• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,079 questions , 2,229 unanswered
5,348 answers , 22,758 comments
1,470 users with positive rep
819 active unimported users
More ...

  Feynman diagrams and Hartree-Fock

+ 4 like - 0 dislike

I am puzzled by some lines I read in Mattuck's book on Feynman diagrams in many-body problems ( http://www.amazon.com/Feynman-Diagrams-Many-Body-Problem-Physics/dp/0486670473 ) Page 21 (1.14) for those who have the book. Basically after representing the full propagator of an electron in an electron gas by expansion of the electron-electron interaction (not specified but QED I guess), it says: "this is the 'Hartree-Fock' approximation for the electron gas", which I still don't understand. The Hartree-Fock method for me is just an iterative tool to calculate the collective wave-function of self-interacting fermions satisfying the correct anti-symmetrized form. This statement is evasive to me, and I'd like to understand in which way it makes sense.

This post imported from StackExchange Physics at 2015-06-15 19:34 (UTC), posted by SE-user Learning is a mess
asked Nov 2, 2012 in Theoretical Physics by Learning is a mess (75 points) [ no revision ]
I don't have this book but I think "Hartree-Fock approximation" means that you consider only states that are slater determinants of one-electron states, i.e. electron-electron correlations are neglected.

This post imported from StackExchange Physics at 2015-06-15 19:34 (UTC), posted by SE-user jjcale
Ok, and on what grounds do we neglect those correlations, it has always perturbed me that the use of a Slater determinant means complete separability between the electrons wave-functions without it being clearly justified.

This post imported from StackExchange Physics at 2015-06-15 19:34 (UTC), posted by SE-user Learning is a mess

1 Answer

+ 3 like - 0 dislike

These Feynman diagrams can be summed by solving the Dyson-Schwinger equation $$ G = G_0 + G_0\Sigma G $$ This is a self-consistency equation for $G$. Now write $G_0$ and $G$ in terms of single particle wave functions, $$ G(x,x';\omega)=\sum_j \phi_j(x)\phi^*_j(x')\left[ \frac{\Theta(E_j-E_F)}{\omega-E_j+i\epsilon} +\frac{\Theta(E_F-E_j)}{\omega-E_j-i\epsilon} \right]. $$ Then the Dyson-Schwinger equation becomes a coupled set of equations for the eigenfunctions $\phi_j$ and the eigenvalues $E_j$. These are the standard Hartee-Fock equations. This is explained in some detail in many text books, for example Negele and Orland, or Fetter and Walecka.

This post imported from StackExchange Physics at 2015-06-15 19:34 (UTC), posted by SE-user Thomas
answered Nov 4, 2012 by tmschaefer (720 points) [ no revision ]

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:
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