Twice of supersymmetric transformation = translation

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1. Feb 16, 2015

wphysics

Hello, I have one conceptual question. I have been working on Supersymmetry.

Now, I understand that twice of supersymmetric transformation is equivalent to translation mathematically(naively).
However, I don't quite understand why this should be the case conceptually. Supersymmetric transformation is that boson to fermion and fermion to boson, meaning that it changes spin by 1/2. How can twice of it be equivalent to just translation?

I just want a conceptual answer, so it does not need to be super consistent.

2. Feb 16, 2015

ChrisVer

Twice a supersymmetric transformation = supersymmetric transformation. I don't know why you say that you have just translation. Otherwise you wouldn't be able to form a group.

What is equivalent to a translation is the commutator of two supersymmetric transformations.

3. Feb 16, 2015

wphysics

In Super Poincare algebra, we know that the anti commutator of two weyl spinors is proportional to translation. This is the equivalence that I am talking.

Thanks

4. Feb 16, 2015

ChrisVer

So you are asking why $\{ Q_a , \bar{Q}_\dot{b} \} = 2 \sigma^\mu_{a\dot{b}} P_\mu$ ?

5. Feb 16, 2015

wphysics

Yes! I am wondering why it is the case conceptually.

6. Feb 16, 2015

ChrisVer

But looking at it, the anticommutator will give you something like $(1/2 , 0) \times (0 ,1/2)$ (the times symbol looks weird here but I hope you understand what I wanted to point out), so it seems natural for me to give you a normal Lorentz repr (and not an irrep). So the $\sigma^\mu$ and the momentum.

7. Feb 16, 2015

wphysics

Yes. That explanation is the most common one. Let me put it in this way. For spin 0 particle, supersymmetric transformation transforms it to spin 1/2 particle. If we do again, it should be back to spin 0 and effectively, its effect is translation.

I cannot draw this picture in my head physically.

8. Feb 16, 2015

ChrisVer

9. Feb 16, 2015

ChrisVer

That's the reason why you can look at supersymmetry as a generalization of spacetime symmetry. And when you write a SUSY element in exponential form, you don't only add the spinor generators Q,Q* but also the momentum. I think it's intrinsic to it for closure reasons. Otherwise you can try to have a look at the Coleman-Mandula theorem.

10. Feb 17, 2015

haushofer

You can regard supertransformations Q as translations in the fermionic direction, and P as translations in spacetime. Both are as such translations in superspace.

11. Feb 25, 2015

Roy_1981

@wphysics

The best way to think about your query, in my opinion is look at the converse of the situation i.e. a susy transformation is DEFINED as a "square root" of a translation. This is exactly analogous to the way we define/picture spinors as "square root" of a vector following Dirac (who as the folklore has it arrived at the Dirac spinor equation by taking a "square root" of Klein-Gordon operator).

There is no simple way to motivate why a square of susy transformation is translation, but as some other members have pointed out, perhaps we need to augment our spacetime with Grassmann directions in addition to usual spacetime i.e. picture a larger spacetime (superspace) . And then the anticommutator of two susy transformation can be seen as a round trip in the Grassman direction, and this round trip does not bring you at the starting point, instead you get a shift along the usual spacetime direction i.e. a translation. This is bit like moving around in a helix every 360 degree doesn't return you to your original position but you arrive a point veritcally shifted over the starting point along the axis.

12. Feb 25, 2015

ChrisVer

Well it is still not enough, because you don't make two fermions... it's a QQ* combination (left/right-handed fermions)...

13. Feb 25, 2015

Roy_1981

You are semi-correct :). You compose two supersymmetry and then undo it by acting in reverse order (for bosons this AB-BA, for fermionic operations, its AB+BA). This set of two operations and then the reveres constitutes the departure from "parallelogram law" with the departure/defect being a translation operation. Two fermions of opposite handedness indeed can be turned into a vector using Pauli matrices ($\sigma^{\mu}$).