 OK. So let's get going. So yesterday we introduced these orbital angular momentum operators, the three of them, the three components of x cross p and divided by h bar to make it dimensionless. And we can make the total angular momentum operator L squared by squaring the individual components and adding. We established these results that the commutator of, for example, Lx with y, is i times z. So if you, similarly, if you do a commutation with Li with, well, Li with pj, I haven't written this down here, you find i times pk times epsilon, sorry. So for example Lx times p, commuted with py would be i times pz. Basically what we're learning here is that L commuted with a component of a vector operator gives you the third vector operator in that set. And from that you can go on very easily to show the demonstration is exactly the same as what we already did with the total angular momentum operator that any of these orbital angular momentum operators commutes with scalars like x squared or p squared or x dot p. Also L squared commutes with any of these things, any of its components. So we have a situation so far which is precisely analogous to the commutation relations to the total angular momentum, what we have so far called the angular momentum operators. So, for example, ji, xj commutator is i sum over k, epsilon i j k x k, et cetera, et cetera. So we have that ji comma jj is equal to i sum over k epsilon i j k j k, which mirrors one which I didn't write down up here and should have done, which is that li comma lj commutator is equal to i sum over k epsilon i j k l k. Which is really a generalisation of this rule, I mean an application of this rule here to the vector lj being put in here so lk appears over there. It just says that l is in fact a vector operator and similarly this is a generalisation. So we have this exact analogy and the question obviously arises, are they the same operator? Is l the operator which, are the operators lx, ly and lz, the operators that generate rotations that we inferred had to exist just by thinking in the abstract about rotations at the end of last term? And so the answer to that question is no and here is the demonstration of it. Let's calculate j squared comma li. Let's work out the commutator between the total angular momentum operator and one of these components. Then this is going to be, this is a commutator for a product and I need to write this, this is a sum of squares so I need to make that explicit so I write that as the sum over i say of ji squared comma, oops, no, no, I won't do because I've got i busy, j li and now each one of these is just a simple product so we use the rule for taking a commutator for a product so this is equal to the sum over j still of, sorry that's a j, what we're summing over. I need to do, I leave one of these standing idly by whilst the l operator commutes with the other one and then I have the l operator commuting with the other one, the first one while the other one stands idly by in his position. Now we know what the result of this is because this is a vector and we know we established last term what the commutator of j with the component of a vector is, it's going to be the third member so this I can replace by an epsilon ji say k so this is going to be a sum over j and then there is going to be a sum over k coming up of jj, there will be an i from the commutator so jj that's this one here, make sure that's clear this is going to be epsilon ji klk so that's the fundamental rule we established that the commutator of this with any vector gave you the third component and then we're here we will have the same thing epsilon ji kl times lk times jj in the back so we can clean this up into the i times the sum of both j and k going from over x, y and z of epsilon ji kl times jjlk plus lkjj and so this is the sort of thing which often vanishes because this is anti symmetric if you swap those two over j and k you get a change of sign and if this was symmetric we would get a change, oh sorry, and if this was symmetric in j and k unless you could swap them over you got no change of sign then you just got the same thing back then we would have, then we would have zero but this is not symmetric if you interchange the order of these, if you swap the indices so under a swap what does this go to it goes to jklj plus ljjk which is not the same as this so this thing is not equal to zero but we know that j squared does compute with all of its components so that sort of suggests that these are not the same thing so let's try and understand what these operators, these orbital angular momentum operators and really understand their relationship to the things that generate rotations so now we need to talk a little bit in abstract about rotation around a path, sorry motion around a path so if we just have a translation, we've already studied translations last term through one displacement A that's generated by the unitary operator which we called u of A which is e to the minus ia dot p on h bar that is the operator which generates out of a state that we would have if our system was shoved along the bench by or was at the different location, a distance, a vector displacement a away so let's consider the following let's make a series of displacements, here's a 1, here's a 2, here's a 3 and so on we're going to make a path by doing a series of displacements then we have that u total is going to be the product of this displacement and this displacement, this displacement, this displacement so it's going to be u A4 operating on the result of using u A3 operating on the result of using u A2, u A1 each of these things is an exponential so this is e to the minus ia4 dot p over h bar e to the minus ia3 dot p over h bar etc and the operators occurring in these exponentials all commute with each other because all momentum operators commute with all other momentum operators they commute with themselves obviously and the px commutes with py etc so when we multiply these exponentials together we have the usual magic of exponential functions we don't have to worry that those are operators so these are operators but we don't have to worry about that because everything commutes so this can be written as e to the minus ia the sum of these vectors summed over i dot it into p over h bar so the operator that generates you is the state that you get as a result of all of these displacements is given by this just a single exponential it's just one of these it's an operator of exactly the original form where the displacement is simply the sum of them so the result of taking it this way and this way and this way and this way we've just shown is the same as the result of just taking it in a straight line over the sum of the ai so that's a general argument and now we say special case so in particular particular for a closed path so if the path carries you all the way around what does that mean? it means the sum of the ai 0 all the vectors add up to nothing and that implies immediately that u total because it's e to the nothing is the identity transformation now that might sound obvious but you'll see in a minute that that's anything but an obvious result so now let's specialise in our paths let our path be made up of the series of so let this be x let this angle in here be delta alpha let n be the unit point out of board then this vector here this displacement vector this is going to be a equals delta alpha n cross x that's just ordinary classical geometry that if I go like this and n is out of the board then that's the displacement that we generate so what does that do? that means that the unitary operator that moves my system from here to here u delta alpha is going to be e to the minus i a which is that delta alpha n cross x dot p upon h bar so that's just applying the standard stuff with the displacement vector of this form this is delta a or a whatever but what is that? this scalar triple product can be reordered standard vector algebra tells us that we can reorder this thing into n dot x cross p so this is equal by vector algebra to i delta alpha n dot x cross p upon h bar so that's the operator and there's the h bar that's here but this is what we define to be l so this is equal to e to the minus i delta alpha n dot l so that's what we've now discovered essentially what l does l is the generator of rotations of movements around circles we can also get something interesting very important by combining these two these two things so for a complete circle so for we can multiply these things together so u for a circle is going to be the sum of this for all sorts of delta alphas e to the minus i delta alpha n dot l which is going to be e to the minus i alpha n dot l where this is the sum sorry, it's going to be the product of these right, if we multiply this it's going to be the product of these right, if we make a series of transformations each one by delta alpha then that product of these exponentials can be written as the exponential of the sum of the arguments the sum of the arguments always at n dot l as the operator and the sum of the alphas of the delta alphas we're going to call alpha so for a circle this is going to be 2 pi so we're going to come to the conclusion that e to the minus 2 pi i n dot l is equal to the identity transformation because we've shown that u going all the way around in any closed path has to be the identity so this thing has to be has to be an identity for any vector any unit vector n and we'll need that result shortly so what do we come on so what is the general picture here we can now understand what the distinction is between orbital angular momentum and angular momentum so here we have a so look at this l and then there's this arrow here if we so we're moving we're moving the point at the base of the arrow around a circle we're just translating it, let's just go back so we're just moving it around a circle but we're not doing anything to its internal structure we're just translating it so if it has a little arrow, it has an orientation as for example the direction of its spin that's not going to change its direction it'll just be carried around now ask what j does what j does is it makes you the system you would have had if you took what you've got and you rotated it on a turntable around the origin if you rotated it on a turntable around the origin this is what happens the orientation of the particle changes as well as its location j is changing locations sorry, l is only changing locations j is doing a complete job by putting the particle on a turntable and rotating it at the same time as translating it that's the difference right, so we've got these operators and we want to know obviously when you have operators you need to know what their spectra are, what the allowed values of their eigenvalues are and we can now immediately say what these are going to be so we know that we can find simultaneous a complete set of simultaneous eigenstates of l squared and any one of its components, l, z that's the same as when we were dealing with the angular momentum operators and we obtained the eigenvalues of the angular momentum operators at j squared and j, z by using by exploiting the commutation relationships between the different components of j that was the only thing we used so we got that the e-values of j squared j, z were they were j j plus one, this is for j squared j squared had eigenvalues of j, j plus one where j got a half one, three halves et cetera right and we had that j, z had eigenvalues m which lay between minus j and plus j and we got these results only using that j i comma j j equals i epsilon i j k j k if you look back at what we did, you'll find that that's the case that nothing else went into this but these commutation relationships here for the l operators we have the same commutation relations so the l satisfy identical commutation relations so the argument we had for j could be repeated line by line there's no point in repeating it, literally but virtually we now repeat that argument line by line with every j replaced by an l and we conclude that the e-values can be l squared has numbers like l, l plus one where l could equal nought a half one blah blah blah and l, z has m lying between l and minus l but then we think, so these are the candidate no other numbers are possible than those numbers right, that's what that argument shows are all those numbers possible, no for the following reason e to the two pi, i, l, z has to be the identity operator we've just shown that, that's this statement with n put equal to the unit vector in the z direction okay, so actually I need a minus sign so this thing has to be the identity operator consequently if we use this operator on one of our states which are going to be called lm right this is the mutual eigenstate of l squared and lz with eigenvalue ll plus one for l squared and m for lz, precisely by analogy with the with what we did with j then this has to be simply lm because this is the identity operator but what actually is this so we're doing an exponential of this operator this operator looks at this and says that's my eigenfunction and therefore it replaces itself with the eigenvalue so this is e to the minus two m pi i times lm comparing this side with this side we have the e to the two pi i m is equal to is equal to the number one and that implies that m is an integer because if it were a half integer this would be e to a certain number of i pi which would be minus one but we've shown it cannot be minus one so it's an integer so this is where there is a difference between what happens between what l does and what j does and why we have to keep track so l is looking very like j but it is not j because we've seen physically that it's not because it merely translates you around a circle it doesn't rotate you around a circle and that has the consequence that m has to be an integer so the eigenvalues of l squared are l plus one where l equals naught one, two, three, four, etc and therefore m is also going to be stuck on integers so what's happening here is that what we learn from this in some sense is that let's just do this again so it's when we have translated our system all the way around a circle it's come back to where it was same orientation, everything the same and the result is that an identity operator is applied as described by applying an identity operator to our state if on the other hand we do translate j all the way around the system you would think it came back to where it comes back to the same place undoubtedly and the little arrow comes back to the same orientation and you think you were back in the same place but quantum mechanics is telling us it's really experimental physics tells us that it's not so that arrow I'm having to describe the orientation of the spin of a particle using an arrow and I've already said this is a hazardous enterprise because quantum mechanics particles of gyros and they in some sense have a spin that point in a direction but you do get into trouble as the Einstein Podolsky Rosen experiment shows if you take this idea of pointing in a direction too seriously and here we have another example of the risks of taking too seriously the idea that you can describe the spin of a particle by pointing in a direction because when it has gone all the way around the state vector has changed sign in the event that this is a half integer and the Stern-Gerlach experiment and other experiments many zillions of experiments show that electrons and protons and so on do have half integer values their angular momentum comes in half integer amounts so when you take a real electron on one of these tours around the origin its state is somehow different from the state it started from it's difficult for us to understand that but that's what the mathematics combined with the experimental physics tells us so the next item on the agenda is the eigenfunctions of L squared and Lz so we already know what the eigenvalues are what we now would like to know is what do these states look like, these LM states look like in the position representation so what we're trying to find so what we want is r theta phi Lm we'd like to find expressions for this which describe these things here and our strategy is basically that we're going to so the strategy is this, the detail is rather tedious so the strategy is this we're going to apply L plus to LL and get nothing we're going to express this abstract so this is Lx plus, I think we talked about this yesterday Lx plus I, Ly so this is the operator which will try and raise the M entry here to one larger but that's not possible because it's already the largest value so it'll kill it so this is an operator equation we will look at this operator equation in the position representation and then it will become a first order differential equation we will solve it, it'll turn out to be dead simple to solve so that will lead us to let me leave off the r because we're not really interested in r for the moment it will lead us to this LL and then we will be able to say that theta phi LL minus one is equal to theta phi this is a horrible square root now LL plus one minus M M minus one times L minus sorry this needs to be to the minus one times L minus so we will having obtained this wave function we will apply L minus to it in the position representation and that will give us essentially the next one down that will lower this by one and by repeatedly doing this we will be able to generate all of the wave functions associated with a particular total anglimentum quantum number L that's the strategy to carry this out what we need is expressions we need expressions in the position representation for these operators so this, let's start by writing down what LZ is what is LZ it's one over H bar of X P Y minus Y P X in the position representation what is that P Y in the position representation is minus I H bar D by D Y so this becomes minus I X D by D Y minus Y D by D X now to find out what that is we could just, we would like to express everything in terms of theta and phi because we know anglimentum is to do with rotations that's why we want to use theta and phi spherical polar coordinates and we could just use the chain rule to turn this into derivatives in theta or phi but it's much easier to go to use the chain rule not going from this to D by D theta D by D phi but to go to write to find out what is D by D phi according to the chain rule it's D X by D phi D by D X plus D Y by D phi D by D Y plus D Z by D phi D by D Z that's just the chain rule from calculus now we put in what these X Y and Z are D X in polar coordinates is R sin theta cos phi we know that Y is R sin theta sin phi and we know that Z is R cos theta so this implies that so take a derivative of this with respect to phi we have that D X by D phi is going to be minus R sin theta sin phi which is the same as minus Y minus is going to be simply minus Y we have that similarly D Y by D phi that becomes a cosine and therefore this becomes X and we have that D Z by D phi so we take these results and stuff them back in here and that tells me that D by D phi is equal to that's a Y sorry that's a minus Y so let's write it down minus Y D by D X plus X D by D Y what is the relationship to what we have up there that is so this is precisely what's in that bracket for LZ so this implies that LZ is minus I D by D phi of course this is no surprise because LZ is the thing that rotates you around the Z axis and everybody knows that the spherical polar coordinates is defined as the angle around the Z axis so it's kind of obvious that this has got to be the case but it's nice to see that the chain rule delivers the goods now the next bit is distinctly more tarsam what we now have to do is what we want to do is express L plus and L minus L X plus I L Y in terms of D by D thetus and D by D phi you could go at this just by brute force but the algebra would be heavy so the algebra is not going to be that wonderful now here is I think this is the way to do it so let's just calculate what D by D theta is using the chain rule it's obviously D X by D theta D by D X plus D Y by D theta D by D Y plus D Z by D theta D by D Z ok we have expressions up there so D X by D theta is going to produce an R cos theta cos phi etc this is going to produce an R cos theta sin phi so I got a common factor of R cos theta open a bracket and then we will be able to write down sin sorry cos phi D by D X plus cos phi sorry plus sin phi and then finally we have D Z by D thingy which is going to be minus R cos we're differentiating cos sin theta D by D Z now let's take our expression for D by D phi and multiply it by theta so we're going to write down cot theta D by D phi just for the fun of it so right so this has to be in doing it we're going to replace well let's take this one first it's going to be R that is R sin theta cos phi when we multiply it by cot which is cos over sin the sin is going to go into a cos so this is going to be R cos theta brackets sin phi excuse me cos phi D by D Y so this this here is R sin theta cos phi which is X times cos over sin which is cot and then that Y is going to be R sin theta sin phi and this cot multiplication will turn the sin theta into cos theta I take out and we will have a sin phi here D by D D by D how much X that's it right so what we now do is just again for the fun of it we take D by D theta and add on I times this so we look at D by D theta plus I times cot theta D by D phi what happens then we have R cos theta as common factors oops quick I should have excuse me yes that's right that's fine so what am I going to be doing I'm going to be adding this and this so we're going to add I times this to that so we're the common factor of D by D X so we're going to have cos phi minus I sin phi which is well known to be written as E to the minus I phi that's how much of D by D X we're going to have and then we're going to have sin phi plus I cos phi and that is going to be plus I times E to the minus I phi D by D Y we haven't quite finished have we so if you expand this out it's going to give you an I cos phi which we want that's this one here times I and it will contain a minus I sin phi times an I which makes it a plus just plain sin phi which is what we want there then finally we've got this in the back so we're from here so we have minus R sin theta D by D Z so supposing we oh and this thing could be written R sin theta is this no no no leave it alone so what we do now is we multiply this by E to the I phi so we have E to the I phi brackets D by D theta plus I cot theta D by D phi which will turn out to be our bottom line is equal to this is going to be cancelled because I'm multiplying through by E to the I phi this will be cancelled so we will have R cos theta brackets D by D X plus I D by D Y and then we will have minus R sin theta times E to the I phi which I choose now to write as cos phi plus I sin phi and now R sin theta cos phi is X and this in the back is going to be Y R sin theta sin phi so we'll be able to write the right hand side as R cos theta D by D X plus I D by D Y minus so this is going to be X plus I Y excuse me this was times D by D Z it always was it was minus R sin theta D by D Z yes it just got lost between this line and this line okay few so I want to now establish that that is actually L plus so in order to do that I write down so what is L plus is L X plus I L Y don't need those brackets which is one upon H bar okay what's P X it's Y P Z minus Z P Y plus I times L Y is Z P X minus X P Z so now I want to turn this into D by D X so this will be so this is minus I H bar D by D Z so we're going to have sorry before we do that let's gather things together because we've got a P Z and a P Z so let's just for the moment write this is one over H bar we're going to have a common factor of Z we're going to have I Z P X plus I P Y I think right this I and that I make the minus sign that we have there Z is the common factor alright and then what about the Z factors the P Z factors well we have a plus Y minus I X P Z now we replace the P's by minus I H bar D by D X et cetera so we'll get a minus H bars are going to go throughout from here we'll have a minus I times an I which will give us a plus so we'll have a Z D by D X here we will have we have two I's making a minus sign we have another minus sign coming in from the minus I so we end up with minus I D by D Y and we'll rather worry by the sign there no the two made a minus and I soaked up sorry these two I's made a minus I brought in another minus from minus I H bar D by D by D the minus is cancelled leaving me only with the I and this is going to bring in a minus I a minus I well so we have a if you propagate this minus I inside here we get a minus sign there so we get a minus X and this is going to be plus I Y because there's a minus I here and there's the minus sign and here's the I D by D Z and that I hope agrees with what we have written it does because our cos theta is also known as Z so we have Z D by D X plus I D by D Y and here we have a minus X plus I Y D by D Z so we have established a very important result that L plus in the position representation is the differential operator e to the I phi brackets D by D theta is plus isn't it plus I cotangent theta D by D phi close brackets there is an analogous calculation which I'm not going to do which tells us that L minus is equal to minus e to the minus I phi D by D theta is it minus yeah it is minus I cot theta D by D phi and if you try and do the two calculations together with plus and minus signs good luck to you it's fantastically confusing right so so at least I got one of them right so those are will be asked well let's no we'll no let's leave it at that that those are going to be our starting points for tomorrow deriving the eigenfunctions of these very important operators