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I'm attending a course on String Theory and AdS-CFT, and my university doesn't have a course dedicated to SUSY, so I only know the basics: for every bosonic field, there is a fermionic one and vice versa. In the mentioned course, we encountered concepts like SUGRA and the ${\cal N}=4$ Super Yang-Mills theory defined on the $D_3$-branes, and I do not understand what the SUSY ${\cal N}$ means. Is it the number of partners each particle has?

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$\mathcal N$ is the number of supersymmetries, i.e there are $\mathcal N$ fermionic generators of supersymmetry transforming in the $(\frac12, 0)$ representation of the Lorentz group, called $Q^I_\alpha$, and another $\mathcal N$ transforming in the conjugate $(0, \frac12)$ representation, called $\bar Q_\dot\alpha^I$. Here $\alpha$ and $\dot\alpha$ are spinor indices, while $I\in{1...\mathcal N}$ is a label.

We use these generators to extend the spacetime Poincare symmetry algebra to the supersymmetry algebra by expressing their commutators with the existing bosonic generators and anticommutation relations between themselves:

$$ I,J\in\{1...\mathcal N\} \\ [P_\mu, Q^I_\alpha]=0 \\ [M_{\mu\nu}, Q^I_\alpha]=i(\sigma_{\mu\nu})_\alpha{}^\beta Q_\beta^I \\ \{Q_\alpha^I, \bar Q_\dot\beta^J\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu\delta^{IJ} \\\{Q_\alpha^I, Q_\beta^J\}=\epsilon_{\alpha\beta} Z^{IJ} $$

and likewise for $\bar Q^I_\dot\alpha$. In particular, the second equation means that $[M_{12}, Q_1^I]=[J_3, Q_1^I]=\frac12 Q_1^I$ and so each $Q_1^I$ maps a state to another state with a helicity/spin greater by $1/2$ a unit. Similar equations hold for the other spinorial indices, as well as the conjugate generators (which lower the spin/helicity by $1/2$ a unit) and so all the supersymmetry generators act generically as $$ Q_\alpha^I|B\rangle=|F\rangle $$ and vice versa.

We then define (fermionic) creation and annihilation operators $a_I, a_I^\dagger$ as certain simple linear combinations of the $Q^I$ and use them to build the supermultiplet. Starting with a "Clifford vacuum" $|E,\lambda_0\rangle$ that is annihilated by all the $a_I$ and furnishes a representation of the Poincaré algebra, we act with combinations of the $a_I^\dagger$ to generate the higher-helicity states in the multiplet $|E,\lambda_0 + \frac k2\rangle, k\in\{0...\mathcal N\}$.

Owing to the fermionic nature of the generators and hence the c/a operators, these supermultiplets contain a finite number of states. This number depends on the dimension of the spacetime, since this determines the existence of Majorana spinors - for instance, in $D=0,4\ \mathrm{mod}\ 8$, the supermultiplet size is $2^\mathcal N$ for the massless case, and between $2^\mathcal N$ and $2^{2\mathcal N}$ states for the massive case, depending on the nature of the "central charge matrix" $Z^{IJ}$ above. See Supergravity by de Wit for the general case.

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  • $\begingroup$ And therefore, if say ${\cal N}$=2, there are 2 $Q_\alpha^I$, and every state $|B\rangle$ is mapped to 2 different states $|F^I\rangle$. But won't these two fermionic states be mapped to 4 bosonic states $|(B^J)^I\rangle$ and so on? Am I missing something? $\endgroup$ May 18 at 9:40
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    $\begingroup$ @MauroGiliberti you probably missed the anticommutativity of the fermionic generators. $Q_1^1 Q_1^1|E,\lambda_0\rangle$ is just 0. So at each level $k$ there are only ${\mathcal N\choose k}$ resp. ${2\mathcal N\choose k}$ states in a generic massless resp. massive multiplet, by simple combinatorics $\endgroup$ May 18 at 10:12
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    $\begingroup$ note that there are some subtleties depending on the number of spacetime dimensions, because the minimal spinor has different properties depending on $d\mod8$. Sometimes they may be real, sometimes complex, so the counting is not as straightforward as $\binom{\mathcal N}{k}$: there might be extra factors of $2$. $\mathcal N$ complex fermions generate a $2^{\mathcal N}$-dimensional module, but $\mathcal N$ real fermions a $2^{\mathcal N/2}$-dimensional one. $\endgroup$ May 18 at 16:52
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    $\begingroup$ @AccidentalFourierTransform good point, I was only thinking of D=4, which is a bit stupid because the OP seems to be talking about supersymmetry in the context of string theory. I will edit it in soon. $\endgroup$ May 18 at 17:00

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