[Wess&Bagger SUSY&SUGRA] 1. Why Supersymmetry?

 This article is one of the posts in the Textbook Commentary Project.

SUSY is good

Supersymmetry is a subject of considerable interest among physicists and mathematicians. Not only is it fascinating in its own right, but there is also a grouwing belief that it may play fundamental role in particle physics. (This book is written in 1982..) This belief is based on an important resulf of Haag, Sohnius, and Lopuszanski, who proved that the supersymmetry algebra is the only graded Lie algebra of symmetries of the $S$-matrix consistent with relativistic quantum field theory. In this chapter, we shall discuss their theorem and its proof. (Readers specificallt interested in supersymmetric theories might prefer to start directly with Chapter II or III.)

SUSY algebra

Before we begin, however, we first present the suppersymmetry algebra: $$\begin{align} \{Q_\alpha^A, \bar{Q}_{\dot{\beta}B}\}_+&= 2\sigma_{\alpha \dot{\beta}}^m P_m \delta^A_B \nonumber\\ \{Q_\alpha^A, Q_\beta^B \}_+&= \{ \bar{Q}_{\dot{\alpha}A},\bar{Q}_{\dot{\beta}B}\}_+= 0\nonumber\\ [P_m,Q_\alpha^A]_-&= [P_m,\bar{Q}_{\dot{\alpha}A}]_-= 0\nonumber\\ [P_m,P_n]_-&= 0.  \end{align}$$ The Greek indices $(\alpha, \beta,\cdots,\dot{\alpha},\dot{\beta},\cdots)$ run from one to two and denote two-component Weyl spinors. The Latin indices $(m,n,\cdots)$ run from one to four and identify Lorentz four-vectors. The capital indices $(A,B,\cdots)$ refer to an internal space; they run from $1$ to some number $N\ge 1$. The algebra with $N=1$ is called the supersymmetry algebra, while those with $N>1$ are called extended supersymmetry algebras. All the notation and conventions used throughout this book are summarized in Appendix A.

Coleman-Mandula Theorem

We are now ready to consider the theorem. Of all the graded Lie algebras, only the suppersymmetry algebras (together with their extensions to include central charges, which we shal discuss at the end of the chapter) generate symmetries of the $S$-matrix consistent with relativistic quantum field theory. The proof of this statement is based on the Coleman-Mandula theorem, the most precise and powerful in a series of no-go theorems abour the possibvle symmetries of the $S$-matrix.

The Coleman-Mandula theorem starts from the following assumptions:

(1) the $S$-matrix is based on a local, relativistic quantum field theory in four-dimensional spacetime;

(2) there are only a finite number of different particles associated with one-particle states of a given mass; and

(3) there is an energy gap between the vacuum and the one particle states.

The theorem concludes that the most general Lie algebra of symmetries of the $S$-matrix contains the energy-momentum operator $P_m$, the Lorentz rotation generator $M_{mn}$, and a finite number of Lorentz scalar operators(charges?) $B_l$. The theorem further asserts that the $B_l$ must belong to the Lie algebra of a compact Lie group.

Another way: Supersymmetry

Supersymmetries avoid the restrictions of the Coleman-Mandula theorem by relaxing one condition. They generalize the notion of a Lie algebra to include algebraic systems whose defining relations involve anticommutators as well as commutators. These new algebras are called superalgebras or graded Lie algebras. Schematically, they take the following form: $$\begin{align} \{Q,Q'\}_+=X\quad [X,X']_-=X''\quad [Q,X]_-=Q''.\end{align}$$ Here $Q$, $Q'$, and $Q''$ represent the odd (anticommuting) part of the algebra, and $X$, $X''$, and $X''$ the even (commuting) part.

The operators $X$ are determined by the Coleman-Mandula theorem. They are either elements of the Poincare algebra $\mathcal{P}=\{P_m,M_{mn}\}$ or elements of a Lorentz-invariant compact Lie algebra $\mathcal{A}$. The algebra $\mathcal{A}$ is a direct sum of a semisimple algebra $\mathcal{A}_1$ and an Abelian algebra $\mathcal{A}_2$, $\mathcal{A}=\mathcal{A}_1\oplus \mathcal{A}_2$. (?? is it useful??)

The generators $Q$ may be decomposed into a sum of representations irreducible under the homogeneous Lorentz group $\mathcal{L}$: $$\begin{align}Q=\sum Q_{\{\alpha_1\cdots\alpha_a\},\{\dot{\alpha}_1\cdots\dot{\alpha}_b\}}.\end{align}$$ The $Q_{\{\alpha_1\cdots\alpha_a\},\{\dot{\alpha}_1\cdots\dot{\alpha}_b\}}.$ are symmetric with respect to the underlined indices $\alpha_1\cdots\alpha_a$ and $\dot{\alpha}_1\cdots\dot{\alpha}_b$. They belong to irreducible spin- $\frac{1}{2}$ $(a+b)$ representations of $\mathcal{L}$. Since the $Q$'s anticommute, the connection between spin and statistics tells us that $a+b$ must be odd.

Restrict $a$ and $b$

We shall now invoke two additional assumptions to prove the $a+b=1$. These assumptions are: 

(1) the operators $Q$ act in a Hilbert space with positive definite metric; and 

(2) both $Q$ and its hermitian conjugate $\bar{Q}$ belong to the algebra.

We start by considering the anticommutator $$\begin{align}\{Q_{\{\alpha_1\cdots\alpha_a\},\{\dot{\alpha}_1\cdots\dot{\alpha}_b\}},\bar{Q}_{\{\dot{\beta}_1\cdots\dot{\beta}_a\},\{\beta_1\cdots\beta_b\}}\},\end{align}$$ where all the indices are assigned the value $1$. The product $$\begin{align}Q_{\{1\cdots 1\},\{\dot{1}\cdots\dot{1}\}},\bar{Q}_{\{\dot{1}\cdots\dot{1}\},\{1\cdots 1\}}\end{align}$$ belongs to a spin-$(a+b)$ representation of $\mathcal{L}$, so $$\begin{align}\{Q_{\{1\cdots 1\},\{\dot{1}\cdots\dot{1}\}},\bar{Q}_{\{\dot{1}\cdots\dot{1}\},\{1\cdots 1\}}\}\end{align}$$ must close into an even element of the algebra with spin $(a+b)$. From the Coleman-Mandula theorem, we know that this element is either zero or a component of $P_m$. For $a+b>1$, it must be zero.

The anticommutator (6) is a positive definite operator in a Hilbert space wit ha positive definite metric. This tells us that $\{Q_{\{1\cdots 1\},\{\dot{1}\cdots\dot{1}\}}=0$ for $a+b>1$. Since the $Q_{\{\alpha_1\cdots\alpha_a\},\{\dot{\alpha}_1\cdots\dot{\alpha}_b\}}$ are irreducible under $\mathcal{L}$, they all must vanish for $a+b>1$. From this we conclude that the odd part of the supersymmetry algebra is composed entirely of the spin-$\frac{1}{2}$ operators $Q_\alpha^L$ and $\bar{Q}_{\dot{\alpha}M}.$

Summarize Commutators

The anticommutator of $Q_\alpha^L$ and $\bar{Q}_{\dot{\alpha}M}$ closes into $P_{\alpha\dot{\alpha}}$, $$\begin{align}\{Q_\alpha^L,\bar{Q}_{\dot{\alpha}M}\}=P_{\alpha\dot{\alpha}}C^L_M,\end{align}$$ where $P_{\alpha\dot{\alpha}}=\sigma_{\alpha\dot{\alpha}}^mP_m$. In Exercise 1(um..) we show that the finite-dimensional matrix $C^L_M$ is hermitian. It may therefore be diagonalized by a unitary transformation. Since $\{Q_1^L,\bar{Q}_{\dot{1}L}\}$ is positive definite, the matrix $C^L_M$ has positive definite eigenvalues. This lets us choose a basis in the odd part of the algebra such that $$\begin{align}\{Q_\alpha^L,\bar{Q}_{\dot{\alpha}M}\}=2P_{\alpha\dot{\alpha}}\delta^L_M.\end{align}$$

We now turn our attention to the anticommutator of two odd elements, both with undotted indices. The right-hand side of this expression may be decomposed into symmetric and antisymmetric parts. The symmetric part has spin $1$. From the Coleman-Mandula theorem, the only possible candidate is the Lorentz generator $M_{\{\alpha\beta\}}$: $$\begin{align}\{Q_\alpha^L,Q_\beta^M\}=\epsilon_{\alpha\beta}X^{[LM]}+M_{\{\alpha\beta\}}Y^{[LM]}.\end{align}$$ From the fact that $P_m$ commutes with $Q_\alpha^L$ (see Exercise 2..), we find that the $Y^{[LM]}$ must vanish. This lets us write the commutator (9) as follows: $$\begin{align}\{Q_\alpha^L,Q_\beta^M\}=\epsilon_{\alpha\beta}a^{l,[LM]}B_l.\end{align}$$ Here $B_l$ is a hermitian element of $\mathcal{A}_1\oplus \mathcal{A}_2$ and $a^{l,[LM]}$ is aytisymmetric in $L$ and $M$. With this result, the supersymmetric algebra takes the following form: $$\begin{align}\{Q_\alpha^L,\bar{Q}_{\dot{\beta}M}\}&=2\sigma_{\alpha\dot{\beta}}^mP_m\delta^L_M\nonumber\\ [P_m,Q_\alpha^L]&=[P_m,\bar{Q}_{\dot{\beta}M}]=0\nonumber\\ \{Q_\alpha^L,Q_\beta^M\}&=\epsilon_{\alpha\beta}a^{l,[LM]}B_l=\epsilon_{\alpha\beta}X^{[LM]}\nonumber\\ \{\bar{Q}_{\dot{\alpha}L},\bar{Q}_{\dot{\beta}M}\}&=\epsilon_{\dot{\alpha}\dot{\beta}}a^*_{l,[LM]}B^l=\epsilon_{\dot{\alpha}\dot{\beta}}X^\dagger_{[LM]}\nonumber\\ [Q_\alpha^L,B_l]&=S_{lM}^LQ_\alpha^M\nonumber\\ [B^l,\bar{Q}_{\dot{\alpha}L}]&=S^{*lM}_L\bar{Q}_{\dot{\alpha}M}\nonumber\\ [B_l,B_m]&=ic_{lm}^kB_k.\end{align}$$

Restrict coefficients

We shall now use the Jacobi identities to further restrict the coefficients $a^{l,[LM]}$ and $S^L_{lM}$ in (11). The ordinary Jacobi identity may be easily extended ot include anticommutators, as is done in Exercise 3: $$\begin{align}\{A,\{B,C]]\pm \{B,\{C,A]]\pm \{C,\{A,B]]=0.\end{align}$$ The bracket structure $\{\ ,\ ]$ signifies either commutator or anticommutator, according to the even or odd character of $A$, $B$, and $C$. The signs are determined by the odd elements. If the odd elements are in a cyclic permutation of the first term, the sign is positive; if not, it is negative. By exploring the Jacobi identities in a certain order, we shall arrive at our results as quickly as possible.

We first consider the identity $$\begin{align} [B_l,\{Q_\alpha^L,\bar{Q}_{\dot{\beta}M}\}]+\{ Q_\alpha^L,[\bar{Q}_{\dot{\beta}M},B_l]\}-\{\bar{Q}_{\dot{\beta}M},[B_l,Q_\alpha^L]\}=0.\end{align}$$ The first term vanishes because $B_l$ and $P_m$ commute. The second and third terms give $$\begin{align} -\{Q_\alpha^L,\bar{Q}_{\dot{\beta}K}\} S^{*lK}_M+\{\bar{Q}_{\dot{\beta}M},Q_\alpha^K\}S_{lK}^L=0,\end{align}$$ or $$\begin{align}2_{\alpha\dot{\beta}}[S^{*lL}_M-S^L_{lM}]=0.\end{align}$$ Equation (15) is true only if $$\begin{align}S^{*lL}_M=S^L_{lM},\end{align}$$ so $S^L_{lM}$ is hermitian.

Next we use the identity ..

We now use the identity...

In Exercise 4..

Conclusion

No further restrictions follow from the other Jacobi identities, as may be proven by checking them all. We have therefore found the most general supersymmetry algebra:

This is the most general graded Lie algebra of symmteries of the $S$-matrix consistent with relativistic quantum field theory. If central charges exist, they must be of the form $X^{[LM]}=a^{l,[LM]}B_l$, where $a^l$ interwines the representations $S_l$ and $-S^{*l}$.