I've read some papers recently that talk about gapped Hamiltonians or gapless systems, but what does it mean?

Edit: Is an XX spin chain in a magnetic field gapped? Why or why not?

  • $\begingroup$ One point to note is that the Mass Gap referred to in an answer below is one of the $1000000 Clay Prizes. So it might be worthwhile understanding this! $\endgroup$ Feb 10, 2011 at 15:57
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    $\begingroup$ You should remind us what an XX spin chain is... $\endgroup$
    – genneth
    Feb 10, 2011 at 17:41
  • $\begingroup$ Good point - $H_{XX} = \frac{J}{2} \sum_l (\sigma_l^x \sigma_{l+1}^x + \sigma_l^y \sigma_{l+1}^y)-B\sum_l \sigma_l^z$ $\endgroup$
    – Jane
    Feb 11, 2011 at 21:17

5 Answers 5


This is actually a very tricky question, mathematically. Physicists may think this question to be trivial. But it takes me one hour in a math summer school to explain the notion of gapped Hamiltonian.

To see why it is tricky, let us consider the following statements. Any physical system have a finite number of degrees of freedom (assuming the universe is finite). Such physical system is described by a Hamiltonian matrix with a finite dimension. Any Hamiltonian matrix with a finite dimension has a discrete spectrum. So all the physical systems (or all the Hamiltonian) are gapped.

Certainly, the above is not what we mean by "gapped Hamiltonian" in physics. But what does it mean for a Hamiltonian to be gapped?

Since a gapped system may have gapless excitations at boundary, so to define gapped Hamiltonian, we need to put the Hamiltonian on a space with no boundary. Also, system with certain sizes may contain non-trivial excitations (such as spin liquid state of spin-1/2 spins on a lattice with an ODD number of sites), so we have to specify that the system have a certain sequence of sizes as we take the thermodynamic limit.

So here is a definition of "gapped Hamiltonian" in physics: Consider a system on a closed space, if there is a sequence of sizes of the system $L_i$, $L_i\to\infty$ as $i \to \infty$, such that the size-$L_i$ system on closed space has the following "gap property", then the system is said to be gapped. Note that the notion of "gapped Hamiltonian" cannot be even defined for a single Hamiltonian. It is a properties of a sequence of Hamiltonian in the large size limit.

Here is the definition of the "gap property": There is a fixed $\Delta$ (ie independent of $L_i$) such that the size-$L_i$ Hamiltonian has no eigenvalue in an energy window of size $\Delta$. The number of eigenstates below the energy window does not depend on $L_i$, the energy splitting of those eigenstates below the energy window approaches zero as $L_i\to \infty$.

The number eigenstates below the energy window becomes the ground state degeneracy of the gapped system. This is how the ground state degeneracy of a topological ordered state is defined. I wonder, if some one had consider the definition of gapped many-body system very carefully, he/she might discovered the notion on topological order mathematically.

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    $\begingroup$ @ Xiao-Gang Wen Dear Prof.Wen, mathematically, is it possible that a Hamiltonian has the "gap property" for both two distinct sequences of sizes of the system as we take the thermodynamic limit, but these two sequences give two different ground state degeneracies ? $\endgroup$
    – Kai Li
    Apr 23, 2014 at 15:26
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    $\begingroup$ Yes. Haah's cubic code is such an example. And one can easily construct many other examples by stack 2D topologically ordered states to form a 3D gapped state. $\endgroup$ Apr 23, 2014 at 15:43
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    $\begingroup$ What if there exists a sequence of sizes such that \delta varies, but always remains above a non-zero threshold? That should be considered gapped too. In particular, why should \delta remain fixed? $\endgroup$
    – Mani Jha
    May 14, 2020 at 15:09

Gapped or gapless is a distinction between continuous and discrete spectra of low energy excitations. For a Hamiltonian $H$ with gapped spectrum, the first excited state has an energy eigenvalue $E_1$ that is separated by a gap $\Delta > 0$ from the ground state $E_0$. For example, a dispersion relation of the form $E = |k|$ is an example of a gapless (continuous) spectrum, whereas $E = \sqrt{k^2 + m^2}$ is an example of a gapped one. $k$ denotes the wave vector and can be any real number. $m$ is the mass which in this case is the cause of the gap.

This distinction leads to a qualitative difference in the physical behavior of gapped and untapped systems - most importantly it determines whether a material is a conductor or an insulator. There are quite fascinating processes that can give rise to a gap such as interactions (interesting examples are the mass gap in Yang-Mills theory, or the gap in BCS superconductivity).

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    $\begingroup$ How is $E = \sqrt{k^2 + m^2}$ a gapped spectrum? Wouldn't the ground state have $E_0=m$ and the spectrum be continuous. Or do we assume $E_0$ to be $0$? $\endgroup$ Jun 5, 2019 at 16:55
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    $\begingroup$ @JoãoBravo I would think $E_0 = 0$. This corresponds to the state of zero particles. But once you go to the next excited state, say with n=1 particles with k = 0 you would find $E_1 = m$; hence a "mass gap". $\endgroup$ Jul 21, 2020 at 19:44
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    $\begingroup$ @AimanAl-Eryani That makes sense, thank you. $\endgroup$ Jul 21, 2020 at 21:23

Gapped and gapless are usually attributes for many-body Hamiltonians. A gapped Hamiltonian is simply one for which there is a non-zero gap between the ground state and the first excited state.

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    $\begingroup$ I would add that often, the difference is physical --- a system with gapless excitations will have its phenomenology dominated by those; in addition, a gapped system is fairly robust against perturbations which might change the phase that the system is in --- it's far easier to mix states which are near each other in energy. So for example, the Fermi liquid is gapless, which makes it unstable towards superconductivity, which is a gapped phase. $\endgroup$
    – genneth
    Feb 10, 2011 at 15:37

A short remark for the "edited" part of your question (whether there is a gap in the XX chain or not). The XX spin chain in a magnetic field, i.e., the model defined by the Hamiltonian

$$ H = \sum_i (\sigma^{x}_i \sigma^{x}_{i+1} + \sigma^{y}_i \sigma^{y}_{i+1} + h \sigma^{z}_i) $$

is gapped when $|h| > 1$. This is not a very difficult results, it comes out immediately if you do the usual Jordan-Wigner and a Fourier transformation a la the famous paper of Lieb, Schultz and Mattis (Ann. Phys. 16, 407, (1961)) (although there the $\sigma^{z}_i$ terms are missing, but they are not hard to incorporate).


I would just like to add a little to these answers in light of the Edit to the question which introduces "XX Spin Chains" as a context for this question. I have found a Tutorial on Spin Chains here. Basically they are N spins on a line. Here is the Hamiltonian from that paper where N=2.

$H_{12}=J/4(\sigma_1^x\sigma_2^x+\sigma_1^y\sigma_2^y+\sigma_1^z\sigma_2^z - I \times I)$

Depending on the sign of J this has either 3 degenerate ground solutions, plus one excited solution or one ground solution. This is a basic model of ferromagnetic/antiferromagnetic states. In this case the solutions have a gap. They will still have a gap for general N.

However many developments of this largely integrable model have happened in recent papers, with an applied continuous magnetic field for example. In some of these cases the model may be gapless. There is also the question of what the model implies in the Thermodynamic limit $N \rightarrow \infty$.


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