Moving to three dimensions we will need new, more complicated, coordinate systems separation of variables is the key method for breaking down a problem into tractable ODEs the classic problems are the infinite rectilinear box, and the spherically symmetric potential (including the Coulomb) in more than one dimension position is a vector r derivatives with respect to position become the gradient second derivatives become the laplacian in cartesian coordinates these are written the TDSE takes the form as usual, separating off the time dependence gives rise to energy eigenstates and a TISE

Expressing this stuff in spherical coordinates (r, ,  ) the volume element is obtained by increasing each coordinate infinitesimally:  (dr) (r d  ) (r sin  d   r 2 sin  dr d  d  r is the distance from the origin  is the polar angle, also called co-latitude: angle down from +z axis; ranges from 0 to   is the azimuthal angle : angle away from +x axis in xy plane in counterclockwise sense; ranges from 0 to 2  x = r sin  cos  y = r sin  sin   z = r cos  with s 2 = x 2 + y 2, we have r 2 = s 2 + z 2 so we get r 2 = x 2 + y 2 + z 2 tan  s/z tan  y/x

Expressing this stuff in spherical coordinates unit vectors defined as an increase in that coordinate only because the unit vectors change when taking space derivatives, gradient and laplacian get all mixed up see Griffith’s E&M text appendix and end-papers for one way to systematically keep track of all the formulas and rules for first and second derivatives to understand a bit more of the form of the Laplacian, note that the unit vectors are NOT CONSTANT! the basic issue is that when one dots the gradient into the gradient, the dot products of the unit vectors are just 1 (three ways) or zero (six ways) as usual, but taking their derivatives is NOT zero (in five cases).

Issues with getting the Laplacian in sphericals I first three terms… of the gradient dotted into itself: simple enough so far… can you visualize it??

Issues with getting the Laplacian in sphericals II second three terms… of the gradient dotted into itself:

Issues with getting the Laplacian in sphericals III third three terms… of the gradient dotted into itself:

insert into TISE; divide by RY; multiply by r 2 ; put r dependence on one side and angular ( ,  dependence on the other side: Separating the TISE into an angular and a radial part assume V(r) =V(r) and separation since left side depends only on r and right side only on ( ,  ), both sides are a constant, which we write (weirdly, for now) l(l+1) we arrive at two distinct DEs (one O and one P), which are…

Processing the angular part of the TISE insert the separated form into the angular equation assume yet another separation into polar and azimuthal factors the Y functions are spherical harmonics calling that constant m 2, we arrive at two ODEs…

Solving the azimuthal equation the  function must be periodic with period 2  :  (  +2  ) =  (  ) the azimuthal equation is easy to handle so this is why it made sense to write the constant as m 2 since we allow for ± m anyway, the ± is superfluous the normalization of this is trivial: all probabilities with a single m eigenvalue are azimuthally (axially) symmetric!

Cracking open the polar equation the polar equation, like the azimuthal equation, contains no physics, and was familiar to the ancients it is the Legendre equation, and its solutions are the associated Legendre polynomials if m = 0, the solutions are the Legendre polynomials (and of course the spherical harmonics have azimuthal symmetry in that case) let x = cos  and rexpress things in this language

Cracking open Legendre’s equation for m = 0 it is even in its variable, so solutions will be even or odd try a power series expansion  (x) =  a n x n, which yields

Cracking open Legendre’s equation for m = 0 there is an even series or an odd series, but not both for large n, this roughly settles down to a n+2 ~ n/n+2 ~ 1 for large 1 thus, the ratio test for successive terms may be applied and we see that the ratio is x 2  a problem at x = ± 1  we must terminate the series so a n+2 = 0  l = n = 1,2,3… to keep the solution finite at  = 0 and  = , l must be a non- negative integer, and m must satisfy the inequality |m|  l because otherwise there is a sign flip in the ODE (see?) and things diverge solutions are associated Legendre polynomials in this, we use the Legendre polynomial, which are given by (another) Rodrigues formula

Graphs of the Legendre polynomials notice how they are ‘normalized’ to be unity at x = cos  =1   = 0 (+z axis) P2P2 P4P4

Normalizing and calculating expectation values azimuthally symmetric solutions, so that’s covered polar solutions are integrated on  angle (from 0 to  often) in x = cos  language,  = 0  x = 1;  =   x =  1, so integration on increasing angle has a sign flip if integrating on increasing x infinitesimal dx gets sign flip too to normalize (or any other integral) we can do it either way: the normalized angular wavefunctions are the spherical harmonics we will shortly see their intimate connection to angular momentum

Example of calculation with the Legendres generate P31; normalize it; find the angles ‘subtended’ by the ‘collar’; find the probability the particle is in that volume

Example of calculation with the Legendres generate P31; normalize it; find the angles ‘subtended’ by the ‘collar’; find the probability the particle is in that volume

Pictures of some orbitals clockwise from upper left: d zz, d yz, d xz, d xy, d x 2 -y 2 Example of the p orbital calculation is to come