 All right, last night, I ended up the hour by introducing some of the formalism of the spatial three-dimensional quantum mechanics working first with a one-dimensional case. So there's the position I can catch, the resolution of the identity, or the normality of the relation to the continuum sense. One of the results of this is the translation table in capital language and wave function language. This first equation was really the definition of the wave function, the scalar product of the position I can catch with a state vector psi. To go to the government, we actually insert a resolution of the identity and we see that the wave function psi acts as the extension coefficients of the state psi with respect to position basis. And I mentioned also the generalization of three dimensions in which there are now three operators, x, y, and z, x1, y1, excuse me, x1, x2, and x3 is the same thing, which form a complete set of commuting observables in a few dimensions. And these examples here were to go over the standard or any other degree of freedom that might be around besides the position variables. I'll remind you today, the complete set of commuting observables is a set of observables such that you can't find any more observables that can be with the ones that you've got. And then the ones that you've got, you may declare that the simultaneous items cases are one-dimensional. It's in fact the way of constructing the open space. So here when I say that, let's say, in a one-dimensional problem that x set by itself forms a complete set of commuting observables, it means there are no others that can find a commuting less that produce any residual degeneracies. That's the idea. I also mentioned that sometimes we choose not to go all the way in completing a final complete set because we're not interested in the physics that further observables might involve. For example, we can ignore the spin in some circumstances, such as when it's not important energetically. All right. Anyway, this is the basic set of the wave functions. Now, there's a simple consequence of this that I want to mention, which is that if you compute the scalar product of two kets, so let's call them phi, excuse me, phi and psi, scalar product of two kets, we can write this in wave function, write it simply by inserting a resolution of the identity of the sequence integral between the phi and psi. This gives us an integral over x of phi scalar product of x times x scalar product of psi. And you'll see that this is integral over x. The first term is, in this output, the definition of the wave function. The first term is the complex conjugate of the phi wave function. The second one is the psi wave function. And thus, you recover a familiar formula for computing scalar products of wave functions. It all comes out as formulas. All right. Now, the next thing I'd like to do, let me start on the board here with some regular topic. I wish there was a third board where I could keep me separate somewhere else for you, but this is the best I can do over the trade. The next thing I'd like to do is introduce translation operators. Basically, the idea of a translation operator is that it takes your physical system in one location and just rigidly moves it over to another place. We'll do this first in one dimension. And we'll call the translation operator T of A where A is the displacement in S. And T of A is defined in the following way. We define it by giving its actual position I can get S. What it does is it maps it into the I can get in a shifted position, which is the position X plus A. If you want to make a picture of that in terms of the X axis, let's say here's a position S0, the I can get which is, I mean, a kind of position located at X0 can be thought of as actually a state, which is highly localized around X0. If we really, of course, apply this to an adult function, a weight function, concentrate it there. If I have another point, X0 plus A, which is further down in axis, then the position I can get at that point is another sharply peak function. And the idea is that T of X moves one and the other. So it's a logical definition of what you mean by a translation operator. We'll take that as a definition. Now, given this definition, the question that arises is what this definition you see is given in terms of the basis sets of position I can get. The question is what does it do to weight functions? Well, let's set this question up like this. Let's say that Psi is an initial state, and it corresponds to a weight function Psi of X, which by the definition of a weight function is a scalar function that's Psi. Now, let's say Phi be a new state, which is a translation operator T of A acting on Psi like this. This corresponds to a new weight function called a Phi of X, which by the definition of weight functions is a scalar product of the I will get X with the state Phi like this. So the question that is, what is the relationship between the old and new weight functions and the weight function language? Well, this is easily dealt with. What we do is we write Phi of X, which is a scalar product here. Let's take this definition of Phi and plug it into that ket in the right hand side, where upon it becomes actually a major solvent, X on the left is a translation operator in the middle of the original state Psi on the right. Now, let's take this expression and let's insert a resolution of the identity between a T of the A and a Psi using the resolution of the left and the normal here. One tricky aspect of this is that the X which appears here is a dummy variable of integration. So let me put primes on it just to distinguish it from the X which appears here. That's the place that the Phi function is evaluated. And if we do this, and this turns into an integral over X prime, the X prime, and then we've got X, make your settlement X in the left T of A in the middle of X prime on the right and then X prime scalar product with Psi like this. Now, the X prime scalar product of Psi by the definition of the weight function of Psi of X prime. And T of A acting on X prime by the definition of the translation operator is the shift of KET. So this is the same thing as the KET X prime plus A. And so what we get is as we get integral the X prime scalar product of X on the left and X prime plus A on the right times Psi of X prime. However, this scalar product by the worth of the melody relation of position eigencats is the same thing as X prime minus A which banishes the less X prime is equal to X minus A as you can see. And so doing the integral is easy and it just gives us Psi of X prime minus A. So there's our answer. The new weight function is equal to the old weight function evaluated with the shifted number shifted position X minus A for the minus sign here. I'll be able to write this over here as another aspect of the translation operator. I'll write it this way as a P of A acting in the function Psi, that's what I'm calling the new function Pi, evaluated with a point X is equal to the old function Psi evaluated with X minus A the inverse shifted function. So that's two expressions for the translation operator. One in KET language and one in weight function language. So this is the difference in sign here plus and again in the minus of the weight function. We have several comments I want to make about this weight function result. In the first place what about this minus sign? Is it being the weight function being moved backwards? The answer is no, the weight function is being moved forwards because if I plot a weight function, let's say the old weight function looks like this let's say this is Psi of X so it's got some loss like this. The weight function which is 5X in this notation over here 5X weighted weight at once is the same since it's been rigidly translated. In the forward direction I take two points on two weight functions that are the same point you'll see the motion forward by the motion A and this is the correct picture in this picture here you must have the minus sign in this formula just a little bit but I'll just mention that there's a useful mnemonic to think about in a formula like this which is to you see X here is really a dummy variable if I replace X by X plus A in both sides of this equation it turns into this it's T of A acting on Psi evaluated in X plus A is equal to Psi evaluated in X and the mnemonic that's sometimes useful is to say the value of the new weight function is equal to the value of the old weight function at the old point if you remember that you'd write this down without the trouble without you to remember where the minus signs go ok so that's one counter these operators move and push forward questions rigidly forward another common is that my use of parentheses here a lot of books would write this equation this way without that X to 7 parentheses and write T of A acting on Psi that's what they say they write it that way and the problem with this notation is the following is that Psi is the weight function Psi of X is the value of the weight function which is just the number for giving the value of X so what does the T of A act on? is it acting the weight function or does it act on the number? well it obviously acts on the weight function and that's why I put the value of X as the old weight function and that's why they say ok so that's a second comment about this alright in any case these are two useful expressions that go with counterweight function language for the translation operator now next I'd like to consider the translation operator in the case of infinitesimal or small translations so let's say that A is small in some sense then we can take T of A and in fact if they extend in the Taylor series of the first quarter turns it becomes T of 0 actually I just realized this is not what I want to do next what I want to do next is to develop the properties of the translation operator so let me do that in all of you do it in the same order since there's probably space for it there's properties, four properties of the translation operators it's all less terrible simple properties the first one is that if you evaluate it in 0 you get the identity operator if the displacement is 0 then cat X just goes into cat X all the cats are left the same so it's obviously the identity the second one is the translation operators T of A are unitary but why is that it's easiest for me to do it over here if I take an old weight function of I of X in a computed square norm that's the integral over X to the absolute value of psi squared that's of course the total probability the new weight function after the translation is psi of X minus A let's compare this to let's say C of X let's compare this to the integral over X of psi of X minus A quantity squared this means that dx prime is equal to dx what we got in there these integrals are from minus infinity to plus infinity I'll just assume that that's the range of integration unless I put something else up on there in any case for this change of variables the new integral becomes equal to the old integral so in fact these two expressions are the same the result is that the translation operator preserves the norm of the weight function this is true for all weight functions now it's one of the homework problems for this week to show that in those circumstances an operator that does this is unitary any operator that preserves the norm for all weight functions is unitary so the translation operator is unitary doesn't change probabilities the third property is the composition property if I take T of A and multiply times T of B that means supply T of B first and then T of A second is some weight function and the answer is it's T of A plus B multiplying translation operators is equivalent to just adding these placements I think that's clear from the picture here as you move it down the axes you can also prove it from the definition of the couple however A plus B is of course the same thing as B plus A so this product of operators is the same thing as the product in the reverse order in other words the translation operator is commute in fact they're called a group which is what you call an A-billion group the word A-billion in the context of groups it really means the same thing as commutative so this is a commutative set of operations of commutative group the fourth property I want to mention is the T of A inverse is equal to T of minus A which is clear because if you advance forward by A and go backwards by A you come back to where you started from however if you want to really prove it then start with a property number three and set B equals to minus A T of A times T of minus A equal to T of zero which by property one is identity that gives you number four also if you have number two then T of A is unitary T of A inverse is also the same thing as T of A banger so I can add that here like this and these are the four properties so the translation operator is what I want to summarize following these are probably definitions now the next thing as I started to say a minute ago let's look at a special case in which the translation is small let's take T of A and effectively expand it in a Taylor series about A equals zero first term is T of zero and the next term is A times the derivative of T with respect to A evaluated in zero and in its higher word terms this is a type of an approximation for the translation operator for small displacements this operator DTDA evaluated in A equals zero no longer depends on A because it's evaluated in A equals zero so it's just an operator without any big parameters at all let's give this operator a name let's call it minus I times K where K is a new operator or if you like I'll do it this way let's define K as an operator is equal to I times DTDA evaluated in A equals zero like this or I'll put it this way it's evaluated in A equals zero let's make this a definition K here is an operator you have to remember that the case of an operator whereas A is just a number if I want to really emphasize that I can put a hat and a K I'm using hats here to indicate operators but usually only in cases where I think there's going to be a confusion where you might forget an operator sometimes you can contrast it with a classical object or a C number which is the number that's a benign value problem alright anyway this is the K operator now as you see I split off a factor of I relating K to the first derivative of the translation operator and why do they do that the reason is to make an operator K and K hat to make it a mission operator this follows from the unitarity of fatigue operators if I take a T here's my expression for T of A underwriting again T of zero of course is the identity by property number one the first direction term now becomes minus I K A plus I word of terms like this now if we take two of A dagger we need to dagger both sides daggering one gives me one daggering minus I gives me plus I daggering K gives me K of dagger and A is the real number because it just goes into itself so this is the expression for T of A dagger let me take the same let me go back to the first equation again and just replace A by minus A to get T of minus A well that doesn't change the first term but the first direction term is the sign change that becomes plus I K A plus I word of terms now because of property number four here two lines have to be equal to each other the first term is already equal and in the second term they're all equal except for the K and K dagger so the equality of these two since these two things are equal as you see this implies K is equal to K dagger and thus we have that this is a permission operator and that was the reason why I do have to remove the I from it to make a permission operator we're particularly interested in permission operators because they correspond to things that are observable and we'll come back to this in the rest of the unit as well as the observable corresponding to K that's the whole point of the fact of these translation operators alright so in any case to summarize then just to repeat this equation here we've got T of A is equal to one minus K A plus I word of terms like this the K is related as you see its definition is related to the first derivative of the translation operator at least higher order terms of course involve higher derivatives it turns out that all these higher derivatives can now only be expressed purely in terms of the first derivative and here's how you do it we're going to have an differential equation for T of A again this reminds us of the first problem where we solve for operators if I use the differential equations this is capital C here I'm going to evaluate this derivative by using the definition of the derivative provided like this is the limit and epsilon goes to zero of T of A plus epsilon minus T of A divided by epsilon now the T of A plus epsilon which appears here if you look at the composition property which is number three in the translation operators this can be written as epsilon times T of A and if we do this both of the terms in the numerator here have a T of A on the right hand side which can be factored out to the right so this turns into the limit is epsilon goes to zero of T of epsilon minus one over epsilon that whole limit multiplied on the right by T of A like this now this remaining limit which appears here by the definition of the derivative let's call it DTDA evaluated in A equals zero that's the meaning of that derivative there however by definition of the k-operator this is the same thing as minus IK or k-coffin that's going to happen and so what we get is is DTDA is it from the minus IK times T of A it's a typical differential equation and it has a solution T of A is equal to B to the minus I KA multiplied times T of zero in the initial condition but T of zero is the identity so I'll just take it off again and the result is that we have a closed form for the translation operator as exponential of the displacement times this operator K hat with this minus sign that appears here alright this means in particular that this Taylor series is started here by getting the first term the w comes easy to write on the higher terms the next term is going to be the minus sign of the 1 hat k squared A squared plus and so on you can now write out all the terms of the Taylor series if you're in trouble it's just an exponential series if I do the same procedure in three dimensions then we have a function called a T of A vector which acts as a white function of X vector and produces a sign of X vector plus A or acting on the cats the main vector acting on the eigen ket X gives us the eigen ket X plus A this is just the obvious three dimensional generalization of what I've done in before for the definition of the T there now there are three K operators as well we can just find K sub I then would be equal to instead of the ordinary derivative of respect to A if we come as equal to I times the partial derivative of T of A vector with respect to A sub I we can evaluate it in A vector we can zero yes should there be a minus sign of X thank you so in my three dimensions we actually get a vector of K operators there's three of them and they're defined this way and we also get a similar argument over here we get an exponential series for T of A which becomes E to the minus I that's right it is K vector but let me write this way K vector dotted and K vector to remind us that it's an operator and this is the three dimensional generalization of the exponential series for the translation operator alright now since we've got three operators here these three K operators there's a question of what is the commutation relations in terms of the commutation relations followed from the properties of the translation operators of cells so let me show you how this works there are a variety of ways of doing this but I'll show you one way and it is the balls of pattern would be displayed in the course so suppose I take T of A vector now times T of B vector a particular product with two translation operators in three dimensions we'll expand both of these out of the Taylor series it turns out I need to go out to the second order so for the A one minus I just write this A dotted with a K I'll leave all the halves here the second order is minus one path A dotted with a K squared minus a squared plus so on the B I swear this is the same it's such a place A by B minus I B dotted with a minus one half B dotted with a K squared and if I do the algebra multiply these two series together the zero order term is one times one the first order term is minus I A dot K dot times one or one times minus I B dot K there's two of them they combine together to give us the minus I the vector is sum of A plus B dotted with a K we need to write a quadratic order to know where I'm going so a quadratic order there's actually three terms let's see the minus a half A dot K squared times one the middle term times the middle term and the left term times the right times three of them we write this right here minus a half A dotted with a K squared minus I A dot K times minus I B dot K is minus A dot K times B dot K and then the last term is minus a half B dot K squared now if I write these in the reverse order T of B times T of A I get the same answer except I just need to swap A's and B's so the one goes into one it just stays the same and this first order term swapping A and B doesn't change that so that's as far as in the quadratic terms the A dot K squared and the B dot K squared just exchange each other so the sum of them is the same we just copy those down the only one that's really different is this the old term here in which the order of these factors gets reversed minus B dot K times A dot K however the translation is huge so T of A times T of B is equal to B times T of A and if I subtract it to a value of zero so everything cancels out except these middle terms zero and so by subtracting these two what I find is the denominator of A dot K of B dot K is equal to zero now this is true for any choices of the vectors A and B so if I choose them to be in the vectors along the x, y, and z directions then I get components of K in the quadratic components x, x, y, y, z, any of them and the resultant is something that maybe I'll put up here the resultant K operators defined in this way satisfy the computational equations that K, I, and K and J commentators equal to zero these are communal operators what does the K operator do in wave functions well maybe I'll say what it does to the cat's person it prevents somewhat easier let's work in three dimensions what does K hat I act when an eigen kind of position what does that equal to by the definition this is a partial of respect to A sub I evaluated in a measure of zero of T of A after one can of x and separately multiplied by I that's equal to I times the derivative of respect to A sub I evaluated in A from zero of x plus A well differentiating respect to A sub I is the same for this thing since it depends on the sum x plus A it's the same for the chain rules it seems to differentiate with respect to x that's what I said A sub zero of A goes away and so the result is this is I times the derivative of respect to x of I on the cat x like this we summarize this over here by writing this in vector form let's say K operator acting like cat x I'll put a half here to remind us that that's an operator it's equal to I times the gradient of cat x this is a strange looking formula it has any field of gradient of x means but remember that these are vectors and the derivative is a limit of difference between vectors so it's meaningful to do these derivatives I'll show you another formula now which is a little easier to interpret I can point and think about that but instead I'd like to go on to in effect a wave function version of this I'm giving both cat and wave functions some of these formulas in this one this is the cat version now let's work on the wave function version so the problem then is if I take K I acting like a wave function of psi and I evaluate that at x what is that equal to if I follow the same procedure as on the line above this is I times the partial of respect to A sub I of the translation operator T of A acting like a psi evaluated at x and this is from set A to 0 again and this is I E partial of respect to A sub I evaluated at A to 0 applied to using the translation identity of the upper board there of psi of x minus A so again by using a chain rule differentiating the respective A with respect to x except you get a minus sign from the chain rule and then you set A to 0 so the A goes away and this becomes minus I derivative of psi of respect to xI and evaluated at x and that's when you close this K operator so allow me to write this this way as K operator and I'll put a hat over again acting on a wave function psi evaluated at x minus I times the gradient of psi maybe it's starting to look familiar with the K operator finally from this expression here it becomes possible to work out the commutation relations already between the K's and the K's and the x's and the x's the x's and the x's got solid up but here they are but those the x's commute with each other now the remaining commutator commutation relations there's the commutation relation between an x and a K and if we work those out the easiest way to do it is use the wave function version of it so if I take x hat I times K hat J acting on a wave function of psi minus K hat J times x hat I acting on a wave function of psi this is all evaluated in some position this is on the left hand side becomes x hat I times minus I well x hat has a setting of multiply by x, so just x hat and alpha hat now and then we get minus I partial with respect to xJ acting on a side minus I partial with respect to xJ times xI for the one side and if you work this out the one that's left over is I delta IJ multiplying psi so the result is you go back up here this is equal to I times delta IJ and now we've got a complete set of commutation relations among position operators in this new operator K which came out of displacements, translations in this space you can no doubt see the operator K as closely related to momentum based on what you know already about momentum operators in quantum mechanics let me make a slight regression into the subject of momentum of classical mechanics and then we'll come back to the question of the momentum of quantum mechanics so let's talk classical mechanics for a few moments in elementary courses in classical mechanics the momentum is defined as the mass times the velocity for a single particle however the more advanced courses the momentum is defined as the derivative of Lagrangian with respect to the velocity virtual light as x dot Lagrangian is a function of x dot and it may also depend on time here I'm speaking of Lagrangian with a single particle moving in three-dimensional space for example now the question is are these two definitions the same and the answer is it depends on the Lagrangian this definition by the way we'll call this the kinetic momentum we'll give it a name and this one we'll call the canonical momentum and the question is is the kinetic and canonical momentum the same or as I say it depends on the Lagrangian the common Lagrangian which applies in the case of in the case of which the forces are derived from the potential the Lagrangian is the kinetic energy it's minus the potential energy it's 2 minus v it's more explicit it's going to have an energy times the velocity x dot square in a way like that minus the potential energy which depends on x this one you may also depend on time so I'll allow that a time dependent theorem in any case this sort of Lagrangian applies when the force on the classical particle can be written as minus the gradient of the potential as you say if there's a potential function such that this is true then this Lagrangian describes the system and as you can see the canonical momentum p which is the lpx dot only comes to the kinetic energy term it's just equal to m times x dot in that case it agrees with the kinetic momentum this is a very common case of practice that the force is derived from the potential it is however not universal and in fact in any case it's always an approximation in this course we'll be interested later on in number one in cases where this is not true for example we have magnetic forces but also we'll be interested in the corrections to this even where this is approximately true this means for example spin argument relativistic effects, spin effects and so on and so it will not be true in this case either although this is a common case it's not universal to take another case where the kinetic and canonical momentum are not the same let's include a magnetic field in the action of this particle if we do then we need to add another term of Lagrangian which is q over c times the velocity x dot the effect of the potential of the general is a function of both position and time so already like that this is really probably a charge particle now moving in a given magnetic field if we do this then you see the canonical momentum requires an extra term if we actually respect x dot here we get a plus q over c and we get a plus q over c so the problem now is this is a case of which the kinetic and canonical momentum are not the same so one question is which one of these is the one right one to do is the one that's talking about momentum in particular in quantum mechanics there are no doubt familiar with the fact that the momentum operator of wave functions anyway is minus by h bar gradient see later exactly what that means but in any case the question arises is a kinetic momentum or is it a canonical momentum that this represents which one is it this is quantum of course that's classical but which is it doesn't include the vector potential the answer is yes this is actually a quantum operator really corresponds to a canonical momentum not the kinetic momentum there are several ways of arguing with the canonical momentum is actually more fundamental than the kinetic momentum just to mention one example is that if you have a system which is invariant of translations which means there's a symmetry a translational symmetry well symmetries in physics get rise to conservation laws in fact it's believed that all the conservation laws of physics, all the important ones but I can all the ones I can think about are always connected with symmetries conservation of energies connected with time translations so if you do a conservation momentum it's one reason why I started talking about translation operators the moment we go but what is the conservation law corresponding to translational invariance the answer is it's the momentum but which momentum is it the answer is it may not make any difference of course it's a simple analogy but in more complicated cases it does make a difference in which one is it the answer is it's an economical momentum that's conserved when you've got translational invariance well translational invariance of a system means that the energy of a particle doesn't depend on where it is in space Hamiltonian is an independent position this applies in particular to free particles and that's why the free particle Hamiltonian depends only on momentum the energy of a particle is also a dependent position in a uniform magnetic field because it's uniform it's the same everywhere and the energy doesn't depend on where you are so in that case you don't have to worry about this is the last term ok so anyway those are some of the aspects of the momentum in classical mechanics and the suggestion is this economical momentum because it's the generators of symmetry it's the momentum which is conserved in translational invariance the momentum is the generator of translational symmetry that's what I mean there goes hand in hand with the fact that it's the quantity that's conserved and the system is translational invariant so this is what should carry over in the quantum mechanics is the generator of translations and the operative that's conserved when the energy is in the dependent position alright so for all these reasons we suspect that the momentum reinterpreted as a quantum mechanical operator remember the measurement of physical observables corresponds to operators in our cut spaces it's one of the postulates but this must somehow be related to the operator k that came out of the translation operators however these two operators can't be equal in one another because they don't have the same dimensions k's got dimensions of inverse length and the momentum's got dimensions of momentum mass times distance over time so to make a reasonable guess we assume that instead of these things being equal they're proportional we'll write the proportionality constant as h bar that has to be the constant of dimensions of action the dimensions are not right now here I'm not talking about operators so I'll put a hat on this this is an assumption alright what is an h bar the constant h bar does not appear in the postulates in quantum mechanics this is the first place where it appears in this development that we're making in fact the logic of the development that I've been presenting leaves open the possibility that different particles might have different h blocks this is kind of a crazy idea it's very hard to believe on any theoretical grounds that this could be true but it is something that can be tested experimentally one reason this in relation to people's h bar k implies for plane waves which are eigenstakes of momentum so I'll mention that the wavelength lambda is 2 pi h bar divided by the magnitude of momentum and so this can be tested experimentally one can find by measuring lambda and p together one can determine h bar these experiments have been done and they show that the h bar is the same for all a whole part of this I'm just kind of comparing this with what I have said I understand how you write everything and I have three mentions how does k depend on the weight of the k and the momentum how did this come out well in a way a jump could have gone I wouldn't have in the Foss equation if you took the h bar k it just missed here we dragged it all on how we just stated that this would have been the weight of the k well this was a jump let's forget this with the broy relation do you just want to look at this I just want to know how that k turned into k ok here's how it happened for the translation operators we discovered this vector of commission operators k and they can leave with each other now permission operators are supposed to correspond to observableness so the question is what is the observable that corresponds to k the k that appears from these translation operators is the generator of translation that just means that the translation operators are exponential like this there's a similar form of classical mechanics because classical translations it's really an exponential iterated Poisson brackets in the momentum I won't go into that because this is in the course of classical mechanics but let's just say that in classical mechanics the momentum is a generator of translations in a sense involving a very similar relation so it's a guess that the operator the observable that corresponds to k is related to the momentum and if it's true that they can't be equal there's got to be a proportionality that for the best type of mentions in action we'll call it HR that's right we're asserting it's a guess that can be tested externally and the experiment is based on involving interference with wave functions which is what it actually measures the land so this is the beginnings of HR several years ago I taught a special course kind of funny course on the history of the Planck radiation law it's turned out to be a huge amount of work I never would have done if I didn't know to give back to the old 19th century physics but it was very, very interesting how it worked out and what Planck did in the 1890s in analyzing the black body spectrum and the experimental data coming in and even before Planck had the right formula for the black body spectrum in fact he had a wrong formula at first there was the there was a formula that was wrong there were two different ways of looking at it but in any case even these wrong formulas had a constant in them which was what we now call HR version of HR and Planck recognized that this was a universal constant that it was going to play the same role in physics with the speed of light and it was constant gravitation and so we started putting two and two together about what you could do with this fundamental constant in terms of finding conventional quantities or fundamental scales for distance, energy and so on it turns out with these fundamental constants HRG and the speed of light you can get a fundamental unit of distance of energy, momentum all these things become fundamental units which are now called the Planck units for example the fundamental unit of by distance is the Planck scale it's about 10 to the minus 33 centimeters this is a enormously much smaller than any distance scale that can be explored in any modern accelerator which gets down to 10 minus 15 centimeters something like that maybe we'll add a little less so it's way beyond anything that's experimentally accessible but nevertheless the belief is that at the Planck scale which quantum mechanics, relativity and general relativity that's the gravitation all become united into some kind of universal quantum gravity version of theory there's a great deal of speculation going on these days what happens at that length scale in any case I just want to say that the fact that HRG was a universal constant was appreciated even before Planck had been done in the correct form of the radiation law alright so that's the HR now given this we can express our operator K and give it a K and express it in terms of momentum and if we do we obtain modified versions of these commutation relations the first one just I multiply my factors of HR the components of momentum can be used with each other and the commutation relations between position and momentum are the connoisseurs we're dealing with IH of our delta IJ like this and the XX commutation relations are the same so in this way we have derived the Heisenberg form of commutation relations connected to position and momentum so if we're going to use the connoisseur momentum which is based on Lagrangian how will we handle a dissipative system because you can't write a Lagrangian to that right you can't able to dissipate the systems in quantum mechanics without but you see dissipated system means an increase in entropy so it can only be done in terms of the density operator anyway there are ways of doing this as these naster equations as long as they involve an evolution of the density operator but there isn't a Schrodinger equation but Schrodinger equation involves pure states that's part of the answer actually we won't deal with dissipation in this course dissipation comes from an interaction with the environment I made in that dependence on the on the notes on classical mechanics where it talks about dissipation we don't even have the Lagrangian description that the system is dissipated but we don't have a Hamiltonian either so you can't quantize if you get a quantum amount of time you know it's an interesting thing if you look at YouTube there's these Feynman discussions he talks about this there's that if you look at the ball the floor of course it bounces but it dissipates and gradually down south goes to a standstill whereas because whereas we know because of thermal notions the individual atoms are jiggling around all the time but they don't damp out so why does the ball damp out but not the atoms and the reason this is the ball has got all these extra degrees of freedom which are the atoms in the floor and so on they get randomized and disappear and it doesn't really disappear but the thermal notions but the atoms don't have any further degrees of freedom they are the degrees of freedom they're down to just a small number of degrees of freedom like the electronic and rotational degrees of freedom in molecules well actually they do have other degrees of freedom they've got for example electronic degrees of freedom they won't get excited at low energies but that's because of the quantum mechanics the energy is too high it organizes the states it organizes them by energy levels and as a result you can't get the dissipation of energy you can't get energy flowing from let's say the ground state of hydrogen and anything else because there's nothing else there's no other energy level discrete energy levels around it can go into if you heated the hydrogen atom high enough gave it enough energy so that we could make other transitions like emitting photons dropping in the lower states then it will do that and that's very similar to the way the bouncing ball organizes energy also this patient is a whole subject we won't do much about dissipation alright okay so so there's so this is the this is the momentum operator now let's say the momentum operator three of them actually in three dimensions is permission let's find a tigant function it's eigenstates if this z is through just one dimension if we look at the momentum operator p hat packed by the wave function psi of x and I really should put parentheses around it like this this will be the eigenvalue p times psi of x this is the same thing as minus i of h of r d psi of d x this gives the obvious solution that psi of x is equal to let's put a p on the side because that's actually the eigenvalue psi of p of x is equal to e to the i of p of x over h of r times the constant, the localization constant and this must be the same thing as the xp if I use the notation kp to stand for the eigenstate of the momentum operator to the eigenvalue p and if we wish this to be these momentum eigenstates would be normalized to p scalar product p prime with delta p minus p prime this is the continuum normalization then this constant that appears to be 1 over the square root of 2 pi h r and so we obtain the matrix only connecting x and p position of momentum representation this turns out to be extremely useful for converting back and forth in certain resolutions the identity this leads to the Fourier transform connecting the position of momentum wave functions and so on I won't go over that again because I actually did that back when we were talking about talking about the mathematical formulas and quantum mechanics I'll just mention one further thing is the trivial generalization of this if we do this in three dimensions the scalar product of x with p is it's an obvious generalization not just becomes a dot product p dot x divided by 2 pi h r now into 3 half power to the right normalization to the left normalization and this is the end of the main elements of the spatial degrees of freedom and position of momentum representations of one device ok, so that's that's all for today