# Why is the magnetic field of a superconductor normally excluded?

1. Sep 23, 2012

### lufc88

Why is the magnetic field of a superconductor normally excluded?

2. Sep 23, 2012

### Dickfore

Re: superconductivity

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

3. Sep 23, 2012

### Dickfore

Re: superconductivity

The flow of Cooper pairs.

4. Sep 23, 2012

### K^2

Re: superconductivity

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.

5. Sep 23, 2012

### Dickfore

Re: superconductivity

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 Hc, 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, Hc1, 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 Hc2, above which the material becomes a normal metal.

6. Sep 23, 2012

### lufc88

Re: superconductivity

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

7. Sep 23, 2012

### lufc88

Re: superconductivity

8. Sep 25, 2012

### K^2

Re: superconductivity

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.

9. Sep 25, 2012

### 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.

10. Sep 25, 2012

### 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.

11. Sep 25, 2012

### 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:
$$\mathbf{J} = -Q \, \mathbf{A}$$
This constitutive relation, coupled with Ampere's Law:
$$\nabla \times \mathbf{H} = \frac{4 \pi}{c} \, \mathbf{J}$$
and the expression for the magnetic field in terms of the vector potential $\mathbf{H} = \nabla \times \mathbf{A}$ 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.

12. Sep 25, 2012

### 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.

13. Sep 25, 2012

### Dickfore

14. Sep 26, 2012

### 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.

15. Sep 26, 2012

### 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 $\Delta \, c^{\dagger}_{\mathbf{k}\uparrow} \, c^{\dagger}_{-\mathbf{k}\downarrow} + h.c.$. 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".

16. Sep 26, 2012

### DrDu

Ok, you are completely right. Where I got confused is here:
The Cooper pairs have zero momentum flow but $0=p=mv-eA$
hence $j=nev=ne^2/m A$ 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
$\sum_{kk'}V_{kk'}c_{k'\uparrow}^*c_{-k'\downarrow}^*c_{-k\downarrow}c_{k\uparrow}$.
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.

17. Sep 26, 2012

### 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:
$$V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k} - \mathbf{k}' \vert^2}$$
which has a singularite for zero momentum transfer, but is something like:
$$V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k} - \mathbf{k}' \vert^2 + k^2_{TF}}$$
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.

18. Sep 26, 2012

### DrDu

No, the finite energy of the plasmons (irrespective of in the superconductor or normal metal) is due to the long range of the Coulomb forces. Screening does not work for this kind of collective modes.
The doubts on the validity of the BCS explanation of the Meissner effect spurred several important developments (I can give you references tomorrow).
For example Nambu wrote in his Nobel lecture ( http://www.nobelprize.org/nobel_prizes/physics/laureates/2008/nambu-lecture.html )
"I will now recall the chain of events which led me to the idea of SSB and
its application to particle physics. One day in 1956 R. Schrieffer gave us a
seminar on what would come to be called the BCS theory [5] of supercon-
ductivity. I was impressed by the boldness of their ansatz for the state vector,
but at the same time I became worried about the fact that it did not appear
to respect gauge invariance. Soon thereafter Bogoliubov [6] and Valatin [7]
independently introduced the concept of quasiparticles as fermionic excita-
tions in the BCS medium. The quasiparticles did not carry a definite charge
as they were a superposition of electron and hole, with their proportion
depending on the momentum. How can one then trust the BCS theory for
discussing the electromagnetic properties like the Meissner effect? It actually
took two years for me to resolve the problem to my satisfaction. There were
a number of people who also addressed the same problem, but I wanted to
understand it in my own way. Essentially it is the presence of a massless col-
lective mode, now known by the generic name of Nambu-Goldstone (NG)
LPN 2008 ekvationer...
boson, that saves charge conservation or gauge invariance."

19. Sep 26, 2012

### Dickfore

I don't understand what you mean here. What collective modes? And how is screening supposed to work for "other modes" that it doesn't work for "these modes"?

Yes, I remember reading his Nobel lecture. Let me emphasize once more that:
1. The flow of a supercurrent is not accompanied with excitations of any kind. Period. In fact, it may be accepted as a condition of superfluidity;
2. I think the BCS Hamiltonian is gauge invariant. It is true that the product:
$$c^{\dagger}_{\mathbf{k} \uparrow} \, c^{\dagger}_{-\mathbf{k}\downarrow}$$
gets a phase after a gauge transformation, but you have to remember that this product is itself multiplied by the gap function $\Delta(\mathbf{k})$, which in the BCS theory is given as the average:
$$\Delta(\mathbf{k}) = -\frac{1}{V} \, \sum_{\mathbf{q}}{\tilde{U}_{\mathbf{q}} \, \langle c_{\mathbf{k} + \mathbf{q}, \uparrow} \, c_{-\mathbf{k} + \mathbf{q},\downarrow} \rangle}$$
so, it acquires an opposite phase than the above product. These two phases cancel one another and the Hamiltonian is left gauge invariant.

20. Sep 27, 2012

### DrDu

Before answering let me say that I mainly wanted to point out the historical perspective, namely how superconductors were found to be the first examples of systems in which the Higgs mechanism is at work.

Let me try to answer this question first: Plasmons are collective modes. They are responsible for the screening in metals and superconductors. However the plasmon cannot screen itself.

Second: In a superconductor global gauge symmetry is broken. Hence one would expect from Goldstones theorem some collective mode which zero frequency in the long wavelength limit, i.e. some collective excitation inside the gap.
It is clear that this may potentially render the analysis of BCS of the Meissner effect invalid, as with collective modes inside the gap there would be no gap and a paramagentic contribution to the current cannot be excluded.

However, there are no excitations inside the gap in a real superconductor. Anderson was (at least one of ) the first who showed that the Goldstone mode has a finite frequency also at k=0 once Coulomb interaction is taken into account and that it becomes a normal plasmonic mode which does not lie in the gap:
http://prola.aps.org/abstract/PR/v112/i6/p1900_1
http://prola.aps.org/abstract/PR/v110/i4/p827_1
Others who analyzed this situation where Rickayzen and Nambu.
Especially Nambu has worked out the mechanism much more clearly making use of the Ward identities: http://prola.aps.org/abstract/PR/v117/i3/p648_1
This whole analysis was spurred by the doubts about the validity of the calculation of the Meissner effect by BCS, let me cite here the abstract from the article by Schafroth
http://prola.aps.org/abstract/PR/v111/i1/p72_1
: "It is shown that a theory of superconductivity which starts from an "effective" Hamiltonian with significantly velocity-dependent interaction between electrons does not possess well-defined magnetic properties. The Meissner effect cannot therefore be established from such a theory. This applies in particular to the Bardeen-Cooper-Schrieffer theory."

At about the same time particle theorist were desperately seeking models of relativistic systems with broken symmetry but not possessing zero mass Goldstone boson.
Anderson wrote a paper in 1963
http://prola.aps.org/abstract/PR/v130/i1/p439_1
where he tried to explain the mechanism to particle theorists.
Higgs developed the first relativistic model showing the Higgs mechanism in response to the article by Anderson.