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 ),(),()(),(2

22

tt

itVtm

rrrr

ˆˆˆ ˆˆˆ 2

2

2

2

2

22

zyxzyxzyx

kjikjir

• as usual, separating off the time dependence gives rise to energy eigenstates and a TISE

)())()()(2

22

rrrr EVm

Expressing this stuff in spherical coordinates (r, , )

• the volume element is obtained by increasing each coordinate infinitesimally: (dr) (r d ) (r sind r2 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 s2 = x2 + y2, we have r2 = s2 + z2 so we get • r2 = x2 + y2 + z2 tans/z tany/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).

2

2

2222

22

23

sin

1 sin

sin

11

sin: sin

1ˆ

1ˆˆ ˆ

rrrr

rr

dddrrddrrr

r rφθrrr

Issues with getting the Laplacian in sphericals I

• first three terms… of the gradient dotted into itself:

• simple enough so far…0ˆ

0ˆ

0ˆ

rrr

φθr

0ˆ

ˆˆ

ˆˆ

φ

rθ

θr • can you visualize it??

rθφ

φθ

φr

ˆsinˆcosˆ

ˆcosˆ

ˆsinˆ

rr

rrrrrrrrrr

rrrrr

rrrrr

rrrrrrrr

222

2

2

2

2

2

2

2

2

2

1210

100

sin

1ˆˆ

sin

1

sin

1ˆˆ

1ˆˆ11ˆˆ

ˆˆˆˆˆ

sin

1ˆ

1ˆˆˆ

rφrφ

rθrθ

rrrrrφθrr

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

sinsin

1sin

cos00

100

sin

1ˆˆ

sin

1ˆˆ

1ˆˆ1ˆˆ

1ˆˆ

11ˆˆ

1ˆsin

1ˆ

1ˆˆ1ˆ

2

22

2

2

2

2

2

22

2

2

2

2

r

rr

rr

rr

rrrrr

rrrrr

θφθφ

θθθθ

θrθr

θφθrθ

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

2

2

2

2

2

2

222

2

22

222

2

2

2

2

sin

1

0sin

10000

sin

1ˆˆ

sin

1ˆˆ

sin

1ˆˆsin

cos

sin

1ˆˆ

sin

1ˆˆ

sin

1

sin

1ˆˆ

sin

1ˆ

sin

1ˆ

1ˆˆsin

1ˆ

r

r

rr

rrr

rrrrr

rrrrr

φφφφ

φθφθ

φrφr

φφθrφ

• insert into TISE; divide by RY; multiply by r2; 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 ),()()( YrRr

known are which arguments, thedrop now ),()(),()()(

),()(sin

1sin

sin

11

2 2

2

2222

2

2

YrERYrRrV

YrRrrr

rrrm

• 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…

RYrERYVRYY

r

RY

r

R

dr

dRr

dr

d

r

Y

m/by multiply

sinsin

sin22

2

2

2222

2

2

rearrange )(sin

1sin

sin

11

222

2

2

22

2

ErrVrY

Y

Y

Ydr

dRr

dr

d

Rm

2

2

22

22

sin

1sin

sin

1 )(

21

Y

Y

Y

YErV

mr

dr

dRr

dr

d

R

Processing the angular part of the TISE

• insert the separated form into the angular equation

• assume yet another separation into polar and azimuthal factors

)()(),( Y• the Y functions are spherical harmonics

• calling that constant m2, we arrive at two ODEs…

rearrange now sin11

sinsin 2

2

2

)l(l

d

d

d

d

d

d

one on this work togo slet' 1sin

1sin

sin

1

nowfor one thisignore )1( )(2

2

2

2

2

22

)Yl(lYY

RllErVRmr

dr

dRr

dr

d

!constant! a equal sidesboth 1

sin1sinsin

2

22

d

d)l(l

d

d

d

d

ΘΦsinbymultiply now 1sin

sinsin 2

2

2θ/ )l(l

d

d

d

d

d

d

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 m2

• since we allow for ± m anyway, the ± is superfluous• the normalization of this is trivial:

2...1,0, 2

1)(

2

1)2(1 2

2

0

22

0

me

AAdAd

im

• all probabilities with a single m eigenvalue are azimuthally (axially) symmetric!

sin1sinsin1 222

2

2

m)l(ld

d

d

dm

d

d

integer or 0 is 122)2(

mimeimAeim

Ae

anyway? , iswhat )(22

2

mimAemd

d

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

0sin)1(sinsin 22

mlld

d

d

d

0sin)1(11

0sin)1(111

1cos1sincos

cos

2222

2221

221

22

21

221

2

mlldx

dx

dx

dx

mlldx

dxx

dx

dx

dx

dx

dx

d

dx

d

d

d

d

d

d

d

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) = anxn , which yields

0)1(101)1(11 2222

lldx

dx

dx

dxll

dx

dx

dx

dx

nn

nnnnnn

nnnn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

ann

llnna

annllannallnaanann

allnaannann

xannllnxaxnnaxnna

xallnxaxnnaxnna

xallnxaxxnnax

12

)1(1

01)1(12)1(12

0)1(2112

01)1(2112

0)1(211

0)1(211

2

22

2

2

00202

012

2

2

0

1

1

2

2

2

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 an+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 x2 a problem at x = ± 1• we must terminate the series so an+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

)(1:)( where)()(cos)( 22 xPdx

dxxPxPAPA l

mmm

lm

lm

l

• in this, we use the Legendre polynomial, which are given by (another) Rodrigues formula

ll

ll xdx

d

lxP 1

!2

1:)( 2

8

324

1448

3542

132

53

2

122

3210 1

xxPxxP

xPxPP

Graphs of the Legendre polynomials

• notice how they are ‘normalized’ to be unity at x = cos =1 = 0 (+z axis)

8

324

1448

3542

132

53

2

122

3210 1

xxPxxP

xPxPP

P2

P4

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

ddd

dd

d

dxdx sin

cos

• to normalize (or any other integral) we can do it either way:

dPAdPAdxxPA ml

ml

ml sin)(cossin)(cos)(1

0

220

221

1-

22

• the normalized angular wavefunctions are the spherical harmonics

0]for )( and 0,for 1[)(cos

!

!

4

12),(

mmimePml

mllY mm

lm

l

• 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

1cos52

sin3)(

11)(1:)(

213

2

322

1521

22

332

521

23

1

21

213

xP

xxxxdx

dxxP

dx

dxxP

xxxx

xxxdx

dx

dx

dxP

2

332

53

2463

623

33

7212048

1

13348

11

!32

1:)(

6

72

127

24221

72

21

3224

9

105

16024

9

105

210770147075024

9

23

22

5

70

7

5024

9

0

24624

9

2

0

2424

9

0

213

2

sin1cos11cos35cos25

sincos11cos10cos25sin)(cos1

AAAAAA

AdA

dAdPA

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

180;0;5.101;5.78

0sinor cos01cos52

sin3)(

0

0 5

10

2213

xP

1351.

sin1cos11cos35cos25

sincos11cos10cos252

sin)(cosyProbabilitCollar

4472.3280.1252.0128.12

7

5

11

5

1

3

11

5

1

5

35

5

1

7

2512

7

2/246

12

7

22/

2424

7

213

2

2/12/32/52/7

0

0

0

0

d

d

dPA

Pictures of some orbitals

• clockwise from upper left: dzz, dyz, dxz, dxy, dx2-y2

• Example of the p orbital calculation is to come