Quantum physics with Green's functions

In summary, Green's functions can be used to express the density of states, and the imaginary part of the Green's function is important for obtaining a more sensible spectral function. Green's functions are advantageous in calculating the density of states in systems with strong interactions, as they take into account factors such as impurities, disorder, temperature, and correlation effects. They are also necessary for including frequency-dependent potentials, which can have a significant impact on the system's properties.
  • #1
Tanja
43
0
The density of states can be expressed in terms of Green's functions and is defined as the imaginary part of the Green function (multiplied by 1/Pi).
Can someone explain me what happens if the imaginary part is just 0? It would be the case if all Eigenvalues of the Schrödinger equation are real, which should happen very often in my understanding of quantum systems.

Thanks
 
Physics news on Phys.org
  • #2
The imaginary part wouldn't really be zero at the eigenvalues of the Hamiltonian, because those will be the poles of the Green's function and it will be undefined at that point.

You get a more sensible spectral function if you add a small imaginary part to the denominator of your Green's function. Starting with a Green's function of the form:

[tex]G(\omega) = \left[ \omega + \mu - H + i\delta \right]^{-1}[/tex]

then the imaginary part is:

[tex]\Im G(\omega) = \frac{-\delta}{ (\omega + \mu - H)^2 + \delta^2}[/tex]

This pushes the poles off the real axis so you don't have singularities there, and also broadens the peaks (important for numerical studies where you are evaluating the Green's function on a regular grid). You will notice is you take the limit as [tex]\delta[/tex] goes to zero you will get a delta functions at the points where [tex]\omega + \mu - H[/tex]
 
  • #3
Thanks. You did help a lot.
I didn't know that there will remain a delta function after delta went to zero. Do you know a derivation of this?

You seem to have a background on Green's functions. Can you help me in one more question, please?
Is it possible to get a solution of the Green's function without solving the Schrödinger equation to get the Eigenvalues?
If it wouldn't be the case I don't really understang why the Green's function method is such an advantage. I mean, solving the Schrödinger equation for Eigenvalues and Eigenvectors, already results in having the necessary components for the density matrix.
Ok, there must be a reason to use Green's functions, otherwise so many people wouldn't use them. But I don't really understand the advantage.
 
  • #4
Off the top of my head, a derivation might go like, as [tex]\delta[/tex] goes to zero, for [tex]x = \omega + \mu - H = 0[/tex] non zero, the denominator stays non-zero and the numerator goes to zero, so you have a zero. For x = 0, then it goes like [tex]1/\delta[/tex] as [tex]\delta[/tex] goes to zero, so this is divergent. The only trick from there would be to show that the integral of [tex]\delta/(x^2 + \delta^2)[/tex] over all x is a constant value (I think it might be pi) independent of [tex]\delta[/tex].

My experience with Green's functions is fairly limited. So the only ways I know of to evaluate Green's functions require knowledge of the full spectrum of the Hamiltonian. But I will list the advantages I can think of, all of which are condensed matter topics.

The density of states you calculate from a non-interacting (or mean-field) theory doesn't correspond very well with experimental photo absorption/emission, because these experiments involve exciting a particle to a state with a finite lifetime that the whole system can have a response to. If the physical system were non-interacting it makes sense that the density of states would be an adequate description, because the excited particle wouldn't affect any of the other particles. But in a system with strong interactions, this isn't the case. The Green's function calculated via [tex]G(t) = \langle c(t) c^\dagger \rangle[/tex] is a direct calculation of particle excitation, so from this argument, one would expect the Green's function to give a more physically realistic picture.

If you calculate the band structure (ie. eigenvalues of H_k) via LDA, you will have a set of energy bands which are sharp delta-function peaks at a specific energy values, all having the weight of one (ignoring degeneracy). This doesn't reflect experiment well because a variety of factors will broaden these peaks, or shift their weight to other bands. Some of these factors are: impurities, disorder, temperature, correlation effects. (In many cases these can be safely neglected, but there are certainly interesting cases where they are relevant). Impurities can be treated with the Coherent Potential Approximation, which uses a Green's function formalism (and that is the extent of my knowledge on CPA).

My knowledge of Green's functions comes from doing Dynamical Mean Field Theory (DMFT) calculations. In the DMFT formalism, we take the Green's function at each k-value:
[tex]G_{k}(\omega) = \frac{1}{\omega + \mu - H_k - \Sigma(\omega)}[/tex]
where you can see a frequency dependent potential has been added to the Hamiltonian (the so-called self-energy). The spectrum of these Green's functions gives the usual k-space band structure, if the self-energy is zero. The self-energy is calculated via some other means, often with some self-consistency condition with the Green's function. The self-energy is non-Hermitian and may have a substantial imaginary part at some frequency. This can broaden the spectrum of H, create additional peaks, shift weight around, etc. to create the kinds of effects that appear in experiment. In DMFT the self-energy is used to include correlation effects. I'd imagine it's difficult / impossible to include any frequency dependent potential without using Green's functions.
 

Related to Quantum physics with Green's functions

1. What are Green's functions?

Green's functions are mathematical tools used in quantum physics to solve for the behavior of a system in response to an external force or disturbance. They represent the response of a system to a point-like particle, also known as an impurity, placed at a specific location within the system.

2. How are Green's functions used in quantum physics?

Green's functions are used to calculate a variety of physical properties in quantum systems, such as energy spectra, scattering amplitudes, and correlation functions. They are also used to study the effects of impurities and perturbations on a system.

3. What is the connection between Green's functions and quantum mechanics?

Green's functions were originally developed in the field of classical mechanics, but they have been adapted for use in quantum mechanics to describe the behavior of particles on a microscopic scale. They are an integral part of many quantum mechanical equations, such as the Schrödinger equation and the Heisenberg equations of motion.

4. How do Green's functions relate to the concept of quasiparticles?

In many systems, particles do not behave as independent entities, but rather interact with each other and their environment to form new collective states known as quasiparticles. Green's functions can be used to study the properties of these quasiparticles and how they emerge from the interactions between particles.

5. Are Green's functions only applicable to quantum systems?

No, Green's functions have applications in many areas of physics, including classical mechanics, electromagnetism, and statistical mechanics. They are also used in other fields such as condensed matter physics, nuclear physics, and cosmology.

Similar threads

  • Quantum Physics
Replies
0
Views
579
Replies
7
Views
807
Replies
1
Views
685
Replies
1
Views
800
Replies
6
Views
864
Replies
2
Views
1K
  • Quantum Physics
Replies
14
Views
2K
  • Quantum Physics
Replies
20
Views
1K
  • Quantum Physics
2
Replies
44
Views
2K
Back
Top