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Difference between gapless excitations and Goldstone bosons

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sam12
#1
Sep4-13, 08:42 AM
P: 3
I have been looking around on the web and in books to clarify this, but can't find a good explanation describing relationship/difference between gapless modes/excitations and Goldsone modes/bosons in Condensed matter physics.

Does the term "gapless modes" mean that no energy is required for these modes/excitations or that only an infinitesimal amount of energy is required for such excitations? If Goldstone modes/excitations require some(small) amount of energy (as mentioned in Ref:http://web.mit.edu/8.334/www/lectures/lec3.pdf), how can they ever be gapless?
Does a previously gapless mode/excitation develop a gap or remain gapless if a continuous symmetry is spontaneously broken in a system?
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fzero
#2
Sep4-13, 06:27 PM
Sci Advisor
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P: 2,606
Quote Quote by sam12 View Post
I have been looking around on the web and in books to clarify this, but can't find a good explanation describing relationship/difference between gapless modes/excitations and Goldsone modes/bosons in Condensed matter physics.

Does the term "gapless modes" mean that no energy is required for these modes/excitations or that only an infinitesimal amount of energy is required for such excitations? If Goldstone modes/excitations require some(small) amount of energy (as mentioned in Ref:http://web.mit.edu/8.334/www/lectures/lec3.pdf), how can they ever be gapless?
Gapless means that it takes an infinitesimal amount of energy to excite some mode. We are comparing energy to the vacuum state, so even if there is no gap, an excitation will still have at least kinetic energy. In a gapless spectrum this energy can be arbitrarily small, as in eq. (II.14) of those notes. In a theory with a mass gap, the lowest-energy excitations have a finite energy (with respect to the vacuum) even at zero momentum. Typically this has the interpretation of either a physical or effective mass of the particle states of the theory.


Does a previously gapless mode/excitation develop a gap or remain gapless if a continuous symmetry is spontaneously broken in a system?
The Goldstone mechanism explains how a gapless mode emerges in a vacuum where a continuous global symmetry is broken. The Landau-Ginzburg model in eq (II.12) is a good example. If we consider the spectrum of fluctuations around ##\psi = 0##, we find a mass gap set by the parameter ##\sqrt{t}##. However, ##\psi = 0## is not the true vacuum of the theory when ##t<0## (and ##\sqrt{t}## becomes imaginary). If we minimize the Hamiltonian and expand around ##\bar{\psi}## we find that the excitations can be described by the gapless mode ##\theta(x)##.

It is possible for a system to develop a mass gap if we allow additional interactions. For instance, if we couple the complex scalar ##\psi(x)## in the LG model to a gauge field, we will find a mass gap. In the false vacuum we have the massive field ##\psi(x)## and a massless gauge field ##A_\mu(x)##. In the true vacuum, we have a massive real scalar ##\bar{\psi}## and a massive gauge field ##B_\mu(x)##, that is actually composed of the 2 degrees of freedom from ##A_\mu## and the single degree of freedom from the would-be Goldstone boson ##\theta(x)##. This model is related to what is usually called the abelian Higgs model.
DrDu
#3
Sep5-13, 01:55 AM
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P: 3,549
The gap mode vanishes in the limit k->0. So a long wavelength exitation looks different than the vacuum over a large extent of space, yet it is energetically practically degenerate, i.e. there are different vacuum states. That is the definition of a broken symmetry, so every gapless mode can be viewed as resulting from symmetry breaking. The other direction does not always work as interactions have to be short range for broken symmetry modes to be gapless.

inempty
#4
Sep5-13, 09:51 AM
P: 29
Difference between gapless excitations and Goldstone bosons

In the context of field theory, a gap in dispersion relation really means a mass. So a gapless excitation, like goldstone boson, doesn't have a mass hence is always relativistic. This is like the situation of photon in relativity.
DrDu
#5
Sep5-13, 10:36 AM
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I would be careful with the word "relativistic". If E is a linear function of k for a gapless mode, then it may be called massless as it is dispersion free and hence a wavepacket will move with the constant speed v= E/hbar k. However, almost never does this speed coincide with the speed of light in vacuo, so it is certainly not relativistically invariant, which is already clear by the solid matrix forming a special system of reference.
inempty
#6
Sep5-13, 01:40 PM
P: 29
Quote Quote by DrDu View Post
I would be careful with the word "relativistic". If E is a linear function of k for a gapless mode, then it may be called massless as it is dispersion free and hence a wavepacket will move with the constant speed v= E/hbar k. However, almost never does this speed coincide with the speed of light in vacuo, so it is certainly not relativistically invariant, which is already clear by the solid matrix forming a special system of reference.
I think this depends on what you mean by "relativistic". Of course in solids most gapless excitations don't move in the speed of light and real relativisitc effects are not present. But for gapless particles, fundamental equations in the "relativistic" form are often used to describe their motion, like slightly modified Dirac equations which is applied to electrons in graphene. This also gives us some analogous "relativistic" phenomenon in CMP.


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