The Spectral Function in Many-Body Physics and its Relation to Quasiparticles

recently, I stumbled accross a concept which might be very helpful understanding quasiparticles and effective theories (and might shed light on an the question How to calculate the properties of Photon-Quasiparticles): the spectral function $$A\left(\mathbf{k},\omega \right) \equiv -2\Im G\left(\mathbf{k},\omega \right)$$

as given e.g. in Quasiparticle spectral function in doped graphene (on arXiv).

It is widely used in many-body physics of interacting systems and contains the information equivalent to the Greens function $G$. For free particles, $A$ has a $\delta$-peaked form and gets broader in the case of interactions.

The physical interesting thing is, as I read, quasiparticles of interacting systems can be found if $A$ is also somehow peaked in this case. I don't understand this relationship, hence my question:

What is the relation of the spectral function's peak to the existence of quasiparticles in interacting systems?

Thank you in advance
Sincerely

Robert

• You'll find the term "spectral function" in nuclear physics as well, were it plays the role of a energy and momentum distributions (as measured in some interaction) for nuclear components. Without being familiar with the application you cite, it appears to have a similar meaning. – dmckee Jan 22 '11 at 21:06
• @dmckee: Thank you for this connection. If I can get my hands on $A$ in this meaning I will have to look if I in turn can understand something from nuclear physics as well :) Greets – Robert Filter Jan 23 '11 at 11:55

Dear Robert, the answer to your question is trivial and your statement holds pretty much by definition.

You know, the Green's functions contain terms such as $$G(\omega) = \frac{K}{\omega-\omega_0+i\epsilon}$$ where $\epsilon$ is an infinitesimal real positive number. The imaginary part of it is $$-2\Im(G) = 2\pi \delta(\omega-\omega_0)$$ So it's the Dirac delta-function located at the same point $\omega$ which determines the frequency or energy of the particle species. At $\omega_0$, that's where the spectrum is localized in my case. If there are many possible objects, the $G$ and its imaginary part will be sums of many terms.

This delta-function was for a particle of a well-defined mass (or frequency - I omitted the momenta). If the particle is unstable, or otherwise quasi-, the sharp delta-function peak will become a smoother bump, but there's still a bump.

Because you didn't describe what you mean by "peak" more accurately, I can't do it, either. It's a qualitative question and I gave you a qualitative answer.

Cheers LM

• Thank you for your response. Could you please comment on why quasiparticles in interacting systems correspond to peaks of $A$ - can one e.g. see cooper pairs and derive BCS theory (for which I am obviously no expert) from this viewpoint? Furthermore, can one apply this to composed systems with different $G$ for different domains along with appropriate boundary conditions? I am sorry for the lengthy comment and the maybe not so sharp question. Greets – Robert Filter Jan 23 '11 at 11:51
• There is a little more to this. The spectral function can be almost directly measured (e.g. with ARPES or surface tunnelling experiments), and so is a massively important link between experiment and theory. The real fun begins when the spectral function doesn't show a dominant peak... – genneth Jan 30 '11 at 10:29
• @genneth: Thank you for the further info! You may want to consider making another nice answer out of your comment with some references and further explanations :) - I assume that most people do not read the comments. Greets – Robert Filter Jan 30 '11 at 12:34
• Just as a mathematical addendum: to see that $-\mathfrak I \left( \lim_{\epsilon \to 0} \frac{1}{x+i\epsilon} \right) = \pi \delta(x)$, note that the left-hand side equals $\lim_{\epsilon \to 0} \frac{\epsilon}{x^2+\epsilon^2}$. This is clearly zero for any $x \neq 0$. However note that it integrates to $\lim_{\epsilon \to 0} \int_{-\infty}^{+\infty} \frac{\epsilon}{x^2+\epsilon^2} \mathrm d x = \lim_{\epsilon \to 0} \int_{-\infty}^{+\infty} \frac{1}{y^2+1} \mathrm d y = \pi$ (where $y = \frac{x}{\epsilon}$). These two properties uniquely characterize the $\delta$-function. – Ruben Verresen Mar 16 '16 at 23:03
• Dear @RubenVerresen - it's approximately right and the conclusion is OK in this case but I think your general statement isn't right. All combinations of derivatives of delta functions are "zero" for nonzero $x$, and their integral over reals is zero, but they're not multiples of the delta function itself. Because the fraction is strictly positive, one may know that no derivatives of delta are included here, however. – Luboš Motl Mar 17 '16 at 15:13

Spectral function gives the number of state(or density of state if you divide volume,...etc), The peak means there's a state or there're several degenerate states there. In single particle system, spectral function are only delta function sets at where eigenstates are. Considering the many-body interaction (for ex: electron-electron interaction, electron-phonon interaction...etc in Condensed Matter) into hamiltonian as a perturb term and calculating the approximate solution in some degree, the new eigenstates ket could be called quasiparticle. Sometimes we called this particle as "dressed electron". It's just a approximation which merge those complicated interaction and electron into a "quasiparticle". Thus, the spectral function couldn't be so simple as a set of delta function in single electron system, but relates with the interaction, which add a so-called "self energy" term in spectral function. The real part of self-energy changes the peak position, the imaginary part changes the life time of the state.

you can see the ch1 & ch2 in this book: Green's Functions and Condensed Matter by G.Rickayzen.