A conceptual question about Green's function's treatment of interaction

Question

Here we have electron gas and some other stuff. We expand the Hamiltonian to the 1st order of one single harmonic oscillator's displacement $\vec{u}$. Its equilibrium position is at the origin. Then we get an effective coupling Hamiltonian $\vec{j}(\vec{r})\times\vec{f}(\vec{r})\cdot \vec{u}$, wherein $\vec{j}(\vec{r})$ is the electron density, $\vec{f}(\vec{r})$ is some effective potential.

I assumed a particular mode (frequency $\Omega$ for x,y,z) of $\vec{u}$. And I tried to work out the above diagram. It's doable.

1. In this electron Green's function calculation, do we take into account of the interaction's influence on the oscillator? Is it damped or not?
2. I'm confused about where the momentum transfer $q$ come from. Potential $\vec{f}(\vec{r})$ or oscillator's motion $\vec{u}$? I only Fourier transform $\vec{j}(\vec{r})$ and $\vec{f}(\vec{r})$ in the Hamiltonian, therefore this $q$ merely appears in terms that come from $\vec{j}(\vec{r})\,,\vec{f}(\vec{r})$. So I suppose this momentum transfer $q$ comes from the potential. Is this correct?

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user huotuichang

Comments

Hi, the diagram shown here is problematic. There is no momentum conservation at the interaction vertex, and therefore the "in" and "out" electrons don't necessarily carry the same momentum. As an exercise, you can think about how you would treat the "static potential" (that is, the oscillator doesn't oscillate and just provides a static potential) problem in terms of diagrams. In addition, the oscillator is apparently damped by the fermion bath. You can compute the decay rate by using the Fermi golden rule.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user Isidore Seville

Answer 1

I am assuming that those are bare Green's function in the diagram, and the dotted line is the harmonic oscillator $<a^\dagger a>$ ?

1) The diagram you drew does not "know" about the damping of the harmonic oscillator, (although it is closely related by the optical theorem and such to the diagrams that would calculate the damping, I suppose that is what Isidore is saying). If it is necessary to include the damping then you would have to think of a way to self consistently include it.

2) I would recommend you write everything down in real space coordinates, get the second order term and then Fourier transform it. That should make it clear.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user BebopButUnsteady

Comments

Thanks! The dashed line means sth. like $<u_i u_j>$. So you mean this calculation doesn't account for the damping? I'm confused. I suppose green's function method is accurate once you count all the perturbative diagrams.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user huotuichang

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@huotuichang: This is only the lowest diagram of the perturbation theory as you note. To include damping you would have to include diagrams like one where the dotted line had an electron bubble in it and so on.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user BebopButUnsteady

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Thanks. I see. Apology for a novice's jabber. As for your 2nd point, admittedly, I suspect the correctness of my calculation, i.e., directly using $\scr{g}(p,\tau)$, which should be valid only for a homogeneous system. How do you think of this? At least, I guess if I included electron interaction, then I must start from coordinate space Green's function.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user huotuichang

Answer 2

1. The electron motion does feed back to the oscillator, but that is another diagram, known as the bubble diagram, in which you calculate the self-energy correction of the oscillator. That self-energy presumably contains imaginary part, which is then interpreted as the damping of the oscillator. You can either calculate the self-energy corrections self-consistently, or you can simply neglect it in the weak coupling limit away from the non-fermi-liquid criticality.

2. The momentum $q$ should be the momentum of the phonons (quanta of the oscillator motion).The potential $f(r)$ also carries a momentum, but this momentum is on the vertex (where the electron emits/absorbs the phonon). More precisely, due to the presence of the potential $f(r)$, momentum is not conserved on the vertex, the non-conserved amount of momentum is provided by the potential scattering.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user Everett You

Comments

Thanks a lot for your clarifying answer. Er, as you may have noticed, the first comment said this diagram is problematic and the in and out electrons don't necessarily carry the same momentum. I also suspect my direct using Matsubara g(k,τ), which should be valid only for a homogeneous system. So is it wrong? Shall I start from coordinate space diagram?

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user huotuichang

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@huotuichang The comment is right. There is no momentum conservation on your vertex in general. But if your potential $f(r)$ is smooth enough, there could be approximate momentum conservation.It is fine to use Matsubara formalism, I do not see any problem. You can still work in the momentum space, just be care of the vertex.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user Everett You

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But I think direct using momentum space Matsubara $g(k,\tau)$ in fact doesn't solve this problem, in which the defect potential is at the origin. Instead, it solves the scenario that the defect position is randomly averaged. Ah, maybe you mean if we had approximate momentum conservation, this calculation could be acceptable?

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user huotuichang

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Fine, if it is a single-impurity problem, real space representation is better.

This post imported from StackExchange Physics at 2014-08-22 05:07 (UCT), posted by SE-user Everett You