[Music] you [Music] you so good morning all of you let us look at the flux quantization which is a very important concept in the context of superconductivity today and in order to understand flux quantization we will have to go back to the London equations the equations that we have learnt a little bit in brief on this effects of electromagnetic fields in superconductors so let's start with london equations and then we'll go over to the flux quantization so to start with london equations so what's the condition for the transport of charges even if they're interacting amongst
each other so if the charges are interacting amongst each other they should create resistance so they should give rise to resistance so how is that that we are encountering a situation in which the current is flowing without any resistance which means that the electrons would stop interacting with each other we of course know partly the answer to this question but here what London and brothers London brothers actually they have suggested is that whenever there is a slowing down or there is a resistance that arises due to the interaction of electrons among the electrons then there
is a external electric field that would overcome the this effect of the interaction and would eventually give rise to a resistance less motion so in order to see that let us write down Newton's laws for a charged particle in an electromagnetic or rather in an electric field here so we write it with M star which is the effective mass of the particles that are contributing to the motion which we know that there are Cooper pairs or we are going to see more elaborately the formation of Cooper pairs and so on and then there is so
this is the Newton's laws which says that so this is nothing but the Newton's laws for a charged particle of charge East our of course you can relate that charge Eastern is nothing but equal to twice of the electronic charge and this electronic charges it's as you know that it's a it's two EE or the Cooper pair but that will come to later let us simply look at this equation and try to understand that what it what it does or what it gives so M star is the effective mass and E star is of course
the effective charge okay and now the current density or the super current density is written as j s is equal to n s e star and V s this is all familiar to you j s is the current density n s is the super electron density or the super fluid density as it's called basically it's a density of super electrons and V s is the velocity of these super electrons and of course Easter is just the electronic charge so that is the the current density equation that one would get and we can write it using
a mod star here of V s and the minus sign is just to make sure that the current density is actually in a direction which is opposite to the direction of the the flow of the super electrons okay so if you call this as equation one and call this as equation 2 so if you put two into one one would get the first London equation which is of the form that mu zero lambda L square a DJ s DT is equal to e where this lambda L is equal to M star divided by mu zero
NS e star square and it's called as the penetration depth and we know that physically what it means is that the depth or the distance are two till in which the magnetic field can penetrate into the sample or more precisely this is the distance over which the magnetic field actually penetrates and falls into a value of 1 over e of the value that it has at the surface so it's called as a penetration depth and it's one of the most important length scales of a superconductor and there's another length scale that we have looked at
which is called as a coherence length and so this is the equation that we get and let's call this as equation maybe three and then of course if I take the curl of both sides of equation three so taking curl of both sides in 3 we have Muse mu zero lambda L square and a del Del T of curl of j s it's equal to curl of e which is nothing but equal to minus of del by Del T by Maxwell's equation and now this requires minute of discussion here that you see there was a
DJ DT or we can simply write it as DDT of that instead of writing so this is a DDT now you see that we need to take a Curl Curl is a space derivative and because there's a time derivative we can interchange the space and time derivative so we have interchanged that and I have written the curl inside and which is taken directly on j s and the DDT has come out so this call it as equation for let us look at what happens to this the right-hand side of the equation by using Maxwell's equation
again we can write down the curl of B this is equal to a mu 0 epsilon 0 del e del T and a plus mu 0 J and if we use this then we can write down this lambda L square del square of del b del T this is equal to a del V del T and this equation since it's true for the rate of change of magnetic field with time it should also be true for the magnetic field itself and we can write this down so this is a special case that we are considering
that where the rate of change of magnetic field is actually taken to be same as the magnetic field so we have ignored or rather we have taken this the value of the magnetic field at the surface to be equal to 0 and in which case we can write down the same equation as for simply B and that is written as so let's write it with this so it's a this is equal to B and since B is equal to MU 0 J s so this B actually looks like it's equal to minus mu 0 lambda
L square curl of J s ok so this is the equation this is one of the london's equations where the magnetic field is expressed in terms of the current density of the curl of the current density so if we introduced introduce the magnetic vector potential which is defined by be equal to curl a and then of course the je s can be written as minus 1 by mu zero lambda L square e because of this of course that your B is equal to curly so that is the relation between the J s and E so
this is basically called as the second London equation equation and just to rectify that we have not written this one was written as the first this one is written as a first London equation all right so these are the two learnin equations that we get but what are they good for what do they convey and can we extract some more information out of them that's the question so of course we know that curl of E is equal to minus del be del T and the curl of so e minus mu 0 lambda L square del
J s del T or we can write it as DJ as DT as well so this is equal to 0 so which means that if a curl of vector is equal to 0 that vector should be written or rather we one should be able to write that vector in terms of a gradient of a scalar quantity so e - e - mu zero lambda L square we'll call it DJ s DT just to make matters simple let's change that as well here and so this is equal to nothing but a Phi where Phi is a
scalar so once that is done one can easily understand that the grad phi dot j s it's equal to zero because of the reason if i take a dot product with the current density the or the super current density that should give me equal to zero so that tells that there is no component of grad phi in the direction of Jas okay and so which means that the J s dotted with the electric field should be equal to zero and which is basically a restatement of nothing but the first London equation okay and this is
basically it looks very reasonable for the reason that that there's no the energy dissipation is zero in a constant magnetic field which should be correct because otherwise it would have violated conservation of energy so this in a way it says that no energy dissipation in a constant magnetic field that is because the magnetic force is is a low-range fortress which goes as V cross B and the energy dissipation is called as a power and the power goes as f dot V so if V cross B is dotted with V then that has to be zero
this is a common wisdom and this is basically saying the same thing so and it says that the conservation of energy is valid or a very important thing at this moment this is strictly true for what is called as a simply connected superconductor okay so let me write that in big and bold if we have a multiple connected superconductor I will just tell you what multiply connected superconductor means in a multiplicative superconductor there could be a component of e that is if we dot it with the super current density it may not still give it
equal to zero which means that the grad Phi perpendicular to J s may still be admissible for a multiplicand acted superconductor so let me write that once again so J s perpendicular to grad Phi which is nothing but the electric field or say that not perpendicular to grad Phi could still be admissible in a multiplicand acted superconductor so this is one of the important statements that we make here and in order to justify that and get a connection with flux quantization we'll define what a multiplicand ected superconductor is so this if this superconducting specimen so
this is a superconductor everywhere okay so it's there is no hole there is nothing so this is called as a simply connected superconductor and there is a hole there so this region where I write hole is not superconducting so I put a hatched area to denote that it's superconducting so this area is superconducting and the whole area is not so this is anything other than superconducting so this is called as a multiplicative superconductor so why did we introduce this in the first place the reason that we introduced it is that we wanted to show flux
quantization that if there is a there's a hole in a superconducting region and the superconductor is called as a multiplicand acted superconductor put it in a magnetic field the flux lines will thread or penetrate through this hole and is cannot take any value it takes only values that are perpendicular sorry that are quantized that are quantized in some known quantum which is called as a flux quantum will tell you the value of that and so there could be one flux quantum or there could be two flux quantum or there could be three flux quantum but
nothing in between this there's no half or one-and-a-half flux quantum and so on so that's one the other thing is that what in line with the preceding discussion we have said that the grad Phi dot J s is equal to zero or rather the e dot J s is equal to zero so basically there is no energy dissipation and E and J s are always perpendicular to each other and that's why the dot product vanishes because you know the dot product comes with of two vectors that come with a cos theta theta is the angle
between the two vectors and if theta is equal to 90 degree then of course the angle goes to so the COS of that angle that becomes zero and so the whole quantity vanishes so here we want to show that also in a multiplicative superconductor it the e and J's they don't have to be necessarily perpendicular now in order to show both of these let us look at the basically we'll just we have looked at enough quantum model of super conductivity and going to see enough of that in the future but however just to formally once
again write it so we'll say quantum model of super conductivity and very interestingly this has relevance with the fact that the quantum mechanics is usually seen in microscopic and sub microscopic scales or rather very small length scales and inside an atom and maybe inside a nucleus and so on however super conductivity is one such example in which quantum phenomena or quantum mechanics manifests itself in macroscopic scales okay so we can have a big superconductor which could be you know extending over many many sort of coherence length or penetration depth and still the quantum formalism or
the formulation of quantum formulation of superconductivity still holds and we can write down a wave function corresponding to that quantum regime and let's call that wave function as psy of X T where it is written as an amplitude and a phase where theta is called as a phase so theta psy is the amplitude and theta is actually called as a phase and it's a real quantity okay but what is the physical significance of that the mod size square is the super current density or the super electron density as we know so in a homogeneous superconductor
NS is does not depend upon the space variable and in an inhomogeneous superconductor it can it can depend upon the space variable or space coordinates so let's write down the canonical momentum for a particle of charge East R and mass M star so that is given by P equal to M star vs plus i.e star a so this is the in an electromagnetic field where a is the vector potential of course we can choose a gauge in which Phi equal to zero and you know also the divergence of a equal to zero so this is
can be written as a minus i h cross del psy by writing P as so this can be you know in a way this can be operated on sy and this m star vs plus e star a sigh and so on and then you write P is equal to minus IH cross del so this becomes equal to minus IH cross del C which is equal to m star vs plus a East are a sigh and so on and now then of course your sy is equal to as written above it's equal to sy to the
power I theta so ad else I is nothing but equal to sy e to the power I theta del theta and and at this moment we are assuming that we are talking about a homogeneous superconductor and in which si is this more amplitude is independent of position and it's only the phase that depends on the special coordinates so if you put it here then it becomes equal to a minus IH cross and there's AI Delta Theta and sigh this is of course equal to a M star vs plus e star a sigh if you do
a simplification then that becomes equal to because your minus I into I will become equal to 1 H Delta Theta so this is equal to M star V S Plus I East are a and this is sy well I mean I've taken off sigh from both the sides and this can still be written as H cross by East R Delta Theta this is equal to a mu 0 lambda L square J s + e that's because the J s is equal to NS e star V s that's the equation for or the defining equation for
the current density so the J s / NS e star into M star is equal to M star V s and then I can introduce this lambda L which was defined earlier by this equation by this lambda L square is equal to M star divided by mu 0 NS s square or East R square and that's is going to give me this and so this is basically so we can write down the so this is an equation that is coming from the London equations and that's why all this exercise of explaining this London equation is
being done let's see what it takes us from there so I can write down the J s it's equal to e star H divided by M star ns and there is a del theta and plus a to e star by H by H cross and a sorry the a has to be in the numerator a and so on I believe this is H cross one can check that so this is the expression for this the current density or the super current density just to remind you that let me also write down the lambda L square
is equal to M star mu 0 NS e star square okay so that's the that's the equation that one gets and so J's becomes equal to this and we're of course all these things are very familiar to you theta is the phase of the wave function so we are getting a gradient of the phase and let us see that how this gives rise to the quantization let us take a multiplicative superconductor like this okay which has got of course this closes smoothly do not so there is a hole there okay so this region is superconducting
and this region is a hole let me take a path I draw it by a dashed line a path okay and this path let me call this as L and of course because it's in the superconducting it falls into the superconducting specimen the J s should be equal to zero here because otherwise of course so this J s should be equal to zero along this line or inside the superconducting sample and that tells that so along a closed loop l so in since we have taken a closed loop the net superconducting current density along the
closed loop would be equal to 0 which means the coefficient cannot be equal to zero so the term that is there on the right inside the bracket should be equal to zero which says that delta theta plus a to e star over h cross a has to be equal to zero which says that delta theta has to be a minus 2 e by H cross a and that is equal to that so this is the variation of the phase and its relationship to the magnetic vector potential so if you integrate let's call this as equation
1 here so if you integrate 1 over L then one can get so it's a closed integral so it's a gradient of theta dotted with DL so we take a small element of length along that capital L or that contour the dashed contour is equal to a DL which is nothing but equal to minus 2 e by H cross and this is equal to a dot DL so I can use Stokes theorem which says that the line integral or the closed line integral of any vector can be written as the curl of the the surface
integral of the curl of the vector which means this is equal to minus 2 e by H cross and this is equal to a curl of a D s where this s is basically the surface area of the I mean where the contour is found by L and s is the corresponding surface area now since you know that a curl of a equal to B so this is equal to minus 2 e by H cross and this is a B dot B s and B dot d s is nothing but the magnetic flux so this
is equal to a minus 2 e by H cross and a Phi which is called as a magnetic flux let's write it with a capital Phi so now remember that our theta is actually the phase of the wave function now we always demand that the wave function is a single valued quantity ok wave function cannot be multiple valued unless we make a full circle of the phase otherwise it will lead to the ambiguity in the probability density if we do not enforce the single valued nests of sy so this is then equal to if you
enforce single valued nests of size so this is equal to 2pi n it's equal to minus e by 2 H cross Phi that is coming from the the top line so that tells that so this means that 2 pi n equal to 2 e over H cross Phi it's equal to Phi over Phi 0 where of course we have written that n Phi 0 it's equal to Phi so the flux through the hole that you are seeing should be quantized in terms of a Phi 0 certain Phi 0 and it can only take values n
equal to 1 and equal to 2 n equal to 3 and so on n equal to 0 is a special case where we have a simply connected superconductor and this of course gives that the Phi 0 value is equal to 2 pi H cross over to e which is equal to H over 2 e and if you put the value of H and E we comes out to be H is of course your the Planck's constant which is six point six ten to the power minus 34 joule-seconds is equal to one point six ten to
the power minus 19 coulomb all these things you will find in any of the data book that one comes across and this is equal to a two point zero six seven nine into ten to the power minus 15 Tesla meter square so this clearly shows that the flux that can pass through the hole is quantized and the quantum of flux which is an important quantity the quantum of flux is this value two point zero six seven nine ten to the power minus 15 Tesla meter square it has of course the magnetic field multiplied by the
area that dimension and so these are experimentally observed and this comprehensively shows that electronic the the East are that takes part in this particular calculation is nothing but equal to 2e so which says that there are two electrons that are responsible and that are relevant for these you know the carrying of the super current and which we know are called as the either you call it them as super electrons we know that they are called as the Cooper pair you [Music] you
