Sunday, August 31, 2014

quantum field theory - 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



Answer



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


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