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I would like to know that what does the nilpotent physically represents? For example, in BRST quantization of point particle, BRST charge is nilpotent means square of this operator gives zero (mathematically) but what does it represents physically?

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The BRST operator, much like some supersymmetry generators that may be nilpotent, is a Grassmannian operator – a fermionic one. So it cannot have any nonzero real eigenvalues.

The discussion of its "meaning" is therefore very special and inseparable from the BRST procedure. The fun is that whenever we have a nilpotent operator $$ Q^2 = 0 ,$$ it is very easy to find some state vectors that obey $Q|\psi\rangle = 0$. How? Simply define $$ |\psi\rangle = Q |\lambda\rangle $$ and you're guaranteed that $$ Q |\psi \rangle = Q^2 |\lambda\rangle = 0. $$ So nilpotent operators are interesting because they allow you to find the states that are "annihilated" by these operators. The interesting solutions to $Q|\psi\rangle = 0$ are those that cannot be written as $Q|\lambda\rangle$, and those are basically called the BRST cohomologies and have the physical meaning of the physical states of non-zero norm.

In this sense, the "physical meaning" of the BRST operator is "maximally physical", indeed – all the physical states are BRST cohomologies of $Q$. To understand why we use $Q$ and nilpotent operators at all, you need to think mathematically, however. It's really a clever mathematical trick to deal with gauge symmetries.

Also, some combinations of supersymmetry generators $Q$ may be nilpotent. In this case, we're not too interested in cohomologies. We're generalizing the Poincaré group by adding some fermionic supersymmetry generators. The nilpotency physically means that such a $Q$ may be imagined as switching you between fermions and their bosonic superpartners. $Q$ may map an electron to a selectron, but if you act with such a $Q$ again, you get zero. Another $\bar Q$ may get you from the selectron to the electron state, but if you act again, you get zero. So the nilpotency means that the second step is zero and it is therefore physically equivalent to the finite number of particle types whose properties are linked to each other by supersymmetry.

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