Wikipedia says:

In particle physics, supersymmetry (often abbreviated SUSY) is a symmetry that relates elementary particles of one spin to other particles that differ by half a unit of spin and are known as superpartners. In a theory with unbroken supersymmetry, for every type of boson there exists a corresponding type of fermion with the same mass and internal quantum numbers (other than spin), and vice-versa. There is only indirect evidence for the existence of supersymmetry [...]

I want a mathematical explanation of SUSY.


2 Answers 2


Mathematically, SUSY begins with the supersymmetry algebra, a Lie superalgebra, which is itself a special case of a more general class of algebras called graded Lie algebras. Of central importance is the supersymmetry algebra referred to as the super-Poincare algebra that extends the Poincare algebra to include supersymmetry "charges" and their anticommutators.

In the context of physics, one studies field theories, both classical and quantum, that exhibit invariance under some action of supersymmetry algebras on fields and Hilbert spaces of these theories. As a result, representations of supersymmetry algebras are especially important in physics.

I would highly recommend that you look at THIS set of notes written by Sohnius, one of the original supersymmetry masters and co-discoverers of THIS famous and important theorem which really motivates why supersymemtry is all the rage in physics. The notes talk about representations of supersymmetry algebras in a lot of detail, and the clarity of the prose is top-notch if you ask me.

Addendum. I almost forgot, you also hear the word "superspace" which is a construction that physicists use to, among other things, make constructing manifestly supersymmetric Lagrangians easier. The mathematics behind this is supermanifolds.

Lastly, there is some discussion of these things on math.SE, see for example

https://math.stackexchange.com/questions/1204/why-are-superalgebras-so-important https://math.stackexchange.com/questions/51274/motivation-for-supermanifolds

  • $\begingroup$ By the way, superspace in the form of super-Minkowski space is right there in the super-Poincare algebra, being the quotient of that by the Lorentz sub-algebra (ncatlab.org/nlab/show/super+Poincare+Lie+algebra). $\endgroup$ Sep 18, 2013 at 17:46
  • $\begingroup$ @UrsSchreiber Interesting stuff. Thanks for the link! $\endgroup$ Sep 18, 2013 at 18:55
  • $\begingroup$ The link to the set of notes written by Sohnius is broken, as far as I can tell. Are the notes you were referencing 'Introducing Supersymmetry' by Martin F. Sohnius, currently available here ? $\endgroup$ Mar 3, 2020 at 1:39
  • $\begingroup$ @perilousGourd Yes that's the one, although it seems one can find versions not behind a paywall as well on Google Scholar. $\endgroup$ Mar 3, 2020 at 16:32

In addition to what Joshua said in his nice answer, may favorite (simplified) point of view is looking at a SUSY transformation as a coordinate transformation (translation) in superspace

$$ x' = x + a + \frac{i}{2}\zeta\sigma^{\mu}\bar{\theta} - \frac{i}{2}\theta\sigma^{\mu}\bar{\zeta}$$

$$ \theta'= \theta + \zeta $$

$$ \bar{\theta}'= \bar{\theta} + \bar{\zeta} $$

with $\theta$ and $\bar{\theta}$ denoting the additional Grassmanian coordinates.

The supersymmetry generators or supercharges, when written down as differential operators, contain momentum operators in both, the "usual" even spacetime coordinates and the odd Grassmann coordinates

$$ Q_a = i\partial_a -\frac{1}{2}(\sigma^{\mu})_{a\dot{b}}\bar{\theta}^{\dot{b}}\partial_{\mu}$$

$$ \bar{Q}^{\dot{a}} = i\bar{\partial}^{\dot{a}} -\frac{1}{2}(\bar{\sigma}^{\mu})^{\dot{a}b}\theta_b\partial_{\mu}$$

Where $\partial_a = \frac{\partial}{\partial\theta^a}$ and $\bar{\partial}^{\dot{a}} = \frac{\partial}{\partial\bar{\theta_{\dot{a}}}}$ are the derivatives along the Grassmanian coordinates.

A nice and very readable introduction to the superspace formalism can for example be found in Ch 11 of this book.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.