Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  How to understand the unitary?

+ 0 like - 0 dislike
927 views

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
asked Oct 10, 2014 in Mathematics by Roger209 (0 points) [ no revision ]
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

1 Answer

+ 2 like - 0 dislike

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 pawel_winzig (70 points) [ no revision ]
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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...