1

Physics 2D Lecture SlidesLecture 27: March 2nd 2005

Vivek SharmaUCSD Physics

Quantum Mechanics In 3D: Particle in 3D BoxExtension of a Particle In a Box with rigid walls 1D → 3D⇒ Box with Rigid Walls (U=∞) in X,Y,Z dimensions

yy=0

y=L

z=L

z

x

Ask same questions:• Location of particle in 3d Box• Momentum • Kinetic Energy, Total Energy• Expectation values in 3D

To find the Wavefunction and variousexpectation values, we must first set upthe appropriate TDSE & TISE

U(r)=0 for (0<x,y,z,<L)

2

The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates

2 2 22

2 2

2 2 2 2 2

22

2

22

2

2 2

Time Dependent Schrodinger Eqn:

( , , , )( , , , ) ( , , ) ( , ) .....In 3D

2

2 22

2

x y z tx y

x y

z t U x y z x t im t

m x m y m

z

mSo

! ! !"

!#$ " # + # =

!

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

+!

*

= +! !

! !

!

!

!

!

x

2

x x [K ] + [K ] + [K ]

[ ] ( , ) [ ] ( , ) is still the Energy Conservation Eq

Stationary states are those for which all proba

[ ]

=

bilities

so H x t E

K

x t

z

% &=

#

(*

=

')

#

-i t

are

and are given by the solution of the TDSE in seperable form:

= (r)e

This statement is simply an ext

constant in time

(

ension of what we

,

derive

, , ) ( , )

d in case of

x y z t r t+,# =#

" "

1D

time-independent potential

y

z

x

Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables

1 2 3

1

2

2 3

2

in 3D:

x,y,z independent of each ( , , ) ( ) ( ) ( )

and substitute in the master TISE, after dividing thruout by = ( ) ( ) (

- ( , , ) ( ,

other , wr

, ) ( , ,

)

and

) ( ,

ite

, )

n

2m

x y z

TISE x y z U x y z x y z E x y

x y z

x y

z

z

! ! ! !! ! !

!

!

! !"

=

+ =!

22

1

2

1

2222

2

2

3

2

3

2

1

2

2

( )1

2 ( )

This can only be true if each term is c

oting that U(r)=0 fo

onstant for all x,y,z

( )1

2 ( )

(

2

r (0<x,y,z,<L)

( )1

2 (

)

zE Const

m z z

y

m

x

m x x

x

m

y y

!!

!!

!

!!# $%

& +' (# $%

+ & =

)

# $%&' (%* +%* +

=

)

&

' (* +

%

%!!

!

!

22

33 32

22

22 21 12 2

1 2 3

)( ) ;

(Total Energy of 3D system)

Each term looks like

( )( ) ;

2

With E

particle in

E E E=Constan

1D box (just a different dimension)

( )( )

2

So wavefunctions

t

zE z

m zy

yE x E

x

y

m!

!!

!!

%& =

%&

%

=

==%

+ +

%!!

3 31 2 21must be like ,( ) sin x ,( ) s ) s nin ( iyy kx k z k z!! !, ,,

3

Particle in 3D Rigid Box : Separation of Orthogonal Variables

1 1 2 2 3 3

i

Wavefunctions are like , ( ) sin

Continuity Conditions for and its fi

( ) sin y

Leads to usual Quantization of Linear Momentum p= k .....in 3D

rst spatial derivative

( )

s

sin x ,

x

i i

z k z

n k

x

L

yk

p

k !

! "

!

"

!# #

$

=

#

=

!!"

"

1 2 3

2

2

1 3 1

2 2 22

2

2 3 ; ;

Note: by usual Uncertainty Principle argumen

(n ,n ,n 1,2,3,.. )

t neither of n , n , n 0! ( ?)

1Particle Energy E = K+U = K +0 = )

2(

m 2(

zy

x y z

n

why

p nL

nmL

p nL L

p p p

"

"

"% & % &= = '( ) (

% &= ( )* +

)* + *

=

+

+ + ="

""

2 2 2

1 2 3

2

1 2 3

2

1

Ei

3

-

3

1

)

Energy is again quantized and brought to you by integers (independent)

and (r)=A sin (A = Overall Normalization Cosin y

(r)

nstant)

(r,t)= e [ si

n , n , n

sin x

sin x ysn in ]t

k

n n

k

A k k

k

k

z

z

!

!

+ +

=, "!

!

!E

-i

et"

Particle in 3D Box :Wave function Normalization Condition

3

*

1

1

2

1

x,y,

E E-i -i

2

E Ei i

*

2

2 2

2

2

3

*

3

z

2

(r) e [ sin y e

(r) e [ s

(r,t)= sin ]

(r,t)= sin ]

(r,t)

sin x

sin x

sin x

in y e

[ si

Normalization Co

(r,t)= sin ]

ndition : 1 = P(r)dx

n y

dyd

1

z

t t

t t

k z

k

k

k

A k

A k

A k zk k

A

z

!

!

"

"

"

#

"

=

=

=

$$$

! !

! !"

"

"

"

""

L L L

2

3 3E

2 2 2

1 2 3

x=0 y=

2 2 -

1

z

3

i

0

2

=0

sin x dx

s

sin y dy sin z dz =

(

2 2 2

2 2 an r,t)=d [ s sinii ex yn ] n

t

L

k

L LA

A kL

k k k

k zL

% &% &% &' (' (' (' (' (' () *) *) *

+ , + ,# = - . - ./ 0 / 0"

$ $ $

!"

4

Particle in 3D Box : Energy Spectrum & Degeneracy

1 2 3

2 22 2 2

n ,n ,n 1 2 3 i

2 2

111 2

2 2

211 121 112 2

2

3Ground State Energy E

2

6Next level 3 Ex

E ( ); n 1,2,3... , 02

s

cited states E = E E2

configurations of (r)= (x,y,z) have Different ame energy d

i

mL

mL

n n n nmL

!

!

" "

!= + + = # $

=

% = =

%

!

!

!

egeneracy

yy=L

z=Lz

xx=L

2 2

211 121 112 2

Degenerate States

6E

= E E2mL

!= =

!

x

y

z

E211 E121 E112ψ

E111

x

y

z

ψ

Ground State

5

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

6

2

( )kZe

U rr

=

The Coulomb Attractive Potential That Bindsthe electron and Nucleus (charge +Ze) into a

Hydrogenic atom

F V

me

+e

r

-e

The Hydrogen Atom In Its Full Quantum Mechanical Glory

2

( )kZe

U rr

=

2 2 2

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

variables(x,y,z) to derive 3 in

1 1

This approach will

dependent d

( ) M

iffer

ore compli

ential. eq

cated form of U than bo

get

x

ns

.

U rr x y z

! = "+ +

2 22

2

To simplify the situation, use appropriate variables

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

very ugly since we have a "conjoined triplet"

Instead of writing Laplacian

)

x y

# $

% %& = +

% %

'

2

2

2 2

2

2 2 2

2

2

2

2

2

2 2 2

2

2

2

, write

1

sin

TISE for (x,y,z)= (r, , ) become

1

r

1 (r, , ) (r, , )

r

s

1 2m+

1= s

(E-U(r))

insi

si

n

1sin

s

(r, , )

n

in

r

r

rr

z

r

r

rr r

r ##

# # #

## #

$

( (

#( # $ ( # $

( # $

# $

# $

% %) *+ ,% %- .

% %)

% %)

*+ ,% %

%+%

%%

%%

*& + ++ ,% %- .

% %) *+ ++ % .-

,% -.

!

!!!! fun!!!

(r, , ) =0 ( # $

r

7

Spherical Polar Coordinate System

2

( sin )

Vol

( )( )

= r si

ume Element dV

n

dV r d rd dr

drd d

! " !

! ! "

=

Don

’t P

anic

: Its

sim

pler

than

you

thin

k ! 2

2

2 2 22

2

2 2

2 2

2

1 2m+ (E-U(r))

sin

Try to free up las

1(r, , ) =0

r

all except

This requires multi

t term fro

plying thruout by sin

si

1sin

si

si si

m

n

n

n

rr

r r

r

r

rr r

! ! !! " #

#

"

!

"" " "

""

"

"

#$ $% &' ($ $) *

+

$ $% &' ($ $) *

$ $% &+ +' ($ $) *

$+

$

$$ !

2

2

2 2 2

2

2m ke+ (E+ )

r

(r, , )=R(r). ( ). ( )

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

sin =0

For Seperation of Variables, Write

( , , )

r

Note tha

(

t :

)

nr

r

#! " # " #

! ! "!

! " #

#

"

"

"

#

"$% &

+' ($) *

$

$$

,

-

, ..

$

!

2 2 22 2

2 2

2

( ). ( )

( , , )( ) ( )

( , , )( ) ( )

s

R(r)

r

( ) when substituted in TI

in sin =0

Rearrange by ta

sin

king the

sin

SE

( )

1 2m ke+ (E+

)r

rR r

rR r

R rr

R r r

" #

" ##

"" #

"

"

"

"##

""

#"

" ""

=, .

$-= .

$$-

= ,$

$ $% $ $,% &+ +' (,

&' $ $)

($ $

$$

$,+

$$.$

$$) .**

.!

2 2

2

2 2 2

2

2

2

2m ke 1+ (E+ )

r

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

LHS= constant = RH

term on RHS

sin s

S =

sinsin

m

in =-

l

R rr

R r r #

" # " #

""

#

"" "

"$ $% &' (

$ .. $

+

$ $,% &+ ' (, $ $)) * *$ $ !

8

2 2 22 2

2

2

2

2

sin sin =m

Divide Thruout by sin and arrange all terms with r aw

Now go break up LHS to seperate the terms...r ..

2m keLHS: + (E+ )

a

&

sinsi

y from

r

1

nl

R r

r

rr

r

R r

R

! !

!

!!

! !

!

!

" "#$ %+ & '# " "( )

" "$

"

%

""

& '" "( )*

!

2

2

2 2

2

m 1sin

sin sin

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

LHS = const =

2m ke(E+ )=

r

What do we have after shuffl

ll r,

= ( in1) g RHS

l

l l

R r

r!

! ! ! !! !

" "#$ %+ & '$ %

+# " "( )

* +

& '"( ) !

22

2

2

2

2 22

2 2 2

!

m1 sin ( 1) ( ) 0.....(2)

sin sin

...............

d ..(1)

1 2m ke ( 1)(E+ )- ( ) 0....(

m 0.

3

.

)r

T

l

l

d R r l lr R r

r dr

d dl

d

d

r

ld

r

!

,

!! ! ! !

- .#$ %+ + + #

- ." +$ %+ =& ' / 0"( ) 1 2

=& ' / 0( ) 1 2

3+ 3 =

!

hese 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

9

Solutions of The S. Eq for Hydrogen Atom2 2

2

2 2 2

2

2

0

1 2m ke ( 1)The Radial Diff. Eqn: (E+ )- ( ) 0

r

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

1. E>0 or has negtive values given by ke 1

E=-2a

d R r l lr R r

r dr r r

Solutions

n

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

*

-

$

!

2

0 2

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

n = principal Quantum # or the "big daddy" qunatu

To

; Bohr

Summa

m #

: The hy

Rad

r drogeize n

ius

atom

a

l

mke

n= .

%= =& '

+!

n = 1,2,3,4,5,....

0,1,2,3,,4....( 1)

m 0, 1, 2, 3,.

Quantum # appear only in Trappe

is brought to you

d systems

The Spati

by the letters

al Wave Function o

f the

Hydrogen Atom

..

l

l n

l

/= .= ± ± ± ±

lm( , , ) ( ) . ( ) . ( ) Y (Spherical Harmonics)

l

l

m

nl lm nl lr R r R0 1 0 12 = 3 4 =

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……..

10

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•n1>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!