Auxiliary fields in supersymmetry

I know that auxiliary fields can be used to close the supersymmetry algebra in case the bosonic and fermionic on-shell degrees of freedom do not match. Could somebody please elaborate on this concept and explain in which context auxiliary fields play a role?

Answer

The existence of auxiliary fields can be motivated in two different (but related) ways that I know about. The first is using superspace. Superfields are functions of position and two Grassman variables. The auxiliary fields are necessary terms to ensure that a superfield remains a superfield under SUSY transformations.

The second way to motivate auxiliary fields is as a method to ensure a classical theory remains supersymmetric when it becomes quantized. To see how this works consider the simplest and most naive SUSY theory (note that most of this discussion follows from the notes by Steve Martin), \begin{equation} {\cal L} = \underbrace{ \partial _\mu \phi ^\dagger \partial ^\mu\phi} _{ {\cal L} _{ scal}} \underbrace{ - i \psi ^\dagger \bar{\sigma} ^\mu \partial _\mu \psi } _{ {\cal L} _{ ferm}} \end{equation} where $ \phi $ is a scalar and $ \psi $ is a Weyl fermion. We assume that we know nothing about but we want to introduce a symmetry that takes bosons into fermions and the inverse.

Consider a SUSY transformation parametrized by $ \epsilon ^\alpha $. Since SUSY transforms bosons into fermions its easy to guess the transformation law, \begin{align} & \delta \phi = \epsilon \psi \\ & \delta \phi ^\dagger = \epsilon ^\dagger \psi ^\dagger \end{align} With this we have, \begin{equation} \delta {\cal L} _{ scal} = - \epsilon \partial ^\mu \psi \partial _\mu \phi ^\dagger - \epsilon ^\dagger \partial _\mu \psi ^\dagger \partial ^\mu \phi \end{equation}

To find the fermion transformations we find the Lagrangian variation under a SUSY transformation and set it equal to zero: \begin{equation} 0 = \delta {\cal L} = \frac{ \partial {\cal L} _{ ferm}}{ \partial \partial _\mu \psi } \delta \partial _\mu \psi + \frac{ \partial {\cal L} _{ ferm} }{ \partial \psi ^\dagger } \delta \psi ^\dagger - \epsilon \partial ^\mu \psi \partial _\mu \phi ^\dagger - \epsilon ^\dagger \partial _\mu \psi ^\dagger \partial ^\mu \phi \end{equation} Its straightfoward to show that in order for this to be cancelled by the change in the fermion Lagrangian, $ \delta {\cal L} _{ ferm} $ we need, \begin{align} & \delta \psi _\alpha = - i ( \sigma ^\mu \epsilon ^\dagger ) _\alpha \partial _\mu \phi \\ & \delta \psi _{ \dot{\alpha} } ^\dagger = i ( \epsilon \sigma ^\mu ) _{\dot{\alpha}} \partial _\mu \phi ^\dagger \end{align}

We designed our theory to be successfully invariant under a single SUSY transformation. If the SUSY generators combined with the Poincare generators form an algebra then a commutator of the SUSY transformations should result in a linear combination of the Poincare+SUSY generators. This just means that they represent a group. Furthermore, the commutator should be independent of which state it acts on.

We first consider the scalar field commutator, \begin{align} \delta _{ \epsilon _2 } \left( \delta _{ \epsilon _1 } \phi \right) - \delta _{ \epsilon _1 } \left( \delta _{ \epsilon _2 } \phi \right) & = \delta _{ \epsilon _2 } \left( \epsilon _1 ^\alpha \psi _\alpha \right) - \delta _{ \epsilon _2 } \left( \epsilon _2 ^\alpha \psi _\alpha \right) \\ & = \epsilon _1 ^\alpha \left( - i ( \sigma ^\mu \epsilon _2 ^\dagger \right) _\alpha \partial _\mu \phi - \epsilon _2 ^\alpha \left( - i \left( \sigma ^\mu \epsilon _1 ^\dagger \right) _\alpha \partial _\mu \phi \right) \\ & = \left( \epsilon _2 \sigma ^\mu \epsilon _1 ^\dagger - \epsilon _1 \sigma ^\mu \epsilon _2 ^\dagger \right) \underbrace{ i \partial _\mu} _{ {\cal P} _\mu } \phi \end{align}

We now need to get the same result acting on a fermion: \begin{align} \delta _{ \epsilon _2 } \left( \delta _{ \epsilon _1 } \psi \right) - \delta _{ \epsilon _1 } \left( \delta _{ \epsilon _2 } \psi \right) & = \delta _{ \epsilon _2 } \left( - i ( \sigma ^\mu \epsilon _1 ^\dagger ) _\alpha \partial _\mu \phi \right) - \delta _{ \epsilon _1 } \left( - i ( \sigma ^\mu \epsilon _2 ^\dagger ) _\alpha \partial _\mu \phi \right) \\ & = \left[ \left( \epsilon _2 \sigma ^\mu \epsilon _1 \right) - \left( \epsilon _1 \sigma ^\mu \epsilon _2 \right) \right] {\cal P} _\mu \psi - i \epsilon _{ 2 \alpha } \left( \partial _\mu \psi \sigma ^\mu \epsilon _1 \right) - \epsilon _{ 1 \alpha } \left( \partial _\mu \psi \sigma ^\mu \epsilon _2 \right) \end{align} This doesn't appear to be equal to the commutator above. However the equations of motion of this theory are, \begin{align} &\partial ^\mu \sigma ^\mu \psi = 0 \\ \Rightarrow &\partial _\mu \psi \sigma ^\mu = 0 \end{align} thus on-shell the commutator is indeed zero. But, we want this to hold everywhere, including off-shell. There is nothing wrong with our previous theory, it really is supersymmetric. However, in order to have a quantum SUSY theory (and hence include off-shell particles) we need a harmless field that will ensure the algebra closes at all momenta. We call this an auxiliary field, $F$. In order to prevent this field from propagatating we can't let it have any derivatives. In order for the field to be zero on-shell we must have, \begin{equation} {\cal L} _{ aux} = F ^\ast F \end{equation} and so the equations of motions are, \begin{equation} F = F ^\ast = 0 \end{equation} as required. Note that this $F$ terms being zero on-shell is a consequence of working with a theory without any interactions. In general a more complicated condition is needed.

Now the full supersymmetry transformations are modified with a new $F $ term for $\psi$ to eat up the extra terms above, \begin{align} & \delta \phi = \epsilon \psi \\ & \delta \psi _\alpha = - i ( \sigma ^\mu \epsilon ^\dagger ) _\alpha \partial _\mu \phi + \epsilon _\alpha F \\ & \delta F = - i \epsilon ^\dagger \bar{\sigma} ^\mu \partial _\mu \psi \end{align} where the $F$ term was artificially chosen such that the algebra closes. The extra auxiliary field fixes the algebra to let the algebra close off-shell. Its straightforward to check that this leaves the Lagrangian invariant.