0
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • $\begingroup$ In doing the hermitean conjugate of $c^\dagger c$ you forgot to revert the order. $\endgroup$
    – Void
    Oct 10, 2014 at 15:07
  • $\begingroup$ Is $c$ a Grassmann-odd operator, and you are worried about the sign convention for Hermitian conjugate of Grassmann-odd operators? $\endgroup$
    – Qmechanic ♦
    Oct 12, 2014 at 15:05

1 Answer 1

Reset to default
2
$\begingroup$

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

Improve this answer
$\endgroup$
5
  • $\begingroup$ Tks. I get it. I misunderstood $^{\dagger}$ as $^{T}$. $\endgroup$
    – Roger209
    Oct 10, 2014 at 8:58
  • $\begingroup$ Tks. I get it. I misunderstood the algebraic properity of ${\dagger}$. $\endgroup$
    – Roger209
    Oct 10, 2014 at 9:05
  • $\begingroup$ @Roger209: Please, be so kind and mark the answer if it helped you. $\endgroup$
    – pawel_winzig
    Oct 10, 2014 at 10:48
  • $\begingroup$ Ah, I have tried to upvote it. But I don't have the power. I will mark it in the future. $\endgroup$
    – Roger209
    Oct 10, 2014 at 11:19
  • $\begingroup$ @Roger209: I meant mark as answer. $\endgroup$
    – pawel_winzig
    Oct 10, 2014 at 11:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.