0
$\begingroup$

The spectral gap of a quantum model or a Hamiltonian, in the context of whether it is a gapped or gapless model, is often defined as the difference between the two lowest distinct eigenvalues of the Hamiltonian in the thermodynamic limit. (1) Is this definition restrictive in the sense that it excludes the case below?

Consider a Hamiltonian in the thermodynamic limit (or even in a finite but large size) has a finite set of distinct eigenvalues below a large spacing characterized by a parameter $\Delta$. That is, its eigenvalues arranged in increasing order are: $$ \{ \lambda_0,\lambda_1,...,\lambda_n,\lambda_n+\Delta,...\}. $$ Now, suppose that $(\lambda_n-\lambda_0) \ll \Delta$. Can $\Delta$ then be consider a spectral gap for which $\{\lambda_0,...,\lambda_n \}$ corresponds to the groundstate manifold?

(2) I would also like to ask if it is important to have the notion of spectral gap tied with the energy associated with local quantum fluctuations. By this I mean, for example, random spin-flip due to quantum fluctuations in spin systems. Suppose $\delta \ll \Delta$ is the energy associated to local quantum fluctuations. Consider $(\lambda_{i+1} - \lambda_i) \leq \delta$ for $i=0,..,n$. Does this make the model gapless even if the system can only be excited up to the $n$th eigenstate by local quantum fluctuations?

$\endgroup$

1 Answer 1

0
$\begingroup$

(1) In the thermodynamic limit (TDL), if the eigenvalues $\lambda_0,...,\lambda_n$ are distinct then there is no degeneracy and the gap is viewed as $\lambda_1 - \lambda_0$. For finite systems, this does not apply, one can have small gaps (due to finite size effects, which may vanish in the TDL). Therefore a gap is essentialy a thermodynamic property. It is a question of whether you have a continuum of modes above your ground state.

(2) Spin-flips are a great example, there are models where a spin-flip 'costs' no energy therefore, $\delta=0$ and this is a gapless mode. However, if there any excitation costs finite amount of energy, your model is gapped.

$\endgroup$

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.