16 - 1 - part i angular momentum - explicit solutions (948) (1)

7/28/2019 16 - 1 - Part I Angular Momentum - Explicit Solutions (948) (1)

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Welcome back everybody.I'm Charles Clark.This is Exploring Quantum Physics.And today we're going to turn to theproblem of angular momentum, one of mostfundamental symmetries in quantumphysics.We say everything revolves around angularmomentum and we're going to start lookingat some of the explicit solutions.We'll try to show you different ways ofcalculating them from first principles,and then with pointers to the literatureso that you can, you can develop betterunderstanding if you so choose.I hope you enjoy it.Now, home works of week six had someproblems of tensor analysis, a lot ofvector algebra.some of you found that quite frustrating,I think.others, I believe, found it enjoyable.in any event, we're going to do a little

bit more of that.And please don't be too alarmed byeverything that's up here.These are really the core, coreidentities that I'm putting forcompleteness.the uh, [INAUDIBLE] tensor.Just the basic notation about, our changein notation in order to simplifycalculations.the definition and the position vector,in terms of the old and the new notation,the radius.

And then the definition of the angularmomentum operator and the square of theangular momentum operator.So we're going to now use some of theseto generate Eigenfunctions of the angularmomentum operator.And I think you'll find this approachvery useful.Especially if you work in other fields soelectrical engineering, or anything thatstudies acoustics or wave motions or o,optics, you'll find a lot of counterpartsin those areas to what we're studying

here.We'll start by showing what I hope youthink is quite an amazing result.If we take any function of the radialcoordinate only.it, it turns out to be an Eigenfunctionof any given complement of the angularmomentum operator and of the squareitself.Okay.


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So let's see how this goes.we start with,Just applying the l, l sub u upon psi.And then that is, very straightforward,expansion of the angular momentumoperator.And now what you see is that the, effectof the derivative on a, a radialfunction, function of the radius only canbe evaluated by the chain rule.And so you see we get an expression thatboils down to this.So we'll go with a little in-line quiz toseeIt get, get, give you a gauge of yourunderstanding of how to deal with anexpression encountered like this.So I hope that you saw that thisvanished, this term vanishes by symmetryrelations because it's just a sum overdummy indices but with a, a sign changeassociated with the permutation of.The two indices, so that term mustnecessarily be zero.

So we've recorded that fact.And then the next one is is, is easybecause if l u of psi of r zero, thenanything else applied to that l u.Of psi of r is also 0, including lusquared, and then the sum of the squaresof the l sub u, which is the totalangular momentum vector is also equal to0.So we've shown that any, any functionthat is a function of the radius alone isan Eigenfunction.Well, it's an Eigenfunction of any

complement of the angular momentumoperator.It is conventional to take the complementalong the third axis.This traditionally the z axis, as thequantization axis.But we'll see that that's not always thecase.Anyway, let's proceed and see what theimplications of this discovery are.Several weeks back, we tried a sort ofsimple constructive technique.It's not a general purpose technique, but

just a way of getting some understandingof the Schrodinger equation by applyingit to functions that we already know, theones that we're familiar with.And seeing what, what type of potentialwould support those as an Eigenfunction.So what we want is to use a function ofthe radius that is, is, is normalized, ornormalizable, that is it can be dividedby another constant to make this


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identity.Correct.And in the previous case, we chose a aradial Gaussian wave function, of this,of this form, which is in fact the thesolution for the ground state of athree-dimensional isotropic Harmonicoscillator.You see that this term here is just equalto r squared, so this is a radialsolution.And it therefore has the anglar momentum,0, that we just described.Well, how many functions do you know thatare radially symmetric and die off ininfinity, in a, in a way that'ssufficient to,To allow the normalization integral to becomputed.Well, of course, one can invent a wholenumber of, contrive functions that dothat.But, to me,This one here sort of stands out.

It's a simple, exponential decay in alldirections.it's easy to, use, and so it's a goodchoice for this constructive approach.So here I show a cut through thatfunction.This is radially symmetric, but I cutthrough the origin, r equals 0, along, itdoesn't matter which axis, any axis looksthe same.And this shows something that you mightnot have appreciated before, which is thewave function has a cusp at the origin.

In fact the second derivative of the wavefunction, you know, with respect to x, isactually a singular there.Well that corresponds, in fact, tosomething else that we find if you justif you take this function and put it intothis equation you will find that the the,the potential that is implied by the useof this function is just the Coulombfunct-, Coulomb potential.Again, Gaussian units, Gaussian units.and this is, in fact, the ground statewave function for the Hydrogen atom.

it's [INAUDIBLE] with energy, given bythis expression.in, whatever the Bohr radius is you'veseen it before, but here it is made moreexplicit.Now there was a, one of the problems in,one of the questions in week fivehomework had to do with approximatingusing a Gaussian approximation to thisfunction in a variational calculation.


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Well, a Gaussian, you see, is going tohave a rather smooth profile of theorigin.So this type of cusp behavior is notsomething that's well represented by aGaussian wave function.As a matter of fact, there were twovariational calculations in that week'shomework.And in some sense, the.Functions that you were being asked toapproximate were the same function.This is actually also, if you think aboutit, the one dimensional cut here thatgives functions the same as a function ofa ground state of an attractive deltafunction potential of one dimension.So this is a rather interestingcorrespondence between the delta functionin one dimension and the Hydrogen atom inthree dimensions.So just for convenience of furthernotation we'll just, we'll write that,this wave function psi sub 1s.

The 1s is the conventional designationfor the quantum numbers that bouncedeight of Hydrogen.Maybe this is also a good time for you torefresh your recollection of the meaningof the Bohr radius.And so that will be the subject of anin-video quiz.Okay.So we're going to break at this point,and you might review this material.See if you can follow the derivations.I also, recommend that you use your time

just to do a few simple calculations,like verifying that this wave function isappropriately normalized.You know, I could make a mistake and I'llgive a nice reward to the first personwho identifies that in the student form.to do this, it requires integration overpolar coordinate angles.A theta in phi.Now we're going to discuss those later.We haven't really dealt with that yet,but you can find pointers on how to dothree dimensional integrals many places

in the open literature.And so this is a sort of exercise forthose of you who want some practice.those of you who had trouble.With the homework problems of week five,maybe this is a good way a good approachto sharpening your skills.Okay, thanks for listening and hope tosee you again.

16 - 1 - part i angular momentum - explicit solutions (948) (1)

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