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