# Opening up of spin gap

1. Apr 28, 2010

### mavipranav

Hi

Can somebody explain what is a "gapless excitation"? If it is an excitation, why does it require no energy (gapless) to excite the system?

Thanks,
Mavi

2. Apr 28, 2010

### stephenhky

Gapless excitations can mean Goldstone modes, gapless modes, massless modes, or soft modes, depending on the context. But they are all the same thing in field theories.

Gapless modes are fluctuation modes that have a spectrum that vanishes at zero momentum. Therefore, it does not require too much energy to excite modes of lower values of momenta (k). On the other hand, a gapless mode means a long-range correlation of fluctuations. An excitation at one point can cause another point far from it to have a chance to fluctuate as well.

By Goldstone theorem, gapless modes arise when the ground state of the system spontaneously breaks a symmetry that the system possesses. Examples of this are phonons in solids, magnons in ferromagnets, photons in EM vacuum, helimagnons in helimagnets etc.

3. Apr 28, 2010

### mavipranav

That makes sense; so isn't the unfortunate nomenclature ("gapless excitation") an oxymoron of some sort? Because if it is multiply-degenerate, why call it an excitation?

4. Apr 28, 2010

### stephenhky

I am not sure if it is oxymoron, because terms come out from different context. In condensed matter physics, it is called Goldstone modes when Goldstone talked about it with his idea of spontaneous symmetry breaking. In high energy physics, this excitation is also called Nambu-Goldstone bosons. In high energy physics, it is called massless modes because such excitations refers to the saddle-point equation of their theories with m=0. I don't know why it is called soft modes. In superconductivity, the term "gapless" is used because this refers to a mode that is excited without a gap? I don't know since I do not know the theory of superconductivity.

5. Apr 29, 2010

### mavipranav

Now I think maybe I misinterpreted your initial statement: " ... it does not require too much energy to excite modes of lower values of momenta (k) ... ". I initially presumed that you were referring to excitations between different degenerate levels with momentum being preserved, but after your second reply I am beginning to think that probably you meant there is a scattering with momentum being exchanged. Can you clarify which one you meant?

6. Apr 29, 2010

### stephenhky

In this case, I would think of there is a scattering with momentum being exchanged. The excitations for any Goldstone bosons is often a result of excitations from other particles in the system.

7. May 24, 2010

### vkroom

In cond mat. gapless modes need not be related to Goldstone bosons. For example within a critical region the system has the full conformal symmetry yet critical excitations are gapless.
The 'gap' of a spectrum can be thought of as the mass of the excitations. Spin gaps usually refer to spin excitations that requires a finite energy. Excitations in an Ising ferromagnet is gapped because the simplest excitation above the ordered state would require one of the spins to flip costing $$2J$$ amount of energy, where $$J$$ is the spin-spin coupling strength.

The 'gap' in superconductor (sc) is the energy required to break Cooper pairs, which is finite. Hence any excitation above the sc state needs a finite amnt of energy. This can in turn be thought of as being the mass of the excitation (similar to the energy that a photon needs to possess in order to produce an electron positron pair).
The s-wave sc is always gapped, while the d-wave ones have certain channels in which one can have gapless excitations.

8. May 25, 2010

### mavipranav

So if there is a continuous band of energies connecting the ground and excited state of the d-wave superconductor (because it is gapless), why should the superconductivity be stable, since arbitrarily small thermal fluctuations can excite it? This should be true even if we consider large correlation lengths.

9. May 26, 2010

### vkroom

the gapless region of a d-wave sc are just 4 points at the 'fermi surface' (these are usually 2-dim systems). the cooper pairs that can be excited out of these points are extremely small compared to the macroscopic no. of pairs that are protected by the d-wave gap. Therefore on whole sc is not destroyed. Also it is important to remember that cooper pairing is a dynamical mechanism, i.e. it happens in the momentum space.