# SUSY multiplets - simple question

Hi,
I have a conceptual problem in understanding the SUSY (N=1) massless supermultiplet.
Using appropriately normalized creation and annihilation operators Q, Q+ (only one component survives in this representation) we have for the quark state:

Q+|p,-1/2>=0 (quark) where the 1/2 labels the eigenvalue of J3 spin operator.

For the gluino we can write
Q+|p,-1/2>=Q+Q|p,-1>=(2E-QQ+)|p,-1> ~ |p,-1> (gluon)

I don't understand why can we go from the gluino to the gluon but for the quark (also spin 1/2 like the gluino) the superpartner is a scalar.

## Answers and Replies

fzero
Homework Helper
Gold Member
Hi,
I have a conceptual problem in understanding the SUSY (N=1) massless supermultiplet.
Using appropriately normalized creation and annihilation operators Q, Q+ (only one component survives in this representation)

I'm not sure why you think that only one component survives. In 4d, you need both spinor components.

we have for the quark state:

Q+|p,-1/2>=0 (quark) where the 1/2 labels the eigenvalue of J3 spin operator.

For the gluino we can write
Q+|p,-1/2>=Q+Q|p,-1>=(2E-QQ+)|p,-1> ~ |p,-1> (gluon)

I don't understand why can we go from the gluino to the gluon but for the quark (also spin 1/2 like the gluino) the superpartner is a scalar.

These are two different representations. To describe a supermultiplet, you also need to describe its representation under the Poincare algebra. So it's best to start with a Poincare multiplet, which is described by a mass $$M$$ and a spin $$s$$. The states are labeled as $$|M,s,m_s\rangle$$. Since you want massless multiplets, we'll take $$M=0$$ and just write the states as $$|s,m_s\rangle$$.

We view $$Q_\alpha$$ as lowering operators and $$Q^\dagger_{\dot{\alpha}}$$ as raising operators. The lowest weight state is

$$|\Omega_s \rangle = Q_1 Q_2 | s, m_s \rangle,$$

since

$$Q_1 | \Omega_s\rangle =Q_2 | \Omega_s\rangle =0.$$

The rest of the states in the supermultiplet are generated by action with $$Q^\dagger_{\dot{\alpha}}$$. So the supermultiplet is

$$|\Omega_s \rangle , ~~ Q^\dagger_{\dot{1}} | \Omega_s\rangle,~ Q^\dagger_{\dot{2}} | \Omega_s\rangle , ~~Q^\dagger_{\dot{1}} Q^\dagger_{\dot{2}} | \Omega_s\rangle .$$

For the scalar, or chiral, supermultiplet, we start with the state

$$|\Omega_0 \rangle = Q_1 Q_2 | 0, 0 \rangle .$$

Acting with $$Q^\dagger_{\dot{\alpha}}$$ gives us a Weyl fermion

$$Q^\dagger_{\dot{1}} | \Omega_s\rangle,~ Q^\dagger_{\dot{2}} | \Omega_s\rangle \sim | \tfrac{1}{2},\pm \tfrac{1}{2} \rangle ,$$

while

$$Q^\dagger_{\dot{1}} Q^\dagger_{\dot{2}} | \Omega_s\rangle \sim | 0,0\rangle$$

pairs with the lowest weight state to form a complex scalar.

For the vector multiplet, we start with a Weyl fermion

$$|\Omega_{1/2} \rangle = Q_1 Q_2 | \tfrac{1}{2}, \pm \tfrac{1}{2} \rangle .$$

Acting with $$Q^\dagger_{\dot{\alpha}}$$ gives us states that fill out a Lorentz vector

$$Q^\dagger_{\dot{\alpha}}|\Omega_{1/2} \rangle \rightarrow | 1 , m_1 \rangle$$,

while acting with $$Q^\dagger_{\dot{1}} Q^\dagger_{\dot{2}}$$ gives us a Weyl fermion of the opposite chirality.

Thank you very much for your reply.
I'm not sure why you think that only one component survives. In 4d, you need both spinor components.
Here I mean that for the massless case $$p_{\mu}=(E,0,0,E)$$ the algebra yields $$\{Q_{a},\bar{Q}_{\dot{b}}\}$$=$$2(\sigma^{\mu})_{a\dot{b}}P_{\mu}$$=$$2E(\sigma^0+\sigma^3)_{a\dot{b}}$$=$${\left(\begin{array}{cc} 1 &0 \\ 0 &0 \end{array}\right)_{a\dot{b}}}$$ implying that $$Q_{2}$$ is zero in this representation.

Apart from that I agree with everything you said in your post. My final question is
For the vector multiplet, we start with a Weyl fermion
why can't we identify this Weyl fermion with the quark? In that case could we raise the spin eigenvalue $$m_{s}$$ by 1/2 and find a massless spin 1 state pairing with the quark?

fzero
Homework Helper
Gold Member
Thank you very much for your reply.

Here I mean that for the massless case $$p_{\mu}=(E,0,0,E)$$ the algebra yields $$\{Q_{a},\bar{Q}_{\dot{b}}\}$$=$$2(\sigma^{\mu})_{a\dot{b}}P_{\mu}$$=$$2E(\sigma^0+\sigma^3)_{a\dot{b}}$$=$${\left(\begin{array}{cc} 1 &0 \\ 0 &0 \end{array}\right)_{a\dot{b}}}$$ implying that $$Q_{2}$$ is zero in this representation.

Apart from that I agree with everything you said in your post.

OK. I've given massive supermultiplets then. I've also missed a scalar in the massive vector multiplet.

My final question is

why can't we identify this Weyl fermion with the quark? In that case could we raise the spin eigenvalue $$m_{s}$$ by 1/2 and find a massless spin 1 state pairing with the quark?

The Weyl fermion in the vector multiplet must transform in the same representation of the gauge group as the vector field, so the adjoint. This explains why the gauginos are in adjoint representations, but it also means that you can't identify them with fundamental quarks. We can put chiral multiplets in arbitrary representations of the gauge group, so they're necessary.