Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It is a very trivial doubt but somehow I am not able to figure it out. While constructing a supersymmetric lagrangian we always even number of fermionic fields.

  1. One reason is of course the product of $\psi_1 \psi_2$ behaves as a bosonic variable, but still I would like to know what are the other reasons?

  2. If we have terms with odd number of fermions will the theory be non-supersymmetric or is there some fundamental property of an action/lagrangian that it restricts such kind of terms?

share|improve this question
add comment

3 Answers

The Lagrangian is Lorentz-invariant, which means it cannot contain fermionic terms.

share|improve this answer
1  
Lorentz-Invariance has to do with space-time symmetries, how come fermionic terms affect it ? The anti-symmetric property of fermions is within it's own algebra. –  Jaswin Nov 22 '12 at 15:49
2  
Fermions, by definition, have spin and transform under the Lorentz group. –  Matt Reece Nov 22 '12 at 16:08
add comment

There is a formulation of supersymmetry which based on odd structures. In mathematics it is called odd symplectic geometry. In this framework one can formulate supersymmetric models with odd Hamiltonians and Lagrandians. Please see the following article by: Soroka, Soroka and Wess. In this work they constructed the supersymmetric harmonic oscillator using an odd Lagrangian. Please see also the following article by Soroka.

share|improve this answer
add comment

I interpret OP's question as

Why does the action $S$ carry definite Grassmann parity, and in particular, why is $S$ Grassmann-even?

This question is actually not restricted to just supersymmetric models, but could be asked for any theory with Grassmann-odd variables.

It is true that the overwhelming majority of published literature have $S$ manifestly Grassmann-even.

  1. At the classical level ($\hbar=0$), there is in principle nothing wrong with an action $S$ with indefinite or odd Grassmann parity, as long as the corresponding Euler-Lagrange equations reproduce the classical equations of motion.

  2. At the quantum mechanical level, the situation changes. For instance, in the path integral formalism, it would be rather weird to consider a Boltzmann-factor $\exp(\frac{i}{\hbar}S)$ where the action $S$ has indefinite or odd Grassmann parity. (If one would like to consider such a construction, one would have to insert appropriate projection operators to ensure that the overlaps and amplitudes remain Grassmann-even.) Recall that no measuring device in an experiment is going to measure a Grassmann-odd number. A measuring device can only produce real outputs $\subseteq\mathbb{R}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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