What is supersymmetry?

Question

1. What is supersymmetry?

2. How is it useful in unifying physics? I have heard that supersymmetry states that every fermion is associated with a boson and vice versa. But I don't get hope this can help resolving problems in unifying physics.

Answer 1

What is supersymmetry (in a big nutshell)?

To explain supersymmetry, let us consider the simplest supersymmetric model which is the Wess-Zumino model with an action,

$$S=-\int d^4x \left( \frac{1}{2}\partial^\mu \phi^\star \partial_\mu \phi + i\psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi + |F|^2 \right)$$

where $\phi$ is a scalar field, $\psi$ a Majorana spinor field and $F$ is known as an auxiliary field, a pseudo-scalar, which arises due to a subtlety explained later. A supersymmetry is a continuous symmetry relating bosons to fermions and vice versa. Most importantly, for it to be a symmetry of the system, we must require that the Lagrangian changes only up to a total derivative. Naively, we might propose:

$$\delta\phi = \epsilon_\alpha \psi^\alpha$$ $$\delta\psi^\alpha = \epsilon^\alpha\phi$$

where we have introduced an infinitesimal parameter $\epsilon^\alpha$ which is required in order to balance the indices on both sides of the transformation. The parameter is fermionic; to deduce the dimensionality, we can do some dimensional analysis.

$$[\partial_\mu\phi^\star\partial_\mu\phi] = 4,\; [\partial_\mu] = 1$$

hence $[\phi] = 1$. Similar reasoning for the fermionic field shows $[\psi] = 3/2$, as expected. Now from the transformation, we may deduce the dimension of the arbitrary parameter:

$$[\epsilon^\alpha] = [\psi]-[\phi] = 1/2$$

and $[\epsilon_\alpha] = -1/2$. Plugging the transformation in, and deducing $\delta\mathcal{L}$, we will find it does not quite cancel to leave only a total derivative left. After some manipulation, we can find:

$$\delta\psi_\alpha = -i(\sigma^\nu\epsilon^\dagger)_\alpha \partial_\nu \phi$$

Now to address the auxiliary field, $F$. If we compute the equations of motion, we find

$$\frac{\partial \mathcal{L}}{\partial F} \sim F \quad \frac{\partial \mathcal{L}}{\partial (\partial F)} = 0$$

which implies $F=0$. Although it may seem adding the field has no effect, remember that $F=0$ is only true on-shell. The purpose of $F$ then is actually off-shell, which you will notice balances the required degrees of freedom of the system for consistency.

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A Historical Note

In 1967, Coleman and Mandula published a paper, All Possible Symmetry of the S-Matrix, which aptly presented a theorem restricting S-matrix symmetries. However, a key assumption was

A symmetry transformation is said to be an internal symmetry transformation if it commutes with P. This implies that it acts only on particle-type indices, and has no matrix elements between particles of different four-momentum or different spin. A group composed of such transformations is called an internal symmetry group.

In other words, the symmetries were bosonic; notice that supersymmetry is in fact fermionic, and hence not subject to the Coleman-Mandula theorem. In some ways, one may attribute the initial development of supersymmetry to a desire to find a loophole in the Coleman-Mandula theorem.

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Resources

• For a simple introduction, see the Beyond the Standard Model course by Professor Veronica Sanz, available here: http://www.perimeterscholars.org/338.html, which includes videos.
• For a modern introduction, complete with superfields, supermanifolds and more, see the text String Theory and M-Theory: A Modern Introduction, by Becker, Becker and Schwarz where it is presented in the context of string theory.