Physics 2D Lecture SlidesLectures Week of March 1, 2010

Sunil SinhaUCSD Physics

Probability Density Functions for Particle in 3D Box

Same Energy Degenerate StatesCant tell by measuring energy if particle is in

211, 121, 112 quantum State

Source of Degeneracy: How to “Lift” Degeneracy• Degeneracy came from the

threefold symmetry of aCUBICAL Box (Lx= Ly= Lz=L)

• To Lift (remove) degeneracy change each dimension such thatCUBICAL box RectangularBox

• (Lx≠ Ly ≠ Lz)• Then

2 22 2 2 2

31 2

2 2 22 2 2

x y z

nn nE

mL mL mL

!! !" #" # " #= + +$ %$ % $ %$ % & '& ' & '

Ener

gy

The Hydrogen Atom In Its Full Quantum Mechanical Glory

2

( )kZe

U rr

=

U (r) !1

r=

1

x2+ y

2+ z

2 " More complicated form of U than box

By example of particle in 3D box, need to use separation of

variables(x,y,z) to derive 3 independent differential. eqns.

This approach will get very ugly since we have a "conjoined triplet"

To simplify the situation, use appropriate variables

Independent Cartesian (x,y,z) # Inde. Spherical Polar (r,$ ,%)

Instead of writing Laplacian &2=

'2

'x2+

'2

'y2+

'2

'z2

, write

&2 = 1

r2

''r

r2 ''r

()*

+,-+

1

r2 sin$

''$

sin$''$

()*

+,-+

1

r2 sin2$

'2

'2%

TISE for . (x,y,z)=. (r,$ ,%) becomes

1

r2

''r

r2 '. (r,$ ,%)

'r

()*

+,-+

1

r2 sin$

''$

sin$'. (r,$ ,%)

'$()*

+,-+

1

r2 sin2$

'2. (r,$ ,%)

'2%+

2m

!2

(E-U(r)). (r,$ ,%) =0

!!!! fun!!!

r

Spherical Polar Coordinate System

2

( sin )

Vol

( )( )

= r si

ume Element dV

n

dV r d rd dr

drd d

! " !

! ! "

=

Don

’t Pa

nic:

Its

sim

pler

than

you

thin

k !

1

r2

!!r

r2 !"!r

#$%

&'(+

1

r2 sin)

!!)

sin)!"!)

#$%

&'(+

1

r2 sin2)

!2"!2*

+2m

!2

(E-U(r))" (r,) ,*) =0

Try to free up second last term from all except *

This requires multiplying thruout by r2 sin2) +

sin2)!!r

r2 !"!r

#$%

&'(+ sin)

!!)

sin)!"!)

#$%

&'(+!2"!2*

+2mr

2 sin2)!

2(E+

ke2

r)" =0

For Separation of Variables, Write " (r,) ,*)=R(r).,()).-(*)

Plug it into the TISE above & divide thruout by " (r,) ,*)=R(r).,()).-(*)

Note that :

!.(r,) ,*)

!r= ,()).-(*)

dR(r)

dr

!.(r,) ,*)

!)= R(r)-(*)

d,())

d)!.(r,) ,*)

!)= R(r),())

d-(*)

d*

+ when substituted in TISE

sin2)R

d

drr

2 dR

dr

#$%

&'(+

sin),

d

d)sin)

d,d)

#$%

&'(+

1

-d

2-d

2*+

2mr2 sin2)!

2(E+

ke2

r) =0

Rearrange by taking the * term on RHS

sin2)R

d

drr

2 dR

dr

#$%

&'(+

sin),

d

d)sin)

d,d)

#$%

&'(

+2mr

2 sin2)!

2(E+

ke2

r) = -

1

-d

2-d

2*

LHS is fn. of r,) & RHS is fn of * only , for equality to be true for all r,) ,*

+ LHS= constant = RHS = ml

2

Now go break up LHS to separate the r & ! terms.....

LHS:sin2!

R

d

drr

2 dR

dr

"#$

%&'+

sin!(

d

d!sin!

d(d!

"#$

%&'

+2mr

2 sin2!!

2(E+

ke2

r) =m

l

2

Divide Throughout by sin2! and arrange all terms with r away from ! )

1

R

d

drr

2 dR

dr

"#$

%&'+

2mr2

!2

(E+ke2

r)=

ml

2

sin2!*

1

(sin!d

d!sin!

d(d!

"#$

%&'

Same argument : LHS is fn of r, RHS is fn of ! , for them to be equal for all r,!

) LHS = const = RHS = l(l +1) What do we have after shuffling?

d2+d, 2

+ ml

2+ = 0...................(1)

1

sin!d

d!sin!

d(d!

"#$

%&'+ l(l +1) *

ml

2

sin2!

-

.//

0

122((!) = 0.....(2)

1

r2

d

drr

2 dR

dr

"#$

%&'+

2m

!2

(E+ke2

r)-

l(l +1)

r2

-

./

0

12R(r) = 0....(3)

These 3 "simple" diff. eqn describe the physics of the Hydrogen atom.

All we need to do now is guess the solutions of the diff. equations

Each of them, clearly, has a different functional form

Solutions of The S. Eq for Hydrogen Atom2

2

2

dThe Azimuthal Diff. Equation : m 0

Solution : ( ) = A e but need to check "Good Wavefunction Condition"

Wave Function must be Single Valued for all ( )= ( 2 )

( ) = A e

l

l

l

im

im

d

!

!

!

!! ! ! "

!

#+ # =

#$ # # +

$# ( 2 )

2

2

A e 0, 1, 2, 3....( )

The Polar Diff. Eq:

Solutions : go by the name of "Associated Legendre Functions"

1 msin

( 1) ( ) 0sin si

Quantum #

n

l

l

im

l

d d

Magnetic

ld

m

ld

! "

% %% % % %

+

& '() *+ + + ( =, - . /0 1

=

2 3

$ = ± ± ±

only exist when the integers and are related as follows

0, 1, 2, 3....

: Orbital Q

;

ua

p

nt

osit

um N

ive numb

umber

er

1For 0, =0 ( ) = ;

2

For

l

l

l

l m

l l

l m

l

m

%

= ± ± ± ± =

= $ (

2

1, =0, 1 Three Possibilities for the Orbital part of wavefunction

6 3[ 1, 0] ( ) = cos [ 1, 1] ( ) = sin

2 2

10[ 2, 0] ( ) = (3cos 1).... so on and so forth (see book)

4

l

l l

l

l m

l m l m

l m and

% % % %

% %

= ± $

= = $( = = ± $(

= = $( +

Φ

Solutions of The S. Eq for Hydrogen Atom

The Radial Diff. Eqn: 1

r2

d

drr

2 dR

dr

!"#

$%&+

2m

!2

(E+ke2

r)-

l(l +1)

r2

'

()

*

+,R(r) = 0

Solutions : Associated Laguerre Functions R(r), Solutions exist only if:

1. E<0 or has negative values given by E=-ke2

2a0

1

n2

!"#

$%&

;a0=!

2

mke2= Bohr Radius

2. And when n = integer such that l = 0,1,2,3,4,,,(n -1)

n = principal Quantum #

To Summarize : The hydrogen atom is brought to you by the numbers

n = 1,2,3,4,5,.....l = 0,1,2,3,,4....(n -1)

ml= 0,±1,±2,±3,...± l

Quantum # appear only in Trapped systems

The Spatial Wave Function of the Hydrogen Atom

/(r,0 ,1) = Rnl

(r) . 2lm

l

(0) . 3m

l

(1) = Rnl

Yl

ml (Spherical Harmonics)

Radial Wave Functions & Radial Prob Distributions

0

0

-r/a

3/2

0

r-2

a

3/200

23

23/20 00

R(r)=

2 e

a

1 r(2

n

1 0 0

2 0 0

3 0 0

- )e a2 2a

2 r(27 18 2 )

a81 3a

l

r

are

a

l m

!

! +

n=1 K shelln=2 L Shelln=3 M shelln=4 N Shell……

l=0 s(harp) sub shelll=1 p(rincipal) sub shelll=2 d(iffuse) sub shelll=3 f(undamental) ssl=4 g sub shell……..

Symbolic Notation of Atomic States in Hydrogen

2 2

4 4

2

( 0) ( 1) ( 2) ( 3) ( 4

3 3 3

) .....

1

4

3

1

s p

s l p l d l f l g l

s

l

s

d

n

p

s p

= = =! = =

"

5 5 5 5

4

5

4

5s p d f g

d f

Note that:•n =1 non-degenerate system•n>1 are all degenerate in l and ml.

All states have same energyBut different spatial configuration

2

2

0

ke 1E=-

2a n

! "# $% &

Facts About Ground State of H Atom-r/a

3/2

0

-r/a

100

0

2 1 1( ) e ; ( ) ; ( )

a 2 2

1 ( , , ) e ......look at it caref

1. Spherically s

1, 0,

ymmetric no , dependence (structure)

2. Probab

0

ully

i

ln l r

ra

m R ! "#

! "#

! "

$ = % = & =

' =

$

= = =

22

100 3

0

Likelihood of finding the electron is same at all , and

depends only on the

radial seperation (r) between elect

1lity Per Unit Volume : ( ,

ron & the nucleus.

3 Energy

,

of Ground ta

)

S

r

ar ea

!#

! "

"(

' =

2

0

kete =- 13.6

2a

Overall The Ground state wavefunction of the hydrogen atom

is quite

Not much chemistry or Biology could develop if there was

only the ground state of the Hydrogen Ato

We ne

m

e

!

boring

eV= (

d structure, we need variety, we need some curves!

Interpreting Orbital Quantum Number (l)

2

RADIAL ORBITAL

RADIAL ORBI

2 22

2 2 2

22

TAL2 2

K

1 2m ke ( 1)Radial part of S.Eq

K ;substitute this form for

n: (E+ )- ( ) 0r

For H Atom: E = E

K K

K + U =

1 2m ( 1)-

2m

d dR r l lr R r

r dr dr r

d dR l lr

r dr dr r

ke

r

! "+# $+ =% & ' () * + ,

+# $+%

+

&

-

+) *

!

!

!

[ ]

ORBITAL

RADIAL

OR

2

2

2

2

B

2

2

( ) 0

( 1)Examine the equation, if we set get a diff. eq. in r

2m

1 2m( ) 0 which

Further, we also kno

depends only on radi

K

K us

w

r of orb

K tha

i

t

t

R r

l lthen

r

d dRr R r

r dr dr

! "=' (

+ ,

+

# $+ =% &

) *

=!

!

22

ITAL ORBITorb

2

AL 2

2

ORBITAL 22

L= r p ; |L| =mv r

( 1

1; K

2 2

Putting it all togather: K magnitude of)

| | ( 1)2m

Since integer=0,1,2,3...(

A

n-1) angular mome

ng.2

nt

Mom

orbit

l lL l l

r

l

Lmv

mr

L

posi

mr

tive

. /

+= = +

=

=

= /

=

/

" " "

!!

um | | ( 1)

| | ( 1) : QUANTIZATION OF Electron's Angular Momentum

L l l discrete values

L l l

= + =

= +

!

!

pr

L

Magnetic Quantum Number ml (Right Hand Rule)

QM: Can/Does L have a definite direction

Classically, direction & Magnitud

? Proof by Negat

ˆSuppose L was precisely known/defined (L || z)

e of L

S

always well defi

n

ed

:

n

io

L r p= !! ! !

!

! !

!

2

z

z z

Electron MUST be in x-y orbit plane

z = 0 ;

, in Hydrogen atom, L can not have precise measurable

ince

Uncertainty Principle & An

p p ; !!!

gular Momentum

value

: L

2

pz E

L r p

So

m

"

= ! #

# $ $ $ # $ %

$

%

$

=

! !

" # " "

!

!

" #

Z

Z

The Z compo

Arbitararily

nent of

L vector spins around Z axis (precesses).

|L | ; 1, 2, 3...

( 1)

It

picking Z axis as a referen

L

:| L |

ca

| | (always)

s

n

in

ce directi

c

o

:

e

n

l l

l

m

Note L

m l

m l l

<

= = ± ± ± ±

< +

!

!

#

# #

Z never be that |L | ( 1)

(breaks Uncertainty Principle)

So you see, the dance has begun !

lm l l= = +# #

L=2, ml=0,±1, ± 2 : Pictorially

Sweeps Conical paths of different ϑ: Cos ϑ = LZ/L and average <LX> = 0 <LY> = 0

What’s So “Magnetic” ?

Precessing electron Current in loop Magnetic Dipole moment µ