5
$\begingroup$

There is something I don't understand about quasiparticles density of states.

I work with the book "Introduction to many body physics" from Coleman.

When he introduces the quasiparticle he does the following.

We consider a Fermi sea of a non interacting system. Now we "enable" the interactions adiabatically. Thus we have an adiabatic correspondance between the non-interacting system and the interacting one. We can thus refer to the interacting system eigenstates by $ | \{n_{p \sigma} \rangle \} $ where $n_{p \sigma}$ is the occupation of electrons in the non interacting system (1-1 correspondance between the eigen states).

Now, he defines a quasiparticle by saying it is the fact to add a single electron above fermi sea and then enabling the interactions. We will end up with a state of the interacting system that has a single excitation. He calls this excitation a quasiparticle.

The energy of a single quasiparticle is then defined by :

$$ E^0_{p \sigma}=E(p,\sigma)-E_0$$

Where $E_0$ : energy of the adiabatic correspondance of the fermi sea of the non interacting system.

$E(p,\sigma)$ : energy of the adiabatic correspondance of the fermi sea + 1e above of the non interacting system.


What I don't understand :

If we have 2 quasiparticles, the energy of them will not simply be : $E^0_{p_1,\sigma_1}+E^0_{p_2,\sigma_2}$ because of the interactions between them.

But in the book, he defines the quasiparticles density of states as :

$$ N^*(E)=2\sum \delta(E-E^0_{p}) $$

For me this would be the density of states if we only had single excitations. Am I right ?

Thus can we really use this quantity as a usual density of states. Like if we want to count the total number of quasiparticles in the system for me we can't simply do :

$$ N=\int N^{*}(E) dE$$

Am I right or can we still do this even if we have the interactions ??

And final question : Imagine I want to compute something like $\chi_c=\frac{1}{V} \frac{\partial N}{\partial \mu} $. Does the $N$ refers to number of quasiparticles or number of electrons ? In fact I am still confused with the notion of quasiparticles, if we want to compute thermodynamic quantities involving the number of particles in the system why should we take the number of quasiparticles instead of the number of particles ?

[edit] : In fact I don't understand if the quasiparticles are excitation above the Fermi sea or just electrons interacting (and thus number of electrons = number of quasiparticles).

PS :

this topic Density of states in a system of interacting electrons says that in an interacting system the DOS doesn't necessarily has a meaning but we can have a meaning in fermi liquids (which is my case). But as I said, for me the quasiparticles are still interacting between themself so I don't know how to give a meaning to it.

$\endgroup$
  • $\begingroup$ Have you looked at Lehnmann representation of retarded greens function. Generally real part of poles of retarded greens function are interpreted as quasiparticle energies and imaginary part as their life time. $\endgroup$ – Sunyam Oct 20 '17 at 11:34
  • $\begingroup$ Thank you for your answer. In fact I am looking for a more physical interpretation. $\endgroup$ – StarBucK Oct 20 '17 at 11:40
  • $\begingroup$ Your second equation is not true even for free fermions. The number of particle is given for free fermions, thanks to the Pauli principle, by $2\int_0^{E_F}N(E) dE$. Your equation counts the total number of accessible states. $\endgroup$ – Adam Oct 20 '17 at 14:49
  • $\begingroup$ The 2 can be put inside the density of states ? That is why I defined my density of states as $2 \delta$. $\endgroup$ – StarBucK Oct 20 '17 at 14:57
  • $\begingroup$ Yes, sure, you can put the 2 in $N(E)$. $\endgroup$ – Adam Oct 20 '17 at 15:07
3
+100
$\begingroup$

I think you misunderstood the concept of a quasiparticle. Take a gas of electrons interacting with each other via photons, or with the cristal lattice via phonons, ... Then the quasiparticle hypothesis consists in approximating this soup of particles as a gas of non-interacting ones, i.e. the electrons+photons+phonons are described in term of a Fermi statistics, have a Fermi surface, ... and the associated particle is called a quasi-particle.

The next approximation is to consider weakly interacting quasi-particles, and to associate a kinetic equation to their evolution. But for the density of state we do not need this complication. Just in passing, the density of state takes the picture of a number occupation and its Keldysh representation as a so-called Keldysh-Green's function in the case of a kinetic equation.

That's perhaps the confusing thing : a weakly interacting particle just have an energy and a lifetime. The lifetime goes to infinity for a bare elementary particle (as electrons are), but at finite temperature in the soup of defects and interactions like a real solid, under averaging and screening conditions, the lifetime might be still pretty large (i.e. larger than the characteristic time of an experiment) and to approximate a bare particle dressed with many interactions by a free quasi-particle is sufficient in many situations. Then the meaning of the density of states for these quasiparticles is the same as for the free electron gas : it represents all possible accessible states at a given energy, as long as the experiment runs for time smaller than the lifetime. For time of experiment longer than the lifetime, sure the quasi-particle will no longer stay in a given energy state.

One can demonstrate many aspects of this quasiparticle construction, and prove it works in many experimentally interesting situations. The standard textbook for this is

Nozières and Pines, Theory of quantum liquids, vol. I and II - Westview Publish.

but I particularly enjoyed the pedagogical introduction in

Mattuck - A guide to Feynman diagrams in the many-body problem, now available at Dover publications

Please read (at least !) https://en.wikipedia.org/wiki/Quasiparticle before continuing, and eventually come back to SE to ask more specific question.

In the meantime, you may close or re-edit this question, since it is poorly related to density of state as far as I can see.


Added after comments :

Please check the references I cited, especially the Wikipedia entry I gave you. I'm just rephrasing what is said there. It's pure time lacking.

In fact, quasi-particle are not free but they are weakly interacting... and so what ? At first they appear as if they were free particles. Then you realise they get, in addition to their free part, a complex part, namely the lifetime. You can see the lifetime as a mean-field approximation of the effect of the other quasi-particle onto the one you study. This effect is weak, so it's just a perturbation of the free particle you got. Call it a $f_{p_{1}p_{2}}^{\sigma_{1}\sigma_{2}}$ if you like, a collision integral in a Boltzmann equation if you prefer, a self-energy term in a kinetic theory if you love so ... it's always small. In fact in many cases it makes the system relaxes to its ground state, by interaction with a bath (be careful, it's not the same as the theory of decoherence, here the bath is macroscopic and it shares only macroscopic quantities with the system, like temperature for instance)

Going back to the lifetime you get in a mean-field approximation, it goes to infinity in case of no interaction, and is finite (but pretty large) in case of (weak) interactions. So quasi-particles are stable objects, up to this lifetime. So you can count quasi-particles, and so the density of state (DOS) makes sense.

A bit of hint of the mathematics behind : The lifetime is related to the self-energy $\Sigma$ of the Green's function. Under peculiar circumstances, you can sum up a full familly of Feynmann diagramms, and your Green's function looks formally like ($\omega$ is the frequency, $H$ the one-body Hamiltonian, $G$ the Green's function)

$$G\sim\dfrac{1}{\omega-H-\mathbf{i}\Sigma}$$

hence you deform the Green's function alongside the DOS by calculating $\Sigma$ and putting it in the Green's function. But $\Sigma$, how large it might be, does not change the topological properties of the Green's function, and a Fermi surface still exists. So you get a mathematical object which formally looks like a free particle, when the interactions with other particles are all encoded in $\Sigma$. $\Sigma$ is not that easy to calculate in realistic situations ... but assuming analytic properties of this function, one can say a lot of interesting things about the system.

$\endgroup$
  • $\begingroup$ Thank you for your answer. I have to admit that I am a lot confused, as I would like to see quasi-particles as a mathematical trick to do everything "as if" there were not interactions (modifying the mass of the electron $m \rightarrow m^*$ and consider them free by doing this for example). But I still have problems. For example you say in your second paragraph that the quasi particles are weakly interacting (I'm ok with this), but then you say it is approximated by a free quasi particle in your 3rd paragraph. For me if we forget about these interactions all the Fermi liquid theory breakdown $\endgroup$ – StarBucK Oct 24 '17 at 7:25
  • $\begingroup$ What I call interactions between quasi particles are the usual coefficients $f_{pp'}$ of the Landau theory. And also, if we indeed have interacting quasi particles then I don't see how we can understand the DOS. Excepted if specifically for the DOS we approximate it by the one of a non interacting case. But then it is an assumption to be made, it is not "obvious" ? Thanks ! $\endgroup$ – StarBucK Oct 24 '17 at 7:26
  • 1
    $\begingroup$ @StarBucK: The mathematical trick is to do everything as if the interactions are weak by transforming to a quasi-particle picture. in a first approximation, the resulting weak interactions can be neglected, but improving the approximation requires their consideration. $\endgroup$ – Arnold Neumaier Oct 24 '17 at 14:14
  • $\begingroup$ Sorry for my late answer. Thus am I right if I summarize as it follows. The quasi particles are submitted to interactions and have a lifetime $\tau(E)$ that is generally high if we are close to Fermi surface. If we consider "short enough" events, the quasiparticle will not be scattered and then we can consider its DOS. In fact I think that my confusion lies in the fact that the quasiparticle doesn't change its energy (thus short time of experiment) is a necessary condition to talk about DOS but not a sufficient one. $\endgroup$ – StarBucK Oct 26 '17 at 9:40
  • 1
    $\begingroup$ You're on the good way. All what you say isn't wrong. Excitations summed up only in the linear approximation (= harmonic oscillator like excitation), which is the first approximation tried indeed. And never forget that, at the end, the experiments confirm or infirm your model. Clearly, quasiparticle is just a model, and in modern experimental situations, it has difficulties. Really you should open Mattuck's book, it's really clear about the lifetime and self-energy, as far as I remember. All the best ! $\endgroup$ – FraSchelle Oct 27 '17 at 8:06

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.