Wednesday, February 04, 2009

To tide you over....

Proposal deadlines are almost done with, and then I'll try to post more. In the meantime, here's a question to tide you over. Superconductors are classified as "type I" or "type II". In type I superconductors, superconductivity is completely destroyed above some critical externally applied magnetic field, Hc. To be more jargon-y, in these materials the coherence length (the typical spatial extent of pair-like correlations between electrons in the superconductor) is much larger than the magnetic penetration depth (the distance that a magnetic field penetrates into a superconductor before it is screened away by circulating supercurrents). In type II superconductors, above a critical field Hc,1 magnetic flux starts to penetrate the superconductor in the form of vortices (localized regions with nonsuperconducting cores through which magnetic flux is threaded, and around which are circulating supercurrents). Above a second, higher critical field, Hc,2, superconductivity is eventually destroyed. In type II superconductors, the coherence length is much shorter than the penetration depth.

So here's the question: why are almost all the pure elemental superconductors type I, and why are essentially all alloys type II? Is there a simple argument that explains this? If there is, I haven't heard it....

13 comments:

CarlBrannen said...

While reading random arXiv articles, I ran into the interesting polemic 0901.4099. So I read a little further. It made sense to me, but this is not my field.

See Charge expulsion, Spin Meissner effect, and charge inhomogeneity in superconductors, "to be published in" Journal of Superconductivity and Novel Magnetism.

From what I gather, it would have to do with a requirement that alloys, being of mixed material, must have lambda_L quite large (see paper). Otherwise the circles whose radii are defined by lambda_L would include differing percentages of the alloy components. I'll send a link to your question to the author, maybe he'll comment.

Doug Natelson said...

The author of that paper is Jorge Hirsch, who is also the person that coined the now famous "h-index" or "h-number" as a means of evaluating scientific productivity and impact. His thoughts on this would, of course, be appreciated, though I should point out that his take on superconductivity is controversial.

Uncle Al said...

The only BCS supercon in a chiral space group AuBe, Tc = 2.64 K (P2(1)3 (#198), a = 4.668 A, Z = 4) is disputed. X-ray powder diffraction cannot see beryllium. A sufficiently large quality crystal has not been grown for neutron diffraction.

What renders chirality and superconductivity incompatible?

Anonymous said...

Interesting paper, the one mentioned in the first comment. It looks to me (a non-specialist) that it addresses more the crossover between type I and type II SCs (Fig. 2, Sec. VII) and not why 2lambda_L > \xi in alloys. It would be interesting to hear from Hirsch on this.

On a related note, after reading his polemic paper comparing BCS phonon-mediated pairing mechanism to Madoff's Ponzi scheme, I wonder whether his ideas have had any traction with experimentalists working in the area.

Doug Natelson said...

Anon. - Yeah. I think the number of physicists who view BCS as "fundamentally flawed" is very small.

Karthik said...

The coherence length of a superconductor depends on the electron mean free path (MFP) in the normal state (L) as
1/c = 1/co + 1/L where co is the coherence length of the pure metal before alloying. All alloys have poor conductivity and small MFP => small coherence length => they become type II.

Doug Natelson said...

Karthik - I know what you're driving at, but I think the statement that all alloys have short mean free paths is not true. Furthermore, single-crystal Nb is type II. I agree that "dirt" points in the right direction, but I don't think that's the whole puzzle. I suppose the real test of this would be to quench-condense an elemental superconductor (lead? tin? mercury?) into a nanocrystalline or amorphous form, and see what happens to the superconductivity.

Doug Natelson said...

Follow-up.... According to this (pdf!), lead-tin solder has a lower resistivity than pure lead, and yet solder is type II while lead is type I....

Anonymous said...

Anon2: Doug, I have very little background in magnetics. What's a typical depth of penetration for a magnetic field into a superconducting solid? Also, does screening of the field with depth imply that superconductivity is a surface phenomenon?

Heumpje said...

My guess:
The important number is kappa=lambda/xi. lambda=penetration depth, xi=coherence length.

lambda is determined by the electrodynamic properties, i.e. the superfluid density. The pentration depth for most materials is on the order of 100 nm (40 nm for Nb).

xi however is determined by crystal properties. Quoting from Tinkham,

coherence length...characterizes the distance over which the wavefunction can vary without undue energy loss (page 11).

I think therefore that Khartik is right in focussing on the coherence length: alloys are generally better described in an atomic picture (i.e. tight binding) while elements are generally itinerant (i.e. free electron picture).

The coherence length will be shorter in the former and longer in the latter, as it is essentially determined by the length scale of variation imposed on the superfluid by the atomic potential. As an aside: Nb is better described in a tight binding model, so this would fit the picture.
Regarding the coherence length in the SC state:

1/xi = 1/xi_0+1/L

with xi_0 proportional to v_f/T_c, and L again the MFP in the normal state.
Lets take L-> infinite. Then xi=xi_0, and the problem reduces to the statement,

Type I: v_f > lambda
Type II: v_f < lambda

At first glance it seems to me I reached the opposite conclusion, as above. On the other hand v_f is inversely proportional to sqrt(m_b). Tight binding gives narrower bands than free electron bands, and hence larger effective masses.

...So, my guess:
free electron : tight binding < - > type I : type II

Further emperical evidence: cuprates have coherencce length on the order of a unit cell and are extreme type II with narrow tight binding bands...

Anonymous said...

Doug Natelson wrote:
I suppose the real test of this would be to quench-condense an elemental superconductor (lead? tin? mercury?) into a nanocrystalline or amorphous form, and see what happens to the superconductivity.
----
Doug, I think this is what Allen Goldman and Bob Dynes have been doing all their career with great success! I do not understand what you think will be new in this now?

Charudatta Galande said...

Dr.Natelson, there have been several attempts to measure nanocrystalline superconductors, and I have worked on one such effort. Nanocrystalline superconductors are all type-2.

A more detailed set of experiments on nano-superconductors can be found in the foll. papers - (both from the same group where I worked)

1. http://link.aps.org/doi/10.1103/PhysRevLett.95.147003

2. http://www.iop.org/EJ/abstract/0953-8984/21/20/205702

-Charudatta.
(MEMS grad student, Rice)

Charudatta Galande said...

From these papers, you would see that Nb and Pb behave very differently in the quantum size regime. Nb is intermediate-coupling and Pb is strong-coupling, so a similar study on a weak-coupling superconductor would make the circle complete and give a complete picture.

-Charu.