# Grassman number

RedX
If you have a Grassman number $$\eta$$ that anticommutes with the creation and annihilation operators, then is the expression:

$$<0|\eta|0>$$

well defined? Because you can write this as:

$$<1|a^{\dagger} \eta a|1>=-<1| \eta a^{\dagger} a|1> =-<1|\eta|1>$$

But if $$\eta$$ is a constant, then shouldn't:

$$<0|\eta|0>=<1|\eta|1>=\eta$$ ?

If you have a Grassman number $$\eta$$ that anticommutes with the creation and annihilation operators, then is the expression:

$$<0|\eta|0>$$

well defined? Because you can write this as:

$$<1|a^{\dagger} \eta a|1>=-<1| \eta a^{\dagger} a|1> =-<1|\eta|1>$$

But if $$\eta$$ is a constant, then shouldn't:

$$<0|\eta|0>=<1|\eta|1>=\eta$$ ?

Grassmann numbers are operators (though they are called numbers).
$$<0|\eta|0>=0$$ is well-defined and vanishes.