 I mean, at this point, the only application we've done is to take the semi-classical limit and now I might take today to give you an application that comes from statistical mechanics. So this is some summary special from that. I'll cover it up here to start the main topic. So for simplicity, let's take a Hamiltonian and a one-dimensional Hamiltonian, a kinetic plus potential, like this. And let's imagine that this is a system which is in contact with the heat bath, which is at a temperature T. And so we introduce the usual thermodynamic parameter, beta, which is 1 over kT, where k is the Boltzmann constant. Then you can know the density operator is given by 1 over the partition function, z of beta, times t to the minus beta h. And it's the normalization of the density operator leads to the condition that the partition function, z of beta, is the price of e to the minus beta h. Like this. Now, let's take the density operator and let's compute its, let's write down its matrix element in the x representation between two x points x0 and x. We can make it x0 as an initial and x as a final point. This is the same thing, obviously. It's 1 over the partition function times the x-page matrix element between x and x0 of the operator e to the minus beta h. This operator e to the minus beta h, by the way, let's give it a name. Let's call this the Boltzmann operator, just to have a name for it. I don't know if anybody else calls it that, but I think it's a reasonable name. In any case, what we've got here in this second factor is the matrix element of the Boltzmann operator. Now, this is obviously similar to the matrix element of the time evolution operator, which is e to the minus it h over h bar, sandwiched between x and x0. The Hamiltonian appears in the x program in the same fashion. This, of course, is the propagator, which is what we've been talking about in the path intervals. In fact, these two matrix elements become identical, but if we make the formal substitution of the time set to e to the minus i beta h bar, as to say we'll introduce an imaginary time. And if we do this, then the propagator goes over to the matrix element of the Boltzmann operator. Now, I have no idea what the physical meaning of this is, this substitution. What it means to talk about imaginary time. But it's a, from a theoretical standpoint, it's certainly convenient because it allows you to convert one into the other and transmit formulas. And in particular, for the propagator, the matrix elements of the propagator, the time evolution operator, we have a path integral. And so this allows us to write down just by making this change of variables to this substitution, right, to the positive right on the path interval for the matrix elements of the Boltzmann operator. So, we'll move that forward again to remind you. Here is our path integral in one dimension for the propagator. You see, it's expressed in terms of this parameter epsilon. Now, I'll remind you that epsilon is essentially a time increment. It's t divided by n. And if we take t and replace it by minus i h bar beta, and this becomes an imaginary increment like this, let's write this as minus i eta just so that eta becomes real. And let's so re-express this propagator in terms of eta instead of epsilon just to get rid of complex numbers. So, one thing you can see is this prefactor is going to be real. It's going to be an eta i epsilon goes to eta. The i epsilon goes to eta here. The epsilon squared over the minus eta squared over there. Hard to know. It'll be the same formula. So, to write this in, we have the matrix elements between x and x zero on the Boltzmann operator. It's now given by a path integral. It's the limit as n goes to infinity of a prefactor, which is the mass divided by 2 pi eta times h bar. eta is a Greek letter next to epsilon in the Greek alphabet. It's just supposed to represent a small quantity just like epsilon does. n over 2. And then we have an integral. The x integration is the same as before. It's the integration of the intermediate points. And then we've got the next exponential. And instead of i epsilon over h bar, i epsilon is the same thing as eta. So, it just turns into eta over h bar. So, we have an integral exponential of the integral of eta divided by h bar. We have the sum j equals 0 to n minus 1. Let's write a graph here. We've got m over 2 xj plus 1 minus xj. The integral in the x is divided by epsilon squared, which turns into minus a squared. So, we'll put a minus sign up front. Minus the potential energy to the derivative of xj. That part doesn't change. And so we get this for a path integral for a matrix element. The Boltzmann operator. Simple substitution. Because of the introducing the imaginary time, the relative sign between what you might call the kinetic and the potential energy has changed. So, we come both to the same signs. Let's take the minus sign out here and put the minus sign out front like this. And if you do this, you see that this exponential is now a real exponential. It's not a cilatory anymore. In fact, insofar as the kinetic protocol, kinetic energy term is concerned, it's really a Gaussian exponential in the therapist s, which is the variable's penetration. What's the potential energy to change that? Yes, squared. Thank you. It should be squared. Yes. All right. So, this is the integral. Now, as in the case of the Feynman path integral, this thermodynamic path integral in the exponent here has a sum, which is formally the same form as a Riemann sum, which looks like an approximation to an ordinary integral. And so if we use that to motivate notation, let's take this thing, everything except the h part here, and let's write this out. This becomes in the limit when n goes to infinity, the a to go to zero. Let's, in the case of the Feynman integral, remember we introduced this variable we called tau in the range between zero and the final time t. Let's let the variable in this, in this modifier, it will be u instead of tau. It will go from zero up to an maximum. What we'll do is we'll let u sub j, and the j times eta is just the j step in the stepping through the summation here. And then x of u sub j, we can think of this going over as x into x of u, just as a function x of u. Also, if we set j equals 10, we get u sub h, which is the last value of u, is equal to n times eta. That's the same thing as beta h bar. So u has a range between zero and beta h bar. And thus this summation, which I circle here, turns into an integral. It's zero, the beta h bar, of the quantity n over 2 dx du squared, plus the potential energy v evaluated at x of u, and the whole thing integrated v u. It turns into this integral like this. And what you see is that there's a theory here of something which is, well, it's not the Lagrangian anymore because it's not kinetic minus potential, it's not kinetic plus potential. You could say this is the energy, that's the point one point of view. Another point of view is this is still Lagrangian, but it's in the Lagrangian of the inverted potential where v has gone into minus v, turned the potential upside down. In any case, that's the integrand that appears in the x column. And so this leads into a compact notation for this path integral, which is that there's first of all a normalization constant which is kind of delineatable. This one's here. And then we have zero over a space of paths we call x of u. And the paths that we integrate over satisfy the condition that x of zero is equal to x zero, and that x of the upper limit, which is beta h, beta h bar, is equal to x, which is the final point. And then what we integrate is e to the exponential of minus 100 h bar, an integral from zero to beta h bar of this thing here. Let me just call this L inverted L i and v v here like this, which is actually the branching in the inverted potential of v went to minus v. Or if you like, you can call this a Hamiltonian. I don't care. But in any case, this is the form or a compact notation for this path integral. All right. Now, this type of path integral has a real exponent instead of a oscillatory one. These real exponents are actually much easier to handle if you want to be mathematically rigorous than they're actually called that. It was given a name, they're called the veneratorals. The history of this is the veneratoral was actually known well before the Feynman path integral from the 1920s, having to do with statistical processes and so on. And it's also much easier to handle mathematically because it's a oscillatory case of one interval. In any case, it's formally just an arbitrary continuation of a Feynman path integral to an imaginary time. Now, let's put this to some use. Let's go back to the partition function, z of beta. This is, as mentioned, is the trace of the Boltzmann operator, v of minus beta h. Let's carry out the trace in the X representation. So it becomes the integral of over some variable x0 of x0 sandwiched around the Boltzmann operator. It's the sum of the diagonal elements that we're doing this at a continuous representation. Well, plugging in for our path integral, this is the same thing as the integral over d of x0. And then for the diagonal matrix element, so the Boltzmann operator, I just take my formula here and just set the final axis to the x0. So there's this constant here. There's an integral over paths that go from x0 back to x0 again. The differential path, the quote-of-quote volume element in path space, this d of x of u is really just an abbreviation for this element of this, and these integrations are the intermediate points on the path. And then we have, let's write it as e to the minus whatever h bar integral 0 to the beta h bar dm over 2, dx dx squared plus d of x of u, the whole thing would be u. Okay? So there's an expression for the partition function. As you know, the partition function is oftentimes a central object of interest because if you can calculate it, then you can get equations of state and all kinds of other information about the system. Yes, because the x0s are the same on both sides. That came from the taking place of the thing. If we were really interested in the full naked cell elements of the multinoxial operator, the initial and final points would be the same. But this is an interesting thing because it means the paths, the bin ones, the paths are, they start at x0, and they wiggle around, and you know what's what. But whatever we know is, when the after in a lapse amount of time, quote-of-quote time, it isn't real time, it's just u time, but when you use Bay of H, the path returns to the initial point. So these are actually closed paths now, where it was only a closed path through the given initial point. All right. Now, this integral, this form, can be subjected to a certain approximation. Let's take the case that the temperature T is large. This would mean that the thermodynamic parameter of Bay is 1 over kT is small. And so let me use the upper limit of integration is small. So this path has to come traverses its excursion in a small amount of the last quote-of-quote time, beta h bar. Now, that in turn means the path cannot deviate very far from the initial conditions because if it does, it has to cover a large amount of distance and a small amount of time. That means that the kinetic energy term here, this is all that quotes kinetic energy, kinetic energy term is large. Because of the minus sign here, it means it's going to damp out the ingrain exponentially. So paths in a large excursion in a short amount of quote-of-quote time, beta h bar, don't contribute very much to the integral. And so this allows us to make an approximation where we take v of x of u and just replace it by v of x of 0, which is an initial point. And so that simplifies the path interval because the second term in the exponent here, we can now do the integral integral from 0 to beta h bar of v of x of u. It's just equal to beta h bar times v of x of 0. And we divide by h bar and that will give us a factor here as you can see if v of the minus beta times v of x of 0, the factor that appears in the integral here. However, this factor depends only on the initial point of the path that does not otherwise depend on the path at all. And so this factor can be actually taken out of the entire path interval and inserted back here. Now, of course, it does depend on x of 0, so we can't bring out that integral, and so the result is that we get an approximation of the partition function of z of beta is equal to integral dx 0 of u of the minus beta v of x evaluated at x 0 times our normalization process, times the integral of the path space x of u. These are still closed paths, x 0 goes to x 0. And then we get u of the minus 1 over h bar integral of 0 to beta h bar. And then we have just the kinetic energy term m dx du squared over 2 du like this. Now, this remaining path integral, the potential's gone away, and so this is the path integral for the free particle with no potential at all. And in fact, it's just an analytic continuation of the propagator for the free particle. Let me just use 0 of t as the propagator for the free particle. 0 means free here, free particle. This was this thing evaluated between x 0 and x 0 of what we said beta, excuse me, what we said t is equal to minus i h bar beta. Well, here I summarized, we worked on the propagator for the free particle, here it is, we did it in several different ways, and here it is. So this is for real time, but I know I first of all said x equals to x 0, and that's going to make the x one go away, so this whole thing disappears. All that's left is the square root, and then setting t equals minus i h bar beta is going to make this into a real square root factor, something 2 pi h bar squared beta. So the result of this is that this entire long rule, let's think of this one here, this entire free particle pattern over which I circle here becomes very simply just the square root of the m divided by 2 pi h bar squared beta. So the result is, finally, we get the partition function, this is an approximation of high temperature limit is equal to the square root of the mass m divided by 2 pi h bar squared beta times integral over x 0 of e to the minus beta times the potential of x 0. And so this is a useful approximation of the high temperature limit. Instead of that course, you always do free particles. You find the partition function for free particles or particles in a box carrying this out explicitly. I'm sure you've done that. What this shows you is that it gives you a correction for the potential in case there's an interaction between the particles. I only did this for one big problem, but this would give you a correction for an external potential for a single particle moving in some potential. But by making a multi-particle system, the potential could be the potential of interactions amongst the particles. And this would give you corrections, for example, and do the ideal gas law taking into account the traction or impulses between the molecules. This result was actually a cluster. It was known to Boltzmann probably in the 1880s if you're not sure what they're after. And it's the easiest way of deriving that's used classical statistical mechanics. However, that doesn't mean a pattern will approach as useless because it's not just a hard way of doing an easy problem because this whole approach leads openly to the possibility of making further corrections to this approximation here, where you're taking into account the finite deviations from the initial point. And if you do this, then you're going to obtain an effect, an expansion of the partition function and the powers of data. It becomes like a high-temperature expansion with lower-temperature corrections, quantum corrections coming into it. This is more of a library calculation. That's all I want to say about half integrals. So are there any questions about them? Yes. Is this equation in the box? Yes. So the factor in front of the integral, that is like the quantum conservation N cubed, that is... Well, also we didn't know about H-bar, so we didn't get the H-bar here. That's certainly true. The 2 pi H-bar is the size of a plug cell in face-to-face. And in order to make classical statistical mechanics make some kind of sense, we've had to have some discretization of face-to-face volume. People knew this even before quantum mechanics was around. But this is coming up with the correct normalization because it really is a quantum calculation. The essential part of it, though, is the integral of this exponential factor over the volume of the container. And that's the part that bolts us into the volume. Okay. So now what I'd like to do is to turn to a new topic, which is that of charged particles in magnetic fields. We'll examine several cases of examples of the dynamics of charged particles in magnetic fields. And look at some of the issues which arise there. To begin with, I'd like to start with just some generalities about charged particles in magnetic fields. Let's say that Q is the charge. Let's say you speak about electromagnetic fields which will allow it to be completely general, time-dependent. In that case, the Hamiltonian is, as we've seen before, 1 over 2 Mp minus Q over C, the vector potential, which in general depends on both position and time. One is squared plus the charge of Q times the scalar potential, which I'll call capital Y, as a function of X and T. We've discussed this before. This is not a relativistic Hamiltonian. You'd like to spin it. If we're doing classical mechanics interpreting this as a classical Hamiltonian, then the X dot in the particle, which is, of course, the same as the velocity, is given by Hamilton's equations as dH and p. And that's the same thing as 1 over the mass times this vector P minus Q over C at A. If you're doing quantum mechanics, you may want to define the velocity operator that's dot by the Heisenberg equation's motion minus I over H bar in commutator, and that's for the Hamiltonian. And if you work that out, you get the same answer. At least it's the same formula with the symbols being interpreted as operators. So let's take both of these as motivation for defining what we call the velocity operator in quantum mechanics as being 1 over M times P minus Q over C A. My point here is to introduce velocity operators. We can expect that these velocity operators correspond to the physical measurement of velocity and velocity of this particle in the system. Now, these velocity operators, so let's explore the property of velocity operators first of all. One thing to say is that if I just take this definition of the velocity operator and plug it into the Hamiltonian, I can eliminate this complicated P minus Q over C A, and the result is the Hamiltonian looks like this. It becomes H is equal to 0 over 2 e to the power 3 plus Q times the potential energy 5. And so what you see is that from a physical, this makes the physical meaning of the Hamiltonian much more clear. It's just the kinetic plus potential energy. We're going to say this in other ways. This first term in this Hamiltonian is just a complicated way of writing the kinetic energy. When you have particles in magnetic fields, the kinetic energy is not P squared over 2 m. It's this entire thing. Okay. All right. Now, let's look at the properties of the velocity operator. First of all, as far as the momentum operators are concerned, we know they commute each second. But what about the velocity operators? Let's take, for example, the commutator of Px with Py. Using this definition, this is just work this out as 1 over m squared times the commutator of Px minus Q over C Ax with Py minus, I'll start Px minus Q over C Ax, commutator with Py minus Q over C Py like this. So there's four terms here. And 1 over m squared. So the commutator of Px with Py is 0. The commutator of Px with the second term is going to be, first of all, minus Q over C. The commutator of Px with Ay is minus Ih bar, the derivative of Ay with respect to x. We're going to take these cross terms. The commutator of the Ax term with Py, again, is minus the factor of Q over C. The commutator of Ax with Py is plus I over h bar partial of Ax with respect to y. And then, finally, the commutator of Ax with Ay is 0 because the A's depend only on the x's. They're functions of the x's and the x's equal. So writing this all out more carefully, it becomes I h bar of Q over h squared of C times partial of Ay with respect to x minus partial of Ax with respect to y. And you see appearing here as the component of the magnetic field, which is B, the magnetic field, Bz, the C component like this. The commutator of the x and y components of the velocity is proportional to the C component of the magnetic field. We can summarize these all together by writing this way as the commutator of P i with dj is equal to I h bar Q divided by m squared of C times x 1 i j k d k in the magnetic field. So the vector potential is dropped out and we have just the magnetic field appearing. We see that the velocity components don't commute with the magnetic field around, but it's impossible to find simultaneous eigenstates of the velocity in the presence of the magnetic field. It's also impossible to make simultaneous measurements within the precision of the different components of the velocity. If the magnetic field is zero, then of course that will change just because then the velocity is proportional to the momentum of the commutator of the commutator of the commutator. While I'm writing down commutators, let me also mention the commutator of the x's with the velocities of x i with dj. These are easy because here's the definition of the velocity. The x commutes with A because A is a function of x so it's only the P. So it's the usual x p commutator divided by m. So this is equal to I h bar divided by m divided by delta i j. Anyway, those are some commutation relations among interesting operators that occur in D. All right. As I said, these are just generalities that apply to any problem involving a charged particle in a magnetic field of quantum mechanics. Yes, I'd like to turn to a specific problem which is that of a charged particle in a uniform magnetic field. This is quite important in practice and it has a lot of applications in mathematics to condense matter physics. So let's take the case in which the electric field is equal to zero. The magnetic field is equal to v times c hat. Let's also take the case of an electron in which q is equal to minus e. The contrast of soccer-wise book, I'm going to adopt a convention called a positive number. It's a charge of a proton. I think it's less confusing to do that that'll write q for the charge of any old particle which is likely to be the case of an electron. This is a quote-unquote spinless electron we'll talk about here. That's actually not too realistic but we'll do it anyway because it's simpler to do it. So the first thing I'd like to do is to deal with the classical problem of the charged particle in the magnetic field. Let's do classical mechanics first. According to classical mechanics the mass times the acceleration is the force which is equal to the charge is minus e times the velocity v over c crossing the magnetic field. However, the magnetic field is the same as v times the magnitude of v times z having the z direction. And so by the mass of m this becomes the acceleration which is equal to the quantity eb over mc times the magnitude of v that crosses the velocity of the equation's motion. The quantity eb over mc turns out its dimensions of frequency so we've given it a name here to find omega is equal to eb over mc and this is called the gyro-frequency and it's the frequency of the assertive motion of the charged particle in the magnetic field. So just substitute it in here for omega for that free factor if you write out the Newton's laws now in three components what you get is the x component of acceleration is minus omega times vy the y component of acceleration is plus omega times vx and the z component of acceleration is equal to zero. The z motion is particularly simple the z gives you it's just a free particle in the z direction so z becomes equal to ez times t plus c0 that's the solution in the c direction and as far as the so the only thing that remains is the x and y direction it's a little bit harder but not too much let's take these equations here and let's integrate them once in time both sides the x and y equation so we get vx in the left hand side is equal to minus omega times y plus the constant of all cx and then dy is equal to plus omega times x minus the constant of all cy cx and cy are just constants of integration it's convenient to say cx is equal to minus omega times capital y and c1 by no plus and cy is equal to minus omega times capital x where capital x and y are new constants new constants we'll interpret these constants physically in a moment but if we do this thing these velocity equations turn into as you can see they'll be vx which is of course the same thing as x dot is equal to it's equal to minus omega of y minus capital y and dy which is y dot is equal to plus omega of x minus capital x equations of motion let's introduce some new terminology c is equal to x minus capital x and let's write eta is equal to y minus capital y the difference between lowercase x and y which is the particle of ordnance and the capital x and y which are the constants of integration and if we do this you can see that c dot is equal to x dot plus capital x is a constant and eta dot is equal to y dot because capital y is a constant and so these equations of motion turn into this they turn into c dot is equal to minus omega times eta and eta dot is equal to plus omega times cc simplifies even further and when these equations are easily solved we find that c of t and eta of t the solution is a function of time is a matrix cosine of omega t minus sine of omega t sine of omega t and cosine of omega t both apply on to the initial condition 2-0 and 8-0 this matrix is a rotation in the x-y plane that goes in the counterclockwise direction so if we have a vector of c and a of 2 components it just rotates in a circle in the counterclockwise direction however from these definitions of c and a you've got x equal to capital x plus c and you've got y equal to capital y plus c plus eta but since capital x and y are constants it means that the particle position moves in a circle around the center of capital x and capital y so we've got the motion of the x-y plane looks like this but there's a certain point in the coordinates of capital x and capital y there's a circle around this if I can draw a circle very well which I can't but something like that and if I have an initial condition here of x-0, y-0 then the final condition is on the same circle and the angle which has been swept up is omega t as I said it's in the counterclockwise direction and so this gives us not only the interpretation of this constant motion x and y they're the center of the circle but it also shows that the motion in the x-y plane is just a circular motion going around and around this is for an electron with a charge of minus c if you've done a proton or a positive charge in the direction it's kind of easier just to do it for one charge and if you need to get a charge for that if you do the equations this will change sides all right if we combine this with the z-motion which is the lead of the free particle moving in z-direction you can see that the motion of the particle is a helix through space this helix can be viewed as there being a circle which is parallel to the x-y plane with a velocity dc and as it does so the particle is moving around and around with a uniform velocity around the circle which itself is moving and it's the center of that circle it's called the guiding center and so you can see that the guided center traces out of line it just follows the z-motion the straight line and so the x and y coordinates are the coordinates the guiding center if I put it in the z-components the guiding center is the point in this coordinates capital X, capital Y and the same z as the particle but the capital X to capital Y and the x and y levels so that capital X to capital Y are the guiding center the guiding center coordinates one further remark is that the guiding center coordinates are introduced as constants of integration and here they are and we can take these equations and solve for x and y as a function of the position of the particle and the velocity and if we do this, well again this is a capital X is equal to the lower case of x minus dy over omega and the capital Y is equal to the lower case of y plus the x over omega and this is all classical equations part of the position x and y are both functions of time and so is the particle velocity but when you compute the right hand side these functions of time inspire in such a way to make the left hand side independent of time that's what you mean by constant emotion so if you just know where the particle was at some point in this velocity you could use this to figure out where the guiding center was and that's all I want to say about the classical problem what about the left one? no, because the pencil dropped it was just starting to rotate and then I had to be in front of that's a perversal view it works as a lot of stuff but that's the way it came out here are the two my daughter has one of these tricks you know that you squeeze a trigger and no thing on the end she loves those things okay so that's the that's the classical problem of the charged particle in the uniform magnetic field relative to one problem okay so let's write the Hamiltonian in terms of, first of all in terms of velocity operators just the kinetic energy in order to be squared this is a quantum Hamiltonian you can of course split this up into what I'll call a perpendicular parts perpendicular to the magnetic field and then the parallel part is parallel in that magnetic field like this we can call these two terms h per and h parallel we can do that now the strategy that I'm going to pursue in solving this problem is first of all work out the energy eigenvalues energy eigenvalues are physically measurable so they have to be gauging married they have to be independent of any choice of gauge convention on the other hand as we've seen wave functions do depend on the choice of vector potential because they're they can change under a gauge transformation so in one of the eigen functions you have to commit yourself to a vector potential so the first step will do the easier part which is to find the energy eigenvalues and we'll do this by working with operators the basic strategy is to borrow definitions of operators from the classical solution as far as the perpendicular motion is concerned there's an obvious set of derivatives from operators there's the particle positions x and y the guiding center positions capital lengths of y and the velocities vx and vy these are the obvious ones as far as the perpendicular motion is concerned it's z and vz or perhaps pz if you want the momentum we'll talk about the velocity operators here so we get this best of interest with operators so the name of the game here is just to x and y are defined classically we'll get to borrow the definitions and convert them into operators so the name of the game is just to explore the properties of these operators let's start with the velocity operators vx and vy and in particular let's work out their commutators velocity operator commutator vx and vy well that's a special case oh gosh I erased it it's a special case of the commutator of the components of the velocity in general it would be high h bar q over n squared c times vz I don't copy that now yet because first of all vz is just b for our problem because we're talking about a uniform magnetic field up at the top of the board there and also q is equal to minus e so I get a minus sign with a charge and then we've got eb over mc is the frequency so this is the same thing as minus i h bar divided by m omega let's say m omega that's the way we'll write this so let me put it in the same place hopefully by itself vx, vy is equal to minus i h bar over m omega for other commutators vx with vz and vy with vz are proportional to the x and y components of the magnetic field which are zero in this case so they commute it's only the x and y velocity operators that don't commute now maybe I'll mention the z and vz in the next because they're particularly easy the commutator of z with vz is the this is a position the velocity commutator I spoke of before this is i h bar divided by mass time is 1 yes indeed it should thank you very much thank you very much yes the big is on top alright let's also compete the commutators of the Geising Center z and y let's do that well as I say we're going to borrow the classical expression so this is classically expressed as the Geising Center positions in terms of particle positions and particle velocities let's just borrow this to make a definition of operators capital x and capital y and the quantum problem so if we do this then I won't put hats on at least not yet but these are now operators so this becomes the commutator of z minus v y over omega with y plus vx over omega well through the calculation there's four terms there's first of all commutator of x with y which is 0 then x then there's plus 1 over omega then there's x with vx then there's minus 1 over omega the commutator of v y with y then there's minus 1 over omega squared the commutator of v y with vx then we'll work up these terms commutator of x with vx is i h bar over m because vx is y p over m so this is i h bar over m omega for the first term for the second term there's commutator of v y with y that's the opposite order so that's minus i h bar over m 2 minus it's cancelled we get another factor of i h bar over m omega for the last term we've got minus 1 over omega squared times the commutator of v y with vx well here's vx with v y we get v y with vx we have to change the sign so it's plus i h bar over m and you see one of these i h bar over m and you see one of these i h bar over m omega and you see one of these i h bar over m omega and you see one of these i h bar over m omega let me summarize that over here on some table y is equal to i h bar over m omega and so those are some commutators now it's pretty clear that all of the perpendicular operators are going to commute with all of the parallel operators because the perpendicular ones only involve x and y and the x and p y and the parallel ones only involve c and v c and it turns out I'm not actually going to need the particle position very much as these remaining four operators and interesting ones if we compute the remaining commutators such as x with vx and x with v y and capital Y with vx and v y we'll find that all of those are 0 we'll leave it for you to do straight forward calculations just like I did so what we can say for the commutators here is that all others are 0 all of the commutators now when we erase this commutator calculation we're going to do the same form as the sum of them over there the result of this is is that by looking at variables of physical interest which are suggested by the classical problem we find what it's something which is almost what you call a canonical transformation that is to say we've found new operators that are almost satisfied the canonical commutation relations these are a pleasant rapport in commutation relations the ordinary x's and p's these commutational relations are the x, y, p, j i, h, r, l, i, j i've got the x's and the x's to be with each other and the p's to be with each other these are the canonical or heisenberg or commutation relations now what we've got is almost the same except for proportionality constants that's what we've got over here the x and the v y are the non-zero commutator the x and the y are the non-zero commutator the z and the v z are the non-zero commutator and all the others are 0 so this is bring out we can bring this out more clearly if I make some definitions here to get rid of the constants let's do this let's say first of all we've got in center position capital X the constant both y and y h bar is 1 over n omega here so let me write capital X is 1 over the square root of n omega times the new operator called q1 capital Y is 1 over the square root of n omega times the new operator called p1 then for the velocities their commutator is i h bar times omega over m with a minus sign so let's write v y to be equal to the square root of it's omega over m like omega over m square root of omega over m times q2 and vx is equal to the square root of omega over m times p2 then finally let's write z is equal to q3 and let's write the velocity vz is equal to capital P3 divided by the mass m like this because if we make these definitions then those commutation relations over there become the canonical commutation relations that I've written here except that I better write q's in these instead of x's in these so let me change them into what they are they will be first of all commuted with each other and all the p's can be with each other but the qi pj commutator is equal to i h bar delta i j so in effect there's a coordinate transformation that tip us from the original x's and p's of the particle over to these variables that preserves the canonical commutation relations now let's go back then and express the Hamiltonian in terms of these new operators the important dig in Hamiltonian is the important dig in the kinetic energy the x squared is the y squared as you see the sum of these is the squared is going to be omega over m times q2 squared plus p2 squared that's multiplied by m over 2 so what we get is for the important dig in the part we get omega over 2 times q2 squared plus p2 squared and as far as the parallel part we use the pz definition as p3 over m and this turns into plus p3 squared over 2 and this is the Hamiltonian now expressed in terms of these new variables you'll see from the perpendicular motion it's a harmonic oscillator with the parallel motion it's a free particle so again I'll call this h perpendicular parallel and so you can write down the energy eigenvalues immediately they're going to be 1 plus a half times h bar omega for the perpendicular energy and then since it's a free particle it's just going to be p3 squared over 2 m for the parallel magnetic energy and this depends on two quantum numbers n and p3 there really ought to be carefully and distinguishing eigenvalue p3 from the operator p3 so let me put a hat on that here I'll use my common notation of putting hats on operators when there's some danger of confusion with a corresponding c number or eigenvalue but in any case this gives us the energy levels for the particle in the uniform magnetic field these harmonic oscillators put in the energy levels for the perpendicular motion on a random level and as you see there's a minimum of energies minimum of energies is 1 half h bar omega in the perpendicular correction p3 is what is usual for a one-dimensional free particle but it's the momentum in the z direction because it reminds me of senator plus senator now there's several remarks to make about this result one of them is that this harmonic oscillator can't tell me it doesn't quite look like the harmonic oscillators you're used to they look like this for a mechanical oscillator it would be a p squared over 2 m plus m omega squared over 2 x squared if I were going to use usual one-dimensional notation this is a question between that harmonic oscillator and this one the answer is this if I make a change of variables and my p is equal to the square root of m omega p prime and x is equal to x prime divided by the square root of m omega this is again a canonical transformation that deserves the commutation relations because it just squeezes the x inscriptions at p but it's the same factor when you see this Hamiltonian above p prime is equal to m omega over 2 times x prime squared plus p prime squared so it takes one in the same form that we see here so by a simple change of variables one harmonic oscillator is converted to the other that's one remark the second remark is how did I know that these energy levels of this thing are given by the standard formula plus a half h bar omega that didn't solve any shorting or equation the answer is that we rely on the on the Dirac algebraic method for solving the harmonic oscillator which as you'll recall involves the introduction of a's and a's daggers and their commutation relations and everything followed from that but the commutation relations of the a's and a's daggers follow the commutation relations of the a's and the q's and the p's and the x's and the p's and since we've got canonical commutation relations they're both the same argument applies exactly here it leads directly to this and also by the way leads to the expressions for the eigenstates in terms of products of creation operators acting like ground state and then finally here's a third remark about this is that this Hamiltonian is unusual because the q1 and p1 variables which are here those are essentially the Geiger center positions x and y do not appear in this Hamiltonian so there's one entire degree of freedom there's three degrees of freedom here one, two, and three you can see the number one degree of freedom is absent from the Hamiltonian it's not that uncommon to find Hamiltonians that are independent of one of the q's so they only know the momentum appears free particles, for example and if that happens it means the momentum is conserved because the h is independent of the q that's true in classical mechanics and also in quantum mechanics but what we have here is more unusual because the Hamiltonian is independent not only of the q but also of p the entire degree of freedom doesn't even occur and what that means is that both x and y are conserved but that's just what we saw in the classical solution is that x and y are constantly emotional but this is how it appears in the quantum context this is a sense of constant emotional both of them are constant emotional even when it appears in this Hamiltonian alright well, the next stage which I don't have time for is to tell you the next time is the energy eigenfunctions so that's what we'll do next time that's fine, yes can you explain all this when you talk about relativity it's the same thing it's the same thing it's the same thing