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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?

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One point to note is that the Mass Gap referred to in an answer below is one of the <en.wikipedia.org/wiki/…; $1000000 Clay Prizes. So it might be worthwhile understanding this! –  Roy Simpson Feb 10 '11 at 15:57
You should remind us what an XX spin chain is... –  genneth Feb 10 '11 at 17:41
Good point - $H_{XX} = \frac{J}{2} \sum_l (\sigma_l^x \sigma_{l+1}^x + \sigma_l^y \sigma_{l+1}^y)-B\sum_l \sigma_l^z$ –  Jane Feb 11 '11 at 21:17
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5 Answers 5

up vote 5 down vote accepted

This is the distinction between continuous and discrete spectrum, but only considering the low energy excitations. For an Hamiltonian with gapped spectrum, the lowest eigenvalue is separated by a gap from $E=0$. For example dispersion relation of the form $E=|k|$ is an example of gapless spectrum, and $E=|k+m|$ is an example of gapped one, where $k$ is the wave vector (which can be any real number) and $m$ is the mass (= gap).

This distinction leads to qualitative difference in the physics - for example the difference between a material being a conductor or an insulator. Many times also, the gap is generated by interesting physics (like the mass gap in Yang-Mills theory, or the gap in BCS superconductivity).

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Thanks. I don't understand - If the ground state and first excited state are different, then the system is gapped. If they aren't different, then they're degenerate.. aren't they? Also, do you mean by "he distinction between continuous and discrete spectrum" that a continuous spectrum cannot be gapped? –  Jane Feb 10 '11 at 15:58
For continuous spectrum there is no "first" excited state, there is a continuum of them with arbitrarily low energies. For example the spectrum $E=k$, where $|k|$ is the wavenumber, which is any real number. –  user566 Feb 10 '11 at 16:05
Good, @Moshe, +1. @Jane, the first excited state, whenever you can find a state that can be called in this way, is always different from the ground state, by definition. ;-) It's like the first U.S. president after George W. Bush. Regardless of gaps or non-gaps, this person can't be George W. Bush himself. You probably didn't want to say that the states were "equal": you wanted to say that they had the same "energy". But it's not the same thing. The very point of degeneracy is that you have different states that have the same energy. Having the same state that have the same energy is easy. –  Luboš Motl Feb 10 '11 at 16:56
For a gapless spectrum, you can find energy eigenstates that are "arbitrarily close" to being degenerate with the ground state (i.e. having the same energy) but if you're really strict, none of those states has exactly the energy of the ground state. The ground state is a unique state with this minimum energy eigenvalue. –  Luboš Motl Feb 10 '11 at 16:59
|k+m| is ungapped in 1d, just biased for left-motion. If you meant ||k|+m| please fix it, but even that is unphysical. You want $\sqrt{k^2 + m^2}$ –  Ron Maimon Sep 4 '11 at 21:04
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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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@ 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 ? –  K-boy Apr 23 at 15:26
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. –  Xiao-Gang Wen Apr 23 at 15:43
@ Xiao-Gang Wen Thanks a lot. Is the spin-1/2 Kitaev honeycomb model also such an example? Where if we take a sequence of sizes of the system $L_i=odd$ (odd number of spins or lattice sites) in the thermodynamic limit, such that there always exist exact Kramer's degeneracy (due to time-reversal symmetry, am I right?) as $L_i$ approaches $\infty $, therefore, this kind of sequence for the thermodynamic limit would give a ground state degeneracy (Kramer's degeneracy), right ? –  K-boy Apr 24 at 10:13
@ Xiao-Gang Wen On the other hand, as pointed by the author, both Lieb's theorem and the author's numerical study suggest that the zero-flux configuration is the unique sector minimizing the ground state (GS) energy, which means that the GS of Kitaev model is nondegenerate or unique. This is in contrary to the above Kramer's degeneracy. It seems that the original paper didn't mention that what kind of sequence of sizes of the system is chosen as taking the thermodynamic limit. –  K-boy Apr 24 at 10:24
Yes, the reasonable sequence should be ensuring the number of lattice sites to be even. –  Xiao-Gang Wen May 9 at 3:55
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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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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. –  genneth Feb 10 '11 at 15:37
Thanks for your answer. Is a gapless Hamiltonian one with a degenerate ground state then? Or have I misunderstood your reply? –  Jane Feb 10 '11 at 15:37
Not necessarily degenerate ground state, just that excitations of arbitrarily low energy exist. For example dispersion relation of the form $\omega =k$ is an example of gapless spectrum, and $\omega = k+m$ is an example of gapped one. Added this to my answer for clarification. –  user566 Feb 10 '11 at 15:45
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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).

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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 $N -> 00$.

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