 In the last lecture, I worked out the energy levels for a particle, a chart particle, moving in the uniform magnetic field. And I summarized some of the calculations here on the board. The main result is the energy levels, which depend on two quantum numbers, M, which is a Landau quantum number, and then P3, which is the momentum parallel to the magnetic field. The particle loses a free particle in that direction. And the Landau level corresponds to a harmonic oscillator with an energy of 1.5 h part omega. Interestingly enough, the Q and the P of the harmonic oscillator are basically the X and Y components of the velocity, which don't commute with each other. So the harmonic oscillator itself is really just the X squared plus the Y squared times M over 2 as kinetic energy. This is the quantized values of the perpendicular kinetic energy of the particle. In fact, the whole energy is kinetic energy. Anyway, that was the main result of the last lecture. And in this lecture, what I'd like to do is to start by working on the, not just the magnetic levels, but also the energy hiding functions. So, let me start with that thing over here. And it worked with that. So the strategy is going to be to choose a complete set of commuting observables, a CSCO. And for starters, we can take the, for starters, I mean, notice that the perpendicular Hamiltonian and the parallel Hamiltonian commute with each other. I want to keep looking at this. I'll have to show you the summary for the last time. The reason they do is that, you know, there's three degrees of freedom. One, which is, basically, the guiding center position X and Y, the Q and the P. The two degree of freedom is the X and Y components of velocity. And the three degree of freedom is the Z and PZ in the parallel direction. So the reason that, the reason that H perpetuates parallel commute with each other is that H perpetually involves number two degrees of freedom whereas H parallel involves number three. They commute with each other because the commutation relations are Q, I, P, J is I, H bar, delta, I, J. So the commutators are zero for different degrees of freedom. So the Hamiltonian is the sum of two terms that commute with each other. And so we can take these two terms H perpetuate parallel with being two of our members in the commuting observables, let's write this set like this. Now, actually, as it turns out, it's still an H per, excuse me, it's still an H parallel. But it's actually better to use P3, the parallel component of momentum. This is really just one dimensional free particle. And the idea here is as it shows, momentum is preferable because an energy is degenerate. You know, free particle of what I mentioned, you've got two waves traveling in opposite directions have the same energy. That's resolved if you use the momentum as a fun number. So allow me to do that and replace H parallel just by P3 like this. That gives us two commuting observables. But we need another one, and it's pretty clear that we do because the number one degree of freedom is not represented by H parallel only per. That's the x and y that might be certain positions. So if we want to find the third observable which commutes with the first two, it obviously can be any function of Q1 and P1. For that matter, it could be called x and y is the same thing apart from these constants. So it's a guiding center position, some function of n. And the simplest choice is just to take it to be x. We can make it y. Let's make it x just to give it a choice. So let's take a complete set of commuting observables, and we'll take it to be x H parallel to P3. And our strategy of them will be to find a set of simultaneous eigenfunctions of those observables. All right. So what we're going to look for then is to solve a set of equations like this. We've got wave functions from these sides, x, y, and z. And we want, first of all, that x acting on this, the operator, brings out an eigenvalue we call x. And you can see that they're distinguished from the operator from the eigenvalues. So let me use my usual trick of putting hats on any operator where I think there's going to be a confusion between the operator and an eigenvalue or some other C number. So anyway, this is the eigenvalue eigenfunction equation for the operator x hat. Likewise, when that h-perf acting on the same side will bring out an e-perf perpendicular energy and it will have a P3 hat acting on the side will bring out a P3 eigenvalue acting on the side, so these are three simultaneous eigenvalue equations that have to be satisfied. Now, so let's work on these. The h-perf is the most complicated of all. And the P3 and x hat are simpler. So let's do the simple ones first. Let's work on P3 hat first. P3 hat is from, it's from, importantly down here, it's the mass times BZ. And BZ, the velocity, is expressed in terms of the canonical momentum and vector potential by this equation. So we'll put those together. P3 hat then is equal to the mass times BZ, I'll put a hat on BZ. And that's the same thing as momentum PZ plus E over C times the vector potential AC, which in general ends at x, y, z. And so this eigenvalue equation, the third eigenvalue equation looks like this now. It replaces PZ by the operator minus i h bar BZ. If we get an equation like this, minus i h bar BZ plus E over C times ACZ, which is the function of x, y, and z, acting on psi is equal to eigenvalue P3 times psi of x, y, and z. Now, what makes this equation hard to solve is the vector potential component ACZ. And so we ask ourselves, is it possible to choose a gauge? Let's try to choose a gauge now to simplify the equations. Can we choose a gauge such that ACZ is equal to zero and then be done? Well, let's write out the equations, the components of B equals del cross A. We take the z component, and we have B, which is the same as the z component. It's at dA and y with respect to x minus partial A to the respect of y, which is right up the full equations. Zero, which is Bx, is equal to partial AC with respect to y minus partial A to y with respect to z. And zero, which is By, which is the third component, is equal to partial A to x with respect to z minus partial AC with respect to x, just writing on the curl. Now, if AC is equal to zero, if we provisionally guess that we can do that, then these two terms here go away. And you can see that what's left is that dA, y with respect to z is equal to zero, and also dA, x with respect to z is equal to zero. So these equations imply that Ax is a function of only a maximum y now, and it also is the same as true for Ay is only a function of x and y. The AC is equal to zero. This means, as you can see, that the entire vector potential is now independent of the z. So partial of the whole vector potential with respect to z is equal to zero. This means that the vector potential is invariant in the translations in the z direction. This is one of the symmetries of the magnetic field. The magnetic field is uniform in the z direction like this. It obviously has a symmetry, a translational symmetry in the z direction. If you move it up and down, it doesn't change anything. You can also move it to the right and left or backwards and forward. And that doesn't change anything either. The magnetic field has a translational symmetry in all three directions. It also has a rotational symmetry about the z-axis. Only about the z-axis, not about the z-axis. Anyway, this magnetic field is uniform in the z direction, and I feel that's four obvious symmetries. So now one of them is reproducing the vector potential, making it the z-translational symmetry. So let's do that. Let's make this assumption, and if we do that, things will suffer because this term goes away. And also, if we do that in this term goes away, we'll see that P3, the operator, is the same as the Pz operator. So let's just change the notation under the same thing. So let's just take P3 hat and just replace it by Pz. That's one of this. And likewise, do the same for the eigenvalue. We'll replace that by Pz. So I'll take this, place this just by Pz. And so now the equation is easy to solve, and gives you the psi of x to y and z is equal to e to the i Pz times z over h bar times an arbitrary function of x and y. Because we're running the x and y derivatives here, this is the general solution for that equation. And so the z dependence is determined, and it turns into a, as you see, a plane wave in the z direction. A logical sense of motion in the z direction is that of a free particle. All right. Anyway, this takes care of the third eigenvalue, which is the P3 equation. Now, the next most easiest one to work with is the x equation. So let's do that. I'll try to start it here. I'm going to shift to my board. So the x, I'll remind you, the x, the guiding center x, is defined as the coordinate x minus dy over omega. And dy is given in terms of the canonical element of the second potential by the standard formula. So in terms of operators, this says that the guiding center operator x hat is equal to the z particle position x hat minus p y hat over omega. And that's the same thing as particle position x hat minus one over little omega times p y hat plus university a y, which we now know is a function only of x and y. And that's an expression for the x operator. I'm putting x at everything where I want to distinguish an operator from the number. And so the equation x hat psi is equal to x psi, the eigenvalue equation, turns into a differential equation, which I can write out like this. It becomes x minus one over little omega. And then we've got minus psi h bar dy, that's equal to p y here, plus university a of x of y. The whole thing multiplies our psi of x, y, z and gives us eigenvalue of x times psi of x, y, z. Now, however, psi is already determined as a phase factor in z direction times an unknown function of an x and y, of x and y variables. We're going to determine this pi function by using these other two eigenvalue conditions. And in particular, by plugging in this expression for the inside, at this point, even to the inside and two sides of this equation, you can see there aren't any z derivatives or anything else that depends on z on either side of this equation. Thus, the phase factor here, the i pz over h bar, pzz over h bar, just cancels. And so, let's look at this up to another board here. Oh, yeah. So it just goes to cancel and ends up being just an equation of y. That's right. I want to do one more thing. Now, having done that, the still is not exactly an easy-looking equation. The main reason is, again, because of this vector potential here. But in addition, there's also this x over here. So before proceeding, let's ask whether we can take this vector potential and use it to cancel out that x. This is there on the other side. What do you think that happens? Is that a supposed to be a x? That's supposed to be a x, yes. A x. No, excuse me. It's a y. Because this is the dy, which came from dy, yes. All right, so can we do that? Well, if we do, if those two terms cancel, then the x here is a minus sign and one over omega and a plus sign here. So what we have to get is a condition that says that x is equal to d over m omega c times ay. Or, obviously, ay is equal to mc over e times omega, which is eb over mc times x. And mc is canceled and eb is canceled. And so this leads to ay is equal to x times b. We do that. Now, if ay is equal to x times b, then this term in the curl equation is just equal to b. So b equals b, and therefore, ea x dy must be equal to zero. And so the logical choice of a x is to take a x equals to zero. And if we do this, of course, these are now, they are functions of x and y, they're really only a function of x. If we do this, we now have a specific to completely determine the age here. And there's a pretty good choice of age. And you can see that the effective potential now has only a y component, and the y component depends only on x. This means that now not only is the effective potential independent with c, but it's also independent of y. So it's translationally invariant also in the y direction, both the y and the z directions. It is not, however, translationally invariant in the x direction. And it better not be because if it were, it would have to be zero. I mean, we're at the constant. If it were constant, it wouldn't the magnetic field would be zero. And the magnetic field was not zero. So there's a moral in this, which is the effective potential can never have all the symmetries of the magnetic field. But anyway, as soon as you set it up, so it has the y and z translationally invariants, but not the x. All right. Anyway, this is not giving us a complete choice of age. And we can now solve this x equation. And so it's going to look like this. It's pretty easy to solve. So it's going to be, I just need to clean this stuff up here. The x in the area will cancel out. And the rest of it is like I get an IH bar over elevator ddy from here to here. And that's almost going to be pi. And that's a y. And that's equal to capital X times pi. That's a y. Capital X is behind the value of the finding certain position of it. And this is easy to solve. It says a pi of x and y is equal to e to the minus i m omega times capital X times y h bar times the arbitrary function of x. So as you see by taking care of the x hat equation we've now gotten the unknown eigenfunction down to just one variable x. And so this leaves us finally with just h per. So let's now write down the h per equation. So h per as an operator is one half m times px plus e over cax plus one over 2m px plus e over cax plus one over 2m py plus e over cay. Of course we've just decided that ax is zero so that goes out. And px here is going to turn into minus i h bar dpx. As far as e over cay y is concerned eay is up here. It's x times b. So this is eb over c times x. Which if I multiply and divide by the mass turns into m omega times x. That's the impartive position x. As far as py is concerned that's minus i h bar ddy. But this is going to act on the full function psi because the z variables don't appear anywhere in h parts. So the z face factor will just drop out. So what we wanted to let it act on is the function phi. The function phi is the face factor of y times f of x. Now the y derivatives do appear. They appear right there. They're going to act on this face factor and bring down a constant. As you can see y is going to be minus i h bar times c. This is the number that's going to turn into a constant factor which is going to be m omega times capital X. Just to multiplication by it. I think it's going to be minus m omega capital X. Right. So after that's been done then there's going to be a common face on the equation h per phi is equal to e per phi. So it's all there's a common face factor. This y dependent face factor is in both sides and we can cancel that. So what's left over when we clean all this up is this. It's minus h bar squared over 2 m times b squared over the x squared. So the equation just in the unknown function f is c. Then what we get is m omega squared over 2 times x minus capital X squared over x. And we get this e squared over x. Now this equation is the eigenvalue equation for a one-dimensional harmonic oscillator. You can see there's a short one-dimensional trigger equation. Here's the kinetic energy. Here's the potential energy. Frequency of omega. The only thing that's a little bit funny about it is the margin has been shifted from x equals 0 over the x equals capital X which is the eigenvalue of the Geithung center operator. So we can write down the energies in the sleuth and the eigenfunctions right away. The energy is e p perpendicular to the distance less than half h bar omega. That's something we do already. And the energy eigenfunctions f of x are going to be able to let me call the harmonic oscillator eigenfunctions just use it in. This is a useful harmony polynomial exponential. But it's evaluated in position x minus x. And so we can now put the whole solution together and we have psi, the wave function which is characterized by the quantum numbers capital X to land on a quantum number in the momentum in the z direction which is a function of x, y and z is equal to e to the i p z z over h bar to the z direction minus i m omega capital X times y over h bar to the y direction times u n of x minus capital X for the X variable. And that's the solution. So those are the wave functions. These are the simultaneous eigenfunctions of these three commuting operators which we chose to set. And because the following isn't a function of those operators in particular just the last two of them p dot energies the total energies are on the level plus the perpendicular energies. Yes? Our argument for why we needed a third of the determinants of the whole thing was that in z I just filled two, three we still had x and y the initial position but we used the pictures of x and now y so we still have another degree of freedom that's our argument to. Well this system has three degrees of freedom. You can see that in the original x, y and z which are just the coordinates of the particle but then we also transformed them to these q's. There was three q, p pairs. It's the same number of degrees of freedom. But in the Hamiltonian there's only two degrees of freedom that appear in number two and number three degree of freedom. The guidance in a position is x and y don't appear because the energy doesn't depend on those parameters. It depends where the plane is from the circle. So in order to find an eigenfunction we need to have a third, we need a third operator between two first two. And it has to be a function of the guidance in the coordinates because the other two would take care of degrees of freedom in number two and three of the guidance in the variables in number one. Does that make sense? Yes. However there was a lot of arbitrariness in the choice of this operator. It could have been any function of the number one variable. It could have been y because it was a function of x and y. It still would have been u to the remaining two. This is a way of saying there's lots of ways of writing down energy eigenfunctions that have the same energy eigenvalue. Energy eigenvalue depends only on the last two fundamental numbers to get into the d z. So this is a way of saying this is a highly degenerate system. Given energy is as many different rates of writing down about energy eigenfunctions. The reason it's so degenerate is because it's highly symmetric. If you include physical effects to break the symmetry, the realistic thing they feel is don't go on forever in x and y. They have to end somewhere in a solid. There's going to be an end of the solid or there's also going to be potentials or electric potentials or something. That will break the symmetry and then the Hamiltonian will defend an x and y. And that will change things. But anyway this is the simplified model. Okay if we take a thought, let's take this way function and sketch it to see what it looks like and try to get some ideas what it means. If I notice the z and y dependence from space factors if you square them and they drop out then all it's left is the only possibility of writing function. If we plot this in the x, y plane and kind of ignore the z direction we can put in a dotted line here for where a particle position is equal to a writing function. Let's also take n equals 0 which is the ground state of the Landau motion then u of n of x and u of n is Gaussian but it's going to be centered on it at x equals x. So you've got a Gaussian that looks like this. It's set to extend out in the y direction. It continues around the y. If I put the z direction it would be a kind of a Gaussian wall that would be out here just in the x, y plane. It's like a Gaussian ground in range which is centered at x equals x. This doesn't look at all like classical motion, classical motion in a circle in the x, y plane situated somewhere like this. What's the relationship between this energy eigenfunction in this class? We might have expected the quantum wave function to be concentrated in a circle. Something of that sort. It doesn't look like that. Why is that? Because we made this an eigenfunction of the operator x hat, the 90 centered position. But the 90 centered y position does not commute with x. In fact, they're like a Q and a B apart from the scaling factor or an x and a B. And so by constructing a state in which the value of x hat from the observable x is known precisely, it means that the value of the course plane y of the observable is completely unknown. And that's why it has to stretch out to the plus and minus infinity in the y direction. It's the uncertainty principle that's forcing this on us. If we had chosen y, not the y, 90 centered y as the set of x as the first observable, then we would have gotten the amount of range going in the other direction. Because then y would have been known that x would be completely unknown. There's a homework problem in which you don't lose either y or x, but you use angular momentum. And if you do that, you find the y function is rotationally varying around the origin. It still doesn't look like the classical motion because the circle in the classical motion doesn't have to be in the origin. It can be anywhere. But anyway, at least it doesn't go to infinity. So anyway, these are all aspects of this problem. There are ways of constructing energy eigenfunctions that do look like the classical motion, but I don't think I want to go into it. Here's another curious thing about this problem. It's that the x-y plane, if I talk about the particle coordinates x and y, of course it's just an ordinary configuration space, and the x-y variables can be used so they can be measured in simultaneous state. If, however, we talk about the guiding center x-y plane, those are operators that don't commute. And in fact, the commentator is i h part times the constant, so they look like an x and a p. In other words, the guiding center x-y plane is really more like a phase space, and it is like a configuration space. And by the way, the same thing is true for the vx and vy variables, which don't commute either. They're like another phase space in a different dimension. In fact, the classical orbit is a circle in velocity space. This is actually like the circular orbit of a harmonic oscillator in the phase space in the classical picture. Well, in any case, to go back to the guiding center variables, x and y, it's like in some respects, it's like a phase space. There's one thing we know about a phase space, which is that there's a concept of a quant cell. Quant cell is a cell that has an area, which is a 2 pi h bar, and it's the area that supports a single quantum state. This is a roughly classical rule. Well, the the x and y variables aren't quite q's and p's. There's one of them, one of the real omega involved in them. So a quant cell in the guiding center plane is going to be a cell in which delta q1 has delta q2. That's the phase space area as equal to 2 pi h bar. So let's talk about quant cells. This is what this would be the condition for it. However, from this formula here, delta q and delta p are square root of omega times delta x delta y. So this is the same thing as m omega times delta x delta y. That's the same thing as m omega is equal to bb over c delta x delta y. And b times delta x delta y is the magnetic flux through the area of the x and y plane. And so what we find is that the flux through a single quant cell v delta x delta y is equal to 2 pi h bar over here times c over e. Which is more usually written this way. It's hc over e. hc over e is the unit of magnetic flux. It's called the flux quantum. And this is actually an important concept in various condensed matter applications. Fun home-effects are where you have not just one particle in magnetic fields. And the quantum states the ground state of the land model fills up the area with one state per quant cell. But the quant cell is an area of the x y plane that captures a single flux of magnetic flux. Which is hc over e. It's a very common concept. The terminology is a little bit bad because the magnetic flux is not quantized. This is just merely a natural unit of magnetic flux that emerges in problems involving magnetic fields. But magnetic flux can take on any value whatsoever. That's the meaning of the flux quantum hc over e. Now that's all I want to say about the charged particle in a uniform magnetic field. I'd like to turn to another problem involving magnetic fields which I guess I'll do here. And this is the hard enough home-effect which is interesting because it challenges our understanding of the global negative-effect potentials in quantum mechanics. The hard enough home-effect may be a fact based on about 1961 so I'm sort of surprising nobody thought of this earlier but there's in a sense there's nothing much to it it's just solving it. Surely nobody invents a new Schrodinger equation or any new physics. It's just solving a Schrodinger equation in some standard way but the results were surprising to some people anyway. Anyway, the situation the situation at all is that of a charged particle moving in the field of a long solenoid a long solenoid a long solenoid is a standard problem in electromagnetic theory so I assume you've seen it already. Let's take the solenoid looking like this it actually is supposed to and ideally it extends to infinity in both directions let's take the z direction just pointing up and put coordinates x and y in this I'll also use I'll also use cylindrical coordinates when it's coming in this is rho, phi, and z where rho is the radius of the x-y plane square root of x squared plus a y squared but in any case the solenoid is supposed to have a current sheet a singular current sheet applied in the azimuthal direction around the solenoid and the result is if you get a magnetic field a constant uniform magnetic field with zero magnetic field outside so the magnetic field is equal to let's just say magnitude b times z hat if the radius is less than the radius of the solenoid let's call that a as the radius of the solenoid in fact while we're making another picture we look at this from above we look at the solenoid from above so the x-y plane is going to be hard to know let's say the radius is a like this so if you're inside the magnetic field is constant if you're outside the magnetic field is zero now in classical mechanics a charged particle moving in the exterior region of the solenoid would not feel the solenoid at all because there's no magnetic field there so there's no force in the particle what's interesting is that there is an effect in the quantum mechanics as we'll see in just a moment to look at this quantum mechanics of course we're running any vector potential because that's what appears in the Schrodinger equation so let's find the vector potential for this magnetic field we'll do this in this manner it turns out that you can find the vector potential which is purely in the 5 direction and we get the vector potential by doing an integral around a closed loop with some radius around the solenoid so let's integrate around some loop a.dl that's equal to the radius the circumference of the loop which is 2 pi a rho times 8 pi however by so stair this is also equal to the magnetic flux it's the integral of b.da over the certainly closed surface and that's the same as the magnetic flux which is enclosed by the loop the magnetic flux which is enclosed by the loop depends on whether the loop is inside the solenoid when it's outside these dotted lines are two examples of loops that we're thinking of here like this if the loop is inside the solenoid then it picks up the flux which is inside the loop so in that case what we get is the area of the loop which is pi over the square times the magnetic field and if the loop is outside the solenoid then we pick up the magnetic flux from the entire solenoid with radius a but nothing beyond that because there's no magnetic field beyond that so in this case it just turns into pi a squared times b for a rho greater than a so dividing by 2 pi rho it's solved that after the 10th hole we find a5 is equal to b over 2 times rho if rho is less than a then b over 2 times a squared over rho if rho is greater than a so we can make a plot it's a function of rho a sub rho we draw in here the radius rho equals a out to in the interior solution the a5 is proportional to rho so it's a straight line like this and then the exterior solution that falls off is 1 over rho so it decays like this if we plot similarly the magnetic field on the same on the same chart same diagram it's constant interior and zero outside so although the magnetic field is zero outside the solenoid the vector potential is non-zero and the vector potential appears in the Schrodinger equation so the question is is that going to affect the quantum mechanics of particles which are outside the solenoid to make this a little more kind of we can imagine that the solenoid is like a hard wall that the particles can't get through so we've got a quantum mechanical problem if I look at it from above we've got a quantum mechanical problem in which the wave function is equal to zero it's a boundary condition in the edge of the solenoid and we're going to be interested in solving the Schrodinger equation in the exterior region we don't care what happens in the interior because the particles don't get in there okay so let's think about doing that well from one point of view you might be tempted to say um in this exterior region we have a non-zero vector potential it's true this is what I just worked out here but this is just one choice of gauge obviously if you have a zero magnetic field the possible choice of gauge is just eight and zero because the current zero is zero so when we do a gauge transformation to get rid of this vector potential end up with eight and zero and then in the exterior region we ought to have just the Schrodinger equation for a free particle and the magnetic field wouldn't make any difference well let's try to do that so the vector potential we've got is this one here this is my component this is B A squared over 2 times rho times pi hat both of my neurons are generated by pi because then pi A squared B is the magnetic flux inside the solenoid so it's pi over 2 pi times rho times pi hat that's what's supposed to be one over rho one over rho times pi hat as we say this vector potential is pure gauge you can get high gauge transformation that can be eliminated that means it must be equal to the gradient of some scalar F what is that scalar F well the scalar F is proportional to the S of the angle phi don't confuse the magnetic flux with the S of the angle phi if you compute the gradient of phi then the coordinates will find it's equal to pi hat divided by rho which is precisely what we have here and so the F here which is the of which A is the gradient to summarize this we can write the vector potential in the exterior region is actually equal to the gradient of the scalar which is the flux of the solenoid divided by 2 pi times the S of the angle phi you'll recall that when you do a gauge transformation on a wave function it goes into psi times e to the i q over h bar c times the gauge scalar F the gauge scalar F is going to be flux over 2 pi times the S of the angle phi and so the result that's not right now so the result is in this attempt to gauge away the vector potential in the exterior region what we end up with is a new wave function which is related to the old wave function by a phase factor that goes like the azimuthal angle specifically it's this the new wave function of the transformation is this e to the i q over h bar c times the function F is where did I write it down the function F is the flux phi over 2 pi times times angle phi h bar c times angle phi flux capital phi times angle phi q over 2 pi h bar c is the same thing as q over h c which is one over what I'll call pi zero where pi zero is the same flux one that we saw in the charge particle in the uniform field this can be written this way psi times e to the i capital phi over phi naught times the azimuthal angle phi like this trouble is you see is that when you go completely around the phi over phi naught is not an injury but this is how we do it in phi yes let me hear that gauge transformation this is the vector bullet inside well I'm glossing over that because I'm saying suppose we just want to solve the Schrodinger equation in the exterior region so I'm supposing let's just try to do a gauge transformation here to get rid of it if you do that what you find is this is the new wave function it's related to the whole wave function times the factor which is the flux in the solenoid divided by the flux prime times the azimuthal angle the new wave function is is discontinuous that's the problem it's discontinuous in the solenoid of its infinite you've got problems with that so let me not go into the problems of the discontinuities of the wave function but this is a reflection of the fact that it's not going to work to gauge away the vector potential in the exterior region even though the magnetic field is zero out there another way of saying it is that if I were able to do a gauge transformation if I were able to do the vector potential in the exterior region and obviously the integral the volume integral of a dot dL would be equal to zero because if a was equal to zero this would certainly be equal to zero but that can't be right because this integral must be equal to the flux which is contained in the solenoid so the result is is you cannot gauge away the vector potential in the exterior region whether it's a hole in its face that's really what makes the difference about being able to do this let me not follow this any further except just to say that the result of this is that there's a phase shift in effect on going around the solenoid compared to the free particle solutions and the phase shift is precisely this that are reflected by two pi going around the solenoid the phase shift is this is the ratio of the flux to the flux one which is in the presence of that vector potential compared to solutions, free particle solutions when there's no vector potential differ by this phase shift going around the solenoid now let me show you how this might have an effect physically it does have an effect physically this is maybe the most dramatic physical effect way of describing the effect of this let's consider a double splitting experiment where we have, let's say, holdways coming here like this two holes and you know what will happen is you get waves radiating from the holes and then you've got a screen down stream from this and as you know you'll get an interference pattern here and the crest are the places where the phase difference coming from the two holes is an an integral of two pi so that the two waves are in phase with one another now let's modify the experiment by putting an arm of bone solenoid right there behind the screen it's part that has never seen the solenoid it's never come closer it's just passing through the holes but because of the presence of the solenoid the solutions in the presence of the solenoid have this extra phase factor a relative phase going around the two sides of the solenoid compared to the three particle solutions and what that means is the relative phase of these two waves is precisely this amount as we say by the flux part of pi naught times 2 pi that's the phase shift the extra phase shift is induced between these two waves and so what happens is the interference pattern moves to the right and the left depending on the flux in the solenoid and if you make this time dependent by changing the current in this we can move this interference pattern back and forth now the point of this is it's not physically observable so this is a physical effect which appears in quantum it's a purely quantum effect it's nothing to know in this classically which occurs in a situation in which charged particles never enter a region where there's a non-zero magnetic field and so this is what Harnoff and Fong pointed out in 1961 and some people said oh it's no big deal it's just a solution to the Schrodinger equation and other people said oh this is the greatest thing so anyway I think there's still kind of a skincey attitude about the Harnoff-Fong effect and it certainly is interesting and this is what it predicts now by the way the Harnoff-Fong effect has been actually experimentally verified as usual in all kind of experiments the real experiments are have to be practical reasons that it's so much different but the Harnoff-Fong effect has been verified experimentally so there's no question that this sort of thing actually really does happen I'm sorry yes would you finish the Harnoff-Fong solution thing you said so this is wrong and now you're saying that it's right no what I said was you see if it were possible to gauge the wave of active potential in the exterior region the Schrodinger equation in the exterior region would be that in a free particle not at all in which case the interference pattern would be just what you always get for what you've always learned but the point is if I put the solenoid in there and turn it on so I've got a magnetic flux then there's an extra phase shift between these two waves so then computing where the peaks occur you have to include an extra shift now it's a relative shift between the two waves it's a relative phase shift and then we might like to think that there's spherical waves coming out of here and these spherical waves of course are the lines of constant phase but who defines the phase it's easy enough that there's no vector potential around you just say for A equals 0 these are the spherical waves where the peaks need to be IKR coming out of here the IKR or R but in the presence of the vector potential then you have to specify a gauge and the phase of the wave function depends on the gauge it changes if you do a gauge transformation so what the wave fronts are depends on your phase convention as weird as it sounds that's true but the phases themselves are not measurable what's measurable are phase differences and the phase difference between these two routes even in the presence of the solenoid depends only on the flux in the solenoid and the flux in the solenoid depends on B not on A so it's a gauge and variant quantity so what's physically measurable is gauge and variant so although the vector potential plays a part of no role in quantum physics and it does in classical physics it still remains true that the gauge convention is still just a convention and that the real physics doesn't depend on that convention the real physics there is a phase shift a relative phase shift between waves only on the flux the flux in the solenoid well so at a time I didn't I didn't really have time to go into what else I wanted to talk about which is how this appears in a pathological formulation let me just say that in a pathological formulation you know you have a phase in the path integral it's either the I over H bar it's a integral over time with a Lagrange let's say T0 and T1 and this is integrated over the path space in the X and T for the real impact integral this is a propagator K and the Lagrangian has two terms it has a kinetic energy which is M times M times X dot 1 over 2 times X dot squared for the path and then it has another term which is Q over C times X dot dot of the vector potential we can call this let's say called 0 which is the free particle contribution of the Lagrangian let's call this LLM which is the magnetic contribution of the Lagrangian so in comparison with the path integral for a free particle of course, let the solenoid in the center and a hard wall on the particles in a sense is only free in the exterior region but we can accommodate that as a path integral just by taking paths that have never interacted with the region so let's look at that here in the path space for integral so then all we've got here is just a free particle of Lagrangian plus a magnetic part of the Lagrangian but the magnetic part of the Lagrangian turns out is is independent of the path up to a it's largely independent of the path I'm sure you want to know about that let's say that here's our hard-on-cold solenoid let's say here's a position x0 here's a position x1 and we've got a path that connects the two of them let's define the magnetic action as being the integral along the path of the magnetic contribution of the Lagrangian let's call this aM it's a magnetic it's a function of the path and it's a t this is the this is q over c times the integral of t0 to t1 of a dotted into the velocity x dot dt well, this turns into a y integral it's q over c the integral from x0 to x1 simply a dot dx and so you see the time where amortization drops out and the magnetic contribution of the action does not depend on how fast a particle is passing actually, it depends on even less than that because if I have two paths let's call them p1 and p2 such that they can be continuously deforming to one another then the magnetic action along these two paths is the same because if I go along one path and back along the other that gives me a loop integral which is equal to the magnetic plus through the enclosed region but this is entirely an exterior region and the magnetic field is 0 so if you stay in the exterior region and you continuously deform one path into another you don't change the magnetic action and so the magnetic action for paths of that sort can be taken out of the path then well, not quite because there are paths for which the magnetic action is not the same it's called as p3 these are paths that relative to the initial path go around the solenoid in fact, if we deform this path continuously we can turn it into this partially follow p1 and make an excursion over here, go around, come back and then follow p1 again so we go backwards and forwards like this if we do this, then the magnetic action along p3 is equal to the magnetic action along p1 plus q over c times the flux in the solenoid and so the result is in the path integral you can break paths up, you can take the homotopy classes these are the classes that indicate how many times it goes around the solenoid and all paths that belong to the same homotopy class give the same magnetic action and that can actually be taken out of the path integral anyway, the result is the path integral becomes a free particle path integral taken over or taken over paths belonging to the single homotopy class multiplied by a phase factor which is precise on this hard-on-bone phase exactly this here times an integer so it becomes e, i, n, 5, or 5, 0 and this is the phase factor that distinguishes one homotopy class from the other anyway, our doing this becomes obvious that the propagators in the presence of the solenoid is not the same as it was in the absence of the solenoid okay, I think that's all so just to remind you that the class this week can all be on code there's a few more homotopes here remember that we're going to say