= {{D α, D α }, D α }. = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ.

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1 PHY 396 T: SUSY Solutions for problem set #1. Problem 2(a): First of all, [D α, D 2 D α D α ] = {D α, D α }D α D α {D α, D α } = {D α, D α }D α + D α {D α, D α } (S.1) = {{D α, D α }, D α }. Second, {D α, D α } = 2iσ µ α α µ, which commutes with all the superderivatives. Consequently, [D α, D 2 ] = {2iσ µ α α µ, D α } = 4iσ µ α α D α µ. (S.2) Similarly, Finally, [D α, D 2 D α D α ] = {D α, {D α, D α }} = 4iD α σ µ α α µ. (S.3) [D 2, D 2 ] = D α [D α, D 2 ] + [D α, D 2 ]D α = D α [D α, D 2 ] [D α, D 2 ]D α = [D α, [D α, D 2 ]] (S.4) = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ. Using the anticommutator {D α, D α } = 2iσ µ α α µ we may rewrite the commutator [D α, D α ] on the last line of eq. (S.4) as 2D α D α 4i σ ααν ν and consequently [D 2, D 2 ] = 8iσ µ α α Dα D α µ + 16σ µ α α µ σ ααν ν = 8iσ µ α α Dα D α µ (S.5) where the second equality follows from σ µ α α σ ααν = tr(σµ σ ν ) = 2g µν. Alternatively, we may use [D α, D α ] = 2D α D α + {D α, D α } on the bottom line of eq. (S.4) to obtain [D 2, D 2 ] = 8iσ µ α α D α D α µ 16σ µ α α µ σ ααν ν = 8i σ ααµ D α D α µ (S.6) 1

2 Problem 2(b): Thanks to eqs. (S.2) and (S.3), we have D α [D 2, D α ] = 4iσ µ α α Dα D α µ = [D α, D 2 ]D α. (S.7) Consequently, D α D 2 D α = D 2 D 2 + D α [D 2, D α ] = D 2 D 2 + [D α, D 2 ]D α = D α D 2 D α. (S.8) Problem 2(c): In light of part (b), 2D α D 2 D α D 2 D 2 D 2 D 2 = D α D 2 D α D 2 D 2 + D α D 2 D α D 2 D 2 = D α [D 2, D α ] + [D α, D 2 ]D α = 4iσ µ α α Dα D α µ 4iσ µ α α Dα D α D α µ = 4iσ µ α α µ{d α, D α } (S.9) = 4iσ µ α α µ 2i σ ααν ν = 8 tr(σ µ σ ν ) µ ν = Problem 2(d): Consider the superderivative product D 2 D 2 D 2. Because the D α operators anticommute with each other and there are only two of them, any product of 3 such operators in a row must vanish, D α D β D γ 0, regardless of indices. In particular, D α D 2 = 0. Consequently, D α D 2 D 2 = [D α, D 2 ]D 2 (S.10) and D 2 D 2 D 2 = D α [D α, D 2 ]D 2 = {D α, [D α, D 2 ]}D 2, (S.11) 2

3 where {D α, [D α, D 2 ]} = 4iσ µ α α {Dα, D α µ } = 4iσ µ α α µ 2i σ ααν ν = (S.12) Thus, D 2 D 2 D 2 = 16 2 D 2 (S.13) and similarly D 2 D 2 D 2 = 16 2 D 2. (S.14) Now let s consider the putative projector operators (1). Thanks to eq. (S.13), (D 2 D 2 ) 2 = 16 2 D 2 D 2 (S.15) and consequently Π 2 C = Π C. Similarly, (D 2 D 2 ) = = 16 2 D 2 D 2 (S.16) and hence Π 2 A = Π A. Next, using D α D β = 1 2 δα β D2 which follows from the anticommutativity of the D operators and the fact that there are only two of them we have (D α D 2 D α ) 2 = D α D 2 D α D β D 2 D β = 1 2 Dα D 2 D 2 D 2 D α = +8 2 D α D 2 D α (S.17) and therefore Π + L 2 = Π L. This proves that all three operators (1) are indeed projector operators. And thanks to part (c), they do add up to the unit operator. All we need to prove now is that the three projectors (1) commute with each other. In fact, all products of these projectors vanish, regardless of the order of multiplication. Indeed, using DDD 0 and DDD 0, we have D 2 D 2 D 2 D 2 = 0 = Π A Π C = 0, (S.18) 3

4 D 2 D 2 D 2 D 2 = 0 = Π C Π A = 0, (S.19) D α D 2 D α D 2 D 2 = 0 = Π L Π C = 0, (S.20) D 2 D 2 D α D 2 D α = 0 = Π C Π L = 0, (S.21) D α D 2 D α D 2 D 2 = 0 = Π L Π A = 0, (S.22) D 2 D 2 D α D 2 D α = 0 = Π A Π L = 0. (S.23) Problem 2(e): Since DDD 0, for any superfield Y (x, θ, θ), D α D 2 Y = 0, which makes D 2 Y a chiral superfield. Consequently, for any general superfield X, Π C X = D 2 D2 X is a chiral superfield (S.24) Likewise, DDD 0 makes D 2 anything an antichiral superfield, hence for any X(x, θ, θ), Π A X = D 2 D2 X is an antichiral superfield (S.25) Finally, Π L X = D α D2 D α 8 2 X = D α D2 D α 8 2 X (S.26) satisfies both D 2 Π L X = D 2 D α = 0 (S.27) and D 2 Π L X = D 2 D α = 0, (S.28) which makes Π L X a linear superfield. 4

5 Now consider Π C Φ for a chiral superfield Φ(y, θ). Since D α Φ = 0, it follows that D 2 D 2 Φ = D α [D α, D 2 ]Φ = {D α, [D α, D 2 ]}Φ (S.29) where {D α, [D α, D 2 ]} = 16 2 (S.30) similar to eq. (S.12). Thus D 2 D 2 Φ = 16 2 Φ = Π C Φ = Φ. (S.31) Similarly, any antichiral superfield Φ satisfies D α Φ = 0 and hence D 2 D 2 Φ = D α [D α, D 2 ]Φ = {D α, [D α, D 2 ]}Φ = 16 2 Φ = Π A Φ = Φ. (S.32) Finally, for a linear superfield L D α D 2 D α L = D α [D 2, D α ]L = 4iσ µ α α Dα D α µ L, D α D 2 D α L = D α [D 2, D α ]L = D α [D 2, D α ]L = 4iσ µ α α D α D α µ L, (S.33) hence D α D 2 D α L D α D 2 D α L cf. part (b) = 1 2 Dα D 2 D α L D αd 2 D α L = 2iσ µ α α µ{d α, D α }L (S.34) = 2iσ µ α α µ 2i σ ααν ν L = +8 2 L and therefore Π L L = L. 5

6 Problem 3(a): Thanks to D 2 D α D 2 D α = 0 and D 2 D α D 2 D α = D 2 D α D 2 D α = 0, equation (3) implies D 2 V = 1 8m 2 D2 D α D 2 D α V = 0 and D 2 V = 1 8m 2 D2 D α D 2 D α V = 0. This makes the on-shell V a linear superfield. And having established that point, we have Π L V = V = 1 8 Dα D 2 D α V = 2 V and hence eq. (3) becomes the Klein Gordon equation (m )V = 0. Note that the equation of motion (3) for the massive vector field V implies both the linear superfield equations D 2 V = D 2 V = 0 and the Klein Gordon equation for all the components. This is similar to what happens to the ordinary (non-susy) massive vector field A µ : its free equation of motion µ F µν + m 2 A ν = 0 (S.35) implies both the 4D-transversality equation µ A µ = 0 and the Klein Gordon equation (m )A µ = 0 for all the components A µ. Problem 3(b): Without gauge fixing, the off-shell vector superfield V has 16 components according to V (x, θ, θ) = C(x) + θ 2 f(x) + θ 2 f (x) + θ α σ µ α α θ α A µ (x) θ2 θ2 (D(x) C(x)) + θ α χ α (x) + θ 2 θ α ( λ α (x) + 2 i σ ααµ µ χ α (x)) + θ α χ α (x) + θ 2 θ α (λ α (x) + 2 i σµ α α µ χ α (x)). (S.36) We need to express the D 2 V = D 2 V = 0 equations in component form. This can be done directly in the (x, θ, θ) coordinates, but it s easier to work with (y, θ, θ) coordinates for the 6

7 D 2 V and (ȳ, θ, θ) for the D 2 V. Indeed, in the (y, θ, θ) coordinates D α = θ and D 2 = 4 α y,θ ( θ 2 ) y,θ (S.37) while in the (ȳ, θ, θ) coordinates D α = θ α ȳ, θ and D 2 = 4 (θ 2 ) ȳ, θ (S.38) Thus, to calculate D 2 V all we need is to change the coordinates from (x, θ, θ) to (y, θ, θ) which gives V (y, θ, θ) = C(y) + θ 2 f(y) + θ 2 f (y) + θ α σ µ α α θ α( A µ (y) i µ C(y) ) θ2 θ2 ( D(y) 2 C(y) i µ A µ (y) ) + θ α χ α (y) + θ 2 θ α λ α (y) + θ α χ α (y) + θ 2 θ α( λ α (y) + iσ µ α α µ χ α (y) ) (S.39) and then drop the θ 2 factors from terms which have them and discard the terms which don t, thus 1 4 D2 V (y, θ) = f (y) + θ α( λ α (y)+iσ µ α α µ χ α (y) ) + θ 2( D(y) 2 C(y) i µ A µ (y) ). (S.40) Similarly, 1 4 D2 V (ȳ, θ) = f(ȳ) + θ α ( λ α (ȳ)+i σ ααµ µ χ α (ȳ) ) + θ 2( D(ȳ) 2 C(ȳ)+i µ A µ (ȳ) ). (S.41) According to these formulae, the D 2 V = D 2 V = 0 equations of motion for the on-shell superfield V expand to component-field equations f = f = 0, (S.42) D 2 C ± i µ A µ = 0, (S.43) 7

8 λ α + iσ µ α α µ χ α = 0, (S.44) λ α + i σ ααµ µ χ α = 0. (S.45) Note that eq. (S.43) is equivalent to D = 2 C and µ A µ = 0. (S.46) Thus, only 4 out of eight bosonic components of V are independent on shell: The 3 component of the A µ (x) (which is proper for the massive vector field in dimensions), plus one real scalar C(x). The remaining scalars either vanish or follow C: f = f = 0 while D = 2 C = m 2 C. Problem 3(c): Finally, consider the fermionic components of V. There are 8 of them χ α, χ α, λ α, and but they give rise to only 4 physical degrees of freedom because they satisfy first-order equations of motion. Specifically, there are 4 two-component Weyl equations λ α = iσ µ α α µ χ α, (S.44) λ α = i σ ααµ µ χ α, (S.45) m 2 χ α = iσ µ α α µ λ α, (S.47) m 2 χ α = i σ ααµ µ λ α. (S.48) The first two equations here follow from the expansion of D 2 V = D 2 V = 0 as we have already seen in part (b). The other two equations follow from the first two and the Klein-Gordon equations (m )(any component) = 0. Indeed, iσ µ α α µ λ α = iσ µ α α µ ( i) σ αβν ν χ β = 2 χ α = +m 2 χ α (S.49) and likewise i σ ααµ µ λ α = i σ ααµ µ ( i)σ ν α β ν χ β = 2 χ β = +m 2 χ β. (S.50) λ α The Weyl equations (S.44) through (S.48) have unconventional powers of the mass m because the χ and the χ fields have a different scaling dimension from the λ and the λ. We 8

9 may remedy that by rescaling χ α mχ α and χ α m χ α, which brings the Weyl equations to conventional form mλ α + iσ µ α α µ χ α = mχ α + iσ µ α α µ λ α = 0, m λ α + i σ ααµ µ χ α = m χ α + i σ ααµ µ λ α = 0. (S.51) We may also package the four Weyl spinor fields into a Dirac spinor fields Ψ(x) and it hermitian conjugate according to Ψ = ( λα m χ α ) and Ψ Ψ γ 0 = (mχ α, λ α ). (S.52) In terms of these Dirac spinors, eqs. (S.44) through (S.48) become the Dirac equations (iγ µ µ + m)ψ = 0 and i µ Ψγ µ + mψ = 0. (S.53) 9