 Good afternoon, as we were still with superconductivity, though this session was supposed to be on optics, we will carry on with superconductivity for some time and if we still have time then we will begin at least the elements of optics. But as before I will start with questions that came in. The point is this that let us, this is a question which came from 1241 NRI Institute of Information Science and Technology at Bhopal. The question is that what is the use of persistent current in daily life? Well let me try to answer that question but I cannot talk about daily life because superconductivity does not appear in daily life but basically the question is you are asking that what good is persistent current? Can you use it for anything? So this is something which one can actually answer. The thing is connected with, so let us just recall what was persistent current. So if you remember that we talked about Meissner effect which is an important tool which says that the magnetic flux cannot enter a sample. If the magnetic flux cannot enter a superconducting sample in case I have a sample in the shape of a doughnut or a ring or tube or whatever then what we said is that the flux lines cannot enter the material of the sample but the problem is that supposing there were flux lines in the semiconductor in the beginning, in the superconductor in the beginning then I we drew the, lowered the temperature finally we drew the magnetic field I said that magnetic flux gets trapped. The magnetic flux gets trapped because the flux lines which had entered the hole cannot get out. Now this Meissner effect and the persistent current is known to have particularly those which are in type 2 superconductors not so much see the full Meissner effect does not have that much of utility. Also primarily it is not too bad a thing because as you know type 1 superconductors the typical critical temperatures are of the order of you know maximum 9 K for niobium can go to about 23 K for a particular alloy. So we are not particularly unhappy about that. So what we are talking about is that if you take a type 2 superconductor as I told you in that there are these vortex tubes which are essentially tubes of normal region. Now there is a thing called flux spinning. Now so what actually you must have heard of something called a magnetic levitation. Now remember what is magnetic levitation and of course here is a lot about you know magnetic levitation many of you might have seen an experiment which is taking a small today it is small but small magnet in the shape of a ring or things like that and putting it above a magnetic field directly opposite a magnetic field and then you can see that it actually stays afloat without anything. And the primary reason is the following that if I have a type 2 superconductors then if there is a magnetic field which you are putting from below then this magnetic field will not be able to enter the specimen of the sample. And so as a result what will happen is that in order to keep those fluxes out the superconductor will generate its own current this is not really called a persistent current but similar to it which will essentially repel these fluxes. So then it is like having so that so that face of the magnet superconductor which is facing the magnetic field pole will behave like a like pole because it is opposing the flux that is coming in. And as you know two similar poles always repel each other so as a result the in order to keep the flux pinned because its pinning occurs because of impurities generally and this flux tubes cannot move around and as a result of that the levitation can take place. And the levitation in principle has a lot of applications later but certainly not a day to day application. Next question 1150 and many others like which properties of superconductor is focused or improved to develop room temperature of superconductor I told you in the morning I told you there is nothing like a given prescription it is essentially like medicine use try out various materials try out various properties and find out that something works. So that question is not going to be answered. Another question though I answered it in the morning but it kept on coming back this was asked in this session and last session Charotter University Nadiag what is red electron? In fact this question was asked even last session with professor Suresh somebody asked what is a super electron? Let me let me make something very clear again and again there is nothing like a red electron there is nothing like a green electron there are only electrons okay. What I told you is that in order to explain some things not on the basis of Cooper pair or the quantum mechanics we sometimes have what is known as a phenomenological theory. So in that phenomenological theory we had assumed that London assumption that supposing I assume there are two types of electrons with different property of transport one type of property which just out of illustration I called it red electron another is green electron I mean I did not really mean that you people will take it very seriously remember I also told you a Cooper pair assume that these two electrons are married now do not tell me that which code did they go to and all that obviously that is not correct sometimes when you are talking giving a simile it does not mean it has to be taken that seriously the what I meant by saying that two electrons are married is to say they remain together of course nowadays not even married people will remain together but that is a different story but what I am trying to tell you is this that supposing there are two categories of electrons with different transport properties one which obeys which dissipates one which dissipates energy the other one flows without resistance okay the one which flows without resistance we will call it we will call it is what I said or rather London call it the superfluid electron or a super electron okay that does not mean that there are electrons called super electrons or red electron or green electron I am sure this question will not come up again alright next question is again we have been talking about it in case of a superconductor this was 1127 Biswakarma Institute Pune why is the current on the surface or the edges well this is this is true for any conductor the now mind you if I am passing a current through it that is a different story but I am trying to say that if you are establishing a current without a battery you say when a battery is connected you know that there is a source of Pmf which is providing the current right now so therefore that is not a problem at all the if I connect a conducting wire to a battery the steady current passes through it that is because there is a source of that current which is providing the charge there is another sink which is removing the charge continuously so that is not what we are talking about we are talking about that when you have a superconductor you have established a flux then we drawn a magnetic field there is neither a battery nor a magnetic field to keep the flux now so there is a magnetic field generated in the hole of the ring which remains I told you also for flux gets trapped but I am coming back to it again so in order that that can happen a current must flow now the such a current cannot flow inside the material it has to be contained only on the surface and you remember we talked about that surfaces discontinuities behave separately when we talked about boundary conditions so inside the bulk material there cannot be a current so current must flow within a very thin sheath of the layer and that is what generates its own magnetic field in order to keep this flux lines so that is and that I said is quantized right 1150 so Sushila had asked how are type 2 conductors helpful type 2 conductors are the ones which are going to become technologically the most important things because any property of superconductor that you can think of for example suppose you want to use somebody asked in the morning what happens if I connect the part of a circuit by means of a superconducting where and I may answer at that time was today nothing happens because your circuits are small but supposing you are looking at a science fiction and you say that I want to transport power without loss from generating station to somewhere else and supposing I had a superconducting material which remained superconductor at normal temperature or even Bombay Delhi temperature then what it means is that it can carry current without any loss at all on the way now that would then be a real practical application now and if it all such a thing can be done whether at a laboratory scale or at a industrial scale then it will have to be for type 2 superconductors because only in type 2 superconductors our chances of getting an increased critical temperature is substantial but type 1 superconductors that way are very useless but the other example that I told you and if you are thinking in terms of for example magnetic levitation you need to have these normal regions inside that where the flux gets picked it is required because the it is only in the normal region in the superconductor that you can have the flux tubes and so the pinned flux can only happen in type 2 superconductors because in type 1 either the magnetic field enters completely or it does not this question 1, 2, 4, 1 what are phonons strictly I it does not it does not fall into the scope of my lectures but since a question has been asked and I use the word phonon let me give a one line answer to that see the thing is that generally any material any matter insulators or anything it has this ion cores and of course there are electrons the ion cores are generally there in a lattice structure now what happens is when you apply when you subjected to a temperature the because of the thermal energy that these ions absorb the ions start vibrating about their main position but the point is this that because of the forces inter atomic forces these ions simply do not move away from each other these are actually like they are connected by a spring and so they can move a little bit from their equilibrium position and again come back to vibrate now so in other words I have ions which can vibrate now if you remember that we did in our we talk about simple harmonic motion of a spring and a mass now we we solve the problem of spin spring and a mass in classical physics we talk about that there will be amplitudes it will so simple harmonic motion and things like that but then you do the same problem in quantum mechanics it becomes a linear harmonic oscillator and in the linear harmonic oscillator your energy states get quantized which was not true in the this here all that happened is that you said that the energy is constant kinetic energy becomes potential energy potential energy becomes kinetic energy and it keeps on continuing there so the when you quantize a thing the the go from classical mechanics to quantum mechanics certain quantizers and takes place usually it is in the form of discretization of the energy energy gets discretized now if these lattice vibrations that you have got you worked in the theory of quantum mechanics instead of having a collection of simple harmonic oscillators what you have is a collection of quantum oscillators and and this vibration which is quantized this is called phonon so in other words to give you a one line answer to that question the quantized vibrations of the ions in a solid are called phonons what is the typical distance between two cooper pairs of electron is it consistent with this electron potential versus distance plot from Netaji Shubash engineering college Gordhya Kolkata let me let me come back to this question a little bit so the point is this that morning I talked about a the concept of a cooper pair and I said that basically the cooper pairs I mean you can imagine that the phonons are providing something like a spring between these two electrons so that these two electrons irrespective of how far they are okay they would still continue to travel together now what actually happens is this that the distance between the two electrons which are connected by the ionic vibrations that is the phonons as we call it that can be reasonably large now incidentally I have to sort of tell you what is meant by a reasonably large in that context but if you take different cooper pairs because these cooper pairs they can come fairly close to each other so there is no problem so in other words the individual members of a cooper pair can be far apart I will explain what is meant by far apart but the different you know you can have a collection of a large number of cooper pairs in the same assembly now in order to understand this you remember I told you that the electrons which participate in this have to remain within a small distance of the Fermi level see in fact this picture I had given you that I have a Fermi surface and I am saying that basically this electron and that electron this is k this is minus k so this must lie within e f plus or minus k t Fermi energy plus or minus k t these are the electrons which are subjected I mean which are going to form pairs and the reason is very simple because otherwise that thermal energy will destroy them the thermal energy realize say I told you already that the gap of the superconductor is that this delta is typically given by 1.76 this is VCS theory 1.76 times k tc and so this is typically 1 milli electron volts or even a fraction of a milli electron 0.1 to 1 milli electron volts so now so therefore the electrons which are close to the near the Fermi surface they are the ones which will participate in it now if you look at the energy uncertainty of such a thing because these electrons were really free electron so I have essentially p square by 2 m as their energy so therefore the energy uncertainty if you like so this is you just differentiate this you get p delta p divided by m now this is approximately an expression for your delta because remember what I told you I said delta is the energy by which the cooper pair is makes the Fermi C unstable in other words the energy of the cooper pair is lower than the energy of the course the metallic things and that is what opens up a gap. So therefore if my gap delta is p delta p and let me use dp because there are too many deltas coming in over m now in that case if you use uncertainty principle then the uncertainty in the position of these pair so in other words what is the distance by which they can be there say if I know the momentum uncertainty then I can find out the I know already what is the maximum of uncertainty of momentum so I know how far these two could be away so this zeta then is given by the momentum position uncertainty which tells me this is equal to again let me use dp instead of delta p okay so zeta not equal to h cross by dp but then this is equal to say the you use this so this quantity I can immediately tell you that this is nothing but h cross okay by dp let me let me multiply this with a velocity Fermi velocity here and denominator also let me multiply with a Fermi velocity so look at what is happening there so here what I have got is h cross vf by dp vf okay and p by m is nothing but the velocity of the electron which is approximately equal to this is a hand waving calculation the velocity of the electrons multiplied by dp so therefore what I got here is essentially h cross divided by delta because this is dp vf and I have got so p by m is so this delta is vf times dp so this is what my h cross by delta is in BCS theory though I have done it with a lot of hand waving it turns out that almost similar expression is obtained by BCS with the change that this works out to this time delta into pi just a small factor is there now you can use this expression or this expression to calculate what should be the value of zeta you can see immediately you have your h cross which is typically of the order of 10 to the power minus 34 or so you have a delta I told you typical delta is about a billy electron volt let us say 10 to the power minus 3 or 10 to the power minus 4 electron volts so but this is an electron volts unit this is in joule second this is in SI unit sorry minus 4 so therefore I need to multiply this with this thing your 1.6 into 10 to the power minus 19 there and I have forgotten a vf to be carried in here okay so so there is vf there all right now what is the Fermi velocity typical velocity of electron corresponding to Fermi energy of a metal okay for 5 electron volts is about you know about electron speed is about 10 to the power 5 meters per second take all these numbers you can see what is happening here so I have a 10 to the power minus 19 and 4 so minus 23 and 5 that is 28 so it is about 10 to the power minus 6 by 1.6 is there let us ignore it because I have taken it exactly as 10 to the power minus 4 so this tells me my zeta is typically 1 micrometer or so of that order so this is what is the typical coherence length in case of the type of theory we are talking about the measurements indicate roughly similar values and of so that is there so that is that is what tells us what it is then having done that why do various superconductors have different TC from South Susheela I do not know these are material properties see as I told you in the morning also the substances have properties depending upon a lot of complicated things there is no great theory to explain them all right. Once again repeated question on from same South Susheela on what do you do to get room temperature superconductors well alchemy as I told you in the morning what are phonons repeated again so I will okay K S Rangeshwamy College of Engineering 1313 asks a question that how long the persistent current will exist how long persistent current will exist thing is there are two issues there one is that there are theoretical prediction on if you have established a flux in a ring how long will it exist and if they they sort of depend upon to what extent you have been able to measure some of the quantities the theoretical estimate seems to suggest that you will have literally billions of years for which it will persist but as I told you one can't wait for a billion year to check whether you are right or not but people have people have tested in the laboratory for a period of up to one year okay and they have not found any reduction in the persistent current so theoretically the persistent current should stay on forever if you want to put a limit on it today's limit is that it will stay for 10 to the power 15 years but as I said that this is the situation there other than mercury and 1128 Biswaswara National Institute of Technology Nagpur other than mercury and liquid helium well liquid helium is a bad example what are the other superconducting materials let me be very clear literally today there are thousands of superconducting materials so I can't obviously give you a list okay but if you are looking at type one see the in elemental superconductor niobium with something like 9.2 niobium with 9.3 degrees is the maximum among the elements there are others for example lead has 7.2 degrees kelvin these are all in kelvins I will have something like now aluminum for example is 1.17 you notice the better a conductor is things are becoming smaller tin has 3.7 mercury of course we have been talking about 4 degrees or so 4.15 degrees similarly it goes on and on in the type to be the new type of superconductors we have for example the first major one yttrium barium copper oxide emko it has 92 kelvin but the we are sort of roughly professor Suresh will be able to tell me tell you better we are still talking in terms of 125 to 130 degrees being about the highest temperature achieved in the laboratories but you know which are steady so therefore there are very large number of them then 1043 it's a good question I will take a bit of a time 1043 which is Sarvajani College of Engineering Technology Surat what will be the magnitude of persons persistent current how big is persistent current good question but let's look at it a little more carefully see this is a very interesting calculation you can do it it depends upon what is it that you know for what geometry you want your persistent current so let me give you a quick example I was saying that let me say the following supposing I have taken a sample in the form of a ring and what I have done is to and and this ring let us suppose I have taken a radius of 1 centimeter now I establish a magnetic field remember the persistent current again how it works you first let the flux lines go in and once and then you know this is at normal temperature reduce the temperature when it becomes a superconductor then I know that this is what will happen this is the expulsion of flux and through the holes of course there would be flux lines so this is when t is less than tc but the magnetic field is still there b not equal to 0 or h not equal to 0 now when you would now withdraw the field then I pointed out that these will go away these will go away because there is no magnetic field to sustain them but these flux lines which were here for example now they get trapped this is the trapping of the flux they get trapped and because there is nowhere to go they just close on themselves now what I also told you is what has been found what has been found is that these flux lines are quantized in units of h by 2e so my phi not my flux that is there has to be a number n times h by 2e now if you calculate what is h by 2e this works out to 2 into 10 to the power minus 5 minus 15 tesla meter square this is my phi not this is simple just take the value of h take the value of e multiply them divide them okay so in other words the flux that is there contained inside that hole it has to be in units of phi not so whatever number of now let me just calculate what is the situation if I want only one flux lines to be enclosed now I have told you that my ring has a radius of 1 centimeter so my area of that ring area of that hole if you like is pi r square so it is pi into 1 centimeter square which is 10 to the power minus 4 meter square now well let us since I am doing a rough calculation forget about this let me take the area to be equal to 10 to the power minus 4 meter square you can put all those pies and things like that okay now if this is your situation that for a single for a single fluxoid my phi not value is 2 into 10 to the power minus 15 tesla meter square and the area of my enclosed is as I said I am taking it as well there is a pie so let me take it as 3 into 10 to the power minus 4 meter square then my the this is my unit of flux this is my area so the magnetic field strength that is there the b itself you remember b is nothing but phi not by the area so therefore this is 2 into 10 to the power minus 15 divided by 2 or 3 it does not matter because calculations are not exact so this is of the order of 10 to the power minus 11 okay your tesla this is my flux okay now the question that I have to ask is what is the current that I must have in order to maintain this strength of the field for that I go back to my electricity magnetism again and look at the expression for field in the center of a circle let us say I mean it does not have to be but this sort of give you rough idea so if you recall your first year physics then you will see that if I radius of my circle is r then the field at the center is given by mu 0 i by 2r this is a circle its center is given like this so this quantity is 10 to the power minus 11 tesla now look at what this means as you know mu 0 is 4 pi into 10 to the power minus 7 i is something I am looking for 2r I have taken 1 centimeter so it is 10 to the power minus 2 is equal to 10 to the power minus 11 so if you look at this i forgetting about these factors what I get is 10 to the power 6 because this 10 to the power minus 2 there minus 11 there so 10 to the power minus 14 13 divided by 10 to the power minus 6 so 10 to the power minus 6 so in other words the due to a single fluxoid the type of this in amperes I type of persistent current that I need is micro ampere but you remember my explanation I said this is the current I require for maintaining a magnetic flux single magnetic flux but of course within that whole there is not going to be a single magnetic flux I will have more right because and typically the if you look at a typical sample of that size we may have as many as 10 to the power 6 you know n may be about 10 to the power 6 fluxes flux lines so in order to get that you the persistent current that you would need would be of the order of an ampere so since the strength of the current depends upon how much of flux that you put in it would also depend upon what was the strength of the magnetic field that you had used to generate these fluxes so supposing originally I had inserted a 1 tesla magnetic field 1 tesla incidentally is a very strong magnetic field normal laboratory magnetic fields that you have will be a gauss or so or maybe 10 gauss 10 gauss is 10 to the power minus 3 tesla so take for example a strength like 10 to the power minus 3 tesla you have now withdrawn the magnetic field and the strength of 10 to the power minus 3 tesla has to be preserved find out the flux now that is very easy area of your sample multiplied by the magnetic field so once you have found out that flux just use the standard formula for magnetic field in a circular domain which is mu 0 i by 2 i this is this is rough way of determining what it is but as I said the answer depends upon what was your original strength of the magnetic field which you had later on withdrawn so we are talking about things which could be small could be milli amperes depending upon if it is a weak field but for the laboratory field it could be as much as it is not a small current it could be as much as an ampere okay let me let me clarify this what is the difference between this is came from some githam university 125 githam university yes the question is what is the difference between a persistent current and an eddy current well you know this question is not as bad as it sounds because the the source of both of them are the same the the how did eddy current arrive the eddy current arose because of changing magnetic field if you have a conductor you need conductors of course if you have a conductor and you are trying to change the flux through it right it will generate its own current on the surfaces and and this direction of the current would be perpendicular to a direction in which you have put in the magnetic field and these are eddy currents which are the differences these eddy currents because of the fact that they are in a normal medium they are also dissipative so they dissipate energy and one has to take care of eddy losses when you talk about the you know electrical parts and so if you like the persistent current arose from similar requirement okay because when we drew the magnetic field but the origin is slightly different when we drew the magnetic field you were changing the flux but that is not the reason why it came in the origin was different the reason why persistent current came in is there was flux which was trapped and because of Meissner effect it could not enter it could not get out okay so so there is a similarity but not much all right I think what I will do is because I have taken my usual quota of half an hour for your questions I will try to you know please don't don't send the same question again and again because this is what is happening because it takes us a lot of time to throw away for example what is the phonon I have received twice so similar questions are coming twice thrice it takes a lot of our time you know and sometimes I have already answered that question like in the morning I answered the question which was asked by Mahakal Institute of Technology Ujjan how is superconductivity related to magnetism I told you why the modern the conventional superconductors exclude the possibility of becoming a ferromagnet because I expect the ground state to be a singlet whereas ferromagnet requires the spins to be aligned okay we proceed with our discussion of in the morning so I told you about the Cooper's pair I would try to finish this session today so what I said is that I have a wave function for the pair of electrons and I sort of proved to told you that look this wave function can be split into a even part and an odd part but since I want the wave function not to be 0 when two electrons hit one over the other okay then I cannot have a sine function there because if r1 is equal to r2 this is 0 and so therefore only the even functions are acceptable so if the function the space part of the function is even then my spin part must be anti-symmetric that is must be a singlet so I made two or three statements about the Cooper pairs one is what we said is these are electrons its origin is phonon mediated attractive interaction so if you like what we are talking about is imagine a picture of two electrons bound to each other by a spring now tomorrow don't ask me that where is that spring see these are all similes being given they are being they are being I have I assume that if there are two masses connected by a spring they remain together okay so it is only in that connection that I said that connected by a spring so what is actually happening is this that the two electrons continually exchange a phonon and it is in this way a better example would be something like this that supposing online on a ground I have two players now and let us suppose they are they are throwing a ball towards each other now so a throws a ball towards b b receives catches it b throws it back now remember during the process the two people a and b could be moving and this you have seen happening if you have gone to any cricket practice or things like that you have seen it happening regularly that a player throws a ball towards another he returns back but in the process they are always moving around it is not that they are just doing that practice staying in one place so in that case these two people that I talked to you who are continuously exchanging the ball they can be considered to be a pair okay and this is the model this is the picture that I use for our super conducting pair where what we are saying is that imagine that electron number one and electron number two they continually continuously exchange a phonon remember I told you phonon is the quantized lattice vibration so they could continuously do a phonon now this pair it remains as a pair it turns out its statistics is Bose Einstein statistics and from what I could understand when I came in towards the end of professor surace's lecture that he told you that it is the bosons which can undergo Einstein condensation the condensation means simply this is not condensation in the sense of the a vapor condensing it is a condensation meaning that these objects the bosons they all come to the same state same momentum state so cooper pairs are collections of bosons and I have said that theoretically the I did a calculation now here is an actual numbers you can see that I did a calculation of 1.6 micrometer I mean I did a calculation of 1 micrometer and a coherence length for aluminum happens to be 1.6 micrometer and it is as low as 40 nanometer for niobium now you can see immediately that there is a relationship because this has a higher critical temperature than that one okay so we also made a statement that the superconducting state is separated from the normal state by an energy gap of the order of about 0.1 milli electron volts okay and this gap you can verify by typical tunneling experiments okay and of course the specific heat is a direct example of the existence of a gap you know that specific heat when it goes exponentially like it happens in case of a semiconductor it is evidence of a gap so and not only that the if you look at remember some time back somebody asked me a question why do you keep on talking about DC resistance what are the resistance is there you see there is something called AC resistance also the thing is that if you are subjecting the sample to an AC having a particular type of frequency then it is possible that it is able to take care of this energy gap so the energy the existence of energy gap can be inferred from various experiments and as I said this is typically of the order of a fraction of a milli electron volts so this is the picture of density of state versus the energy and here is the Fermi energy now remember somebody was asking me asking professor Suresh towards the end where is the Fermi energy for an insulator or a semiconductor well there is a I mean in one line I will tell you that Fermi energy at zero temperature for a metal has the interpretation that it is the energy of the highest occupied level this is not true this definition is not true for the case of a semiconductor or an insulator now what we do in case of a semiconductor insulator is to define the Fermi energy or a chemical potential to be that energy at which the probability of occupation becomes half and if you look at the expression for the Fermi energy in the Fermi Dirac statistics you will find it is 1 over 1 plus e to the power epsilon minus mu by kT so if your epsilon is equal to mu that is Fermi energy the denominator becomes equal to 2 so that the fraction is 1 by 2 so it does not have that interpretation and for an intrinsic semiconductor the Fermi energy lies exactly in the middle of the gap so I told you just before we conclude at last session that the wave function of the Cooper pair can be considered to satisfy the Schrodinger equation so I have written it down there is a minus sign problem here but otherwise all that I have done is to write down p square by 2m plus well e star at this moment is the Cooper pair charge which is 2e phi is a potential scalar potential a is the vector potential and this is minus ih cross del by 2m is p by 2m essentially p square by 2m because there is square but p becomes p minus ea in this model so this is nothing you can write down this equation now only difference that I am doing now is the following I am saying that my Cooper pairs are bosons and so therefore the in the lowest state all the pairs must have the same momentum and that momentum is 0 because you see had it been fermions I will not be allowed to do it because once a particular momentum state is occupied the next pair cannot occupy that because so what we are trying to say now is but that is not true now because I can have any number of Cooper pairs in the same state of momentum so let us put all of them there and that must be the lowest so typically it is this since the amplitude is square root of the probability density so the lowest level if its occupancy is n it is you know higher by a factor of square root of n with respect to any other so the wave function now this is not a wave function of a single electron this is not wave function of a Cooper pair this is the type of wave function where I am saying this is a macroscopic wave function this is a wave function of let us say the whole assembly of my Cooper pairs so I write this because I know a wave function is an amplitude which can be in principle a complex number so I write it as square root of rho so that psi star psi is essentially rho rho has the interpretation of density times a phase factor which is e to the power i theta r and both of them are real by definition so if I substitute this into the expression for the probability current density which is psi star psi plus you know i h cross minus i h cross del psi star psi plus complex conduit you get this request a little bit of an algebra you get an expression like this h cross by m delta theta minus q by h a this now inside the superconductor well inside the superconductor I do not have a current so therefore this is 0 current density is 0 which means by delta theta is q by h a now this is what I am trying to do is remember in the morning I worked out the flux quantization assuming Bohr-Sommerfeld model we said that if a a charge is moving accelerating in a circle I must use Bohr condition because otherwise my charge will radiate and finally slow down and stop now here I am doing the same thing but using my ideas of quantum mechanics so I am saying that this wave function that is there this wave function that is there has to be single value because at the same point I cannot have two values for the wave function now what is what is meant by single value supposing you start here now when you start here now go around the circle you come back when you come back you have increased theta by 2 pi so since it is still the same point then your wave function in order that it is single value it tells me if you change theta by 2 pi 4 pi whatever multiple of 2 pi 6 pi then your you must come back to the same point right so theta when it changes by 2 and pi this is what it is now since my inside the material my current density is 0 so probability current is 0 so my delta theta is q by h a which is what I did here this quantity is 0 so therefore this what happens okay so write that now what you do is you realize by stokes theorem in line integral of a dot dl is nothing but surface integral of the magnetic field which is my flux so that is the confirmation that my flux now how did it come from here when I did the integral over d theta I said my theta can change the because of the single value condition the line integral of d theta in a closed contour must be 2 and pi so my phi must be m h by q I told you I am using q here so that I do not confuse you with that star which is a complex conjugate this q happens to be equal to 2e and this is the flux quantization condition okay with that let me spend the remaining 15 minutes if you like by decide by telling you a very interesting effect which incidentally probably has a very huge potential of an industrial application namely the Josephson junction you have all heard of squid and things like that namely the Josephson junction okay so let me let me continue with this so what is a Josephson junction in this situation I have this is typical experiments of tunneling that you have seen in your this is a provides a barrier and all that but there is a difference I have on two sides of an insulator insulator the a superconductor on this side S1 superconductor on that side S2 now what I do is this that the two superconductors are separated by a thin layer of insulator now remember normally that if I now connect the two ends of the superconductor with the battery now if this insulator is thick I do not expect any current to go through because you know I mean the insulator does not allow the motion of charges so let me look at what is happening what is when this insulator is thin so that a tunneling effect is expected but you will find something very interesting here supposing the left hand side I denote by Psi1 it is its bulk wave function and for the right hand side I have a Psi2 let us assume there is no magnetic field now what are these two amplitudes these two amplitudes are I have on the left hand side Psi1 I write it as square root of rho1 e to the power i theta1 on the right hand side I have got Psi2 which I write at square root of rho2 e to the power i theta2 now using this let me write down a very simple Schrodinger equation remember that supposing there were no connectivity between the two supposing I take this by itself then my Schrodinger equation for Psi1 is ih cross d Psi1 by dt is e Psi and so let us say e1 Psi or e1 Psi does not matter now the equation for this one is ih cross d Psi2 by dt is u2 Psi now what I now have so I have got this forget about this for a moment so I have got these two superconductors but there is a coupling of Psi1 to Psi2 because of this reason that this one has an effect because of the fact that on the other side there is another wave function okay let me take a simple model in simple model I say that look if whenever writing the equation of motion for Psi1 that is Schrodinger equation for Psi1 there is add to it a potential term which is due to Psi2 and vice versa now let us call it a coupling term and I will assume coupling to be symmetric so ih cross d Psi1 by dt is u1 Psi1 u1 is the energy of the left hand side in the absence of any coupling plus k times Psi2 so equation for Psi1 has a contribution from Psi2 likewise the symmetric equation for Psi2 has a contribution from Psi1 so this is my pair of equations which govern this situation incidentally I must point out that this beautiful derivation without very rigorous mathematics is due to Richard Feynman and I would urge all of you that please read a very nice article on Josephson junction which I am following for this particular part of my lecture which is in Feynman lectures volume 3 chapter 21. So as I said k represents the coupling between two junctions now if k is equal to 0 then the equations describe each superconductor with lowest energy and these are u1 and u2 so what I am doing now is this suppose we connect look at it suppose we connect the two ends of the superconductor to terminals of a battery so that there is a potential difference of v between them remember a battery will maintain the energy difference as v which means u1 minus u2 is equal to v I can take the 0 of the energy halfway through now if I do that then my first equation look at look at what I am doing I said u1 minus u2 is the battery potential that is what I want to keep it but I measure the energy from the middle of this if I do that then one of them is higher by v the other one is lower by v so that is what I have done I have said ih cross d psi 1 by dt is q which is again your charge q v by 2 psi 1 plus k psi 2 and ih cross d psi 2 by dt is minus q v by 2 because I have taken the origin to be at the center plus k psi 1 now I need to solve this pair of equation I need to solve this pair of equation now what do I do so what I do in this case is I substitute rho 1 equal to psi 1 equal to square root of rho 1 e to the power i theta 1 this is the form I gave you earlier psi 2 is equal to square root of rho 2 e to the power i theta 2 now I need to do a little bit of an algebra to tell you how to get the equation that is there so for a moment I will go over to this and I will not work out the complete equation but let me sort of at least do a part of that equation now I am copying this equation on my pad so that I can tell you what to do with these equations yeah so these are the pairs of equation I had and I said that let us do the following let us write psi 1 equal to square root of rho 1 this is the for the left hand side e to the power i theta 1 this is trivial algebra but I will not have it in my notes so therefore I am telling you how to do it so look at what do I get let me do just one equation first so because the second equation can then be written down from symmetry so I have ih cross I need d psi 1 by dt now both rho 1 and theta 1 can vary with time so I have got the first one is d by dt of square root of rho 1 which is half denominator rho 1 rho 1 dot because d by dt of general differentiation then e to the power i theta 1 plus differentiation of this one which is root rho 1 i times e to the power i theta 1 that is differentiation of an exponential times theta 1 dot so this is my left hand side this is equal to q v by 2 root rho 1 e to the power i theta 1 plus k times root rho 2 e to the power i theta 2 now I could repeat that so this is my equation number one I could repeat the same without having really to do any calculation because the equations are very symmetric so I have rho 1 going to rho 2 this is equal to now there was a minus sign on the second equation and rho 1 becomes equal to rho 2 this is my second equation now what do I do with this equation so one of the things that you do with this equation is that take the real and the imaginary part you have two equations here so notice this e to the power i theta 1 as right as cosine theta 1 plus i sin theta 1 etc now these two equations separate their real and imaginary part separate their real and imaginary part so total since I have got two equations and each I will get a real and imaginary part so it is something like this let me write down one equation here so I have got ih cross 1 by 2 square root of rho 1 rho 1 dot cos theta 1 plus i sin theta 1 that is why e to the power i theta 1 plus root rho 1 i times cos theta 1 plus i sin theta 1 is important that you keep track of all the i's properly because otherwise you will miss out that this is actually a real this is equal to key v by 2 root rho 1 again cos theta 1 plus i sin theta 1 plus k square root of rho 2 cos theta 2 plus i sin theta and you can immediately do that for example let us look at some some part let us say the imaginary part here this is an imaginary part because there is an i outside so I get 1 by h cross minus sin because i into i is minus 1 2 root rho 1 cos theta 1 rho 1 dot another imaginary part is this one root rho 1 cos theta 1 like that you can do that q v by 2 square root of rho 1 sin theta 1 plus k square root of rho 2 sin theta 2 so this is one equation write down the second equation okay and do some algebraic manipulation come back to this point this is this is absolutely trivial and and this can be very easily done so my equations which are wrote down and once you separate them so what you have to do is to you know manipulate these four to get an expression for rho 1 dot rho 2 dot theta 1 dot and theta 2 dot you will find that you can write them in terms of theta 1 minus theta 2 and you can see why because you have you have to eliminate see from this equation you need to eliminate equations which are connected with theta 2 and things like that so you get four equations as I told you but written in a particular fashion so 2 by h cross k square root of rho 1 dot 2 sin delta this one I write down blindly because I know there is a minus sign difference and similarly theta 1 dot and theta 2 now what is actually happening is this that look at this equation rho 1 dot is this rho 2 dot is this so this is my all the four equations now if you compare these two expressions you notice rho 1 dot is equal to minus rho 2 dot now the thing is you should have actually expected this relationship because rho 1 dot is the changing I have connected them by a something is the changing density if you like of the left hand side rho 2 dot is the changing density of the right hand side so what is actually happening is this that now if you now say that look what is this quantity now this is some constant let us say 2 by h cross k square root of rho 1 rho 2 dot I have called that j 0 so rho 1 dot equal to minus rho 2 dot because this is minus of this equal to j 0 sin delta where delta is theta 2 minus theta 1 now you would immediately say well this is obvious because you know each one must remain the same so the thing is this that what is interesting about is this that I need to worry about what is the consequence there but before I do that say remember this equation is extremely simple rho 1 dot is nothing but j because that is the current which is coming from the left hand side that must be the current which is leaving the right hand side which is must be minus rho 2 dot so both of them being equal and opposite and understandable and this current density j is some j 0 which is given by this coupling constants times sin delta so my equation for the Josephson junction is at this moment j equal to j 0 sin delta which is an extremely simple equation now if you did the same thing with theta 1 dot and theta 2 dot they are absolutely trivial equation so you can solve them like this now let us look at what is my delta because I define delta is theta 2 minus theta 1 so delta dot is theta 2 dot minus theta 1 dot and if you look at this when you subtract them the complicated term cancels out and you are left with theta 2 dot minus theta 1 dot is equal to q v by h cross which tells me delta of t is some constant of integration plus q by h cross vt dt now comes the interesting part supposing I have applied a dc voltage constant voltage across the junction v is constant if v is constant delta of t has the solution it is delta 0 plus q by h cross into vt notice one thing vt sorry into v into t because v is constant then this is v times t now if I apply such a voltage this tells me that I have a q by h cross there I have a q by h cross there but h cross as you know is a very small quantity so since h cross is small quantity and q which is the effective charge of the cooper pair is 2 times e what you get when you write j naught sin delta this is the expression that we just now derived so suppose I look at this expression that if I have a voltage constant voltage then the argument which comes in with t multiplicative factor of t is extremely large because h cross is small even e by h cross is small now sorry e by h cross is very large in which case the argument of sin will fluctuate very rapidly and the average value of this sin function will become 0 what is surprising about it what is surprising about it is that if I apply a dc voltage to the ends of a Josephson junction I find that the current is 0 what is more interesting is supposing you do not apply any voltage this term v is 0 then you find my constant current will be constant because this delta 0 is the phase difference between the two sides and so this can take any value between plus j naught to minus j naught depending upon what the value of that. Now this is very surprising because what we are saying is that in the junction if I apply a dc voltage I do not get a current but if I do not apply a voltage I have a current the last thing that I want to talk about is there following that suppose instead of a dc voltage I were to apply an ac voltage that is there is a dc part which we have seen will become 0 and there is a small amplitude but cosine omega t type of thing. Now I can do that integration of delta t delta dot t because it is a cosine function and you get a term like this then I find out what is sin delta by the usual sin a plus b equal to sin a cos b cos a sin b formula and find a situation like this one I have just now interpreted but what is what is additional here is a term like this. So now what happens is this that if you were to control your value of omega in such a way that the if my omega happens to be q v 0 by h cross then if you expand that cosine here you will find a term which will be also b sin cosine a plus b is cos b minus sin a sin b. So in the second term minus sin a sin b there would be another term which is a sin omega t. Now if that happens the I get a sin square omega t whose average as we know is not 0 but is half. So therefore the second term gives me an alternating current provided my omega is suitably adjusted to give me a resonance condition and such signals have actually been experimentally verified. I will stop it because we have exceeded the time please send me the questions because I see a lot of questions but I have actually already taken one extra class on superconductivity than I originally meant to but I will see that we will shorten the optics a little bit and try to answer your questions. Thank you very much.