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Do electrons exist within a Cooper Pair?

  1. Nov 4, 2008 #1
    Hi guys I hope you can help.
    The other day I tried to explain to someone the mechanism behind 'stripes' in lightly doped cuprates that gives rise to the Mott insulator/ anti-ferromagnetic stuff.
    It quickly became clear that my understanding of super-conductivity differs considerably from the orthodox.
    I hope this forum will be kind enough to explain my heresies to me.

    While a Cooper Pair exists, do the electrons that formed it (and that it will decay into) cease to exist?

    I say: Yes, the CPB is a single entity and thus has its own bosonic wave-function, which may be described as a Bose-Einstein condensate of two electrons. So the constituent electrons are no longer independent wave-functions, and hence cease to exist.
    The CPB pays no attention to Pauli and hence may take any energy 'in' the local density of states, including ones forbidden to electrons.
    Plausible?
     
    Last edited: Nov 4, 2008
  2. jcsd
  3. Nov 4, 2008 #2
    http://arxiv.org/abs/0801.2500

    I think section 4.4 of this article can be helpful. It deals with the general many-body wavefunction when pairing occurs, although it is cold atoms, rather than electrons, which are pairing.
     
  4. Nov 5, 2008 #3
    Ok, that helped me focus my problem with the electrons at a distance model of Cooper Pairs. The maths merely stipulates a binding energy gap and is undeniably successful against observation. The distant electron model does not seem to be intrinsic to the BCS maths??
    Since a bosonic wave-function is symmetric it seems to me that indistinguishibility must apply over the whole wave-function. I like to think of it as a "Distinguishability exclusion principle" or DEP as a nod to Pauli. Since the constituents of the boson are anti-spin any magnetic field in the wave-function would cause Zeeman splitting of the energies were it to be seen as 2 separate electrons the DEP forbids it in the same way as Pauli forbids 2 fermions in the same state. So the Meissner effect is a macroscopic quantum effect of the DEP as atomic shells (chemistry?) are an effect of the Pauli Exclusion Principle.
    Hope that's not too simplistic??
    Anyway 2 electrons at a Cooper Pair distance are extremely distinguishable so can't have a symmetric wave-function in that configuration and must have a linear orientation, which would make them rather the ugly duckling of the boson family.
    Applying strict DEP suggests there can be no spatial distinguishability between the components (which would show up by spin localisation) which would suggest that each constituent is equally represented everywhere within the wave-function. Within this view the 'distance between the electrons' becomes the size of the wave-function. Josephson junctions show that bosonic wave-functions reach beyond the conductors that house them.
    So I'm really asking if there has been any definitive proof of electron structure within a super-conducting boson??
     
  5. Nov 5, 2008 #4

    ZapperZ

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    I find this to be highly confusing and puzzling.

    So what do you think is the composite "boson" that constitutes a Cooper pair that has (i) a charge of 2e, (ii) a coupling energy equal to the energy gap, (iii) a "single particle spectral function" that corresponds very well to the quasiparticle spectral function? The s-wave and d-wave order parameters that have been observed at NOT the order parameter of the Cooper pair, but rather the order parameter of the electrons that make up the Cooper pair, i.e. single-particle order parameter. Are you proposing a new "elementary particle" within a superconductor that has ALL of those properties that we have currently observed?

    Furthermore, maybe you should look at the pairing interaction one more time. You'll notice that while the composite particle is a boson and obeys the bosonic rules, the 2 electrons (or holes as in the cuprate superconductors) still obey the FD statistics whereby each of them occupy a unique momentum value within the electron gas states. Why do you think we talk about the single state for most superconductors and the triplet states for the ruthenates superconductors? These ARE the pairing symmetries of the electrons within the Cooper pairs themselves!

    Zz.
     
  6. Nov 5, 2008 #5
    Ok, so you seem to be saying a spin up electron can be 'symmetrical' with a spin down electron?
    <very confused>
    I thought the basic definition of a bosonic wavefunction was that exchange of particles made no difference.
    psi(r1, r2) = psi(r2, r1)
    ??
     
  7. Nov 5, 2008 #6

    ZapperZ

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    You are confusing the composite boson itself with the CONSTITUENT of the boson. Within the composite boson, the constituent (the 2 electrons) still obey the general FD statistics of ALL the electrons in the Fermi sea!

    I think you may need to get back to the BCS theory again and look at the pairing state.

    Zz.
     
  8. Nov 6, 2008 #7
    So the constituent electrons still have anti-symmetric fermion wave-functions but they are part of a boson. Does this mean the boson doesn't have a its own wave-function??
    As I understand quantum mechanics each entity within a system has its own wave-function.
    Also Cooper Pairs seem to have a total energy that defines their wave-functions and then a binding energy that doesn't contribute to their wave-function total energy (were it to do so then the net total energy would require a state in the local density of states that is either already taken by a fermion, or forbidden to fermions). Are there any other cases in quantum mechanics where one uses something other than the total energy when considering the wave-function and its behaviour?

    Addendum: Sorry, still puzzling this, are there any other bosons which have distinguishable constituents, ie. disobey the exchange invariance?
     
    Last edited: Nov 6, 2008
  9. Nov 6, 2008 #8
    Any boson made from constituent fermions will have the problems that seem to be confusing you here. For example helium (which has many constituent fermions). Which can bose condense into a super fluid.

    It sounds like your question can be rephrased:
    How can multiple bosons occupy the same state, if their constituents do not?


    My understanding (which may be incorrect):
    Take ANY two body system. You can write it in terms of the positions of the two bodies, or you can rewrite it in terms of the positions of the separation and the center of mass. If the two bodies involved were identical fermions, you can view this as two fermions, or one boson (the 'center of mass' coordinate) and one explicit combination of fermions (the reduced mass, difference of positions coordinate).

    It is only the "center of mass" bosonic wavefunction that has the same state during boson condensation. Exchanging bosons here, would require exchanging both fermions between two bosons... which will not yield a sign change to the overall wavefunction, since you can write this as two swaps of fermions.

    Make sense?
    (Can someone else please comment as well, I'd like to know if this is the correct understanding, or if I am somehow incorrectly oversimplifying.)
     
  10. Nov 6, 2008 #9
    Technically we can call an electron spin 0 Cooper Pair a Bose-Einstein Condensate of two electrons. A condition of a condensate is exchange invariance. The electrons have opposite spins. I can only see one rather elegant solution to those constraints.
     
  11. Nov 6, 2008 #10

    ZapperZ

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    But I've just told you about the existence of the triplet spin states for the ruthenates superconductor! So yes, there can be more than just the singlet state that you just mentioned. This is, in fact, analogous to the He3 condensation.

    I don't know what the issue is here anymore. In case you are still not clear about this:

    Composite bosons - obey BE statistics
    Fermions that make up the composite bosons - still obey the FD statistics.

    Not sure where the confusion is here.

    Zz.
     
  12. Nov 6, 2008 #11
    How can a spin up fermion be exchange invariant with a spin down fermion within a composite boson?
     
  13. Nov 6, 2008 #12

    ZapperZ

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    Can you write down the singlet state? Where is it invariant there?

    Zz.
     
  14. Nov 7, 2008 #13
    While it is a singlet boson it must have a symmetric wave-function and its constituents must be exchange invariant. In order for a pair of anti-spin electrons to form a bosonic condensate there can be no net spin at any part of the wave-function since that would distinguish one constituent from the other. Thus within the boson the contribution from each constituent must be the same at all points. If the constituents are effectively on top of each other spatially then one may expect the major contribution to internal energy for the boson to be from electrostatic repulsion with a small term for magnetic attraction. Thus the internal energy of the singlet state is inversely related to the volume taken by the boson which would try to get as big as possible to minimise its energy.
    For the boson to have a QM wave-function its constituents must cease from being independent particles.
    The boson decays when its energy is sufficient to create two electrons to fill available states in the local density of states.
    As I understand it the tools used to measure 'the boson' have actually been measuring electrons. May I suggest that a lot of current thinking about the electron Cooper Pair is ascribing properties of the local decay products to the boson itself.
    If you accept (hypothetically ;) ) that the boson's wave-function cannot contain fermion wave-functions and hence can take whatever energy it likes independent of the constraints of the LDOS, then may I share with you my qualitative explanations for:
    Binding energy/decay condition,
    Type I & II super-conductivity,
    The non-BCS behaviour of cuprates (requires postulate of a semi-conductor bandgap)
    The stripes/anti-ferromagnetism/Mott insulator behaviour around critically doped cuprates.

    I'll come clean, all I read on the current thinking on Cooper Pairs seems deeply dubious since it requires exceptions to basic quantum mechanical principles, furthermore it is limiting our ability to explain current experimental results.
    The Holy Grail of super-conductivity is room-temperature SC. I'm sure we already have the technology if we look in the right place.
    (Go on, hear me out. Worst that can happen is you die laughing :D:D )
     
    Last edited: Nov 7, 2008
  15. Nov 7, 2008 #14

    ZapperZ

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    Huh?

    Before you proceed any further with your "thinking", you need to re-read the PF Guidelines and our policy on unverified, personal theory.

    Secondly, you seem to not be able to comprehend that the constituents of a cooper pair are fermions and STILL obey the FD rules, while the cooper pair itself as a whole is a composite boson that obey the BE rules. I have no idea why this is that difficult to accept especially based on ALL of our experimental observations. None of these are contradictory (exceptions?) to anything in QM, especially when QM was used to arrive at these conclusion (seen the BCS ground state lately?)

    I'm a condensed matter physicist by training, and did all my graduate and post-doctoral work in high-Tc superconductors. So far, nothing that you have said here made a lot of sense, especially your objection to the existence of cooper pairs and the constituents. When I asked you way in the beginning whether you are actually proposing another "fundamental particle" that has all the observed property of a cooper pair, you never replied.

    BTW, the "LDOS" that one sees when one does a spectroscopy measurement is a SINGLE-PARTICLE density of state, not a 2-particle density of state. So you do NOT get the "boson" DOS but rather the DOS of one of the electrons that make up the consituents! That's why you get the energy gap! There's no energy gap for the composite boson (the bosons do not couple to anything - that's why they can move freely!), but there is an energy gap for the electrons in the single-particle spectrum! This is why I clearly mentioned about the "single-particle spectral function" way in the beginning of this thread, but obviously, you missed it or have no idea of the significance of that term.

    Zz.
     
    Last edited: Nov 7, 2008
  16. Nov 8, 2008 #15
    I came here because I wanted to float a potential mechanism behind anti-ferromagnetic stripes and the associated Mott insulator (and to predict that the stripes will super-conduct above a back-voltage of the order of volts).

    I see now that my understanding that a Cooper Pair is a bosonic condensate wave-function that has the electrons as constituents is very different from the idea of two conduction electrons connected by a negative energy boson that doesn't lower the electrons energy.
    We agree that the addition of energy to the boson results in its destruction and 2 electrons in the LDOS (yup, single particle only). Thus it seemed to me that my viewpoint would also give the "single-particle spectral function" (I'd appreciate a link to a clear explanation since I've only found references presuming an understanding) on being hit by ARPES photons.
    I should have smelt a rat when I read about 'heavy fermion' super-conductors showing Andreev reflection like normal SCs and 'science would like to know the mechanism'. To me although the decay states are odd the actual Cooper Pair is much as normal, so I was confused at the confusion.
    My "thinking" is based on basic solid state and quantum mechanics and reported experimental phenomena. It seems basically self-consistent to me and explains things for which 'science would like to know the mechanism'.
    I suppose it doesn't come up for discussion on these forums until, like Bose, I can find an Einstein to prepare it for publication.
    Thank you very much for explaining where I depart from SC orthodoxy, it defines my next barrier if not my next step. :D:D
     
  17. Nov 8, 2008 #16

    ZapperZ

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    Haven't people like Steve Kivelson done that already?

    You may want to look at He3 while you're at it, which I had also mentioned earlier.

    I don't see how. The only "convincing" argument for any signature of stripes from ARPES experiment (and I did ARPES on high Tc superconductors as well - my avatar is a raw ARPES spectra from an overdoped Bi2212) is a theoretical paper by Orgad et al. (Phys. Rev. Lett. 86, 4362(2001)), and that is still being debated as far as experimental verification is concerned.

    Zz.
     
  18. Nov 9, 2008 #17
    Blither, could only get the abstract. Have a way of getting an academic log on, so I'll try that later.
    In the (relatively few) cuprate field/temp phase diagrams I've seen there seems to be a distinct kink at ~15K for high doping reducing (in temp.) as the doping reduces.
    Is this kink real enough to have been noticed? What would work on it be called, so I can search for it?
    Is there a collection of cuprate field/temp phase diagrams for various doping & configurations anywhere? I'm interested in consistencies and trends for things like 0K crit. field, and particularly getting a handle on the non-BCS part of the curve (after the kink so > 20K).
    Finally, what level of heresy would it be to suggest that there is a semi-conductor bandgap just below the fermi level which would manifest in k-space as a surface just below the fermi surface?
    May I also express my immense gratitude for your patience so far, google, wikipaedia and stubborn curiosity can get you so far, but unless you stumble on a keyword and guess its meaning some doors remain closed. Your 'tutoring' has focused my ideas and shown me where my understanding differs.
     
    Last edited: Nov 9, 2008
  19. Nov 9, 2008 #18

    ZapperZ

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    I have no idea what you just said here. It would help if you give an exact citation to the source on where you read about these things.

    There are "kinks" in the band dispersion based on ARPES spectra of the cuprate superconductors, but this appears to not be what you are asking. There is a "break" in the phase diagram of LSCO where the material's Tc is suppressed at around 1/3 doping, but this is, so far, only a characteristics of that compound and not of other cuprate superconductor. But I have no idea if this is what you are referring to.

    It would make no sense. A semiconductor gap will not produce a "pile-up" of states at the bandgap edge in the density of states, which is observed from tunneling spectroscopy. Furthermore, it would be puzzling to explain why the gap would suddenly kick in at Tc, and then cause an abrupt drop in DC resistivity and a drop in magnetic susceptibility as temperature drops. A semiconductor's resistivity goes UP with decreasing temperature - that is a characteristic of a semiconductor!

    Zz.
     
    Last edited: Nov 9, 2008
  20. Nov 10, 2008 #19
    I'm right up against the veils of academia and my remote logon thingy is suffering from incompetence. I'd be grateful for a reference on the pile-up of states from STM (maybe e-mail the pdf if you'd be so kind, no worries if not).
    I'm still having trouble picturing this composite Cooper Pair. Everything I read says the CP has neglible momentum, but for the electrons to still have fermi level k vectors then the electrons momentum must be equal and opposite, blowing any spatial constraints on the composite structure. Conversely electrons keeping a constant distance apart would have to have parallel k vectors and hence the composite would have a large aggregate momentum. Constant distance and zero overall momentum could only be met by each electron having zero momentum which would be forbidden by the LDOS for single independent particles. I read an interpretation of BCS that seemed to say that BCS presumes the electrons momentum falls to zero, if so why doesn't its energy drop to a level that's taken or forbidden in the LDOS?
    Also, I'm interested in the idea that conduction becomes more and more anisotropic as temperature drops, ie. draws back to 45deg nodes. Is there an official mechanism for this? If not may I offer mine? :D
     
  21. Nov 10, 2008 #20

    ZapperZ

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    You are mixing conventional superconductors with the cuprate superconductors. Considering that you still have problems in understanding the basic idea of BCS superconductors, I'm sure you'll understand if I'm skeptical of your "official mechanism" to explain the order parameter of high-Tc superconductors. It would also be your own personal, unverified theory, which I've already noted to be against our PF Guidelines.

    This discussion appears to be going nowhere, because we are back at "Cooper pairs" once more. There are plenty of texts on this as it is already, considering that BCS superconductivity is one of the most verified theory in all of physics. I am only willing to take steps back only so many times before it gets exasperating. So I'm done.

    Zz.
     
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