# How to understand the unitary? [closed]

In the page 219 of Mahan's Many Particle Physics(3ed), there exists a transform $$S=c^{\dagger}c\sum_q\frac{M_q}{\omega_q}(a_q^{\dagger}-a_q)$$ In order to prove that the transformation relating to $e^{S}$ is $\textit{unitary}$, we should prove that $$(e^S)^{\dagger}(e^S)=I$$ or equivalently, $$S^{\dagger}=-S$$

However, in my opinion, $$S^{\dagger}=\big(c^{\dagger}c\big)^{\dagger}\sum_q\frac{M_q}{\omega_q}(a_q^{\dagger}-a_q)^{\dagger}=\big(cc^{\dagger}\big)\sum_q\frac{M_q}{\omega_q}(a_q-a_q^{\dagger})=\big(-c^{\dagger}c\big)\sum_q\frac{M_q}{\omega_q}\big(-(a_q^{\dagger}-a_q)\big)=S$$ What's wrong?

• In doing the hermitean conjugate of $c^\dagger c$ you forgot to revert the order. – Void Oct 10 '14 at 15:07
• Is $c$ a Grassmann-odd operator, and you are worried about the sign convention for Hermitian conjugate of Grassmann-odd operators? – Qmechanic Oct 12 '14 at 15:05

The problem is: $(AB)^\dagger=B^\dagger A^\dagger$. Look how you treat $c^\dagger c$.
• Tks. I get it. I misunderstood $^{\dagger}$ as $^{T}$. – Roger209 Oct 10 '14 at 8:58
• Tks. I get it. I misunderstood the algebraic properity of ${\dagger}$. – Roger209 Oct 10 '14 at 9:05