Why is the magnetic field of a superconductor normally excluded?

lufc88

Why is the magnetic field of a superconductor normally excluded?

Dickfore

Because of the screening effect due to the dissipationless supercurrent flowing in the superconductor.

Dickfore

The flow of Cooper pairs.

K^2

It's not really excluded. It's held constant. If you try to change the magnetic field, dB/dt induces an electric field, which cause currents through superconductor that cancel out the change in the magnetic field. This happens in any conductor, but in a normal conductor, there is resistance, so the screening currents die out, and eventually the field changes. In superconductor, these currents keep going, so the field inside never changes.

Naturally, the actual physics of how and why is a bit more complex and involves quantum mechanics. If you understand at least the Shroedinger's Equation, I can elaborate a bit more.

Dickfore

No, B actually is zero in the bulk of the SC - Meissner effect. This is why perfect conductors differ from superconductors.

It drops to zero in a very thin region on the surface of the SC, called London magnetic penetration depth.

But, this expulsion of external magnetic field can only persist up to a point - critical magnetic field H_c, after which the SC goes into a normal metal state. This behavior is characteristic for so called type I SCs.

For type II SCs, among which are all the high Tc SCs, the behavior is different. After the value of the external field exceeds a value, H_c1, there is the possibility of some of the magnetic field lines to penetrate the bulk of the superconductor in vortices which have a quantized magnitude of the magnetic flux passing through them. This is a mixed state, and persists up to a higher critical field H_c2, above which the material becomes a normal metal.

lufc88

Does the cooper pairs spins (up and down) cancelling out have anything to do with anything?

lufc88

cheers for replying as well

K^2

I see. I was somewhat confused over how the sueprconducting magnet works, leading me to believe that the field is "frozen in" rather than expelled. I'll have to take a closer look at the QM involved, though, I think I'm starting to see why it actually has to be zero rather than just constant.

M Quack

There are type-I and type-II superconductors, depending on the ratio of the London penetration depth and the coherence length.

The magnetic field inside type-I superconductors is exactly zero.

In type-II SC, the magnetic field can penetrate the SC, but it gets bundled into quantized "flux lines"

http://en.wikipedia.org/wiki/Type-II_superconductor

All superconducting magnets are built from type-II superconductors.

DrDu

On a more abstract level, the expulsion of the magnetic field from a superconductor is due to the breaking of the U(1) gauge symmetry. Due to the Anderson-Higgs mechanism, the electromagnetic field becomes massive and therefore cannot enter far into the superconductor.
The Anderson Higgs mechanism was just recently confirmed in elementary particle physics with the detection of the Higgs boson. In a superconductor, the Cooper pairs correspond to the Higgs boson.

Dickfore

Yes, this is the explanation following from the Ginzburg-Landau phenomenological theory.

In the microscopic BCS theory, if one calculates the response of the current density J, to the vector potential A, one obtains:
[tex]
\mathbf{J} = -Q \, \mathbf{A}
[/tex]
This constitutive relation, coupled with Ampere's Law:
[tex]
\nabla \times \mathbf{H} = \frac{4 \pi}{c} \, \mathbf{J}
[/tex]
and the expression for the magnetic field in terms of the vector potential [itex]\mathbf{H} = \nabla \times \mathbf{A}[/itex] leads to a Proca type equation.

One can immediately see the breaking of gauge invariance, because a current is expressed in terms of a gauge-dependent quantitiy, namely the vector potential.

DrDu

I wouldn't say so. Anderson derived the Anderson-Higgs mechanism in superconductors using random phase approximation to derive the current response to electromagnetic fields. The BCS hamiltonian is not gauge invariant and hence J=-QA cannot be derived from it. That's why Anderson (and others) was looking for a more elaborate microscopic description.

Dickfore

I'm afraid I don't follow your logic.

DrDu

Neither do I, sometimes :-)

In post #2 and #3 you ascribe the Meissner effect to the screening effect of dissipationless supercurrent of Cooper pairs. However, the classical BCS explanation of the Meissner effect involves only single particle excitations, the reduced BCS hamiltonian does not even have collective excitations which could be described as a flow of Cooper pairs.

Dickfore

But, that is the whole point. The flow of a suppercurrent does not excite any excitations, and that is why it is dissipationless. If you go through the calculation, you will see the Q kernel has a Δ² in the numerator, and must vanish in the normal state of the metal. After the integral over all momenta and the sum over Matsubara frequencies is performed, you see that at zero T, the dependence on Δ cancels, and the Q parameter is proportional to the total density of the electrons. This is understandable, since all the electrons are in a "BCS condensate" at zero temperature. As the temperature is increased, the magnitude of the Q parameter decreases, mainly because the density of electrons in the "condensate" decreases due to thermal excitations and breaking of Cooper pairs.

It is true that the BCS Hamiltonian is a one-body Hamiltonian which may be diagonalized in terms of the so called Bogoliubov fermionic excitations (which have a gap Δ in their spectrum).

However, it is precisely this gap that it is the effect of the Cooper pairs. That is why we have the anomalous vertex [itex]\Delta \, c^{\dagger}_{\mathbf{k}\uparrow} \, c^{\dagger}_{-\mathbf{k}\downarrow} + h.c.[/itex]. Obviously, this term does not conserve the number of electrons. The explanation is that you may get two or destroy two electrons by breaking up a Cooper pair or binding them up into one from the "condensate".

DrDu

Ok, you are completely right. Where I got confused is here:
The Cooper pairs have zero momentum flow but [itex] 0=p=mv-eA[/itex]
hence [itex] j=nev=ne^2/m A [/itex] so there is still a diamagnetic current.
The point I wanted to make is the following:
The appearance of the gap (which is responsible for the absence of the paramagnetic current) is not physical in the model of BCS, but is due to the artificial long range nature of the reduced interaction
[itex]\sum_{kk'}V_{kk'}c_{k'\uparrow}^*c_{-k'\downarrow}^*c_{-k\downarrow}c_{k\uparrow}[/itex].
In a real superconductor there is a long range repulsive interaction between the electrons, namely the Coulomb interaction which gets collective excitations out of the gap.
That's why many people didn't believe the explanation of the Meisner effect of BCS.

Dickfore

Actually, there is a Coulomb interaction in any metal, regardless whether its SC or normal. But, this interaction is screened and is no longer of infinite range (see Thomas-Fermi screening). Translated into k-space, the potential Coulomb potential energy is no longer:
[tex]
V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k} - \mathbf{k}' \vert^2}
[/tex]
which has a singularite for zero momentum transfer, but is something like:
[tex]
V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k} - \mathbf{k}' \vert^2 + k^2_{TF}}
[/tex]
and saturates to a constant for small momentum transfers.

Additionally, there an electron-phonon interaction, which becomes attractive in the same region and overcompensates the screened Coulomb interaction.

I think you mean that some people were puzzled what overcomes the Coulomb repulsion of the electrons to bind into a Cooper pair, but I find it pretty unbelievable that people suspected the BCS explanation of the Meissner effect.