I would like to understand an apparently rather simple calculation which checks the closure of the Supersymmetry algebra via the commutator of 2 supersymmetric variations of the type:

$$\delta \phi = i(\epsilon^\alpha Q_\alpha + \overline{Q}_\dot{\alpha} \overline{\epsilon}^\dot{\alpha})\phi$$

where $Q_\alpha$ and $Q_\dot{\alpha}$ are the supercharges and $\phi$ is a complex scalar field. In the following the commutator $[\delta_1,\delta_2]\phi$ is evaluated:

$$[\delta_1,\delta_2]\phi =-[\epsilon^\alpha_1 Q_\alpha + \overline{Q}_\dot{\alpha} \overline{\epsilon}_1^\dot{\alpha},\epsilon_2^\beta Q_\beta + \overline{Q}_\dot{\beta} \overline{\epsilon}^\dot{\beta}_2]\phi =$$

$$-\left( [\epsilon^\alpha_1 Q_\alpha, \epsilon^\beta_2 Q_\beta] +[\epsilon^\alpha_1 Q_\alpha, \overline{Q}_\dot{\beta} \overline{\epsilon}^\dot{\beta}_2] + [\overline{Q}_\dot{\alpha} \overline{\epsilon}^\dot{\alpha}_1, \epsilon_2^\beta Q_\beta ] + [\overline{Q}_\dot{\alpha} \overline{\epsilon}^\dot{\alpha}_1,\overline{Q}_\dot{\beta} \overline{\epsilon}^\dot{\beta}_2]\right)\phi$$

Now I assume that the commutators $[\epsilon^\alpha_1 Q_\alpha, \epsilon^\beta_2 Q_\beta]$ and $[\overline{Q}_\dot{\alpha} \overline{\epsilon}^\dot{\alpha}_1,\overline{Q}_\dot{\beta} \overline{\epsilon}^\dot{\beta}_2]$ are zero.

In the paper (I regret I no longer have the reference) I've seen the calculation, however, the 2 remaining commutators turn into anti-commutators :

$$[\delta_1,\delta_2]\phi =-\left(\epsilon_1^\alpha \left\{Q_\alpha, \overline{Q}_\dot{\beta}\right\}\overline{\epsilon}_2^\dot{\beta}-\epsilon_2^\beta \left\{Q_\beta, \overline{Q}_\dot{\alpha}\right\}\overline{\epsilon}_1^\dot{\alpha}\right)\phi$$

and with simplified notation the following is obtained($P_\mu$ is the momentum 4-vector):

$$[\delta_1,\delta_2]\phi =-2\left(\epsilon_1 \sigma^\mu\overline{\epsilon}_2 P_\mu- \epsilon_2 \sigma^\mu\overline{\epsilon}_1 P_\mu\right)\phi = 2i\left( \epsilon_1 \sigma^\mu\overline{\epsilon}_2 - \epsilon_2 \sigma^\mu\overline{\epsilon}_1\right)\partial_\mu\phi$$

So in this calculation it is shown that the supersymmetric algebra is closed, however I have no clue at all how the commutator $[\epsilon^\alpha_1 Q_\alpha, \overline{Q}_\dot{\beta} \overline{\epsilon}^\dot{\beta}_2]$ respectively $[\overline{Q}_\dot{\alpha} \overline{\epsilon}^\dot{\alpha}_1,\epsilon^\beta_2 Q_\beta ]$ is changed to $\epsilon_1^\alpha \left\{Q_\alpha, \overline{Q}_\dot{\beta}\right\}\epsilon_2^\dot{\beta}$ respectively $-\epsilon_2^\beta \left\{Q_\beta, \overline{Q}_\dot{\alpha}\right\}\overline{\epsilon}_1^\dot{\alpha}$.

Actually it is only an algebraic problem(however, SUSY algebra is really complicated), however, for me it is particularly curious how a commutator can become an anticommutator.

  • $\begingroup$ Hi Frederic, are you sure your first equation is correct? If $Q$ denotes a generic Noether (super)charge, then $\delta\phi\equiv i\theta[Q,\phi]$, with a (super)commutator. Did you forget to write the brackets? Or is $Q$ not a (super)charge? If it is the super covariant derivative, the standard notation is $D,\bar D$. $\endgroup$ Commented Jul 21, 2018 at 14:40

2 Answers 2


Actually in SUSY, there is only one commutator and it is called graded commutator or super-commutator, which is denoted and defined as follows: $$ \{ A, B ] = AB - (-1)^{ab} BA $$ where $a$ and $b$ are the degrees of $A$ and $B$, respectively, which is $1$ if the operator is fermionic and $0$ if it is bosonic. Some textbooks or papers use only [,] notation for graded commutators.

So, if even one of $A$ or $B$ is bosonic, then graded commutator reduces to the usual commutator, otherwise it becomes anti-commutator. This means that the algebra is closed under graded commutation relations of the generators.

Note that $\delta_i$ are bosonic since terms like $\epsilon_i^\alpha Q_\alpha$ are not fermionic as a whole. However, graded commutators satisfy the commutation axioms regardless their arguments. So for the distribution property of a graded commutator over multiplication, the result will be again in terms of graded commutators, even if the arguments were bosonic operators. One may switch graded ones into usual in the end of the calculation, of course.


Oktay Dogangun's answer is exactly right:

  1. The only relevant (graded) Lie-bracket is the supercommutator $$[A,B]_{SC}~:=~AB-(-1)^{|A||B|}BA ,\tag{1}$$ where $|A|$ and $|B|$ denote the Grassmann parities of $A$ and $B$, respectively.

  2. It is easy to check that the supercommutator satisfies a graded skew/antisymmetry $$ [A,B]_{SC}~=~-(-1)^{|A||B|}[B,A]_{SC}, \tag{2}$$ a graded Jacobi identity, $$ \sum_{A,B,C~{\rm cycl.}} (-1)^{|A||C|}[[A,B]_{SC},C]_{SC}~=~0, \tag{3}$$ and a graded Leibniz rule $$[A,BC]_{SC}~=~[A,B]_{SC}C+ (-1)^{|A||B|}B[A,C]_{SC}, $$ $$[AB,C]_{SC}~=~A[B,C]_{SC}+ [A,C]_{SC}B(-1)^{|B||C|}. \tag{4}$$

  3. Example: When we reduce a supercommutator $[A_1\ldots A_n,B_1 \ldots B_m]_{SC}$ with the help of the graded Leibniz rule (4) we again get $nm$ terms involving supercommutators $[A_i,B_j]_{SC}$. This explains OP's observation.


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