# How to understand the unitary?

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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?

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Roger209
In doing the hermitean conjugate of $c^\dagger c$ you forgot to revert the order.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Void

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The problem is: $(AB)^\dagger=B^\dagger A^\dagger$. Look how you treat $c^\dagger c$.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user pawel_winzig
answered Oct 10, 2014 by (70 points)
Tks. I get it. I misunderstood $^{\dagger}$ as $^{T}$.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Roger209
Tks. I get it. I misunderstood the algebraic properity of ${\dagger}$.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Roger209
@Roger209: Please, be so kind and mark the answer if it helped you.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user pawel_winzig
Ah, I have tried to upvote it. But I don't have the power. I will mark it in the future.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user Roger209
@Roger209: I meant mark as answer.

This post imported from StackExchange Physics at 2014-10-11 09:51 (UTC), posted by SE-user pawel_winzig

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