ̄∂̂Φ−
Ψ̄µ∂̂Ψµ + i(Ψ̄γ)(∂Ψ)−
(Ψ̄γ)∂̂(γΨ) +
ψ̄∂̂ψ −
Ω̄µ∂̂Ωµ + i(Ω̄γ)(∂Ω)−
(Ω̄γ)∂̂(γΩ) +
χ̄∂̂χ (2)
it is not hard to check that the most general supertransformations leaving sum of massless
Lagrangians invariant have the form (round brackets denote symmetrization):
δΦµν = −
σαβ∂α(cos(θ2)hβ(µγν) − sin(θ2)γ5fβ(µγν))η
δhµν =
2 cos(θ2)(Φ̄µνη) + i sin(θ2)(Ψ̄(µγν)η) (3)
δfµν =
2 sin(θ2)(Φ̄µνγ5η) + i cos(θ2)(Ψ̄(µγν)γ5η)
δΨµ = −σαβ∂α(sin(θ2)hβµ + cos(θ2)fβµγ5)η
for the supermultiplets containing spin 2 fields and
δΩµ = −
σαβ(cos(θ1)Aαβ − sin(θ1)γ5Bαβ)γµη
δAµ =
2 cos(θ1)(Ω̄µη) + i sin(θ1)(ψ̄γµη) (4)
δBµ =
2 sin(θ1)(Ω̄µγ5η) + i cos(θ1)(ψ̄γµγ5η)
δψ = −
σαβ(sin(θ1)Aαβ + cos(θ1)γ5Bαβ)η
for those with (axial)vector ones. The last supermultiplets is simple:
δχ = −iγµ∂µ(ϕ+ γ5π)η, δϕ = (χ̄η), δπ = (χ̄γ5η)
To construct massive supermultiplet we have to add mass terms for all fields as well as
appropriate corrections to fermionic supertransformations. In this, the most important ques-
tion is which lower spin fields play the roles of Goldstone ones and have to be eaten to make
main gauge fields massive. For the bosonic fields (taking into account parity conservation)
the choice is unambiguous: vector Aµ and scalar ϕ fields for tensor field hµν and axial-vector
Bµ and pseudo-scalar π — for pseudo-tensor fµν . Thus bosonic mass terms will be:
2[hµν∂µAν − h(∂A)] −
3Aµ∂µϕ+
2[fµν∂µBν − f(∂B)]−
3Bµ∂µπ
L2 = −
(hµνhµν − h2)−
hϕ+ ϕ2 −
(fµνhµν − f 2)−
fπ + π2 (5)
But for fermions we have two spin 3/2 and two spin 1/2 fields and there is no evident choice.
Thus we introduce the most general mass terms for the fermions:
Lm = −
Φ̄µνΦµν + (Φ̄γ)
µ(γΦ)µ +
−iα1[Φ̄µνγµΨν −
Φ̄(γΨ)]− iα2[Φ̄µνγµΩν −
Φ̄(γΩ)] +
+a1[Ψ̄
µΨµ − (Ψ̄γ)(γΨ)] + a2[Ω̄µΩµ − (Ω̄γ)(γΩ)] + a3[Ψ̄µΩµ − (Ψ̄γ)(γΩ)] +
+ia4(Ψ̄γ)ψ + ia5(Ψ̄γ)χ+ ia6(Ω̄γ)ψ + ia7(Ω̄γ)χ+
+a8ψ̄ψ + a9ψ̄χ+ a10χ̄χ (6)
and proceed with calculations. Cancellation of variations with one derivative gives:
sin(θ2) = cos(θ2) =
sin(θ1) = cos(θ1) =
, α2 =
a1 = −
, a3 = 1, a4 =
2, a5 = 0
a6 = a2
2, a7 =
3, a8 = 0, a9 = −
while variations without derivatives give:
, a10 = −
Resulting fermionic mass terms:
Lm = −
Φ̄µνΦµν + (Φ̄γ)
µ(γΦ)µ +
Φ̄µνγµΨν +
Φ̄(γΨ)− i
2Φ̄µνγµΩν +
Φ̄(γΩ) +
Ψ̄µΨµ +
(Ψ̄γ)(γΨ) +
Ω̄µΩµ −
(Ω̄γ)(γΩ) + Ψ̄µΩµ − (Ψ̄γ)(γΩ) +
2(Ψ̄γ)ψ +
(Ω̄γ)ψ + i
3(Ω̄γ)χ−
6ψ̄χ−
χ̄χ (7)
correspond to invariance of the Lagrangian (besides global supertransformations) under three
local spinor gauge transformations:
δΦµν = ∂(µξν) +
γ(µξν) +
gµνξ1 +
gµνξ2
δΨµ = ∂µξ1 +
γµξ1 +
δΩµ = ∂µξ2 +m
2ξµ +
γµξ1 +
γµξ2 (8)
δψ = m
2ξ1 +
ξ2 δχ = m
From these formula one can easily determine which combination of spin 3/2 fields Ψµ and
Ωµ plays the role of Goldstone field for spin 5/2 field Φµν . Indeed, if one introduces two
orthogonal combinations:
Ψ̃µ =
Ωµ, Ω̃µ = −
then after diagonalization of mass terms one finds that Ψ̃µ is a Goldstone field, while Ω̃µ —
physical field with the same mass as Φµν .
Similar to the case of massive supermultiplet with superspin 1, both mixing angles have
been fixed: θ1 = θ2 = π/4 and this, in turn, means that all bosonic fields enter through the
complex combinations only:
Hµν = hµν + γ5fµν , Cµ = Aµ + γ5Bµ, z = ϕ+ γ5π
Introducing gauge covariant derivatives:
∇µHαβ = ∂µHαβ −
Cµgαβ, ∇µz = ∂µz −m
we can write final form of fermionic supertransformations as:
δΦµν = [−
σαβ∇αH̄β(µγν) −mHµν +
γ(µ(γH)ν) +
gµνz]η
δΨµ = [−
σαβ∇αHβµ −
(γH)µ +
∇µz +
γµz]η
δΩµ = [−
σαβC̄αβγµ +
(γH)µ +
∇µz −
γµz]η (9)
δψ = −
σαβCαβη δχ = −iγµ∇µzη
Note also that due to complexification of bosonic fields the Lagrangian and supertrans-
formations are invariant under global axial U(1)A symmetry, axial charges for all fields being:
field η Φµν , Ψµ, Ωµ, ψ, χ Hµν , Cµ, z
qA +1 0 –1
2 Superspin 5/2
Our next example — massive supermultiplet with superpin 5/2. It also contains four massive
fields: with spin 3, 5/2, 5/2’ and 2 and in the massless limit it should reduce to six massless
supermultiplets:
5/2 5/2
0, 0′
By analogy with all previous cases we will take into account possible mixing for bosonic tensor
and vector fields, so we will start with the following structure of massless supermultiplets:
hµν fµν
Aµ Bµ
So we introduce sum of the massless Lagrangians for bosonic fields:
L0 = −
∂ρΦµνλ∂ρΦµνλ +
(∂Φ)µν(∂Φ)µν − 3(∂Φ)µν∂µΦν +
∂µΦν∂µΦν +
(∂Φ)2
∂αhµν∂αhµν − (∂h)µ(∂h)µ + (∂h)µ∂µh−
∂µh∂µh−
(∂µϕ)
∂αfµν∂αfµν − (∂f)µ(∂f)µ + (∂f)µ∂µf −
∂µf∂µf −
(∂µπ)
2 (10)
as well as sum of the massless Lagrangians for fermionic fields:
Ψ̄µν ∂̂Ψµν − 2i(Ψ̄γ)µ(∂Ψ)µ + i(Ψ̄γ)µ∂̂(γΨ)µ + i(Ψ̄γ∂)Ψ−
Ψ̄∂̂Ψ+
Ω̄µν ∂̂Ωµν − 2i(Ω̄γ)µ(∂Ω)µ + i(Ω̄γ)µ∂̂(γΩ)µ + i(Ω̄γ∂)Ω−
Ω̄∂̂Ω−
Ψ̄µ∂̂Ψµ + i(Ψ̄γ)(∂Ψ)−
(Ψ̄γ)∂̂(γΨ) +
ψ̄∂̂ψ −
Ω̄µ∂̂Ωµ + i(Ω̄γ)(∂Ω) −
(Ω̄γ)∂̂(γΩ) +
χ̄∂̂χ (11)
and start with the following global supertransformations:
δΦµνλ = i(Ψ̄(µνγλ)η) δΨµν = [−σαβ∂αΦβµν +
∂(µγν)(γΦ)]η (12)
for the supermultiplet (3, 5/2),
δΩµν = −
σαβ∂α(cos(θ2)hβ(µγν) − sin(θ2)γ5fβ(µγν))η
δhµν =
2 cos(θ2)(Ω̄µνη) + i sin(θ2)(Ψ̄(µγν)η) (13)
δfµν =
2 sin(θ2)(Ω̄µνγ5η) + i cos(θ2)(Ψ̄(µγν)γ5η)
δΨµ = −σαβ∂α(sin(θ2)hβµ + cos(θ2)fβµγ5)η
for the mixed (5/2, 2) and (2, 3/2) supermultiplets,
δΩµ = −
σαβ(cos(θ1)Aαβ − sin(θ1)γ5Bαβ)γµη
δAµ =
2 cos(θ1)(Ω̄µη) + i sin(θ1)(ψ̄γµη) (14)
δBµ =
2 sin(θ1)(Ω̄µγ5η) + i cos(θ1)(ψ̄γµγ5η)
δψ = −
σαβ(sin(θ1)Aαβ + cos(θ1)γ5Bαβ)η
for the mixed (3/2, 1) and (1, 1/2) supermultiplets and
δχ = −iγµ∂µ(ϕ+ γ5π)η, δϕ = (χ̄η), δπ = (χ̄γ5η)
for the last one.
By analogy with the superspin 3/2 case, we will assume that fermionic mass terms are
Dirac ones:
Lm = −Ψ̄µνΩµν + 2(Ψ̄γ)µ(γΩ)µ +
Ψ̄Ω +
[−Ψ̄µνγµΨν +
Ψ̄(γΨ)− Ω̄µνγµΩν +
Ω̄(γΩ)] +
Ψ̄µΩµ −
(Ψ̄γ)(γΩ) + 2i(Ψ̄γ)ψ + 2i(Ω̄γ)χ− 3ψ̄χ (15)
where all coefficients are completely fixed by the requirement that the Lagrangian has to be
invariant not only under the global supertransformations, but under four (by the number of
fermionic gauge fields) spinor gauge transformations:
δΨµν = ∂(µξν) +
γ(µην) +
gµνξ1, δΩµν = ∂(µην) +
γ(µξν) +
gµνξ2,
δΨµ = ∂µξ1 +m
ξµ + im
γµξ2, δΩµ = ∂µξ2 +m
ηµ + im
γµξ1,
δψ = 2mξ1, δχ = 2mξ2
As for the bosonic fields, here the roles of the fields are evident (again taking into account
parity conservation): we need tensor hµν , vector Aµ and scalar ϕ fields to make spin 3 field
Φµνλ massive, while pseudo-tensor fµν field needs to eat axial-vector Bµ and pseudo-scalar
π fields. So the bosonic mass terms are also completely fixed:
3[−Φµνλ∂µhνλ + 2Φµ(∂h)µ −
Φµ∂µh] +
5[hµν∂µAν − h(∂A)]−
6Aµ∂µϕ +
2[fµν∂µBν − f(∂B)]−
3Bµ∂µπ (16)
ΦµνλΦµνλ −
ΦµΦµ +
ΦµAµ −
(fµνfµν − f 2)−
fπ + π2 (17)
Now we require that the whole Lagrangian be invariant under global supertransformations
with appropriate corrections to fermionic transformations. This fixes both mixing angles:
sin(θ2) =
, cos(θ2) =
, sin(θ1) =
, cos(θ1) =
and gives the following form of additional terms for fermionic supertransformations:
δΨµν = [−
γ(µ(γh)ν) −
fµνγ5 +
γ(µ(γf)ν)γ5]η
δΩµν = i[(γΦ)µν +
gµν(γΦ)−
γ(µAν) +
gµνÂ−
γ(µBν)γ5 +
gµνB̂γ5]η
δΨµ = [
γµ(γΦ)−
γµÂ−
Bµγ5 −
γµB̂γ5]η (18)
δΩµ = i[
(γh)µ +
(γf)µγ5 − γµϕ− γµγ5π]η
δψ = [−ϕ− 2γ5π]η
δχ = i[
6Â +
3B̂γ5]η
The complete supertransformations for fermionic fields could be simplified by introduction
of gauge invariant derivatives;
∇µhαβ = ∂µhαβ −
Aµgαβ, ∇µϕ = ∂µϕ−m
∇µfαβ = ∂µfαβ −
Bµgαβ, ∇µπ = ∂µπ −m
This time bosonic fields do not combine into complex combinations, but due to the fact that
fermionic mass terms are Dirac ones the Lagrangian and supertransformations are invariant
under global axial U(1)A transformations, provided axial charges of all fields are assigned as
follows:
field η, Ψµν , Ψµ, ψ hµν , fµν , Aµ, Bµ, ϕ, π Ωµν , Ωµ, χ
qA +1 0 –1
3 Massive particles
All our previous and subsequent calculations heavily depend on the gauge invariant descrip-
tion of massive high spin particles. For reader convenience and to make paper self-contained
we will give here gauge invariant formulations for massive particles with arbitrary integer
[2, 3] and half-integer [10] spins. We restrict ourselves to flat d = 4 Minkowski space but all
results could be easily generalized to the case of (A)dS space with arbitrary dimension d.
3.1 Integer spin
The simplest way to describe massless bosonic field with arbitrary spin s is to use completely
symmetric rank s tensor Φ(α1α2...αs) which is double traceless. In what follows we will use
condensed notations where index denotes just number of free indices and not the indices
themselves. For example, the tensor field itself will be denoted as Φs, it’s contraction with
derivative as (∂Φ)s−1, it’s trace as Φ̃s−2 and so on. As we will see this does not lead to any
ambiguities then working with free Lagrangians quadratic in fields. In these notations the
Lagrangian for massless particles of arbitrary spin s could be written as:
L0 = (−1)s[
∂µΦs∂µΦ
(∂Φ)s−1(∂Φ)s−1 −
s(s− 1)
∂µΦ̃s−2∂µΦ̃
s−2 +
s(s− 1)
(∂Φ)s−1∂(1Φ̃s−2) −
s(s− 1)(s− 2)
(∂Φ̃)s−3(∂Φ̃)s−3] (19)
where
Φ = 0. This Lagrangian is invariant under the following gauge transformations:
δ0Φs = ∂(1ξs−1), ξ̃s−3 = 0
where parameter ξs−1 is completely symmetric traceless tensor of rank s− 1.
To construct gauge invariant Lagrangian for massive particle which has correct (i.e. with
right number of physical degrees of freedom) massless limit, we start with the sum of massless
Lagrangians with 0 ≤ k ≤ s:
(−1)k[
∂µΦk∂µΦ
(∂Φ)k−1(∂Φ)k−1 −
k(k − 1)
∂µΦ̃k−2∂µΦ̃
k(k − 1)
(∂Φ)k−1∂(1Φ̃k−2) −
k(k − 1)(k − 2)
(∂Φ̃)k−3(∂Φ̃)k−3] (20)
Then we add the following cross terms with one derivative as well as mass terms without
derivatives:
(−1)kak[(∂Φ)k−1Φk−1 + (k − 1)Φ̃k−2(∂Φ)k−2 +
(k − 1)(k − 2)
(∂Φ̃)k−3Φ̃k−3]
(−1)k[dkΦkΦk + ekΦ̃k−2Φ̃k−2 + fkΦ̃k−2Φk−2] (21)
and try to achieve gauge invariance with the help of appropriate corrections to gauge trans-
formations:
δΦk = αkξk + βkg(2ξk−2)
Straightforward but lengthy calculations give a number of algebraic equations on the un-
known coefficients which could be solved (and this is non-trivial because we obtain overde-
termined system of equations) and give us:
(s− k)(s+ k + 1)
2(k + 1)2
, βk+1 =
k + 1
αk, 0 ≤ k ≤ s− 1
ak = −
(s− k + 1)(s+ k)
, dk =
(s− k − 1)(s+ k + 2)
4(k + 1)
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
3.2 Half-integer spin
For the description of massless spin s+1/2 particles we will use completely symmetric rank
s tensor-spinor Ψs such that (γΨ̃)s−3 = 0 (in the same condensed notations as before). Then
Lagrangain for such field could be written as:
L0 = i(−1)s[
Ψ̄s∂̂Ψs − s(Ψ̄γ)s−1(∂Ψ)s−1 +
(Ψ̄γ)s−1∂̂(γΨ)s−1 +
s(s− 1)
(Ψ̄γ∂)s−2Ψ̃s−2 −
s(s− 1)
∂̂Ψ̃s−2] (22)
and is invariant under the following gauge transformations:
δ0Ψs = ∂(1ξs−1), (γξ) = 0,
where gauge parameter ξs−1 is a γ-traceless tensor-spinor of rank s− 1.
Once again we start with the sum of massless Lagrangians with 0 ≤ k ≤ s:
i(−1)k[
Ψ̄k∂̂Ψk − k(Ψ̄γ)k−1(∂Ψ)k−1 +
(Ψ̄γ)k−1∂̂(γΨ)k−1 +
k(k − 1)
(Ψ̄γ∂)k−2Ψ̃k−2 −
k(k − 1)
∂̂Ψ̃k−2] (23)
To combine all these massless fields into one massive particle we have to add the following
mass terms:
(−1)k
2(k + 1)
[Ψ̄kΨk − k(Ψ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2]+
−ick[(Ψ̄γ)k−1Ψk−1 −
k − 1
(γΨ)k−2]
and corresponding corrections to gauge transformations:
δΨk = αkξk + iβkγ(1ξk−1) + ρkg(2ξk−2)
Then total Lagrangian will be gauge invariant provided:
(s+ 1)2 − k2
, αk =
k + 1
, βk =
2k(k + 1)
, ρk =
4 Massless supermultiplets
It is not easy to find in the recent literature the explicit component form of massless su-
permultiplets with arbitrary superspin [22], so for completeness we will give their short
description here. As we have already seen on the lower superspin cases, supermultiplets
with integer and half-integer superspins have different structure and have to be considered
separately.
(s, s+1/2). Supermultiplet with integer superspin s contains bosonic spin s field and
fermionic spin s+1/2 one. In this and in two subsequent sections we will use the same con-
densed notations as in the previous one. By analogy with superspins 1 and 2 supermultiplets,
we start with the following ansatz for the supertransformations:
δΨs = iα1σ
µν∂µΦν(s−1γ1)η δΦs = β(Ψ̄sη)
Indeed, calculating variations of the sum of two massless Lagrangians one can see that most
of variations cancel, provided one set α1 = −β2 . The residue:
δL = −(−1)sβ
(s− 1)(s− 2)
[2(Ψ̄∂∂)s−2Φ̃s−2 − ˜̄Ψ
∂2Φ̃s−2 − (s− 2)( ˜̄Ψ∂)s−3(∂Φ̃)s−3 −
−2(Ψ̄γ∂)s−2∂̂Φ̃s−2 + 2(Ψ̄γ∂∂)s−3(γΦ̃)s−3 − ( ˜̄Ψ∂)s−3∂̂(γΦ̃)s−3]
contains terms with Φ̃s−2 only, so we proceed by adding to fermionic supertransformations
one more term:
δ′Ψs = iα2∂(1γ1Φ̃s−2)
Then the choice α2 =
(s−1)(s−2)β
leaves us with:
δL = −(−1)sβ
(s− 1)(s− 2)
[2(Ψ̄γ∂∂)s−3(γΦ̃)s−3 − ( ˜̄Ψ∂)s−3∂̂(γΦ̃)s−3]
where the only terms are whose with (γΦ̃)s−3. So we make one more (last) correction to
supertransformations:
δ′′Ψs = iα3g(2∂1(γΦ̃)s−3)
and obtain full invariance with α3 = − (s−1)(s−2)β4s . To fix concrete normalization we will use
closure of the superalgebra. Calculating the commutator of two supertransformations we
obtain:
[δ1, δ2]Φs = −iβ2(η̄2γνη1)∂µΦs + . . .
where dots mean “up to gauge transformation”. So we set β =
2 and our final result looks
like:
δΨs = −
σµν∂µΦ̄ν(s−1γ1)η +
i(s− 1)(s− 2)
[∂(1γ1Φ̃s−2) − g(2∂1(γΦ̃)s−3)]η
δΦs =
2(Ψ̄sη) (25)
(s+1/2, s+1). Half-integer superspin multiplet contains fermionic spin s + 1/2 fields
and bosonic spin s + 1 one. Again by analogy with lower superspin case we will make the
following ansatz for supertransformations:
δΨs = α1σ
µν∂µΦν(s)η, δΦs+1 = iβ(Ψ̄(sγ1)η)
This time most of the variations cancel if one set α1 = −β leaving us with:
δL = i(−1)sβ
s(s− 1)
[2(Ψ̄∂∂)s−2(γΦ̃)s−2 − ˜̄Ψ
∂2(γΦ̃)s−2 −
−2(Ψ̄γ∂)s−2∂̂(γΦ̃)s−2 − (s− 2)( ˜̄Ψ∂)s−3(γ∂Φ̃)s−3]
Then the full invariance could be achieved with the following correction to supertransforma-
tions:
δ′Ψs = α2∂(1γ1(γΦ̃)s−2)
provided α2 =
. To check the closure of superalgebra and to choose normalization we
calculate commutator of two supertransformations:
[δ1, δ2]Φs+1 = −2iβ2(η̄2γµη1)∂µΦs+1 + . . .
Then our choice will be β = 1 and our final form:
δΨs = −σµν∂µΦν(s)η +
∂(1γ1(γΦ̃)s−2)η, δΦs+1 = i(Ψ̄(sγ1)η)
Note that starting with superspin 2 the structure of supertransformations are defined up to
possible field dependent gauge transformations and our choice differs from that of [22]. It
makes no difference for massless theories but for massive case the structure of corrections
for fermionic supertransformations depends on the choice made.
5 Integer superspin
Now, having in our disposal gauge invariant description of massive particles with arbitrary
(half-)integer spins, known form of supertransformations for massless arbitrary superspin
supermultiplets and concrete examples of massive supermultiplets with lower superspins, we
are ready to construct massive arbitrary superspin supermultiplets. As we have seen, integer
and half-integer cases have different structures and have to be considered separately.
In this section we consider massive supermultiplet with integer superspin. Such super-
multiplet also contains four massive fields: two bosonic spin s fields (with opposite parity)
and fermionic spin (s+1/2) and (s-1/2) ones. Calculating total number of physical degrees
of freedom and taking into account possible mixing of supermultiplets containing bosonic
fields with equal spins and opposite parity, we start with the following structure of massless
supermultiplets:
As Bs
Ak Bk
By analogy with superspin 1 and 2 cases, we will assume that all bosonic fields enter through
the complex combinations Ck = Ak + iBk only (so that all possible mixing angles are fixed
and equal π/4). Thus we choose the following form of supertransformations for massless
supermultiplets with 1 ≤ k ≤ s:
δΦk = −
σµν∂µC̄ν(k−1γ1)η +
i(k − 1)(k − 2)
[∂(1γ1C̃k−2) − g(2∂1(γC̃)k−3)]η
δC̄k = 2(Φ̄kη) + i
2(Ψ̄(k−1γ1)η) (26)
δΨk−1 = −
σµν∂µCν(k−1)η +
k − 2
∂(1γ1(γC̃)k−3)η
and also
δΦ0 = −i∂̂zη, δz̄ = 2(Φ̄0η)
As a result of our assumption mass terms for bosonic fields are completely fixed:
(−1)kck[C̄k∂(1Ck−1) − (k − 1) ˜̄C
(∂C)k−2 +
(k − 1)(k − 2)
( ˜̄C
∂(1C̃k−3) + h.c.)]
(−1)k[dkC̄kCk + ek ˜̄C
C̃k−2 + fk(
Ck−2 + h.c.)] (27)
where
(s+ k)(s− k + 1)
, dk =
(s− k − 1)(s+ k + 2)
4(k + 1)
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
As for the fermionic mass terms, apriori we don’t have any restrictions on them so we have
to consider the most general possible form:
(−1)k
a1k[Φ̄
kΦk − k(Φ̄γ)k−1(γΦ)k−1 −
k(k − 1)
Φ̃k−2]+
+a2k[Φ̄
kΨk − k(Φ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2] +
+a3k[Ψ̄
kΨk − k(Ψ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2] +
+ib1k[(Φ̄γ)
k−1Φk−1 −
k − 1
(γΦ)k−2] +
+ib2k[(Φ̄γ)
k−1Ψk−1 −
k − 1
(γΨ)k−2] +
+ib3k[(Ψ̄γ)
k−1Φk−1 −
k − 1
(γΦ)k−2] +
+ ib4k[(Ψ̄γ)
k−1Ψk−1 −
k − 1
(γΨ)k−2]
Where:
a1s = −
, a2s = a3s = b3s = b4s = 0
The requirement that total Lagrangian be invariant under (appropriately corrected) super-
transformations gives:
a1k = −
, a2k = −
k + 1
ck+1, a3k =
2(k + 1)
b1k = −2ck, b2k = −
, b3k = 0, b4k = −2ck+1
In this, additional terms for fermionic supertransformations look like:
δ′Φk =
2ick+1
k + 1
[(γC)k +
k(k − 1)
4(k + 1)
γ(1C̃k−1) −
(k − 1)2(2k + 1)
8k(k + 1)
g(2(γC̃)k−2)]−
−Ck +
k − 1
γ(1(γC)k−1 −
(k − 1)(k − 2)
g(2C̃k−2) −
[γ(1Ck−1) − g(2(γC)k−2)] (29)
δ′Ψk =
kck+2
2(k + 1)
γ(1(γC̃)k−1) −
2(k + 1)
[(γC)k −
k(k − 1)
4(k + 1)
γ(1C̃k−1) +
(k − 1)(3k + 1)
8k(k + 1)
g(2(γC̃)k−2)]−
2ck+1
k + 1
[Ck +
γ(1(γC)k−1) +
(k − 1)(k − 2)
g(2C̃k−2)] (30)
Here the supertransformations for Φk field contain terms with Ck+1, Ck and Ck−1 fields in
the first, second and third lines correspondingly, while that of Ψk contain terms with Ck+2,
Ck+1 and Ck fields.
6 Half-integer superspin
Next we turn to the half-integer superspin case. This time we have two fermionic spin (s+1/2)
fields and bosonic ones with spins (s+1) and s. Usual reasoning on physical degrees of freedom
and possible mixings leads us to the following structure of massless supermultiplets we will
start with:
Φs Ψs
Ak Bk
We see that this structure is rather similar to that of integer superspin case. The main
difference (besides the presence of As+1, Ψs supermultiplet) comes from the mixing of bosonic
fields. We have no reasons to suggest that all mixing angles could be fixed from the very
beginning so we have to consider the most general possibility here. Let us denote:
Ck = cos(θk)Ak + γ5 sin(θk)Bk, Dk = sin(θk)Ak + γ5 cos(θk)Bk
In these notations supertransformations for massless supermultiplets could be written as
follows. Highest supermultiplet:
δAs+1 = i(Ψ̄(sγ1)η) δΨs = −σµν∂µAν(s)η +
∂(1γ1(γÃ)s−2)η
Main set (1 ≤ k ≤ s):
δΦk = −
σµν∂µC̄ν(k−1γ1)η +
i(k − 1)(k − 2)
[∂(1γ1C̃k−2) − g(2∂1(γC̃)k−3)]η
δAk =
2 cos(θk)(Φ̄kη) + i sin(θk)(Ψ̄(k−1γ1)η) (31)
δBk =
2 sin(θk)(Φ̄kγ5η) + i cos(θk)(Ψ̄(k−1γ1)γ5η)
δΨk−1 = −σµν∂µDν(k−1)η +
k − 2
∂(1γ1(γD̃)k−3η
and the last supermultiplet:
δΦ0 = −i∂̂zη, δz̄ = 2(Φ̄0η)
By analogy with superspins 3/2 and 5/2 cases we will assume that fermionic mass terms are
Dirac ones. This immediately gives:
(−1)k
s + 1
k + 1
[Ψ̄kΦk − k(Ψ̄γ)k−1(γΦ)k−1 −
k(k − 1)
Φ̃k−2]−
− ick[(Ψ̄γ)k−1Ψk−1 −
k − 1
(γΨ)k−2 + (Ψ → Φ)]
where:
(s+ k + 1)(s− k + 1)
The choice for the bosonic mass terms (taking into account parity) is also unambiguous:
(−1)kak[Ak∂(1Ak−1) − (k − 1)Ãk−2(∂A)k−2 +
(k − 1)(k − 2)
Ãk−2∂(1Ãk−3)] +
(−1)kbk[Bk∂(1Bk−1) − (k − 1)B̃k−2(∂B)k−2 +
(k − 1)(k − 2)
B̃k−2∂(1B̃k−3)] (33)
for the terms with one derivative, where:
(s+ k + 1)(s− k + 2)
, bk =
(s+ k)(s− k + 1)
and the following terms without derivatives:
(−1)k[d̂kAkAk + êkÃk−2Ãk−2 + f̂kÃk−2Ak−2] +
(−1)k[dkBkBk + ekB̃k−2B̃k−2 + fkB̃k−2Bk−2] (34)
Here:
d̂k =
(s− k)(s+ k + 3)
4(k + 1)
, êk =
k(k − 1)
16(k + 1)
[(s− k + 3)(s+ k) + 6]
f̂k = −
(s− k + 3)(s+ k)(s− k + 2)(s+ k + 1)
(s− k − 1)(s+ k + 2)
4(k + 1)
, ek =
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
Note that hatted coefficients differ from the unhatted ones by replacement s→ s+ 1.
Now we require that total Lagrangian be invariant under the supertransformations. First
of all this fixes all mixing angles:
sin(θk) =
s+ k + 1
2(s+ 1)
, cos(θk) =
s− k + 1
2(s+ 1)
and gives us additional terms for fermionic supertransformations:
δ′Ψk = α1Ak + α2γ(1(γA)k−1) + α3g(2Ãk−2) +
+β1Bk + β2γ(1(γB)k−1) + β3g(2B̃k−2) +
kck+1
4(k + 1)
[sin(θk+2)γ(1(γÃ)k−1) + cos(θk+1)γ(1(γB̃)k−1)]
δ′Φk = α4(γA)k + α5γ(1Ãk−1) + α6g(2(γÃ)k−2) +
+β4(γB)k + β5γ(1B̃k−1) + β6g(2(γB̃)k−2) −
cos(θk)[γ(1Ak−1) − g(2(γA)k−2]−
sin(θk)[γ(1Bk−1) − g(2(γB)k−2)]
Where:
α1 = −
s− k√
2(k + 1)
cos(θk), α2 = −
k2 + s+ k + 1√
2k(k + 1)
cos(θk)
α3 = −
(s + 1)(k − 1)(k − 2)
2k2(k + 1)
cos(θk)
β1 = −
s+ k + 2√
2(k + 1)
sin(θk), β2 =
k2 − s+ k − 1√
2k(k + 1)
sin(θk)
β3 = −
(s + 1)(k − 1)(k − 2)
2k2(k + 1)
sin(θk)
k + 1
sin(θk+1), α5 =
k(k − 1)(s− k)
4(k + 1)2
sin(θk+1)
(k − 1)[(k + 1)(s+ 1)− 2k2(s− k)]
8k(k + 1)2
sin(θk+1)
k + 1
cos(θk+1), β5 =
k(k − 1)(s+ k + 2)
4(k + 1)2
cos(θk+1)
(k − 1)[(k + 1)(s+ 1)− 2k2(s+ k + 2)]
8k(k + 1)2
cos(θk+1)
We have explicitely checked that (rather complicated) formulas from this and previous
sections correctly reproduce all lower superspins results.
Conclusion
Thus, using supersymmetric generalization of gauge invariant description for massive parti-
cles, we managed to show that all massive N = 1 supermultiplets could be constructed out
of appropriate set of massless ones. In this, in spite of large number of fields involved, all
calculations are pretty straightforward and mainly combinatorical. Certainly, using gauge
invariance one can fix the gauge where all but four physical massive fields are equal to
zero. But in this case all supertransformations must be supplemented with field dependent
gauge transformations restoring the gauge. So the structure of resulting supertransformation
become very complicated and will contain higher derivative terms.
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	Superspin 2
	Superspin 5/2
	Massive particles
	Integer spin
	Half-integer spin
	Massless supermultiplets
	Integer superspin
	Half-integer superspin

  In this paper we give explicit construction of massive N=1 supermultiplets in
flat d=4 Minkowski space-time. We work in a component on-shell formalism based
on gauge invariant description of massive integer and half-integer spin
particles where massive supermultiplets are constructed out of appropriate set
of massless ones.

Introduction
In a flat space-time massive spin s particles in a massless limit decompose into massless spin
s, s− 1, . . . ones. This, in particular, leads to the possibility of gauge invariant description
of massive spin s particles, e.g. [1]-[13]. In this, two different approaches could be used.
From one hand, one can start with usual non gauge invariant description of massive particle
and achieve gauge invariance through the introduction of additional fields (thus promoting
second class constraints into the first class ones). From the other hand, one can start with
the appropriate set of massless particles having gauge invariance from the very beginning and
obtain massive particle description as a deformation of massless theory. This last approach
closely mimic situation in spontaneous gauge symmetry breaking where gauge field has to
eat some Goldstone field(s) to become massive.
In the supersymmetric theories all particles must belong to some supermultiplet, massive
or massless. Till now most of investigations in supersymmetric theories where bounded to
massless supermultiplets. Quite a few results on massive supermultiplets mainly devoted
to superspins 1 and 3/2 exist [14]-[21]. The aim of this paper is to extend these results to
include massive N = 1 supermultiplets with arbitrary superspins. Certainly, it would be
nice to have superfield off-shell description of such supermultiplets, but as previous results
clearly show it is a highly non-trivial task. So in this paper we restrict ourselves with
component on-shell formalism in terms of physical fields. The same reasoning on the massless
limit means that massive supermultiplets could (should) be constructed out of the massless
ones in the same way as massive particles out of the massless ones. So our approach will
be supersymmetric generalization of the second approach to massive particle description
mentioned above. Namely, we will start with appropriate set of massless supermultiplets
and obtain massive one as a smooth deformation.
The paper is organized as follows. Though our previous examples on massive superspin
1 [21] and superspin 3/2 [15] supermultiplets already give important hints on how general
case of arbitrary superspin could looks like, due to peculiarities of lower spin fields they are
not enough to achieve such generalization. Thus, in the first two sections we give two more
concrete examples, namely massive supermultiplets with superspin 2 and 5/2, correspond-
ingly. All these and subsequent results heavily depend on the gauge invariant description of
massive particles with integer [2, 3] and half-integer [10] spins as well as on the known form
of massless supermultiplets [22]. For reader convenience and to make paper self-contained,
in the next two sections we give all necessary formulas in compact condensed notations. One
of the lessons from previous investigations is that the structures of massive supermultiplets
with integer and half-integer superspins are different, so in the last two sections we consider
these two cases separately. We will see that, in spite of large number of fields, all calculations
are pretty straightforward and mainly combinatorical.
1 Superspin 2
Massive superspin 2 supermultiplet contains four massive particles with spins 5/2, 2, 2’ and
3/2, correspondingly. In the massless limit massive supermultilets must decompose into the
appropriate set of massless ones in the same way as massive spin s particles — into massless
spin s, s-1, ... ones. Simple counting of physical degrees of freedom immediately gives:
0, 0′
So we will start with five massless supermultiplets (Φµν , hµν), (fµν ,Ψµ), (Ωµ, Aµ), (Bµ, ψ)
and (χ, z). From our previous experience with massive superspin 1 and superspin 3/2 su-
permultiplets we know that it is crucial for the construction of massive supermultiplets to
make dual rotation of vector Aµ and axial-vector Bµ fields mixing massless supermultiplets
containing these fields. But now we have two tensor fields hµν and fµν as well, moreover they
necessarily must be tensor and pseudo-tensor ones. Thus we have to consider the possibility
to mix massless supermultiplets with these fields as well and the real structure of massless
supermultiplets we are going to work with looks like:
hµν fµν
Aµ Bµ
Then, introducing a sum of the massless Lagrangians for bosonic fields:
∂αhµν∂αhµν − (∂h)µ(∂h)µ + (∂h)µ∂µh−
∂µh∂µh+
∂αfµν∂αfµν − (∂f)µ(∂f)µ + (∂f)µ∂µf −
∂µf∂µf −
(∂µϕ)
(∂µπ)
2 (1)
as well as sum of the massless Lagrangians for fermionic fields:
Φ̄µν ∂̂Φµν − 2i(Φ̄γ)µ(∂Φ)µ + i(Φ̄γ)µ∂̂(γΦ)µ + i(Φ̄γ∂)Φ−
Φ̄∂̂Φ−
Ψ̄µ∂̂Ψµ + i(Ψ̄γ)(∂Ψ)−
(Ψ̄γ)∂̂(γΨ) +
ψ̄∂̂ψ −
Ω̄µ∂̂Ωµ + i(Ω̄γ)(∂Ω)−
(Ω̄γ)∂̂(γΩ) +
χ̄∂̂χ (2)
it is not hard to check that the most general supertransformations leaving sum of massless
Lagrangians invariant have the form (round brackets denote symmetrization):
δΦµν = −
σαβ∂α(cos(θ2)hβ(µγν) − sin(θ2)γ5fβ(µγν))η
δhµν =
2 cos(θ2)(Φ̄µνη) + i sin(θ2)(Ψ̄(µγν)η) (3)
δfµν =
2 sin(θ2)(Φ̄µνγ5η) + i cos(θ2)(Ψ̄(µγν)γ5η)
δΨµ = −σαβ∂α(sin(θ2)hβµ + cos(θ2)fβµγ5)η
for the supermultiplets containing spin 2 fields and
δΩµ = −
σαβ(cos(θ1)Aαβ − sin(θ1)γ5Bαβ)γµη
δAµ =
2 cos(θ1)(Ω̄µη) + i sin(θ1)(ψ̄γµη) (4)
δBµ =
2 sin(θ1)(Ω̄µγ5η) + i cos(θ1)(ψ̄γµγ5η)
δψ = −
σαβ(sin(θ1)Aαβ + cos(θ1)γ5Bαβ)η
for those with (axial)vector ones. The last supermultiplets is simple:
δχ = −iγµ∂µ(ϕ+ γ5π)η, δϕ = (χ̄η), δπ = (χ̄γ5η)
To construct massive supermultiplet we have to add mass terms for all fields as well as
appropriate corrections to fermionic supertransformations. In this, the most important ques-
tion is which lower spin fields play the roles of Goldstone ones and have to be eaten to make
main gauge fields massive. For the bosonic fields (taking into account parity conservation)
the choice is unambiguous: vector Aµ and scalar ϕ fields for tensor field hµν and axial-vector
Bµ and pseudo-scalar π — for pseudo-tensor fµν . Thus bosonic mass terms will be:
2[hµν∂µAν − h(∂A)] −
3Aµ∂µϕ+
2[fµν∂µBν − f(∂B)]−
3Bµ∂µπ
L2 = −
(hµνhµν − h2)−
hϕ+ ϕ2 −
(fµνhµν − f 2)−
fπ + π2 (5)
But for fermions we have two spin 3/2 and two spin 1/2 fields and there is no evident choice.
Thus we introduce the most general mass terms for the fermions:
Lm = −
Φ̄µνΦµν + (Φ̄γ)
µ(γΦ)µ +
−iα1[Φ̄µνγµΨν −
Φ̄(γΨ)]− iα2[Φ̄µνγµΩν −
Φ̄(γΩ)] +
+a1[Ψ̄
µΨµ − (Ψ̄γ)(γΨ)] + a2[Ω̄µΩµ − (Ω̄γ)(γΩ)] + a3[Ψ̄µΩµ − (Ψ̄γ)(γΩ)] +
+ia4(Ψ̄γ)ψ + ia5(Ψ̄γ)χ+ ia6(Ω̄γ)ψ + ia7(Ω̄γ)χ+
+a8ψ̄ψ + a9ψ̄χ+ a10χ̄χ (6)
and proceed with calculations. Cancellation of variations with one derivative gives:
sin(θ2) = cos(θ2) =
sin(θ1) = cos(θ1) =
, α2 =
a1 = −
, a3 = 1, a4 =
2, a5 = 0
a6 = a2
2, a7 =
3, a8 = 0, a9 = −
while variations without derivatives give:
, a10 = −
Resulting fermionic mass terms:
Lm = −
Φ̄µνΦµν + (Φ̄γ)
µ(γΦ)µ +
Φ̄µνγµΨν +
Φ̄(γΨ)− i
2Φ̄µνγµΩν +
Φ̄(γΩ) +
Ψ̄µΨµ +
(Ψ̄γ)(γΨ) +
Ω̄µΩµ −
(Ω̄γ)(γΩ) + Ψ̄µΩµ − (Ψ̄γ)(γΩ) +
2(Ψ̄γ)ψ +
(Ω̄γ)ψ + i
3(Ω̄γ)χ−
6ψ̄χ−
χ̄χ (7)
correspond to invariance of the Lagrangian (besides global supertransformations) under three
local spinor gauge transformations:
δΦµν = ∂(µξν) +
γ(µξν) +
gµνξ1 +
gµνξ2
δΨµ = ∂µξ1 +
γµξ1 +
δΩµ = ∂µξ2 +m
2ξµ +
γµξ1 +
γµξ2 (8)
δψ = m
2ξ1 +
ξ2 δχ = m
From these formula one can easily determine which combination of spin 3/2 fields Ψµ and
Ωµ plays the role of Goldstone field for spin 5/2 field Φµν . Indeed, if one introduces two
orthogonal combinations:
Ψ̃µ =
Ωµ, Ω̃µ = −
then after diagonalization of mass terms one finds that Ψ̃µ is a Goldstone field, while Ω̃µ —
physical field with the same mass as Φµν .
Similar to the case of massive supermultiplet with superspin 1, both mixing angles have
been fixed: θ1 = θ2 = π/4 and this, in turn, means that all bosonic fields enter through the
complex combinations only:
Hµν = hµν + γ5fµν , Cµ = Aµ + γ5Bµ, z = ϕ+ γ5π
Introducing gauge covariant derivatives:
∇µHαβ = ∂µHαβ −
Cµgαβ, ∇µz = ∂µz −m
we can write final form of fermionic supertransformations as:
δΦµν = [−
σαβ∇αH̄β(µγν) −mHµν +
γ(µ(γH)ν) +
gµνz]η
δΨµ = [−
σαβ∇αHβµ −
(γH)µ +
∇µz +
γµz]η
δΩµ = [−
σαβC̄αβγµ +
(γH)µ +
∇µz −
γµz]η (9)
δψ = −
σαβCαβη δχ = −iγµ∇µzη
Note also that due to complexification of bosonic fields the Lagrangian and supertrans-
formations are invariant under global axial U(1)A symmetry, axial charges for all fields being:
field η Φµν , Ψµ, Ωµ, ψ, χ Hµν , Cµ, z
qA +1 0 –1
2 Superspin 5/2
Our next example — massive supermultiplet with superpin 5/2. It also contains four massive
fields: with spin 3, 5/2, 5/2’ and 2 and in the massless limit it should reduce to six massless
supermultiplets:
5/2 5/2
0, 0′
By analogy with all previous cases we will take into account possible mixing for bosonic tensor
and vector fields, so we will start with the following structure of massless supermultiplets:
hµν fµν
Aµ Bµ
So we introduce sum of the massless Lagrangians for bosonic fields:
L0 = −
∂ρΦµνλ∂ρΦµνλ +
(∂Φ)µν(∂Φ)µν − 3(∂Φ)µν∂µΦν +
∂µΦν∂µΦν +
(∂Φ)2
∂αhµν∂αhµν − (∂h)µ(∂h)µ + (∂h)µ∂µh−
∂µh∂µh−
(∂µϕ)
∂αfµν∂αfµν − (∂f)µ(∂f)µ + (∂f)µ∂µf −
∂µf∂µf −
(∂µπ)
2 (10)
as well as sum of the massless Lagrangians for fermionic fields:
Ψ̄µν ∂̂Ψµν − 2i(Ψ̄γ)µ(∂Ψ)µ + i(Ψ̄γ)µ∂̂(γΨ)µ + i(Ψ̄γ∂)Ψ−
Ψ̄∂̂Ψ+
Ω̄µν ∂̂Ωµν − 2i(Ω̄γ)µ(∂Ω)µ + i(Ω̄γ)µ∂̂(γΩ)µ + i(Ω̄γ∂)Ω−
Ω̄∂̂Ω−
Ψ̄µ∂̂Ψµ + i(Ψ̄γ)(∂Ψ)−
(Ψ̄γ)∂̂(γΨ) +
ψ̄∂̂ψ −
Ω̄µ∂̂Ωµ + i(Ω̄γ)(∂Ω) −
(Ω̄γ)∂̂(γΩ) +
χ̄∂̂χ (11)
and start with the following global supertransformations:
δΦµνλ = i(Ψ̄(µνγλ)η) δΨµν = [−σαβ∂αΦβµν +
∂(µγν)(γΦ)]η (12)
for the supermultiplet (3, 5/2),
δΩµν = −
σαβ∂α(cos(θ2)hβ(µγν) − sin(θ2)γ5fβ(µγν))η
δhµν =
2 cos(θ2)(Ω̄µνη) + i sin(θ2)(Ψ̄(µγν)η) (13)
δfµν =
2 sin(θ2)(Ω̄µνγ5η) + i cos(θ2)(Ψ̄(µγν)γ5η)
δΨµ = −σαβ∂α(sin(θ2)hβµ + cos(θ2)fβµγ5)η
for the mixed (5/2, 2) and (2, 3/2) supermultiplets,
δΩµ = −
σαβ(cos(θ1)Aαβ − sin(θ1)γ5Bαβ)γµη
δAµ =
2 cos(θ1)(Ω̄µη) + i sin(θ1)(ψ̄γµη) (14)
δBµ =
2 sin(θ1)(Ω̄µγ5η) + i cos(θ1)(ψ̄γµγ5η)
δψ = −
σαβ(sin(θ1)Aαβ + cos(θ1)γ5Bαβ)η
for the mixed (3/2, 1) and (1, 1/2) supermultiplets and
δχ = −iγµ∂µ(ϕ+ γ5π)η, δϕ = (χ̄η), δπ = (χ̄γ5η)
for the last one.
By analogy with the superspin 3/2 case, we will assume that fermionic mass terms are
Dirac ones:
Lm = −Ψ̄µνΩµν + 2(Ψ̄γ)µ(γΩ)µ +
Ψ̄Ω +
[−Ψ̄µνγµΨν +
Ψ̄(γΨ)− Ω̄µνγµΩν +
Ω̄(γΩ)] +
Ψ̄µΩµ −
(Ψ̄γ)(γΩ) + 2i(Ψ̄γ)ψ + 2i(Ω̄γ)χ− 3ψ̄χ (15)
where all coefficients are completely fixed by the requirement that the Lagrangian has to be
invariant not only under the global supertransformations, but under four (by the number of
fermionic gauge fields) spinor gauge transformations:
δΨµν = ∂(µξν) +
γ(µην) +
gµνξ1, δΩµν = ∂(µην) +
γ(µξν) +
gµνξ2,
δΨµ = ∂µξ1 +m
ξµ + im
γµξ2, δΩµ = ∂µξ2 +m
ηµ + im
γµξ1,
δψ = 2mξ1, δχ = 2mξ2
As for the bosonic fields, here the roles of the fields are evident (again taking into account
parity conservation): we need tensor hµν , vector Aµ and scalar ϕ fields to make spin 3 field
Φµνλ massive, while pseudo-tensor fµν field needs to eat axial-vector Bµ and pseudo-scalar
π fields. So the bosonic mass terms are also completely fixed:
3[−Φµνλ∂µhνλ + 2Φµ(∂h)µ −
Φµ∂µh] +
5[hµν∂µAν − h(∂A)]−
6Aµ∂µϕ +
2[fµν∂µBν − f(∂B)]−
3Bµ∂µπ (16)
ΦµνλΦµνλ −
ΦµΦµ +
ΦµAµ −
(fµνfµν − f 2)−
fπ + π2 (17)
Now we require that the whole Lagrangian be invariant under global supertransformations
with appropriate corrections to fermionic transformations. This fixes both mixing angles:
sin(θ2) =
, cos(θ2) =
, sin(θ1) =
, cos(θ1) =
and gives the following form of additional terms for fermionic supertransformations:
δΨµν = [−
γ(µ(γh)ν) −
fµνγ5 +
γ(µ(γf)ν)γ5]η
δΩµν = i[(γΦ)µν +
gµν(γΦ)−
γ(µAν) +
gµνÂ−
γ(µBν)γ5 +
gµνB̂γ5]η
δΨµ = [
γµ(γΦ)−
γµÂ−
Bµγ5 −
γµB̂γ5]η (18)
δΩµ = i[
(γh)µ +
(γf)µγ5 − γµϕ− γµγ5π]η
δψ = [−ϕ− 2γ5π]η
δχ = i[
6Â +
3B̂γ5]η
The complete supertransformations for fermionic fields could be simplified by introduction
of gauge invariant derivatives;
∇µhαβ = ∂µhαβ −
Aµgαβ, ∇µϕ = ∂µϕ−m
∇µfαβ = ∂µfαβ −
Bµgαβ, ∇µπ = ∂µπ −m
This time bosonic fields do not combine into complex combinations, but due to the fact that
fermionic mass terms are Dirac ones the Lagrangian and supertransformations are invariant
under global axial U(1)A transformations, provided axial charges of all fields are assigned as
follows:
field η, Ψµν , Ψµ, ψ hµν , fµν , Aµ, Bµ, ϕ, π Ωµν , Ωµ, χ
qA +1 0 –1
3 Massive particles
All our previous and subsequent calculations heavily depend on the gauge invariant descrip-
tion of massive high spin particles. For reader convenience and to make paper self-contained
we will give here gauge invariant formulations for massive particles with arbitrary integer
[2, 3] and half-integer [10] spins. We restrict ourselves to flat d = 4 Minkowski space but all
results could be easily generalized to the case of (A)dS space with arbitrary dimension d.
3.1 Integer spin
The simplest way to describe massless bosonic field with arbitrary spin s is to use completely
symmetric rank s tensor Φ(α1α2...αs) which is double traceless. In what follows we will use
condensed notations where index denotes just number of free indices and not the indices
themselves. For example, the tensor field itself will be denoted as Φs, it’s contraction with
derivative as (∂Φ)s−1, it’s trace as Φ̃s−2 and so on. As we will see this does not lead to any
ambiguities then working with free Lagrangians quadratic in fields. In these notations the
Lagrangian for massless particles of arbitrary spin s could be written as:
L0 = (−1)s[
∂µΦs∂µΦ
(∂Φ)s−1(∂Φ)s−1 −
s(s− 1)
∂µΦ̃s−2∂µΦ̃
s−2 +
s(s− 1)
(∂Φ)s−1∂(1Φ̃s−2) −
s(s− 1)(s− 2)
(∂Φ̃)s−3(∂Φ̃)s−3] (19)
where
Φ = 0. This Lagrangian is invariant under the following gauge transformations:
δ0Φs = ∂(1ξs−1), ξ̃s−3 = 0
where parameter ξs−1 is completely symmetric traceless tensor of rank s− 1.
To construct gauge invariant Lagrangian for massive particle which has correct (i.e. with
right number of physical degrees of freedom) massless limit, we start with the sum of massless
Lagrangians with 0 ≤ k ≤ s:
(−1)k[
∂µΦk∂µΦ
(∂Φ)k−1(∂Φ)k−1 −
k(k − 1)
∂µΦ̃k−2∂µΦ̃
k(k − 1)
(∂Φ)k−1∂(1Φ̃k−2) −
k(k − 1)(k − 2)
(∂Φ̃)k−3(∂Φ̃)k−3] (20)
Then we add the following cross terms with one derivative as well as mass terms without
derivatives:
(−1)kak[(∂Φ)k−1Φk−1 + (k − 1)Φ̃k−2(∂Φ)k−2 +
(k − 1)(k − 2)
(∂Φ̃)k−3Φ̃k−3]
(−1)k[dkΦkΦk + ekΦ̃k−2Φ̃k−2 + fkΦ̃k−2Φk−2] (21)
and try to achieve gauge invariance with the help of appropriate corrections to gauge trans-
formations:
δΦk = αkξk + βkg(2ξk−2)
Straightforward but lengthy calculations give a number of algebraic equations on the un-
known coefficients which could be solved (and this is non-trivial because we obtain overde-
termined system of equations) and give us:
(s− k)(s+ k + 1)
2(k + 1)2
, βk+1 =
k + 1
αk, 0 ≤ k ≤ s− 1
ak = −
(s− k + 1)(s+ k)
, dk =
(s− k − 1)(s+ k + 2)
4(k + 1)
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
3.2 Half-integer spin
For the description of massless spin s+1/2 particles we will use completely symmetric rank
s tensor-spinor Ψs such that (γΨ̃)s−3 = 0 (in the same condensed notations as before). Then
Lagrangain for such field could be written as:
L0 = i(−1)s[
Ψ̄s∂̂Ψs − s(Ψ̄γ)s−1(∂Ψ)s−1 +
(Ψ̄γ)s−1∂̂(γΨ)s−1 +
s(s− 1)
(Ψ̄γ∂)s−2Ψ̃s−2 −
s(s− 1)
∂̂Ψ̃s−2] (22)
and is invariant under the following gauge transformations:
δ0Ψs = ∂(1ξs−1), (γξ) = 0,
where gauge parameter ξs−1 is a γ-traceless tensor-spinor of rank s− 1.
Once again we start with the sum of massless Lagrangians with 0 ≤ k ≤ s:
i(−1)k[
Ψ̄k∂̂Ψk − k(Ψ̄γ)k−1(∂Ψ)k−1 +
(Ψ̄γ)k−1∂̂(γΨ)k−1 +
k(k − 1)
(Ψ̄γ∂)k−2Ψ̃k−2 −
k(k − 1)
∂̂Ψ̃k−2] (23)
To combine all these massless fields into one massive particle we have to add the following
mass terms:
(−1)k
2(k + 1)
[Ψ̄kΨk − k(Ψ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2]+
−ick[(Ψ̄γ)k−1Ψk−1 −
k − 1
(γΨ)k−2]
and corresponding corrections to gauge transformations:
δΨk = αkξk + iβkγ(1ξk−1) + ρkg(2ξk−2)
Then total Lagrangian will be gauge invariant provided:
(s+ 1)2 − k2
, αk =
k + 1
, βk =
2k(k + 1)
, ρk =
4 Massless supermultiplets
It is not easy to find in the recent literature the explicit component form of massless su-
permultiplets with arbitrary superspin [22], so for completeness we will give their short
description here. As we have already seen on the lower superspin cases, supermultiplets
with integer and half-integer superspins have different structure and have to be considered
separately.
(s, s+1/2). Supermultiplet with integer superspin s contains bosonic spin s field and
fermionic spin s+1/2 one. In this and in two subsequent sections we will use the same con-
densed notations as in the previous one. By analogy with superspins 1 and 2 supermultiplets,
we start with the following ansatz for the supertransformations:
δΨs = iα1σ
µν∂µΦν(s−1γ1)η δΦs = β(Ψ̄sη)
Indeed, calculating variations of the sum of two massless Lagrangians one can see that most
of variations cancel, provided one set α1 = −β2 . The residue:
δL = −(−1)sβ
(s− 1)(s− 2)
[2(Ψ̄∂∂)s−2Φ̃s−2 − ˜̄Ψ
∂2Φ̃s−2 − (s− 2)( ˜̄Ψ∂)s−3(∂Φ̃)s−3 −
−2(Ψ̄γ∂)s−2∂̂Φ̃s−2 + 2(Ψ̄γ∂∂)s−3(γΦ̃)s−3 − ( ˜̄Ψ∂)s−3∂̂(γΦ̃)s−3]
contains terms with Φ̃s−2 only, so we proceed by adding to fermionic supertransformations
one more term:
δ′Ψs = iα2∂(1γ1Φ̃s−2)
Then the choice α2 =
(s−1)(s−2)β
leaves us with:
δL = −(−1)sβ
(s− 1)(s− 2)
[2(Ψ̄γ∂∂)s−3(γΦ̃)s−3 − ( ˜̄Ψ∂)s−3∂̂(γΦ̃)s−3]
where the only terms are whose with (γΦ̃)s−3. So we make one more (last) correction to
supertransformations:
δ′′Ψs = iα3g(2∂1(γΦ̃)s−3)
and obtain full invariance with α3 = − (s−1)(s−2)β4s . To fix concrete normalization we will use
closure of the superalgebra. Calculating the commutator of two supertransformations we
obtain:
[δ1, δ2]Φs = −iβ2(η̄2γνη1)∂µΦs + . . .
where dots mean “up to gauge transformation”. So we set β =
2 and our final result looks
like:
δΨs = −
σµν∂µΦ̄ν(s−1γ1)η +
i(s− 1)(s− 2)
[∂(1γ1Φ̃s−2) − g(2∂1(γΦ̃)s−3)]η
δΦs =
2(Ψ̄sη) (25)
(s+1/2, s+1). Half-integer superspin multiplet contains fermionic spin s + 1/2 fields
and bosonic spin s + 1 one. Again by analogy with lower superspin case we will make the
following ansatz for supertransformations:
δΨs = α1σ
µν∂µΦν(s)η, δΦs+1 = iβ(Ψ̄(sγ1)η)
This time most of the variations cancel if one set α1 = −β leaving us with:
δL = i(−1)sβ
s(s− 1)
[2(Ψ̄∂∂)s−2(γΦ̃)s−2 − ˜̄Ψ
∂2(γΦ̃)s−2 −
−2(Ψ̄γ∂)s−2∂̂(γΦ̃)s−2 − (s− 2)( ˜̄Ψ∂)s−3(γ∂Φ̃)s−3]
Then the full invariance could be achieved with the following correction to supertransforma-
tions:
δ′Ψs = α2∂(1γ1(γΦ̃)s−2)
provided α2 =
. To check the closure of superalgebra and to choose normalization we
calculate commutator of two supertransformations:
[δ1, δ2]Φs+1 = −2iβ2(η̄2γµη1)∂µΦs+1 + . . .
Then our choice will be β = 1 and our final form:
δΨs = −σµν∂µΦν(s)η +
∂(1γ1(γΦ̃)s−2)η, δΦs+1 = i(Ψ̄(sγ1)η)
Note that starting with superspin 2 the structure of supertransformations are defined up to
possible field dependent gauge transformations and our choice differs from that of [22]. It
makes no difference for massless theories but for massive case the structure of corrections
for fermionic supertransformations depends on the choice made.
5 Integer superspin
Now, having in our disposal gauge invariant description of massive particles with arbitrary
(half-)integer spins, known form of supertransformations for massless arbitrary superspin
supermultiplets and concrete examples of massive supermultiplets with lower superspins, we
are ready to construct massive arbitrary superspin supermultiplets. As we have seen, integer
and half-integer cases have different structures and have to be considered separately.
In this section we consider massive supermultiplet with integer superspin. Such super-
multiplet also contains four massive fields: two bosonic spin s fields (with opposite parity)
and fermionic spin (s+1/2) and (s-1/2) ones. Calculating total number of physical degrees
of freedom and taking into account possible mixing of supermultiplets containing bosonic
fields with equal spins and opposite parity, we start with the following structure of massless
supermultiplets:
As Bs
Ak Bk
By analogy with superspin 1 and 2 cases, we will assume that all bosonic fields enter through
the complex combinations Ck = Ak + iBk only (so that all possible mixing angles are fixed
and equal π/4). Thus we choose the following form of supertransformations for massless
supermultiplets with 1 ≤ k ≤ s:
δΦk = −
σµν∂µC̄ν(k−1γ1)η +
i(k − 1)(k − 2)
[∂(1γ1C̃k−2) − g(2∂1(γC̃)k−3)]η
δC̄k = 2(Φ̄kη) + i
2(Ψ̄(k−1γ1)η) (26)
δΨk−1 = −
σµν∂µCν(k−1)η +
k − 2
∂(1γ1(γC̃)k−3)η
and also
δΦ0 = −i∂̂zη, δz̄ = 2(Φ̄0η)
As a result of our assumption mass terms for bosonic fields are completely fixed:
(−1)kck[C̄k∂(1Ck−1) − (k − 1) ˜̄C
(∂C)k−2 +
(k − 1)(k − 2)
( ˜̄C
∂(1C̃k−3) + h.c.)]
(−1)k[dkC̄kCk + ek ˜̄C
C̃k−2 + fk(
Ck−2 + h.c.)] (27)
where
(s+ k)(s− k + 1)
, dk =
(s− k − 1)(s+ k + 2)
4(k + 1)
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
As for the fermionic mass terms, apriori we don’t have any restrictions on them so we have
to consider the most general possible form:
(−1)k
a1k[Φ̄
kΦk − k(Φ̄γ)k−1(γΦ)k−1 −
k(k − 1)
Φ̃k−2]+
+a2k[Φ̄
kΨk − k(Φ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2] +
+a3k[Ψ̄
kΨk − k(Ψ̄γ)k−1(γΨ)k−1 −
k(k − 1)
Ψ̃k−2] +
+ib1k[(Φ̄γ)
k−1Φk−1 −
k − 1
(γΦ)k−2] +
+ib2k[(Φ̄γ)
k−1Ψk−1 −
k − 1
(γΨ)k−2] +
+ib3k[(Ψ̄γ)
k−1Φk−1 −
k − 1
(γΦ)k−2] +
+ ib4k[(Ψ̄γ)
k−1Ψk−1 −
k − 1
(γΨ)k−2]
Where:
a1s = −
, a2s = a3s = b3s = b4s = 0
The requirement that total Lagrangian be invariant under (appropriately corrected) super-
transformations gives:
a1k = −
, a2k = −
k + 1
ck+1, a3k =
2(k + 1)
b1k = −2ck, b2k = −
, b3k = 0, b4k = −2ck+1
In this, additional terms for fermionic supertransformations look like:
δ′Φk =
2ick+1
k + 1
[(γC)k +
k(k − 1)
4(k + 1)
γ(1C̃k−1) −
(k − 1)2(2k + 1)
8k(k + 1)
g(2(γC̃)k−2)]−
−Ck +
k − 1
γ(1(γC)k−1 −
(k − 1)(k − 2)
g(2C̃k−2) −
[γ(1Ck−1) − g(2(γC)k−2)] (29)
δ′Ψk =
kck+2
2(k + 1)
γ(1(γC̃)k−1) −
2(k + 1)
[(γC)k −
k(k − 1)
4(k + 1)
γ(1C̃k−1) +
(k − 1)(3k + 1)
8k(k + 1)
g(2(γC̃)k−2)]−
2ck+1
k + 1
[Ck +
γ(1(γC)k−1) +
(k − 1)(k − 2)
g(2C̃k−2)] (30)
Here the supertransformations for Φk field contain terms with Ck+1, Ck and Ck−1 fields in
the first, second and third lines correspondingly, while that of Ψk contain terms with Ck+2,
Ck+1 and Ck fields.
6 Half-integer superspin
Next we turn to the half-integer superspin case. This time we have two fermionic spin (s+1/2)
fields and bosonic ones with spins (s+1) and s. Usual reasoning on physical degrees of freedom
and possible mixings leads us to the following structure of massless supermultiplets we will
start with:
Φs Ψs
Ak Bk
We see that this structure is rather similar to that of integer superspin case. The main
difference (besides the presence of As+1, Ψs supermultiplet) comes from the mixing of bosonic
fields. We have no reasons to suggest that all mixing angles could be fixed from the very
beginning so we have to consider the most general possibility here. Let us denote:
Ck = cos(θk)Ak + γ5 sin(θk)Bk, Dk = sin(θk)Ak + γ5 cos(θk)Bk
In these notations supertransformations for massless supermultiplets could be written as
follows. Highest supermultiplet:
δAs+1 = i(Ψ̄(sγ1)η) δΨs = −σµν∂µAν(s)η +
∂(1γ1(γÃ)s−2)η
Main set (1 ≤ k ≤ s):
δΦk = −
σµν∂µC̄ν(k−1γ1)η +
i(k − 1)(k − 2)
[∂(1γ1C̃k−2) − g(2∂1(γC̃)k−3)]η
δAk =
2 cos(θk)(Φ̄kη) + i sin(θk)(Ψ̄(k−1γ1)η) (31)
δBk =
2 sin(θk)(Φ̄kγ5η) + i cos(θk)(Ψ̄(k−1γ1)γ5η)
δΨk−1 = −σµν∂µDν(k−1)η +
k − 2
∂(1γ1(γD̃)k−3η
and the last supermultiplet:
δΦ0 = −i∂̂zη, δz̄ = 2(Φ̄0η)
By analogy with superspins 3/2 and 5/2 cases we will assume that fermionic mass terms are
Dirac ones. This immediately gives:
(−1)k
s + 1
k + 1
[Ψ̄kΦk − k(Ψ̄γ)k−1(γΦ)k−1 −
k(k − 1)
Φ̃k−2]−
− ick[(Ψ̄γ)k−1Ψk−1 −
k − 1
(γΨ)k−2 + (Ψ → Φ)]
where:
(s+ k + 1)(s− k + 1)
The choice for the bosonic mass terms (taking into account parity) is also unambiguous:
(−1)kak[Ak∂(1Ak−1) − (k − 1)Ãk−2(∂A)k−2 +
(k − 1)(k − 2)
Ãk−2∂(1Ãk−3)] +
(−1)kbk[Bk∂(1Bk−1) − (k − 1)B̃k−2(∂B)k−2 +
(k − 1)(k − 2)
B̃k−2∂(1B̃k−3)] (33)
for the terms with one derivative, where:
(s+ k + 1)(s− k + 2)
, bk =
(s+ k)(s− k + 1)
and the following terms without derivatives:
(−1)k[d̂kAkAk + êkÃk−2Ãk−2 + f̂kÃk−2Ak−2] +
(−1)k[dkBkBk + ekB̃k−2B̃k−2 + fkB̃k−2Bk−2] (34)
Here:
d̂k =
(s− k)(s+ k + 3)
4(k + 1)
, êk =
k(k − 1)
16(k + 1)
[(s− k + 3)(s+ k) + 6]
f̂k = −
(s− k + 3)(s+ k)(s− k + 2)(s+ k + 1)
(s− k − 1)(s+ k + 2)
4(k + 1)
, ek =
k(k − 1)
16(k + 1)
[(s− k + 2)(s+ k − 1) + 6]
fk = −
(s− k + 2)(s+ k − 1)(s− k + 1)(s+ k)
Note that hatted coefficients differ from the unhatted ones by replacement s→ s+ 1.
Now we require that total Lagrangian be invariant under the supertransformations. First
of all this fixes all mixing angles:
sin(θk) =
s+ k + 1
2(s+ 1)
, cos(θk) =
s− k + 1
2(s+ 1)
and gives us additional terms for fermionic supertransformations:
δ′Ψk = α1Ak + α2γ(1(γA)k−1) + α3g(2Ãk−2) +
+β1Bk + β2γ(1(γB)k−1) + β3g(2B̃k−2) +
kck+1
4(k + 1)
[sin(θk+2)γ(1(γÃ)k−1) + cos(θk+1)γ(1(γB̃)k−1)]
δ′Φk = α4(γA)k + α5γ(1Ãk−1) + α6g(2(γÃ)k−2) +
+β4(γB)k + β5γ(1B̃k−1) + β6g(2(γB̃)k−2) −
cos(θk)[γ(1Ak−1) − g(2(γA)k−2]−
sin(θk)[γ(1Bk−1) − g(2(γB)k−2)]
Where:
α1 = −
s− k√
2(k + 1)
cos(θk), α2 = −
k2 + s+ k + 1√
2k(k + 1)
cos(θk)
α3 = −
(s + 1)(k − 1)(k − 2)
2k2(k + 1)
cos(θk)
β1 = −
s+ k + 2√
2(k + 1)
sin(θk), β2 =
k2 − s+ k − 1√
2k(k + 1)
sin(θk)
β3 = −
(s + 1)(k − 1)(k − 2)
2k2(k + 1)
sin(θk)
k + 1
sin(θk+1), α5 =
k(k − 1)(s− k)
4(k + 1)2
sin(θk+1)
(k − 1)[(k + 1)(s+ 1)− 2k2(s− k)]
8k(k + 1)2
sin(θk+1)
k + 1
cos(θk+1), β5 =
k(k − 1)(s+ k + 2)
4(k + 1)2
cos(θk+1)
(k − 1)[(k + 1)(s+ 1)− 2k2(s+ k + 2)]
8k(k + 1)2
cos(θk+1)
We have explicitely checked that (rather complicated) formulas from this and previous
sections correctly reproduce all lower superspins results.
Conclusion
Thus, using supersymmetric generalization of gauge invariant description for massive parti-
cles, we managed to show that all massive N = 1 supermultiplets could be constructed out
of appropriate set of massless ones. In this, in spite of large number of fields involved, all
calculations are pretty straightforward and mainly combinatorical. Certainly, using gauge
invariance one can fix the gauge where all but four physical massive fields are equal to
zero. But in this case all supertransformations must be supplemented with field dependent
gauge transformations restoring the gauge. So the structure of resulting supertransformation
become very complicated and will contain higher derivative terms.
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	Superspin 2
	Superspin 5/2
	Massive particles
	Integer spin
	Half-integer spin
	Massless supermultiplets
	Integer superspin
	Half-integer superspin
