What happens in the Hartree and Fock diagrams? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-22T23:15:25Z https://physics.stackexchange.com/feeds/question/457988 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/457988 2 What happens in the Hartree and Fock diagrams? B. Brekke https://physics.stackexchange.com/users/142976 2019-01-31T08:55:04Z 2019-01-31T12:32:33Z <p>I am trying to understand the Hartree and Fock diagram shown in the picture.<a href="https://i.stack.imgur.com/27q3E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/27q3E.png" alt="enter image description here"></a></p> <p>To understand it a assume there is an electron entering and leaving at the tail of the tadpole (Hartree diagram) and an electron entering and leaving at the edges of the Fock diagram.</p> <p>If I were to guess the meaning of the diagrams I would say:</p> <p>Hartree: An electron interacts with another electron, described by the circle, which first appears with some momentum and the suddenly disappear giving the momentum back to the first electron, and in the end no real change happened.</p> <p>Fock: An electron enters, interacts with itself, (I do not understand the result of this self interacting) and then moves on, and in the end no real change happened.</p> <p>I don't find it to suspicious that no real change is happening, this is just the way a particle is propagating.</p> <p>I have no clue if these explanations are even close to the correct interpretation and would appreciate any help and input!</p> https://physics.stackexchange.com/questions/457988/-/457999#457999 4 Answer by Alex for What happens in the Hartree and Fock diagrams? Alex https://physics.stackexchange.com/users/220271 2019-01-31T09:53:47Z 2019-01-31T12:32:33Z <ol> <li><p>In Hartree term the time and the spatial position of the ends of circle line (representing Green's function) coincide. i.e. equal to <span class="math-container">$\langle \psi^\dagger (t,\mathbf{r}) \psi (t,\mathbf{r})=n(t,\mathbf{r})\rangle$</span>, which is nothing else as an electron density. Therefore, Hartree diagram is basically the potential created by all other electrons that an incoming electron feels: <span class="math-container">$\int d^3 \mathbf{r}'V(\mathbf{r}-\mathbf{r}')n(\mathbf{r}')$</span>. That's why is is called mean-free field approximation. This is classical term and neither diagrams nor quantum mechanics is needed to understand it.</p></li> <li><p>Fock, meanwhile, is a quantum correction to the mean-free field description. Since electron is a "probability cloud", i.e. not located in a single point of the space, the Green's function <span class="math-container">$\langle \psi^\dagger (t,\mathbf{r}) \psi (t,\mathbf{r}')\rangle$</span> in general is not zero even if <span class="math-container">$\mathbf{r}\ne \mathbf{r}'$</span>, like in second diagram. It looks like, electron can interact with itself, but I think it's wrong. Accurate calculation shows that taking <span class="math-container">$t=t'$</span> the Green's function of a single electron (<span class="math-container">$+i0$</span> in denominator) is <span class="math-container">$$\int d\omega \frac{e^{+i0\omega}}{\omega-\epsilon_p+i0} =0$$</span> where exponent comes from accurate work with time-ordering in Green's function: <span class="math-container">$\langle \psi^\dagger (t) \psi (t)\rangle=-\left.\langle T \psi (t)\psi^\dagger (t') \rangle\right|_{t'-t\to+0}$</span>. Therefore the term is non zero only in the presence of other electrons. Practically it leads to the correction of the electron dispersion relation, rather then to a new potential. But this is another story :)</p></li> </ol> <p> L. S. Levitov and A. V. Shytov, Green's functions. Theory and practice.<br/>  A. A. Abrikosov , L. P. Gorkov , I. E. Dzyaloshinski, Methods of Quantum Field Theory in Statistical Physics</p>