Another step toward understanding of high-temperature superconductivity

Apr 04, 2013 by Nik Papageorgiou
Another step towards free electricity
Credit: Wikipedia

(Phys.org) —Superconductors can radically change energy management as we know it, but most are commercially unusable because they only work close to absolute zero. A research group at EPFL has now published an innovative approach that may help us understand and use superconductivity at more realistic temperatures.

are materials that allow electrical current to flow with no , a phenomenon that can lead to a vastly energy-efficient future (imagine computers that never overheat). Although most superconductors work close to absolute zero (0°K or -273.15°C), some can operate at higher temperatures (around -135°C) – but how that happens is something of a mystery. Publishing in a recent PNAS article, Fabrizio Carbone's Laboratory for Ultrafast Microscopy and Electron Scattering (LUMES) at EPFL has developed a method that can shed light on "high-temperature" superconductivity.

How conventional superconductivity works

When electricity passes through a conductor, e.g. a wire, some energy is lost because of resistance. This is not always a bad thing, since it can be used for heat (radiators) or light (). But when it comes to things like national energy grids and high-voltage cables, electrical resistance (up to 7% in some grids) can mean money and constant wear.

This is where superconductors come in. These are materials that, when cooled down enough, with no resistance – and therefore no loss. How? As superconductors cool below a certain temperature, their atoms fall in line and "nudge" charge-carrying electrons together to make new particles called . These observe and form an unusual (a Bose-Einstein Condensate) that is not affected by .

Superconductors can revolutionize the way we use and distribute energy, change modes of transportation (e.g. Japan's magnetic levitation trains) and give us 100% energy-efficient technology. So why hasn't that happened yet? The problem is temperature: Most superconductors only work when they are cooled close to the forbidding . The solution lies with those that work at higher temperatures.

A new method for exploring high-temperature superconductivity

Recently, the lab of Fabrizio Carbone at EPFL addressed the issue by developing a novel method that can advance our understanding of high-temperature superconductivity. High-temperature superconductors (HTS) show promise because they can operate at temperatures around -135°C – still low, but considerably cheaper and more feasible than for conventional superconductors. However, progress in HTS is limited because, even though we know that Cooper pairs are involved in high-temperature superconductivity, there is no consensus as to how they are formed.

Carbone's group was able, for the first time, to directly observe the formation of Cooper pairs in real time in a superconducting HTS and determine how the process affects the optical properties of the superconductor. Using a novel approach, the scientists cooled an HTS to its superconducting temperature and then repeatedly fired laser pulses on it to break up the Cooper pairs back into single electrons. As the Cooper pairs broke and re-formed, they caused a periodical change in the color spectrum of the superconductor. By measuring the color change, the researchers were able to directly study what happens in a superconducting HTS. What they discovered was that Cooper pair formation follows a completely different path than in conventional superconductors.

Carbone's findings mark the first direct observation of Cooper pair formation in HTS superconductivity. They also provide scientists with a powerful tool to observe the phenomenon in real time. The hope is that by extending this innovative approach to different materials, we can begin to understand how high-temperature superconductivity really works.

Explore further: Creating optical cables out of thin air

More information: Coupling of a high-energy excitation to superconducting quasiparticles in a cuprate from coherent charge fluctuation spectroscopy, www.pnas.org/content/110/12/4539.abstract

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BrettHarris
2.7 / 5 (3) Apr 04, 2013
Interesting article, but I'd like to point out that in paragraph one, absolute zero is noted as 0 degrees Kelvin. There are no degrees of Kelvin. That's why it's the absolute temperature scale.
johanfprins
1 / 5 (2) Apr 05, 2013
There are no Cooper pairs in any of the metal and any of the ceramic superconductors: Such entities are probably totally fictional.

Furthermore a superconducting condensate is NOT a Bose-Einstein Condensate since the charge-carriers within a superconducting material follows plain old Botzmann statistics: This is so since they are separate distinguishable entities. If they are not, they cannot be charge-carriers.
johanfprins
1 / 5 (3) Apr 05, 2013
separate distinguishable entities
You mean particles?


Nope! There are no particles: Charge-carriers within a normally conducting material are superposed waves: Within an energy band consisting of delocalised electron waves, the latter waves superpose to form wave-packets as soon as you apply an electric-field. That this is so is modeled in all decent textbooks on Solid State Physics.

In the case of superconduction, the charge-carriers are localized anchored waves, which move by cooperate tunneling through the insulating material that separate them.

Charge-carriers within a conducting material are NEVER particles: They are ALWAYS localised waves. Since these localised waves are distinguishable they are modeled by Boltzmann statistics. How many times do I still have to explain this simple physics to you?
ValeriaT
1 / 5 (2) Apr 08, 2013
How many times do I still have to explain this simple physics to you?
You're not explaining it - you're preaching it. This is indeed a difference. Localized wave is an oxymoron: wave is concept, which is delocalized by its very definition. The "separate distinguishable entities", which you're talking about, are quantum wave packets, which do follow particle-wave duality. Their behavior is as similar to wave, as similar is the Dirac quantum equation to d'Alambert wave equation (a hint: well, not too much).
johanfprins
1 / 5 (2) Apr 08, 2013
How many times do I still have to explain this simple physics to you?
You're not explaining it - you're preaching it. This is indeed a difference.


When you preach you ask people to believe what you cannot prove: This is what YOU do with your ducks paddling within a non-existent aether. I base my conclusions on experiment, logic and quantitative mathematical deductions. You do not have the skills to do the same.

Localized wave is an oxymoron: wave is concept, which is delocalized by its very definition.
Whose definition? This is another one of your hallucinations! When you emit an coherent laser wave of length L and cross-sectional area S, its EM-wave energy is localized within the volume SL.

The "separate distinguishable entities", which you're talking about, are quantum wave packets, which do follow particle-wave duality.


No they do not since each wave packet is a superposition of ALL the delocalized valence electron-waves: i.e. N>>>1 electrons.
johanfprins
1 / 5 (2) Apr 08, 2013
These waves are obtained by solving Schroedinger's many-body equation. When you solve for a single free electron within the vacuum, you only have a solution for a SINGLE electron which cannot superpose with other electron-waves (since they are not there) to form a wave-packet.

Thus "wave-particle" duality for a single free electron has NOTHING to do with such an electron being a wave-packet. In fact Schroedinger's equation (as well as Dirac's abomination which is miscalled as the relativistic equation for an electron)cannot model a free electron in the vacuum at all. Since a free electron's center-of-mass MUST be stationary within an inertial reference frame that is unique to the electron, the position and momentum of a free electron MUST manifest simultaneously with 100% accuracy; also when the electron moves with a speed v. There is no uncertainty in the position and momentum of the center-of-mass of an electron-wave whatsoever!
johanfprins
1 / 5 (1) Apr 08, 2013
BTW If you believe, that all particles are "waves only", you should derive the Schrodinger/Dirac equation of photon or electron


Firstly, Dirac's equation has no physics merit whatsoever. Schroedinger's equation is only valid for stationary solutions where the boundary conditions are defined by the potential energy-field that the electron finds itself in. It cannot model a freely moving electron.

from d'Alambert's wave equation...


Photon: These normal EM wave equations for the electric potential and the magnetic vector potential (Lorentz gauge) give the solution for any coherent light wave with a frequency omega and with any energy E>(hbar)*(omega). The energy density is proportional to the sum of point product of the electric-field and the point product of the magnetic-field: Integration over the volume of the wave gives E. When the energy E of such a wave is equal to (hbar)*(omega) you have a free photon-wave. Coherent EM waves with less energy are not possible.
johanfprins
1 / 5 (1) Apr 08, 2013
with adding of our "boundary conditions" (special relativity, Lorentz transform, etc)


Freely moving electron: Assume the freely moving electron has a stationary cross-section A(e) and it is moving past at a speed v. When you now apply the Lorentz transformation, the free electron will have a length L(e) along the direction in which it moves where one has that: L(e)=(gamma)*A(e). Remember that

(gamma)= (1-(v/c)^2)^(-1/2).

It also has a phase time difference across it given by

T=(v/c^2)*L(e)

Using this change in phase time to derive the wavelength of the passing electron one obtains the de Broglie wavelength. Thus, the passing electron is a coherent EM wave which is moving slower than the coherent photon-wave. It is the easy to verify that the solution of this wave is also given by Maxwell's equations for the electric potential and the magnetic vector potential (the latter models the "spin" of the electron".

The energy of the wave is E=m*c^2 where m=(gamma)*m(e).
johanfprins
1 / 5 (1) Apr 08, 2013
It would be rigorous proof, that your "all is the wave" approach can replace the Copenhagen quantum mechanics, which would be acceptable with mainstream physics. Until you do it, you will just serve as a source of fun.


I have done this, and much more, but cannot get it published; since there are people like YOU who consider the Voodoo concept of "wave-particle duality" as having been written in stone by God himself.

The derivation of the de Broglie wavelength of a freely moving electron, by using the Lorentz transformation, can be seen here:

de http://www.cathod...tion.pdf

If only YOU would learn to first do your homework before putting both of your feet into your mouth!
johanfprins
1 / 5 (1) Apr 08, 2013
But such a EM-energy is delimited with material resonator. The photon has no such resonator built in, or its mass couldn't be zero.


The mass of a laser pulse IS also zero! So what delimiter are you talking about?

each wave packet is a superposition of ALL the delocalized valence electron-waves: i.e. N>>>1 electrons.
This is easy to say, but where is the formal and experimental proof?


As I have pointed out already many times, the proof is within ALL good textbooks on Solid State physics. The charge-carriers have to be modeled as wave-packets formed by superposition of the delocalized electron waves.

In addition, this model is wrong, as the electron is apparently localized particle - only its wake wave in vacuum is not localized (too much).


WAKE UP!! There is NO WAKE WAVE!!!!! The Lorentz transformation proves that the electron itself is the wave when it moves past with a speed v: There are no "wakes" around it whatsoever!
johanfprins
1 / 5 (1) Apr 08, 2013
If you have the equations and all - try to draw your model of electron/photon in similar way, like this http://www.neti.n...Sim.html (the black line corresponds the particle character, the colored one the wave-like behavior in Copenhagen quantum mechanics).


This applet is pure BS. The electron is a wave with mass energy and it therefore has a center-of-mass. It is not a particle with a wake wave! IT IS A WAVE and only a wave! Its energy is pure electromagnetic energy as one would expect that it must be!
johanfprins
1 / 5 (1) Apr 08, 2013
LOL.. Apparently, for you everything is a WAVE - no matter whether it's wave packet of photon, "delocalized electron wave" or common wave at the water surface - i.e. the objects of very different shape, geometry and character.


I have not talked about water waves: I am not so demented to think that ducks paddling in water have any relationship with light or matter waves.

A photon is NOT a wave packet since a wave-packet is NOT a coherent wave.

That electrons do form both delocalised wave states and localized Mott phases within a material is so well demonstrated experimentally that only a moron will question the existence of a delocalized electron that fills thye whole volume of a perfect crystal.

arver Mead has stated that he measured an electron that is 5 meters long! I will rather believe Carver Mead who has an impeccable track record in physics and EE than a troll like you who hides behind numerous aliases since you know that you are a NOTHING!

johanfprins
1 / 5 (1) Apr 08, 2013
It's not surprising, which such sloppy thinking you're finding your ideas self-consistent under all circumstances. You just replaced the "particle" concept with "boundary conditions" word.


Anybody who thinks that he can model a real wave without finding the applicable boundary conditions is not just a sloppy thinker but a total demented fool: Such fool has to then fudge his results by invoking fictitious particles like Dirac had to do with his nonsensical, un-physical "relativistic equation".

OK - if you don't think so, try to demonstrate some feasible experiment, in which your interpretation/paradigm/theory/whatever can be distinguished from standard Copenhagen interpretation of quantum mechanics. If not, then... you know what...


According to the Copenhagen interpretation a photon is a "probability wave" with an uncertainty in its energy and momentum. Thus when you send single photons through a double slit diffractometer, they must be detected with different ...
johanfprins
1 / 5 (1) Apr 08, 2013
energies (frequencies) and wave-vectors as determined by the probability distribution of the wave. This, in turn, demands that the diffraction pattern that they form will not have clearly defined lines since the latter only occurs when each photon-wave has the exact same frequency.

Lines are measured: This proves that the Copenhagen interpretation is not working. The intensity of the photon-waves are thus not "probability-distributions" but must be that of a single coherent wave.

The latter is also proved by the fact that a photon is modeled by Maxwell's equations for light: This can also be seen as follows: According to Einstein's relativity the relationship between a photon's energy E and is momentum is E=pc: Take the square and replace E with -i(hbar)*(d^2/dt^2)(PHI) and the momentum with -i(del)(PHI) and you get:

(del)^2(PHI)=(1/c^2)(d^2/dt^2)(PHI)

which is Maxwell's equation for the electric-potential of a light-wave: (PHI)^2 IS THUS NOT a "probability density".

QED!
johanfprins
1 / 5 (1) Apr 10, 2013
A photon is NOT a wave packet since a wave-packet is NOT a coherent wave
This is incoherent objection, since the photon is not a coherent wave.


Why does one obtain a diffraction pattern that can ONLY be the result of a coherent wave when sending individual photons with the same frequency through double slits?

The rest of BS I'll ignore for not to pollute PO with neverending discussions with people


Why do you then post neverending garbage under different aliases based on a single Voodoo-mantra: AWT, AWT, AWT! You are really very boring since you NEVER answer any question based on real physics.

,who cannot think consistently even in single short sentence.


A very good description of you.

You're a shame of scientific community.


I agree that I cannot blame you of this, since YOU have NEVER been a a member of the scientific community and NEVER will be one: YOU just have too little grey matter within the hollow cavity between you ears!
ValeriaT
1 / 5 (2) Apr 10, 2013
Why does one obtain a diffraction pattern that can ONLY be the result of a coherent wave when sending individual photons with the same frequency through double slits?
This is probably the very basic misunderstanding of yours - the individual photons are doing dots at target like the particles - newer waves. [url=http://www.youtube.com/watch?v=MbLzh1Y9POQ]Watch .
just have too little grey matter within the hollow cavity between you ears
Do you mean that grey neuron, which is connecting the ears to head?
johanfprins
1 / 5 (1) Apr 11, 2013
This is probably the very basic misunderstanding of yours - the individual photons are doing dots at target like the particles - newer waves.


What are the individual photons when they JUST reach the screen BEFORE collapsing to be observed OWING TO THEIR INDIVIDUAL INTERACTIONS WITH ATOMIC SIZED DETECTORS IN THE SCREEN. In your normal stupid manner YOU assume that each photon is a "localized entity" BEFORE it interacts with the screen.

You can just as well conclude that a falling egg is already broken while it falls because when it collides with the floor you observe a broken egg. Can you not see how stupid you really are?

just have too little grey matter within the hollow cavity between you ears
Do you mean that grey neuron, which is connecting the ears to head?


NO, I mean the neurons required to logical interpret what one sees and measures! You have none of them!
johanfprins
1 / 5 (1) Apr 11, 2013
..YOU assume that each photon is a "localized entity" BEFORE it interacts with the screen.
Negative. This is the picture of gamma ray photons, as they travel through http://www.ep.ph....red.png. each photon is a "localized entity" BEFORE it interacts with the screen


Only a moron like you will conclude that a photon that is suffering multiple interactions (measurements) when it moves through space towards a screen is the same as a photon moving through space to the screen while NOT not suffering ANY interactions!! The one thing that QM teaches us is that every interaction (measurement) can change the state: Thus within the spark chamber you collapse the photon LOOOONG before it reaches the screen. While when NOT moving through the spark chamber the photon ONLY collapses when it reaches the screen! For God's sake: Try and WAKE UP out of your STUPOR!!

johanfprins
1 / 5 (1) Apr 11, 2013
Of course it is. As you can even see, every photon can interact with multiple wires repeatedly. It means, it existed both before, both after it.


When it hits the first wire it collapses and it collapses with every subsequent wire it strikes. When the wires are not there it collapses only when it reaches and interacts with the screen!

LOL, it's just YOU, who believes, that the egg had a different form before impact - not me...:-)


If there are wires with which the egg can collide before reaching the floor (as in the case of a photon within a spark chamber), it will obviously break before reaching the floor. When not, it will only break when hitting the floor.

This argument applies to YOU instead. I'm just saying, the photon is a particle both before, both during impact.


Impact with what? The first wire? Why do you not put your dummy back into your mouth, and spend the rest of your life in the foetus position? It is your only natural position!
ValeriaT
3 / 5 (2) Apr 11, 2013
When it hits the first wire it collapses and it collapses with every subsequent wire it strikes
It must be very tiny after then...
Why do you not put your dummy back into your mouth, and spend the rest of your life in the foetus position? It is your only natural position!
You're sorta cute, when you're getting upset. Do you really believe, that the atom, electron or photon gets collapsed again and again when it travels across material full of particles and interacts with each of them?
johanfprins
1 / 5 (1) Apr 12, 2013
When it hits the first wire it collapses and it collapses with every subsequent wire it strikes
It must be very tiny after then...


At last an intelligent remark! When the boundary conditions change the wave must adjust to the new conditions. When a photon of any shape and size strikes the first wire it collapses: If there is no change in boundary conditions while travelling to the next wire it will maintain its collapsed state. The next wire presents the SAME boundary conditions as the first wire, so that the photon collapses to maintain the same wave-size and shape that it collapsed into at the first wire: Therefore it maintains its localized state all the way.

If at the end of its path you place a double slit, the photon will morph by inflating in size to simultaneously move through both slits. It cannot be a localized wave when it moves through the slits. If it remains localized it will not develop a diffracted intensity distribution as we know that it does!
johanfprins
1 / 5 (1) Apr 12, 2013
Do you really believe, that the atom, electron or photon gets collapsed again and again when it travels across material full of particles and interacts with each of them?


When the boundary conditions change at different positions along their paths, they will collapse or inflate to adjust to each change in boundary conditions that they encounter. The amount of change depends on the wavelength. A moving atom obviously has a very short wavelength compared to a photon and an electron, so that the collapse or inflation of the atom-wave is not as large.
johanfprins
1 / 5 (1) Apr 12, 2013
Why not? Do you believe, the electron or atom transmutes, when it moves through Wilson cloud chamber and it collides with billions of particles there?


What do you mean by transmute? Transmute is when one thing changes into something different. When a wave changes shape and size it does not transmute since it is still the same wave!

Within a Wilson cloud chamber, every interaction with a gas molecule, in the chamber, is the same so that the boundary conditions do NOT change every time. Thus the atom or electron-wave need only morph during the first "collision": There-after it maintains its shape and size when colliding.

Schroedinger's equation tells you that this must be so, since the Hamilton-operator acting on one of its own eigen-states do not change the eigen-state. Only when it operates on a wave that is not an eigen-state must this wave morph into an eigen-state, If the same operator again acts on this now morphed eigen-state the wave does not again have to morph.
johanfprins
1 / 5 (1) Apr 12, 2013
So its particle along whole its path there? Why it couldn't collapse before, after then?

It could have done so, but it could also have moved through a small hole and then spread out into space so that it is not a localized wave when it interacts with the first wire; whence it only then becomes localized.

Why it should collapse at all?


Even according to the Voodoo of "wave-particle duality" it is claimed that the wave can be spread out in space and that it then has to collapse in order to miraculously "create" a "particle".

You're essentially saying, during radioactive decay the gamma ray photon released travels like (spherical) wave, until it's not observed.


Until it "is not" observed? You probably mean until it "is" observed. Can you prove experimentally that the gamma-ray photon does NOT form a spherical wave which only collapses into a localized wave when it is observed? No you cannot. The shape and size of a light wave is ..(continued)
johanfprins
1 / 5 (1) Apr 12, 2013
determined by the emitter and need not stay the same when it is observed.

If the light can be generated in form of directional beam, then it's apparent the photons must be localized particles as well.


Not at all: Whether the light is generated as a directional beam or a spherical wave or what have you depends on the boundary conditions defined by the emitter. There is no reason why different emitters which emit photons with the same energy must emit waves with the same shape or size.

We can only determine the shape and size when measuring the emitted photon. If by doing this we have to change the boundary conditions from the one determined by the emitter, the wave has to morph into a different shape and size.

To observe a single photon we always have to change the boundary conditions so that it becomes a localized wave. If you are stupid, you will conclude that the wave must also have been a localized entity before you made the measurement.

johanfprins
1 / 5 (1) Apr 12, 2013
Well, I know about it very well.


Another blatant lie! When it comes to physics you do not know ANYTHING about ANYTHING! You have NEVER passed ANY examination and NEVER contributed anything in physics: But you are so arrogant that you anonymously parade as if you are the last word on physics!

You just replaced the "particle" word with "boundary conditions" word,


I have NOT replaced "particle" by boundary conditions since photon- and electron-"particles" do not exist. One cannot replace something which does not exist with something which does exist.

not but in fact you're adhering to all concepts of Copenhagen interpretation of quantum mechanics firmly


Where do I adhere to to the "probability interpretation" and the "uncertainty" claptrap? Examples please!

I'm perfectly aware of it. It's typical behavior of confused crackpots.


You are the actual crackpot who believes that with no knowledge you know enough to criticize physics anonymously.
johanfprins
1 / 5 (1) Apr 12, 2013
Funnily enough, just the phenomena, in which the photons violate the Copenhagen interpretation of quantum mechanics


Which phenomena. Examples please! I am getting tired of your hand-waving arguments which you cannot back up experimentally in any manner.

and which do manifest their particle nature in most apparent way are those, which you're fighting against most obstinately....;-)


The examples you gave of "particle" behavior does not prove that the photon is NOT a wave. The fact that a coherent wave can move in a straight line as if it has a center-of-mass, does NOT prove that it is NOT a wave.
johanfprins
1 / 5 (1) Apr 12, 2013
All objects moving with a constant speed v are each a coherent wave, even when the object is a cannon ball. If you had the competence, which you obviously do not have and NEVER will have, you could have derived this result for yourself from Einstein's Special Theory of Relativity!

If the object has rest mass its wave-length is given by (lambda)=h/p where p=m*v with m the sum of the rest mass and the dynamic mass.

If it is a coherent light-wave the momentum can be calculated Maxwell's equations for the different possible wave-shapes and energies. In the special case of a photon-wave moving along a single direction (like a laser wave does), the relationship is also (lambda)=h/p, where p=m*c: m is now only the dynamic mass given by m=E/c^2, where E=h*(nu).

Thus, all entities moving along a single direction are propelled along this direction since in each case the moving entity is a coherent wave. If this would not be so, nothing would have been able to move relative to anything else!
johanfprins
1 / 5 (1) Apr 12, 2013
The fact that a coherent wave can move in a straight line as if it has a center-of-mass, does NOT prove that it is NOT a wave.
And what else should prove it? Einstein defined the particles just as a "objects with center of mass".


I do not think Einstein would have done this, since only an idiotic fool will use such a daffy-nition: Einstein, although he made mistakes was a genius! For example, Jupiter is an "object" with a center-of-mass, so is the earth and so is the Sun: I do not think this is what is implied by "particle" when the morons advocate "wave-particle duality".

Photon is not coherent wave - it's a wave packet, being localized both in time, both in space.


It cannot be a wave packet since each photon has a DEFINITE frequency and a DEFINITE wave vector k such that they fulfill the coherency condition given by (omega)/c=k. This can NEVER be valid for a wave-packet!! My God! You are REALLY more stupid than I EVER imagined that a human being can be!