Ginzberg-Landau and the Cooper Pair

[Zymandia]

It seems to me that the Ginzburg-Landau equations are derived from considering the particle of superconduction to be behaving in a superfluid way. This required the particle of superconduction to be a boson so that B-E statistics might apply. The use of Bose's presumptions surely tell us that the superconducting particle is an indistinguishable boson and hence can have no structure, certainly not fermionic. The use of two electrons' charge and mass in the Ginzburg-Landau equations requires that the observables of two electrons are in the bosonic wave-function, and hence have no separate existence as fermions. On another thread I notice that Landau was most unhappy that BCS did not provide the creation and annihilation operator required by B-E.
I think it may be instructive to consider the implication that a super-conducting Thingy is created from two electrons. This Thingy is a bosonic wave-function, which annihilates/decays back into two electrons. This has R(360) symmetry, whereas the electrons remaining fermions in the Cooper Pair requires R(720) symmetry.
There seems surely to be a basic conflict between Ginzburg-Landau and perturbative electron coupling theories?

 

[nbo10]

The GL equations are based on Landaus theory of second order phase transitions. They are not a microscopic solution.

 

[Zymandia]

They are not a microscopic solution, of course. However they impose the constraint on any microscopic solution that the Cooper Pair electrons have been consumed by a boson creation operator and will be re-created on the annihilation of the boson. Lev Landau realized the necessity of this.

 

[0xDEADBEEF]

I don't think that the 720° symmetry is necessary. The combination of the two electrons will carry a spin zero representation of the Lorenz group which will behave bosonic so 360° symmetry. I am not sure about the details there is a theorem by Hermann Weyl that shows, that this representation exists and how it connects space and spin part of the wavefunction.

 

[Zymandia]

"The combination of the two electrons will carry a spin zero representation of the Lorenz group which will behave bosonic so 360° symmetry"
Only if the combination is a single wave-function with no distinguishable structure. If, however, the Cooper Pair is two electrons bound by a gauge boson (phonon) then R(360) leaves the boson phase unchanged but the constituent electrons are now anti-phase.

 

[kanato]

The reference Physics Letters A, 373, 269 (2009) might be relevant. They seem to show that a superposition of Cooper pairs obeys boson commutation relations. I think a superposition like this might be thought of as the quasiparticles used in Ginzberg-Landau.

 

[Zymandia]

If they are bosonic quasi-particles then they may not have fermionic constituents. It seems to me from the abstract that these guys are trying to have their cake and eat it. They are suggesting that a collection of Cooper Pairs maybe considered a BEC while the electrons remain fermions. If the electrons remain fermions then their mass and charge are not part of the bosonic wave-function and the Ginzburg-Landau equations wouldn't apply.
Dirac, Pauli et al. built QED as we know it on basic mathematical rules of which, surely, one is that no wave-function maybe both symmetric and anti-symmetric.
Intriguingly if one does accept Landau's view that the Cooper Pair is a single bosonic wave-function which contains the observables of two electrons, it is quite simple to explain the non-BCS behaviour of HTSCs and the Nernst effect.

 

[kanato]

Well I haven't checked their math. But they construct creation and annihilation operators out of Cooper pair creation/annihilation operators and seem to show that these new operators have the proper bosonic commutation relation. So, if the math is sound I don't see what the problem of having their cake and eating it too is. The constituent particles are still the electrons, so the electron charge and mass are still the fundamental quantities.

Surely a wavefunction can't be both symmetric and antisymmetric with respect to the same particles, but you definitely can't show that a wavefunction can't be antisymmetric with respect to the fundamental particles and have different symmetry with respect to quasiparticles (eg. A Cooper pair wavefunction is not antisymmetric under exchange of Cooper pairs). So I don't see a compelling reason to believe that it's impossible to have a wavefunction where the quasiparticles are symmetric and the constituent particles are antisymmetric. Obviously getting the right symmetry in your quasiparticles is not automatic, but the statement that it is impossible is a rather strong statement that needs justification, particularly since this article seems to show that it is in fact possible.

 

[0xDEADBEEF]

If they are bosonic quasi-particles then they may not have fermionic constituents. [...]

Why not? Would you care to explain why there are bosonic atoms full of electrons, and protons, if bosons cannot have fermionic constituents.?

 

[Zymandia]

While they have indentifiably fermionic constituents they cannot have an overall symmetric wave-function surely. Only down at the milliKelvins can whole atoms become so indistinct that bosonic behaviour is seen, indeed it seems to me that the bosonic / fermionic phase change at critical temperature is precisely about indistinctness. I've been trying to find some discussion about precisely what the Bose condition means.
If I may, I'll try and explain. Take a hypothetical N particles at Absolute Zero, increase temp till one particle goes into the lowest excited state. Maxwell Boltzmann says there N ways this can happen, but, as I understand it, Bose said there is only one way this can happen. One way of envisioning this is to say each particle is a mixed state of 1/N excited and (N-1)/N ground states, thus each particle is in an identical state and Bose is obviously right. Another view is that quantum mechanics lays a cloak of unobservability on the whole system so that the number N loses all physical significance other than as a quantum number of the bosonic function. Are there any texts on these ideas?

 

[0xDEADBEEF]

I know that on the mathematical bottom level these things are far from being solved. But it is generally accepted that two fermions can team up to form a boson. This happens with hadrons, where the quarks join and also with atoms. Fermionic atoms can pair and then form fermionic BECs. If you check out the fermionic BEC papers I am sure there must be word or two about this.

 

[kanato]

What exactly do you mean when you say "overall symmetric"? Of course if you have a wavefunction built from fermionic particles, if you exchange two fermion operators (i /= j) you will pick up a negative sign. But you can construct a wavefunction out of bosonic quasiparticle operators, and that wave function will be symmetric under the exchange of the quasiparticles, ie. commuting two boson operators (i /= j) does not change the sign. Since it's built out of fermion operators swapping two fermions will still result in the anti-symmetry, but it is symmetric with respect to the boson operators. With regards to your original question, I think since Ginzberg-Landau is phenomenological it works for Cooper pairs as long as it's possible to construct these bosonic operators.

With regards to the second paragraph in #10, I take the view that quantum mechanics says N is just a quantum number of the bose field. Mathematically, we would write the ground state as |N,0,0,...\rangle and the excited state as |N-1,1,0,0,...\rangle = N^{-1/2} a_1^\dagger a_0|N,0,0,...\rangle (if I have the normalization right). AFAICT there is no way to construct a superposition which involves each particle, since the operators are labeled by their states. So this view is consistent with the mathematics of second quantization. But you question maybe is a bit more philosophical, and I don't think I can shed any light on that aspect.

 

[Zymandia]

But it is generally accepted that two fermions can team up to form a boson"
The 'team up' is the crux. A pion is a boson formed of two fermions that are so heavily interacting that it would be meaningless to ask about its constituents before it decayed. However if we talk about a bosonic quasi-particle with fermionic constituents then on swapping the quasi-particles the constituents of both quasi-particles, being fermionic, would change sign. Saying 'we can make bosonic style creation and annihililation operators for this entity, so it is a boson' seems like 'lions have tails, this animal in my hand has a tail so it must be a lion'. A quasi-particle, to be bosonic, must show R(360) symmetry which it can't do if it is a perturbation of two distinct fermions because the fermions by definition will have R(720) symmetry.
For a Cooper Pair to be bosonic the electrons must be interacting so strongly that coupling interaction cannot be regarded as a perturbation on a pair of underlying fermionic wave-functions, because if the wave-functions were distinct enough to be called electrons they would be R(720) anti-symmetric and hence the overall Cooper Pair would be fermionic.
'Boseness' seems to be about indistinctness of the particles, at least that's where the maths points, so it seems absurd to suggest that an indistinct particle may yet have distinct constituents.
For Physicists Maths can be a harsh mistress, and it sometimes tells us things we simply don't want to hear. In this case the presumptions of the maths of Bose-Einstein statistics requires that the Cooper Pair is a single bosonic wave-function with no constituents. Ginzburg-Landau comes directly from the presumption that the superconducting wave-function is bosonic ie. too heavily mixed an interaction to meaningfully talk about the electrons it will decay into.
Attempting to talk about electron-electron coupling in Cooper Pairs seems to me to be avoiding what the maths says. Add in that Tate et al. found the Cooper Pair in Niobium to be about 43eV per e- too heavy for a superconducting gap in the milli eV and the idea of a perturbative interaction between electrons looks even more dubious.
The Cooper Pair is far too odd and has too many weirdities associated with it to be seen as two electrons in an unusual dance, how it should be seen is what I'm desparate to discuss.

 

[kanato]

But it is generally accepted that two fermions can team up to form a boson"
Saying 'we can make bosonic style creation and annihililation operators for this entity, so it is a boson' seems like 'lions have tails, this animal in my hand has a tail so it must be a lion'. A quasi-particle, to be bosonic, must show R(360) symmetry which it can't do if it is a perturbation of two distinct fermions because the fermions by definition will have R(720) symmetry.

You say some things that aren't clear. For instance, you say "A quasi-particle, to be bosonic, must show R(360) symmetry," but the symmetry of the wavefunction is with respect to the exchange of particles, and has no bearing on a single particle wavefunction. For quasiparticles to be bosonic, you need the wavefunction to symmetric under exchange of the quasiparticles, not the constituent particles. This is true regardless of whether they have fermionic or bosonic constituents. Also, what exactly do you mean by R(360) and R(720) symmetry? I am not familiar with this notation except in reference to rotations in 3D space, but this is not the general symmetry property of multi-particle wavefunctions.

Boson operators are defined by their commutation relations. If you have the proper commutation relations for an operator, then it is a boson operator. Your analogy with lions doesn't hold. These operators aren't "bosonic style", they are boson operators.

For a Cooper Pair to be bosonic the electrons must be interacting so strongly that coupling interaction cannot be regarded as a perturbation on a pair of underlying fermionic wave-functions, because if the wave-functions were distinct enough to be called electrons they would be R(720) anti-symmetric and hence the overall Cooper Pair would be fermionic.
'Boseness' seems to be about indistinctness of the particles, at least that's where the maths points, so it seems absurd to suggest that an indistinct particle may yet have distinct constituents.

Why does it seem absurd?

Have you read the paper I posted? It seems to mathematically address exactly the concern that you raise. If you haven't read it, it's rather short and clearly written so there's no excuse for not reading it. Can you show that their boson operators don't have the required symmetry? (If so, you should publish a response to their article.) You say math is a harsh mistress, but your arguments seem light on the math and heavy on the hand-waviness.

For Physicists Maths can be a harsh mistress, and it sometimes tells us things we simply don't want to hear. In this case the presumptions of the maths of Bose-Einstein statistics requires that the Cooper Pair is a single bosonic wave-function with no constituents.

Where does the "no constituents" part come from? I've never heard of that being required for Bose-Einstein statistics to work. Obviously this will be required if you want to examine excitations that break up the quasiparticles, but for low energy properties I don't see that this is required.

Ginzburg-Landau comes directly from the presumption that the superconducting wave-function is bosonic ie. too heavily mixed an interaction to meaningfully talk about the electrons it will decay into.

The superconducting state is gapped.

Add in that Tate et al. found the Cooper Pair in Niobium to be about 43eV per e- too heavy for a superconducting gap in the milli eV and the idea of a perturbative interaction between electrons looks even more dubious.

Do you have a complete reference for the Tate paper?

 

[Zymandia]

Tate Paper at:
http://prola.aps.org/abstract/PRL/v62/i8/p845_1

 

[Zymandia]

"I take the view that quantum mechanics says N is just a quantum number of the bose field"

"For quasiparticles to be bosonic, you need the wavefunction to symmetric under exchange of the quasiparticles, not the constituent particles. This is true regardless of whether they have fermionic or bosonic constituents"

What I find odd is that a condensate makes its constituents so indistinct that only the total number is conserved as a quantum number for the state, yet within that it is plausible to talk of constituent electrons. I'm sorry if my argument seems hand-wavy, but as I see it we are talking about the root definitions on which QM is based. The idea of swapping bosonic quasiparticles surely requires their fermionic constituents to be swapped with them, and they must go anti-symmetric. This can only work if each pair is anti-symmetric in themselves and occupying the same physical co-ordinates, otherwise the swap can't be bosonic. I wish we could ask Dirac or Pauli about this.

 

[kanato]

Well if you swap two cooper pairs, you are swapping two pairs of electrons, so you pick up two negative signs. This brings it back to a symmetric exchange of particles (with caveats about the quantum numbers, since Cooper pairs are hard core bosons).

 

[Zymandia]

Run that by me again. How does one swap pick up two negative signs?

 

[Minich]

If we read original Ginzburg-Landau paper carefully, then we don't find any fermion nature in ORDER PARAMETER.

GL order parameter equation does not have any fermion and/or boson nature.

You can consder it as an equation for SINGLE LARGE COMPLEX PARTICLE/WAVE.

 

[Zymandia]

If it is a 'SINGLE LARGE COMPLEX PARTICLE/WAVE' then it wouldn't obey the Pauli Exclusion Principle with regard to the surrounding lattice electrons/Brouillin Zone??

 

[Zymandia]

"If we read original Ginzburg-Landau paper carefully,"
I can't find an english language link (and the russian link didn't work), is it up anywhere?
Also can anyone point me at some cuprate critical field/temp plots for various samples?
There is a kink at < 20K for the higher Tc samples (> 70K) which I think shows the cuprate leaving the BCS regime. The regime it enters is simply described if the single bosonic wave-function of a Cooper Pair has taken an energy that is forbidden to electrons by a semi-conductor band-gap. Thus this regime is telling us about the B-E chemical energy of the Cooper Pair wave-function.
I think.

 

[kanato]

Run that by me again. How does one swap pick up two negative signs?

Consider a fermionic wave function with four particles, \psi(x_1,x_2,x_3,x_4), and we consider two bosonic quasiparticles, the first constructed from fermions 1 and 2, and the second from fermions 3 and 4. To swap the quasiparticles, we swap all of the constituents, so x_1 gets swapped with x_3 and x_2 gets swapped with x_4. The single swap of the pair of quasiparticles results in swapping two pairs of fermions, picking up two negative signs which cancel.

Consider a state with two Cooper pairs:
b_k^\dagger b_{k&#039;}^\dagger |0\rangle (k /= k' of course)
In fermion operators this is
c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger c_{k&#039;\uparrow}^\dagger c_{-k&#039;\downarrow}^\dagger |0\rangle

Now commute the fermion operators until you get b_{k&#039;}^\dagger b_k^\dagger |0\rangle. You will find this takes four commutations, picking up four negative signs, yielding a result with no additional sign.

 

[Zymandia]

Hmm, two swaps on a wave-function will always be symmetric even for an anti-symmetric wave-function. Your original 4 electron state is plainly fermionic since swapping any single pair of particles will be anti-symmetric. However you are saying that we can declare two fermions to be a single 'quasi-particle' and now the swapping rules may apply to two pairs of fermions at once. Could you please prove to me that the swapping rule may be applied to composite particles, since it seems if they can that any system can be bosonic with the appropriate choice of quasi-particle.

 

[kanato]

Hmm, two swaps on a wave-function will always be symmetric even for an anti-symmetric wave-function. Your original 4 electron state is plainly fermionic since swapping any single pair of particles will be anti-symmetric. However you are saying that we can declare two fermions to be a single 'quasi-particle' and now the swapping rules may apply to two pairs of fermions at once. Could you please prove to me that the swapping rule may be applied to composite particles, since it seems if they can that any system can be bosonic with the appropriate choice of quasi-particle.

You should work through this example (with the assumption that k is not equal to k') to see it for yourself:

Consider a state with two Cooper pairs:
b_k^\dagger b_{k&#039;}^\dagger |0\rangle (k /= k' of course)
In fermion operators this is
c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger c_{k&#039;\uparrow}^\dagger c_{-k&#039;\downarrow}^\dagger |0\rangle

Now commute the fermion operators until you get b_{k&#039;}^\dagger b_k^\dagger |0\rangle. You will find this takes four commutations, picking up four negative signs, yielding a result with no additional sign.

The question of whether or not any system could use bosonic statistics depends on its excitations. If the excitations of the system are bosonic quasiparticles then bose statistics. If the excitations are electron-like or fermionic quasiparticles, then the boson quasiparticles won't be able to represent those excitations. For a superconductor above the critical temperature, it's possible to easily excite individual electrons (there are excitations with arbitrarily small energy), but below the critical temperature, since the superconducting state is gapped it takes a minimum finite amount of energy to break the pairs. So there, Fermi-Dirac statistics don't apply because average thermal energy is not enough to break pairs.

 

[Zymandia]

In normal BCS compliant superconductivity the phonon that breaks the Pair (not the weird thing that holds the pair together) is surely a normal F-D thermal phonon?
In the HTSCs the problem is that they survive to temperatures where kT is much bigger than the superconducting gap. Currently this is not understood, as I understand it. I'd suggest it is because the energy (per decay electron) of the single wave-function Cooper Pair has hidden itself in the band structure.

 

[Zymandia]

I've thought hard on whether one can take a group of fermions and by considering them as pairs declare them bosonic. It seems to me that the pairing to a quasi-particle requires that the quasi-particle wave-function be unique for a given point which seems unlikely since its substructure (the 2 electrons) will have extra degrees of freedom. I am sure there are more theoretical restrictions as to when one may declare pairs of fermions to be bosonic. Instinct says two electrons can only be considered bosonic when their interaction is sufficiently beyond a perturbation that there are no underlying fermionic wave-functions, but I can't find that explicitly stated anywhere.
Certainly all the texts are very clear that wave-functions are one or the other (until you hit anyons and FQHE) and the idea of a "bosonic wave-function with fermionic constituents" where both are distinct enough to have an effect seems wrong. Atoms with fermionic constituents can become bosonic at milliKelvins but I'd argue the fermionic nature of the constituents must be indistinct in order for the whole atom to be considered as a bosonic wave-function with a single point of origin.
Aside from the theory there is phenomenology that suggests that there is something more to the Cooper Pair than lightly bound electrons. As I understand it Josephson Junctions have had Cooper Pairs crossing distances > 20nm which is much more than we've ever seen an electron tunnel, yet apparently two highly repulsive electrons choose to jump many lattice cell sizes together. Another anomaly is that which Tate found, namely that in Niobium the Cooper Pair appeared to be 43eV heavier per electron with a binding energy of milli eV. Both these phenomena strongly suggest that the Cooper Pair is more than a perturbative interaction between two distinct electrons.

 

[ZapperZ]

Certainly all the texts are very clear that wave-functions are one or the other (until you hit anyons and FQHE) and the idea of a "bosonic wave-function with fermionic constituents" where both are distinct enough to have an effect seems wrong.

No, this statement is wrong.

That whole experimental discovery of finding the BCS-BEC crossover clearly shows not only the theoretical connection between BCS and BEC, but also that it is experimentally observable via the fermionic condensates. There are tons of theoretical description now on such a crossover[1]. Which part of the ground state wavefunction that went from one to the other that you could not comprehend? It would be helpful, rather than this handwaving criticism, for you to clearly point out where exactly in the formalism that you found "wrong".

Zz.

[1] http://jfi.uchicago.edu/~qchen/Papers/PhysRep412p1-88.pdf

 

[ZapperZ]

In normal BCS compliant superconductivity the phonon that breaks the Pair (not the weird thing that holds the pair together) is surely a normal F-D thermal phonon?
In the HTSCs the problem is that they survive to temperatures where kT is much bigger than the superconducting gap. Currently this is not understood, as I understand it. I'd suggest it is because the energy (per decay electron) of the single wave-function Cooper Pair has hidden itself in the band structure.

This is where the problem lies.

In BCS theory, it really doesn't matter what the pairing "glue" is. It just happens that in conventional superconductors, it is phonons. So why this is being brought into this thread is a mystery. It does not affect the pairing symmetry. In fermionic condensates, it is the external magnetic fields. And in high-Tc cuprates, some believe it is the spin-fluctuation. It really doesn't matter and this is simply a red-herring to the fact that one CAN get fermionic paring to form a composite boson.

And to add another wrinkle to this, your understanding of the current progress in high-Tc superconductor is severely lacking. The phonon picture has not been ruled out even in this family of superconductor. Look at the photoemission papers coming out of Stanford the past few years (ref: Lanzara's Nature paper) that clearly advocated the apical oxygen phonon breathing mode as having sufficient characteristics to sustained such high Tc.

The more I read this thread, the more I think it is based on a series of false understanding. And from the responses by others that I've read, everyone else seems to think so as well.

Zz.

 

[ZapperZ]

Aside from the theory there is phenomenology that suggests that there is something more to the Cooper Pair than lightly bound electrons. As I understand it Josephson Junctions have had Cooper Pairs crossing distances > 20nm which is much more than we've ever seen an electron tunnel, yet apparently two highly repulsive electrons choose to jump many lattice cell sizes together. Another anomaly is that which Tate found, namely that in Niobium the Cooper Pair appeared to be 43eV heavier per electron with a binding energy of milli eV. Both these phenomena strongly suggest that the Cooper Pair is more than a perturbative interaction between two distinct electrons.

OK, here's another faulty piece of deduction.

There's nothing to prevent the cooper pairs in a josephson junction to tunnel across that distance. The coherence length in conventional superconductor can range from 30 to 1600 nm! {ref: R. Meservey and B. B. Schwartz:Superconductivity, edited by R. D. Parks (New York, N.Y., 1969)). This gives the scale length of the order parameter in GL theory. So there's NOTHING out of the ordinary here as far as the experimental observation of tunneling across such distances.

Secondly, the EFFECTIVE mass of a charge carrier in a material isn't really that big of a deal, is it? I mean, if you want to be impressed by such heavy effective mass, look at the heavy fermion superconductors that have an effective mass up to 200 times the bare mass of electrons! So what's the big deal?

Or is it that you do not understand the Fermi Liquid theory of metals and why these are called "quasiparticles" instead, and that it is out of these quasiparticle that conventional superconductivity emerged! This is NOT a mystery and it is certainly not inconsistent or contradictory to any existing theory of superconductivity. It certainly does NOT present a problem with the formation of Cooper Pairs!

Zz.

 

[Zymandia]

I do understand current theory but I think there is a lot of phenomenological evidence to show that the two electron Cooper Pair is wrong.
Long-distance JJ tunnelling is simply explained if we have a single very long wave-length, ie. low |k|, wave-function performing normal QM tunneling through a gap of a few wavelengths.
I think it is fair to associate the coherence length with the wave-length of the boson G-L postulates.
If the Cooper Pair has high k fermion constituents then the tunnelling potential will attenuate their wave-functions in a few wavelengths of the fermions. So for a two electron Cooper Pair to cross the gap standard QM won't do.
What is official thinking on how a two electron Cooper Pair crosses such a gap?

Incidentally has anyone ever measured the weight of a Cooper Pair in a heavy fermion conductor?
I say it will be within 1% the weight of two electrons, presumably current thinking would predict the effective mass multiplier would carry through to the Cooper Pair ie. 100s times?
An easy test?

 

[ZapperZ]

You were given the BCS ground state wavefunction somewhere in this thread. Look at all the available momentum for each fermion pair and look how they are summed in that series! If you have "understood" the current theory, you would have noticed this. So it is puzzling why you still continue to ask this question.

There is nothing "wrong" with the observation of Josephson tunneling, especially when the phenomena was PREDICTED by Josephson based on what was understood to be the supercurrrent.

And here's something to consider. The CODATA standard for the accepted values of "e" and "h" were ALL derived out of measurement in the superconducting state. In fact, the value of "e" came out of the VERY certain measurement of "2e" in a superconducting flux quanta. If "cooper pair is wrong", so will a lot of our electronics, AND, many precise measurement will not make sense. In fact, you probably should never get out of the house, because we use SQUID detectors in making many non-destructive tests on materials.

There are many aspects of science that has varying degree of certainty. Superconductivity in conventional material AND the formation of Cooper Pair has one of, if not THE, highest degree of certainty as far as knowledge is concerned.

Your next "objection" must contain quantitative values to show where theory doesn't match observation. These hand-waving objection is no longer sufficient or accepted.

Zz.

 

[Zymandia]

You have not answered the question.
How do high k short-wavelength constituents of a Cooper Pair tunnel over > 1000 wave-lengths without the barrier potential attenuating them to nothing. Simple basic quantum mechanics is all that is needed for a v. low k single wave-function.

 

[ZapperZ]

You have not answered the question.
How do high k short-wavelength constituents of a Cooper Pair tunnel over > 1000 wave-lengths without the barrier potential attenuating them to nothing. Simple basic quantum mechanics is all that is needed for a v. low k single wave-function.

It depends on the barrier widths!

The supercurrent has a WHOLE RANGE OF k! That was how I answered your question, but you didn't have a clue what that meant, obviously. The BCS ground state summed up all the plane wave states from 0 to the Fermi momentum! The tunneling current doesn't involved ALL of the Cooper Pairs! If they do, the tunneling current would be identical to the supercurrent, which we do NOT see. All the cooper pairs do not have the same probability of tunneling!

So here's your turn to answer MY question. http://arxiv.org/PS_cache/cond-mat/pdf/0410/0410184v1.pdf" , without any Cooper pair tunneling, show me a detailed theoretical explanation for the energy gap formation at 2Delta in the tunneling conductance, AND the presence of the Josephson current, even at zero bias. Remember, there have been ZERO other explanation for this other than the current conventional explanation which has worked very well, AND, what was predicted by Josephson using the BCS formulation. This is the non-handwaving argument that I'm expecting and it will be the last time I will ask this.

Zz.

 

[f95toli]

"Direct" tunnelling of Cooper pairs is quite rare in real junctions; in most cases the Josephson current is mediated by Andreev states in the barrier; i.e there is no tunnelling over "long distances" involved.
All you need in order to get a Josephson current (i.e. a NET transport of Cooper pairs) is two electrodes connected via a weak link of some sort. The SIS junction is just one extreme, at the other end you have SNS junctions and everything in-between (usually specified using the BTK parameter, 0 being an insulating barrier and 1 a normal metal) not to mention junctions that do not really fit that classification such as SS'S, ScS, SFS etc.
In junctions made from hight-Tc superconductors the situation becomes even more complicated due to the d-wave symmetry and the nature of the interfaces.

The original papers by Josephson just deal with a special case; the Josephson effect is much more general than even he ever realized.

 

[Zymandia]

"The supercurrent has a WHOLE RANGE OF k! That was how I answered your question, but you didn't have a clue what that meant, obviously."

No I don't have a clue, and attempts to find-out just finds high-phalutin' physicists giving hand-waving explanations of the 'super-current'.
So let's use logic. A low k long-wavelength 'supercurrent' is required to tunnel the distances observed, but it can't be a low-energy electron because that would be forbidden by Pauli. However we know that current flows across the gap so mass and charge are transferred.
No, I give up. What has an extremely long wavelength so presumably low kinetic energy, yet carries charge and mass in the ratio of the electron, what type of wave is a 'supercurrent'?

 

[ZapperZ]

"The supercurrent has a WHOLE RANGE OF k! That was how I answered your question, but you didn't have a clue what that meant, obviously."

No I don't have a clue, and attempts to find-out just finds high-phalutin' physicists giving hand-waving explanations of the 'super-current'.
So let's use logic. A low k long-wavelength 'supercurrent' is required to tunnel the distances observed, but it can't be a low-energy electron because that would be forbidden by Pauli. However we know that current flows across the gap so mass and charge are transferred.
No, I give up. What has an extremely long wavelength so presumably low kinetic energy, yet carries charge and mass in the ratio of the electron, what type of wave is a 'supercurrent'?

You didn't answer MY question.

And you are more than welcome to look at Josephson's original theory. See if THAT is "hand-waving". This thread is done.

Zz.