CAR-Algebra: Nontriviality?

Question

Given a Hilbert space $\mathcal{h}$.

Consider the abstract CAR-algebra $a:\mathcal{h}\to\mathcal{A}_\text{CAR}$.

Then their actually isometries: $$a:=a(f):\quad a^*\|f\|a=a^*\{a,a^*\}a=(a^*a)^*\\\implies\|f\|^2\|a\|^2=\|f\|^2\|a^*a\|=\|(a^*a)^2\|=\|a^*a\|^2=\|a\|^4\implies\|a\|=\|f\|$$ What if before the last step: $$f_0\neq0:\quad a(f_0)=0$$ Obviously, it holds: $$f=0\implies a(f)=0$$ But that is not enough...

Answer 1

Aah, sometimes things are so simple. :)

Suppose it vanishes: $a(f_0)=0$

Then one has by the CAR relations: $$0=\{a(f),a(f)^*\}=\|f\|^2\neq0$$ That is a contradiction!