Thermal average of operators

[Niles]

Hi

Please take a look at equation 8.60 in the following link: http://books.google.com/books?id=v5vhg1tYLC8C&pg=PA131&dq=%22where+the+last+equality+follows+from+the+fer mi%22&hl=da&ei=i6v2TP2sLYnoOaSxueUI&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwAA#v=onepage&q=%22where%20the%20last%20equality%20follows%20fro m%20the%20fermi%22&f=false

My problem is that I cannot see how they go from second last line to \left\langle {c_\nu ^\dag c_\nu } \right\rangle + \left\langle {c_\nu c_\nu ^\dag } \right\rangle. It has something to do with the thermal average of operators (see here equation 1.118, e.g.: http://books.google.com/books?id=v5vhg1tYLC8C&pg=PA27&dq=%22likewise+the+thermal+average&hl=da&ei=Iaz2TPnqEc-aOsuo3OgI&sa=X&oi=book_result&ct=result&resnum=3&ved=0CC4Q6AEwAg#v=onepage&q=%22likewise%20the%20thermal%20average&f=false). I would appreciate a hint very much.

Best,
Niles.

[xepma]

First off all, you know that the thermal average is given by:
\langle c c^\dag\rangle = \frac{1}{Z}\sum_n \langle n|c c^\dag\rangle e^{-\beta n}

You can link to this the expression for the spectral density by insterting the identity operator:

\mathbf{1} = \sum_{n'}|n'\rangle\langle n'|

This is a common trick, so it's good to be familiar with it:

\langle c c^\dag\rangle =\frac{1}{Z}\sum_{n,n'} \langle n|c| n'\rangle\langle n'| c^\dag\rangle e^{-\beta n}

Can you solve it from there?

[Niles]

Yeah, I believe I can take it from here. Thanks for replying so fast.

Best wishes,
Niles.

[Niles]

Just for clarification: When the authors e.g. write something like (eq. 8.61 in http://books.google.com/books?id=v5vhg1tYLC8C&pg=PA131&dq=%22where+the+last+equality+follows+from+the+fer mi%22&hl=da&ei=i6v2TP2sLYnoOaSxueUI&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwAA#v=onepage&q=%22where%20the%20last%20equality%20follows%20fro m%20the%20fermi%22&f=false) the Greens function G(v, ω), then the v is a set of quantum numbers. By v, is it correct that we are refering to e.g. the many-particle state given by


\left| { \uparrow \,\, \downarrow } \right\rangle


where (in this specific case) spin up and spin down denote single particle states?

Is it furthermore correct that v can also denote just a single particle state like


\left| { \uparrow } \right\rangle


?

Best,
Niles.