# Physics 2D Lecture Slides Lectures Week of June 1,2009 Sunil Sinha ```Physics 2D Lecture Slides
Lectures Week of June 1,2009
Sunil Sinha
UCSD Physics
Probability Density Functions for Particle in 3D Box
Same Energy  Degenerate States
Cant tell by measuring energy if particle is in
211, 121, 112 quantum State
• Degeneracy came from the
threefold symmetry of a
CUBICAL Box (Lx= Ly= Lz=L)
• To Lift (remove) degeneracy 
change each dimension such that
CUBICAL box  Rectangular
Box
• (Lx≠ Ly ≠ Lz)
• Then
&quot; n12! 2
E =\$
2
2
mL
x
&amp;
# &quot; n22! 2
% + \$\$
2
2
mL
y
' &amp;
# &quot; n32! 2 #
%% + \$
2 %
2
mL
z '
' &amp;
Energy
Source of Degeneracy: How to “Lift” Degeneracy
The Hydrogen Atom In Its Full Quantum Mechanical Glory
U (r ) !
1
=
r
1
x +y +z
2
2
2
&quot; More complicated form of U than box
By example of particle in 3D box, need to use seperation of
r
variables(x,y,z) to derive 3 independent differential. eqns.
This approach will get very ugly since we have a &quot;conjoined triplet&quot;
To simplify the situation, use appropriate variables
Independent Cartesian (x,y,z) ' Inde. Spherical Polar (r,# ,\$ )
%2
%2
%2
Instead of writing Laplacian &amp; = 2 + 2 + 2 , write
%x %y %z
2
1 % ) 2 % *
1
% )
% *
1
%2
&amp; = 2 +r
,+ 2
+ sin #
,+ 2 2
r %r - %r . r sin # %# %# . r sin # % 2\$
2
TISE for ( (x,y,z)=( (r,# ,\$ ) becomes
kZe 2
U (r ) =
r
1 % ) 2 %( (r,# ,\$ ) *
1
% )
%( (r,# ,\$ ) *
r
+
sin
#
+
, 2
+
,+
2
r %r %r
%#
. r sin # %# .
1
% 2( (r,# ,\$ ) 2m
+ 2 (E-U(r))( (r,# ,\$ ) =0
2
2
2
r sin #
%\$
!
!!!! fun!!!
Spherical Polar Coordinate System
Volume Element dV
dV = (r sin ! d&quot; )(rd! )(dr )
= r 2 sin ! drd! d&quot;
Don’t Panic: Its simpler than you think !
1 ! # 2 !&quot; &amp;
1
! #
!&quot; &amp;
1
! 2&quot; 2m
r
+
sin )
+
+
(E-U(r))&quot; (r,) ,* ) =0
!) (' r 2 sin 2 ) ! 2* ! 2
r 2 !r %\$ !r (' r 2 sin ) !) %\$
Try to free up last term from all except *
This requires multiplying thruout by r 2 sin 2 ) +
! # 2 !&quot; &amp;
! #
!&quot; &amp; ! 2&quot; 2mr 2 sin 2 )
ke 2
sin ) % r
+ sin )
sin )
+
+
(E+
)&quot; =0
!r \$ !r ('
!) %\$
!) (' ! 2*
r
!2
2
For Separation of Variables, Write &quot; (r,) ,* )=R(r).,() ).-(* )
Plug it into the TISE above &amp; divide thruout by &quot; (r,) ,* )=R(r).,() ).-(* )
Note that :
!.(r,) ,* )
!R(r)
= ,() ).-(* )
!r
!r
!.(r,) ,* )
!,() )
= R(r)-(* )
+ when substituted in TISE
!)
!)
!.(r,) ,* )
!-(* )
= R(r),() )
!)
!*
sin 2 ) ! # 2 !R &amp; sin ) ! #
!, &amp; 1 ! 2 - 2mr 2 sin 2 )
ke 2
r
+
sin )
+
+
(E+
) =0
R !r %\$ !r ('
, !) %\$
!) (' - ! 2*
r
!2
Rearrange by taking the * term on RHS
sin 2 ) ! # 2 !R &amp; sin ) ! #
!, &amp; 2mr 2 sin 2 )
ke 2
1 !2r
+
sin )
+
(E+
)=R !r %\$ !r ('
, !) %\$
!) ('
r
- ! 2*
!2
LHS is fn. of r,) &amp; RHS is fn of * only , for equality to be true for all r,) ,*
+ LHS= constant = RHS = m 2l
Now go break up LHS to separate the r &amp; ! terms.....
sin 2 ! &quot; # 2 &quot;R &amp; sin ! &quot; #
&quot;) &amp; 2mr 2 sin 2 !
ke 2
2
LHS:
r
+
sin
!
+
(E+
)
=m
l
R &quot;r %\$ &quot;r ('
) &quot;! %\$
&quot;! ('
r
!2
Divide Throughout by sin 2! and arrange all terms with r away from ! *
m 2l
1 &quot; # 2 &quot;R &amp; 2mr 2
ke 2
1
&quot; #
&quot;) &amp;
r
+
(E+
)= 2 +
sin !
2
%
R &quot;r %\$ &quot;r ('
r
)sin
!
&quot;
!
&quot;! ('
\$
!
sin !
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!
d 2,
2
+
m
, = 0...................(1)
l
2
dm 2l 1
1 d #
d) &amp; .
sin !
+ 0 l(l + 1) +
3 )(! ) = 0.....(2)
2
sin ! d! %\$
d! (' 0/
sin ! 32
1 d # 2 &quot;R &amp; . 2m
ke 2 l(l + 1) 1
r
+ 0 2 (E+
)3 R(r) = 0....(3)
2
2
%
(
r
r dr \$ &quot;r ' / !
r
2
These 3 &quot;simple&quot; 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 Atom
The Azimuthal Diff. Equation :
d 2#
+ ml2# = 0
2
d!
Φ
Solution : # (! ) = A eiml! but need to check &quot;Good Wavefunction Condition&quot;
Wave Function must be Single Valued for all ! \$ # (! )=# (! + 2&quot; )
\$ # (! ) = A eiml! = A eiml (! + 2&quot; ) \$ ml = 0, &plusmn;1, &plusmn;2, &plusmn;3....( Magnetic Quantum # )
1 d )
d( * &amp;
ml2 '
The Polar Diff. Eq:
, sin %
- + l (l + 1) + 2 / ((% ) = 0
sin % d% 0
d% 1 .2
sin % 3
Solutions : go by the name of &quot;Associated Legendre Functions&quot;
only exist when the integers l and ml are related as follows
ml = 0, &plusmn;1, &plusmn;2, &plusmn;3.... &plusmn; l ; l = positive number
l : Orbital Quantum Number
For l = 0, ml =0 \$ ((% ) =
1
2
;
For l = 1, ml =0, &plusmn; 1 \$ Three Possibilities for the Orbital part of wavefunction
[l = 1, ml = 0] \$ ((% ) =
[l = 2, ml = 0] \$ ((% ) =
6
cos%
2
[l = 1, ml = &plusmn;1] \$ ((% ) =
3
sin%
2
10
(3cos2% + 1)....and so on and so forth (see book)
4
Solutions of The S. Eq for Hydrogen Atom
1 d &quot; 2 !R % ( 2m
ke 2 l(l + 1) +
r
+ * 2 (E+
)- R(r) = 0
2
\$
'
r
r dr # !r &amp; ) !
r
,
Solutions : Associated Laguerre Functions R(r), Solutions exist only if:
ke 2 &quot; 1 %
!2
1. E&lt;0 or has negative values given by E=;a0 =
2'
2
\$
2a 0 # n &amp;
mke
2. And when n = integer such that l = 0,1,2,3,4,,,(n . 1)
n = principal Quantum # or the &quot;big daddy&quot; quantum #
To Summarize : The hydrogen atom is brought to you by the letters
n = 1,2,3,4,5,..../
l = 0,1,2,3,,4....(n . 1) Quantum # appear only in Trapped systems
m l = 0, &plusmn;1, &plusmn;2, &plusmn;3,... &plusmn; l
The Spatial Wave Function of the Hydrogen Atom
0(r,1 ,2 ) = Rnl (r) . 3 lm (1 ) . 4 m (2 ) = Rnl Yl l (Spherical Harmonics)
m
l
l
n l ml R(r)=
2 -r/a
1 0 0 3/2 e
a0
1
r
2 0 0
(2- )e
3/2
a0
2 2a 0
-
r
2a 0
2
2
r
r
3 0 0
(27 ! 18 + 2 2 )e
3/2
a0
a0
81 3a 0
n=1  K shell
n=2  L Shell
n=3  M shell
n=4  N Shell
……
!
r
3a0
l=0 s(harp) sub shell
l=1 p(rincipal) sub shell
l=2 d(iffuse) sub shell
l=3 f(undamental) ss
l=4 g sub shell
……..
Symbolic Notation of Atomic States in Hydrogen
l!
n
s (l = 0)
p (l = 1)
d (l = 2) f (l = 3) g (l = 4)
1
2
3
1s
2s
3s
2p
3p
3d
4
5
4s
5s
4p
5p
4d
5d
.....
&quot;
4f
5f
Note that:
•n =1 non-degenerate system
•n1&gt;1 are all degenerate in l and ml.
All states have same energy
But different spatial configuration
5g
ke 2 ! 1 &quot;
E=# \$
2a 0 % n 2 &amp;
Facts About Ground State of H Atom
n = 1, l = 0, ml = 0 \$ R(r ) =
2 -r/a
1
1
e
;
%
(
!
)
=
;
&amp;
(
&quot;
)
=
a 3/2
2#
2
0
'100 (r , ! , &quot; ) =
1
a0 #
e-r/a ......look at it carefully
1. Spherically symmetric \$ no ! ,&quot; dependence (structure)
1 ( 2ar
2. Probability Per Unit Volume : '100 (r , ! , &quot; ) = 3 e
# a0
2
Likelihood of finding the electron is same at all ! ,&quot; and
depends only on the
radial seperation (r) between electron &amp; the nucleus.
ke 2
3 Energy of Ground State == (13.6eV
2a 0
Overall The Ground state wavefunction of the hydrogen atom
is quite boring
Not much chemistry or Biology could develop if there was
only the ground state of the Hydrogen Atom!
We need structure, we need variety, we need some curves!
Interpreting Orbital Quantum Number (l)
1 d # 2 dR \$ ! 2mr 2
ke2 l (l + 1) &quot;
Radial part of S.Eqn: 2 % r
(E+
)- 2 ( R( r ) = 0
&amp;+
r dr ) dr * '+ ! 2
r
r ,
ke
;substitute this form for E
r
1 d # 2 dR \$ 2m !
! 2 l (l + 1) &quot;
K RADIAL + K ORBITAL R( r ) = 0
%r
&amp;+
r 2 dr ) dr * ! 2 '+
2m r 2 (,
For H Atom: E = K + U = K RADIAL + K ORBITAL -
Examine the equation, if we set K ORBITAL
L
2
p
r
! 2 l (l + 1)
=
then get a diff. eq. in r
2
2m r
1 d # 2 dR \$ 2m
[K RADIAL ]R( r ) = 0 which depends only on radius r of orbit
%r
&amp;+
r 2 dr ) dr * ! 2
1 2 &quot; &quot; &quot;
L2
Further, we also know that K ORBITAL = mvorbit ; L= r . p ; |L| =mv orb r / K ORBITAL =
2
2mr 2
Putting it all togather: K ORBITAL
! 2 l (l + 1)
L2
=
=
/ magnitude of Ang. Mom | L |= l (l + 1)!
2m r 2
2mr 2
Since l = positive integer=0,1,2,3...(n-1) / angular momentum | L |= l (l + 1)! = discrete values
| L |= l (l + 1)! : QUANTIZATION OF Electron's Angular Momentum
! ! !
L = r ! p (Right Hand Rule)
Magnetic Quantum Number ml
!
Classically, direction &amp; Magnitude of L always well defined
!
QM: Can/Does L have a definite direction ? Proof by Negation:
!
!
ˆ
Suppose L was precisely known/defined (L || z)
! ! !
Since L = r ! p # Electron MUST be in x-y orbit plane
p2
# \$z = 0 ; \$pz \$z &quot; # # \$pz &quot; %; E =
&quot; % !!!
2m
!
So, in Hydrogen atom, L can not have precise measurable value
Uncertainty Principle &amp; Angular Momentum : \$Lz \$&quot; &quot; #
Arbitararily picking Z axis as a reference direction:
!
L vector spins around Z axis (precesses).
!
The Z component of L
|L Z |= ml #;
ml = &plusmn;1, &plusmn;2, &plusmn;3... &plusmn; l
Note :| L Z | &lt; | L | (always)
since ml # &lt; l (l + 1)#
It can never be that |L Z |= ml # = l (l + 1)#
(breaks Uncertainty Principle)
So you see, the dance has begun !
L=2, ml=0,&plusmn;1, &plusmn; 2 : Pictorially
Sweeps Conical paths of different
ϑ: Cos ϑ = LZ/L and average
&lt;LX&gt; = 0
&lt;LY&gt; = 0
What’s So “Magnetic” ?
Precessing electron Current in loop  Magnetic Dipole moment &micro;
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