It's not clear to me what are the elementary excitations of Luttinger liquids. Quoting from Giamarchi's book Quantum Physics in One Dimension:

In one dimension, [...], an electron that tries to propagate has to push its neighbours because of electron-electron interactions. No individual motion in possible. Any individual excitation has to become a collective one.

Judging by this, I would say that in Luttinger liquids the elementary excitations can only be a collective one, regarding the system as a whole. But then, talking about the bosonization method for solving such systems, it is said:

It means that in one dimension the particle-hole excitations are well-defined "particles" [...]. These bosonic quasiparticles will just be the key in solving our one-dimensional problem.

Along these lines is this quote from Sénéchal's notes An introduction to bosonization:

The basic idea behind bosonization is that particle-hole excitations are bosonic in character, and that somewhat the greatest part (if not the totality) of the electron gas spectrum might be exhausted by these excitations.

And judging by this I would say that particle-hole excitations, which are bosonic in nature, are the elementary excitations of the system.

What is, exactly, going on physically in a Luttinger liquid?

  • 1
    $\begingroup$ Localized particle-hole excitations are not eigenstates of the system, but you get (approximate) eigenstates by using a superposition of particle-hole excitations (e.g. as a variational ansatz). $\endgroup$ Mar 6, 2021 at 15:20
  • $\begingroup$ Thank you. If you don't mind I still have a question: knowing that density fluctuations, which are a superposition of particle-hole excitations, are eigenstates, how do I deduce that the excitation must be a collective one? I mean how does this emerge formally. $\endgroup$ Mar 6, 2021 at 16:48

1 Answer 1


The boson fluctions are the "sound waves" in the electron gas. As with any density fluctuation, they are a collective linear superposition of many electron-hole pair states, and can be made by acting on the undisturbed gas by exponential operators $$ U[\varphi]=\exp\left\{i \int \hat \rho(x) \varphi (x) dx\right\}, $$ where $\hat \rho(x)= \hat \psi^\dagger(x) \hat \psi(x)$ is the electron density operator and $\varphi(x)$ is a c-number. This operator has the effect of multiplying each one-particle wavefunction by a phase $e^{i\varphi(x)}$ because (I have not checked the sign) $$ \hat \psi(x) \to U[\varphi]\hat \psi(x)U^{-1}[\varphi] = e^{i\varphi(x)} \hat \psi(x) $$

  • $\begingroup$ Thanks! What's now not entirely clear to me is how the "collectiveness" emerges. For a linear dispersion model, density fluctuations are exact eigenstates, but how does this formally lead to a collective excitation? $\endgroup$ Mar 6, 2021 at 16:51
  • 1
    $\begingroup$ "Collective" is an imprecise term. In general it just means that many interacting particles are involved. A density fluctuation with wavelength longer that the mean interparticle spacing is thus "collective." $\endgroup$
    – mike stone
    Mar 6, 2021 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.