I am getting introduced into supersymmetry reading Ryder "Quantum Field Theory". I have taken an introductory course on QFT last semester so I am far from being an expert. Sorry if this is a silly question.

Now I have a question: why do we want that the transformations we use to exploit the symmetries of theories, such as Poincaré transformations, to form a group?

I ask this question because in Ryder chapter 11 page 448 he introduces two auxiliary fields $F$ and $G$ in order that SUSY transformations to form a group. So it seems important that the transformations to form a group, otherwise we are doing things wrong... Why?

  • $\begingroup$ related: Quantum symmetries which are not described by groups. $\endgroup$ – AccidentalFourierTransform Aug 26 '18 at 19:43
  • $\begingroup$ Okay, so not all symmetries come from group transformations. But why is it so important that SUSY transformations to form a group? It is such an important fact that we need to postulate two hypothetical fields for this to be true? $\endgroup$ – user171780 Aug 26 '18 at 19:50
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    $\begingroup$ Well, if doing X doesn’t change anything, and doing Y doesn’t change anything, then shouldn’t doing X and then doing Y not change anything either? That’s the only idea Ryder used. $\endgroup$ – knzhou Aug 26 '18 at 20:51

Consider three observers Alice, Bob and Charles. They have different views of the world on account of them being moved in some way relative to one another. For example Bob might be rotated with respect to Alice whilst Charles might be boosted with respect to Bob. Alice encodes her view of the world by an element $a$ belonging to some set of "states" $A$. So, $a\in A$ is Alice's current state of the world as she measures it. Similarly $b\in B$ is Bob's state of the world as he measures it and Charles sees the world as $c\in C$. Now, since Alice, Bob and Charles are students of physics and not English literature, they believe that they can come up with a consistent description of the world. So, Alice's set of states $A$ and Bob's set of states $B$ must be related by some map $T_{g_{1}}$ such that $T_{g_{1}} : A\rightarrow B$. Here $g_{1}$ is just a label that picks the particular map. Similarly the sets of states of Bob and Charles are related by another map $T_{g_{2}}:B\rightarrow C$. There must also be a map connecting the states of Alice and Charles $T_{g_{3}}:A\rightarrow C$. So, there are two ways of relating the sets of Alice and Charles. We can go $C=T_{g_{2}}B=T_{g_{2}}T_{g_{1}}A$ and $C=T_{g_{3}}A$. Since they believe they can achieve a consistent description of the world, then $T_{g_{3}}=T_{g_{2}}T_{g_{1}}$. This relation suggests that the maps $T_{g}$ are a group representation. This is confirmed by some further investigation. Suppose Bob is not moved with respect to Alice. Then Bob and Alice are the same observer and $A=B$ and so, in this case $T_{g_{1}}$ is the identity map. So, we know that there is an identity. Furthermore, it must be possible to invert the map to get $A=T_{g_{1}'}B$ so every map has an inverse. So, we've got all the rules for a group representation. So, the answer to the question, is that the transformations form a group representation because the physicists believe there is a consistent description of the world for all observers.


In a more general way, the observers could be replaced by the freedom in physical description. Alice's set of states $A$ could have some coordinates to label each $a\in A$. The particular coordinates chosen should not be important as each coordinate system should be equivalent. The various coordinate systems would also form a representation of a group.

Regarding the supersymmetry transformations. The physics is described by some fields. The values which the fields can take correspond to Alice's set $A$ in my original explanation. The simplest supersymmetric field theory is the massless, non-interacting Wess-Zumino model. This has a scalar field $\phi$ and a chiral fermion field $\psi^{i}$ with components $i=1,2$. I've only studied supersymmetry a little in order to try to decide if it is well-motivated or not. My understanding is that the scalar field and the chiral field are to be regarded as coordinates of a single object which describes the physics. The supersymmetry transformation is a rotation of these coordinates. The small rotation causes the scalar to change by $\delta \phi$ and the chiral spinor to change by $\delta\psi^{i}$. The physics should be the same regardless of how we choose to rotate the coordinates. This means that the Lagrangian of the theory should be invariant under the rotation up to a total derivative. In the supersymmetric theory, the condition that the rotations form a group $T_{g_{3}}=T_{g_{2}}T_{g_{1}}$ has to be studied for infinitesimal rotations for the sake of simplicity. For infinitesimal rotations this condition becomes the requirement that the commutator of two small rotations is another small rotation (closure). It turns out that the supersymmetry transformations are not closed under the commutator unless the fields are on-shell. However, by inventing an auxiliary scalar field $F$ with trivial dynamics and with a different transformation rule to the original scalar field, the supersymmetry transformations close off shell - they form a group. So we end up with an object with coordinates $\phi,\psi^{i},F$ which describes the physics. The physics should not depend on the how we rotate these coordinates. These rotations are the supersymmetry transformations.

In summary, we have some set $A$ which describes the physics. We have to have some coordinates to label states $a\in A$. There must be a freedom to describe the physics with different coordinates. The transformations (rotations) between different coordinates must be a group.

  • $\begingroup$ I cannot see how SUSY transformations are related to different observers. I mean, if Alice is at $x=0$ and Bob at $x=27$ then they are related by a translation in space. If Bob's view of the universe is related to Alice's one by a SUSY transformation, what does this mean? $\endgroup$ – user171780 Aug 27 '18 at 11:15
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    $\begingroup$ @user171780 : I've edited my post to try to address why the supersymmetry transformations must form a group. $\endgroup$ – Stephen Blake Aug 27 '18 at 18:12
  • $\begingroup$ Thanks for your answer. It is similar to what happens with the electric and magnetic fields in electromagnetism and the boost transformations. They are in fact part of the one entity $F^{\mu\nu}$ we call the electromagnetic field. $\endgroup$ – user171780 Aug 27 '18 at 21:06
  1. On the bottom of p. 443 Ryder asks if the supersymmetry transformations (11.96) (on the Wess-Zumino fields $A$, $B$ & $\Psi$) together with the Poincare transformations "form a group." In the beginning of Section 11.4 on p. 446 he re-asks this question, and state that they don't "form a group."

  2. In practice Ryder only considers a (representation of a super) Lie algebra rather than a (super Lie) group. Since the supertransformations (11.96) (together with the Poincare transformations) do indeed realize an on-shell representation of the super-Poincare algebra, it becomes clear that Ryder is asking for an off-shell representation by extending with 2 auxiliary fields $F$ & $G$.

  3. Returning to OP's title question (v4) Ryder himself answers on the bottom of p. 447 that the off-shell realization may be needed/useful in the path integral formulation, where off-shell contributions play a role.


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