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!