# Supersymmetry algebra

• Let the $i=x,y,z$ components of the angular momentum have the commutation relations with the supersymmetry generators (also called supercharges?) $Q_a$ ($a = \pm \frac{1}{2}$) as, $[J_i , Q_a] = -\frac{1}{2} (\sigma _i )_{ab}Q_b$. Now if I want to calculate the $J^2$ eigenvalue of $Q_a$ then I might have thought of doing the following calculation using properties of Lie bracket, (using the summation convention)

$[J_i^2,Q_a] = J_i[J_i,Q_a] + [J_i,Q_a]J_i = -\frac{1}{2} (\sigma_i)_{ab}\{ J_i, Q_b \}$

But this is apparently the wrong answer!

The right answer is gotten by doing the following calculation where by the action of $J^2$ on $Q_a$ is defined as (using the summation convention)

$[J_i,[J_i,Q_a]] = [J_i,-\frac{1}{2} (\sigma _i )_{ab}Q_b] = -\frac{1}{2} (\sigma _i )_{ab} [J_i, Q_b] = \frac{1}{4} (\sigma _i)_{ab} (\sigma _i)_{bc} Q_c = \frac{3}{4}Q_a$

I would like to know why is the above the correct action and not the former.

• But when lets say one wants to evaluate the action of a supercharge on the square of say the scalar component of a chiral superfield then the right action is of the former kind i.e,

$[Q,\phi ^2] = \phi [Q,\phi] + [Q,\phi]\phi$

I don't understand why here what was "wrong" earlier is now the right thing.

• When the supercharge acts on the fermion say $\psi$ then in terms of the auxiliary field in the same multiplet (say $F$) the action as derived by Weinberg in his book is,

$\{ Q_b, \psi_a^*\} =2i \delta _{ab}F^*$

Some people tell me that the above can't be right because the RHS does not have the same spinorial symmetry as the LHS. Also they say that the above implies that irrespective of the nature of $F$ the RHS will go to $0$ if $a\neq b$. But apparently there are theories with superpotential known where even if the spinor indices don't match the supercharge has non-trivial anti-commutation relation with the fermion.

I can't see anything wrong about Weinberg's argument and nor do I see Weinberg using any assumption about the superpotential in deriving the above. I would like to know what is the right way to think about the above.

• If $A$ is the gauge field component of a vector superfield and the chiral superfield to which it couples has $\phi$ and $\psi$ as scalar and spinor components then the following seems to be true,

$[Q,A] = [\phi, \psi]$

I would like to know if there is a general argument for the above and if it is always true. And also what is the space in which the above commutators on the both sides are defined.

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The answer for the first question is that you are using wrong definition of the Casimir operator. Casimir operator is defined as $\sum_i X_i^2$ but the $X_i$ have to be understood as generators of the given representation. In your case $J_i \neq X_i$. That's because you are acting on operators, not just vectors. The generators of the action on operators are $X_i = [J_i, \cdot]$. This is the adjoint representation of the usual $J_i$ (which acts the way you suggested only on vectors). So you see that the correct definition of the Casimir operator in this case is the one with doubled commutator.

Now, I don't see what relation $[Q,\phi^2]$ has to do with the above. It's completely different computation.

I can't say anything about the last two questions but they seem completely disconnected from the first two. I suggest you ask these things separately for the next time.

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@Marek Thanks for your answer. Can you give some references to this point ou are making that the Casimir operator should be thought of as not the square of the generators but the generator acted twice? I haven't come across this difference earlier and hence am a bit puzzled. It would be great if you can give some reference. –  user6818 Feb 22 '11 at 15:42
@Anirbit: well, it's still a square. But you have to remember that it is an operator, so the multiplication is given by composition. So $X_i^2 = [J_i, [J_i, \cdot]]$. It takes some getting used to but it is a part of standard group theory treatment of adjoint representations. Every Lie algebra $\mathfrak g$ has a faithful adjoint representation ${\rm ad}: {\mathfrak g} \to End({\mathfrak g}), X \mapsto [X, \cdot]$ which provides a representation of the elements (thought of as vectors) in terms of operators acting on the underlying algebra. –  Marek Feb 22 '11 at 16:37
@Marek The angular momentum Lie algebra if acting on itself by the adjoint representation then I can easily see what you are saying. But why should the angular momentum Lie algebra act on the supersymmetry algebra by the adjoint action? Why is the double commutator the right thing to do? Is there a derivation for this action or is this a definition put in by hand? It would be helpful if you can state the action of this Casimir on the supersymmetry operator in the language of maps between spaces. –  user6818 Feb 23 '11 at 7:50
@Anirbit: because there's simply no other way to act. You could suggest just pure multiplication $J_i Q$ but this obviously isn't action of a Lie algebra because $[J_i,J_j] Q = J_i J_j - J_j J_i Q \neq J_i J_j Q$. The only way you can get a real action is by using $[J_i, \cdot]$ (which works thanks to Jacobi identity). Are you at least familiar with tensor operators? It's the very same thing. E.g. $\mathbf X$ (position operator in QM) transforms as a vector under $SO(3)$ because $[J_i, X_j] = i \varepsilon_{ijk} X_k$. This is again action by commutator. (cont.) –  Marek Feb 23 '11 at 8:29
Perhaps the group action should look more familiar: $U(g)^{-1}X_kU(g) = \sum_l R(g)_{lk} X_l$. Again, there is no other way for the group to act on operators than just by conjugation (unless there is more structure present in the group; but this is the only generic action). When you differentiate this action, you'll get commutators. –  Marek Feb 23 '11 at 8:32

Dear Anirbit, the action of symmetry generators - such as $Q$ - on operators is always given by the commutator (or anticommutator, with both Grassmann-odd) of the symmetry generator with the operator.

So $Q$ and $J_i$ act on states as $$|\psi\rangle \to Q|\psi\rangle, \quad |\psi\rangle \to J_i|\psi\rangle, \quad$$ but they act on operators $M$ as $$M \to [Q,M]_{\pm}, \quad M \to [J_i,M]. \quad$$ When you ask how $J^2=J_i J_i$ acts on $Q$, you must appreciate that $Q$ is an operator. So the right factor of $J_i$ acts on it as a commutator. The result is $[J_i,Q_a]$ which is still an operator, so the left factor $J_i$ still acts on it as a commutator, so the result is a double commutator summed over $i$.

$[Q,\phi^2]$ may be interpreted as an action of symmetry generator $Q$ on the operator $\phi^2$. Note that in this case, it is the "generic operator" - the right player - that is squared while the symmetry generator, $Q$, is not squared. So no double commutators will occur in this case. After all, the identity for $[Q,\phi^2]$ that you wrote is nothing else than the Leibniz rule. The Leibniz rule is always right and it doesn't depend on any definitions "what we mean by the transformation of something by something".

In Penrose's formalism, $\delta_{ab}$ doesn't exist as an invariant. The correct Penrose version of the equation you mention would either have one of the indices $a,b$ raised - so there would be $\delta^a_b$ or vice versa - or it would have $\epsilon_{ab}$ if both indices were kept as subscripts (raising and lowering of 2-valued indices is done by $\epsilon_{ab}$ which is antisymmetric). There is no contradiction when it comes to the antisymmetry in $ab$. In particular, it is not true that $\{Q_b,\psi^*_a\}$ is $ab$-symmetric because $Q$ is not the same letter as $\psi$, so by exchanging them, you don't get the same thing back!
However, at the schematic level, the supercharge, when written as an operator in terms of fields (integral of its density), contains terms of the type $\int \psi \partial_0 A \phi$ from the covariant derivatives, and when commuted with $A$, $\partial_0 A$ cancels and we're left with the $\psi\phi$ terms - with a dependence on the color indices that was written as a commutator (it only works for the adjoint representation).