2
$\begingroup$

I am trying to understand the Hartree and Fock diagram shown in the picture.enter image description here

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.

If I were to guess the meaning of the diagrams I would say:

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.

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.

I don't find it to suspicious that no real change is happening, this is just the way a particle is propagating.

I have no clue if these explanations are even close to the correct interpretation and would appreciate any help and input!

$\endgroup$
4
$\begingroup$
  1. In Hartree term the time and the spatial position of the ends of circle line (representing Green's function) coincide. i.e. equal to $\langle \psi^\dagger (t,\mathbf{r}) \psi (t,\mathbf{r})=n(t,\mathbf{r})\rangle$, 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: $\int d^3 \mathbf{r}'V(\mathbf{r}-\mathbf{r}')n(\mathbf{r}')$. 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.

  2. 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 $\langle \psi^\dagger (t,\mathbf{r}) \psi (t,\mathbf{r}')\rangle$ in general is not zero even if $\mathbf{r}\ne \mathbf{r}'$, like in second diagram. It looks like, electron can interact with itself, but I think it's wrong. Accurate calculation shows that taking $t=t'$ the Green's function of a single electron ($+i0$ in denominator) is $$ \int d\omega \frac{e^{+i0\omega}}{\omega-\epsilon_p+i0} =0 $$ where exponent comes from accurate work with time-ordering in Green's function: $\langle \psi^\dagger (t) \psi (t)\rangle=-\left.\langle T \psi (t)\psi^\dagger (t') \rangle\right|_{t'-t\to+0}$. 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 :)

[1] L. S. Levitov and A. V. Shytov, Green's functions. Theory and practice.
[2] A. A. Abrikosov , L. P. Gorkov , I. E. Dzyaloshinski, Methods of Quantum Field Theory in Statistical Physics

$\endgroup$
  • $\begingroup$ If I understood it correctly, this means that for a single electron propagating alone, it can not have any Hartree interacting, because there are no other electrons producing a mean field. However, it can have a Fock interactions where it interacts with itself. $\endgroup$ – B. Brekke Jan 31 at 10:04
  • $\begingroup$ Now however, I struggle to understand why, in the Hartree interaction, the particle has the same momentum as before the interaction? I understand that this can be the case, but not why it has to be the case. $\endgroup$ – B. Brekke Jan 31 at 10:10
  • 1
    $\begingroup$ 1.2. In any vertex the net momentum is conserved. Try to draw the momenta, you'll see there's the only way: same in and out momentum. $\endgroup$ – Alex Jan 31 at 12:13
  • $\begingroup$ (1). For Hartree you're surely correct. For Fock to be honest I need to correct myself. The interaction is generally written in such way that the term is explicitly zero for single particle. Let me correct the answer.... $\endgroup$ – Alex Jan 31 at 13:00
  • 1
    $\begingroup$ @LewisMiller, yes, and no. Formally you're right. If we would calculate the loop for quantum dots, i.e. with finite number of electrons, then you're fully correct: the interaction is build as $\psi^+_1\psi^+_2\psi_2\psi_1$ so that no self-interaction is there, but the single electron would give a finite density of charge. Therefore the Fock term will be cancelling it out. But for continuous empty system, the integral over $\omega$ will be zero (for both Fock and Hartree, as I showed in answer). This is while in bulk system one electron gives an infinitesimal density of charge. $\endgroup$ – Alex Feb 1 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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