Transcript 3D Schrodinger Equation

3D Schrodinger Equation
• Simply substitute momentum operator
• do particle in box and H atom
• added dimensions give more quantum numbers.
Can have degeneracies (more than 1 state with
same energy). Added complexity.


p  i or
 2
2m
2
x 2
 
2
  ( x, y, z, t )  V ( x, y, z, t )  it
2
• Solve by separating variables
( x, y, z, t )   ( x, y, z ) (t )
 2
2m
 2  V ( x, y, z )  E
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• If V well-behaved can separate further: V(r) or
Vx(x)+Vy(y)+Vz(z). Looking at second one:
2
2m
 2  (Vx ( x)  V y ( y )  Vz ( z ))  E
assum e ( x, y, z )   x ( x) y ( y ) z ( z )
(
2
x 2

)  (Vx  V y )
2
y 2
 x y z
(
2
x 2

) x y
2
y 2
 x y

1
2
x 2
 z dz 2
 x y
(
 ( E  Vz )
 x y z
 2 z
 (Vx  V y ) 
• LHS depends on x,y
 d 2 z


 2
z 2

z z
2
 E  Vz
RHS depends on z
 ( E  Vz )  S

2
y 2
) x y  Vx  V y  S
• S = separation constant. Repeat for x and y
P460 - 3D S.E.
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d 2 x
 x dx 2
d 2 y
 y dy 2
d 2 z
 z dz 2
 Vx  S '  E x
 Vy  S  S '  E y
 Vz  E  S  E z
E x  E y  E z  S ' ( S  S ' )  ( E  S )  E
• Example: 2D (~same as 3D) particle in a Square
Box
V 
V 0
x  0, x  a , y  0, y  a
inside box
satisfies V  Vx ( x )  V y ( y )
 ( x, y )   x ( x ) y ( y )
• solve 2 differential equations and get
E  Ex  E y 
 2 2
2 ma 2
(n  n )
2
x
2
y
• symmetry as square. “broken” if rectangle
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E  Ex  E y 
 ( x, y)  A sin
 2 2
2 ma 2
n x x
a
(n  n )
2
x
sin
2
y
n y y
a
nx , n y  1,2..
2
|

|
 dxdy  1 norm alization
• 2D gives 2 quantum numbers.
Level
nx
ny
Energy
1-1
1
1
2E0
1-2
1
2
5E0
2-1
2
1
5E0
2-2
2
2
8E0
• for degenerate levels, wave functions can mix
(unless “something” breaks degeneracy: external or
internal B/E field, deformation….)
 12  A sin ax sin 2ay
 21  A sin 2ax sin ay
 mix  12   21  2   2  1
• this still satisfies S.E. with E=5E0
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Spherical Coordinates
• Can solve S.E. if V(r) function only of radial
coordinate
2
2M
   V ( r )  E ( r , ,  )
 2
2M
[ r 2r ( r 2
2
2
1
r 2 sin 2   2

r
)

1
r 2 sin  
(sin


)
] ( r , ,  )  V ( r )  E
• volume element is
d (vol)  dr(rd )(r sin d )
• solve by separation of variables
 (r , ,  )  R(r )( )( )
( E V ) R 2 M


2
2
1
r 2 sin 2   2
(
 r 2
r 2 r r
 r 2 sin1 2 
 sin  


) R
R
• multiply each side by
 r sin 
R
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2
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Spherical Coordinates-Phi
• Look at phi equation first
1 d2
 d 2
 ( )  f (r , )  ml2
• constant (knowing answer allows form)
( )  e
iml
• must be single valued
(  2 )  ( )
e
iml (  2 )
e
iml
 ml  0,1,2.......
• the theta equation will add a constraint on the m
quantum number
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Spherical Coordinates-Theta
• Take phi equation, plug into (theta,r) and rearrange
1 d
R dr
(
ml2
 sin 
2
r 2 dR
dr

)
2 M 2r 2
2
d
1
 sin  d
(
[ E  V ( r )] 
sin  d
d
)  l (l  1)
• knowing answer gives form of constant. Gives
theta equation which depends on 2 quantum
numbers.
2
m
sin

d

l
1 d
sin  d
d
sin 2 
)
(
 l (l  1)
• Associated Legendre equation. Can use either
analytical (calculus) or algebraic (group theory) to
solve. Do analytical. Start with Legendre equation
(1  z )
2
d 2 Pl
dz 2
z  cos
 2z
dPl
dz
 l (l  1) Pl  0
Pl  Legendre function
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Spherical Coordinates-Theta
• Get associated Legendre functions by taking the
derivative of the Legendre function. Prove by
substitution into Legendre equation
|m |
2 |ml | / 2 d l Pl
 lml  (1  z )
dz |ml |
 20  P2
 21  (1  z )
2
 22  (1  z )
2
1
2
dPz
dz
  2 1
d 2 Pz
dz
• Note that power of P determines how many
derivatives one can do.
• Solve Legendre equation by series solution
(1  z )
2
d 2 Pl
dz 2
 2z
dPl
dz

Pl   ak z k
k 0
2
d P
dz 2
 l (l  1) Pl  0

dP
dz
  ak kz k 1
k 1

  ak k ( k  1) z k  2
k 2
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Solving Legendre Equation
• Plug series terms into Legendre equation
k 2
k
{
k
(
k

1
)
a
z

[
k
(
k

1
)

l
(
l

1
)]
a
z
} 0

k
k
• let k-1=j+2 in first part and k=j in second (think of
it as having two independent sums). Combine all
terms with same power
j
{(
j

2
)(
j

1
)
a

[
j
(
j

1
)

l
(
l

1
)]
a
}
z
0

j 2
j
• gives recursion relationship
j ( j 1)l (l 1)
j 2
( j 2)( j 1)
a

aj
• series ends if a value equals 0  L=j=integer
a j 2  0  j( j 1)  l (l 1)
• end up with odd/even (Parity) series
a1  0, aeven  0 or a0  0, aodd  0
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Solving Legendre Equation
• Can start making Legendre polynomials. Be in
ascending power order
l  0, a0  1, a1  0  P0  1
l  1, a0  0, a1  1  P1  z
l  2, a0  1, a1  0, a2 

06
21
j ( j 1) l ( l 1)
( j  2 )( j 1)
 3  P2  1  3z 2
• can now form associated Legendre polynomials.
Can only have l derivatives of each Legendre
polynomial. Gives constraint on m (theta solution
constrains phi solution)
lml  (1  z )
2 |ml | / 2 d |ml |
dz |ml |
Pl
 | ml | l
P460 - 3D S.E.
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Spherical Harmonics
 00  1
10  z
1, 1  (1  z )
2
1
2
 20  1  3 z 2
1
2
 2, 1  (1  z ) z
2
 2, 2  (1  z 2 )
• The product of the theta and phi terms are called
Spherical Harmonics. Also occur in E&M.
• They hold whenever V is function of only r. Seen
related to angular momentum
Ylm  lm m
 spherical harm onics
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3D Schr. Eqn.-Radial Eqn.
• For V function of radius only. Look at radial
equation
1 d  r 2dR  2
l (l  1)

  2 V ( r )  E ) R 
R
2
r dr  dr  
r
• can be rewritten as (usually much better...)
 2 d 2u
2 l (l  1)

 (V 
)u  Eu
2
2
2 dr
2 r
P ( r )  4R 2 r 2 dr
u( r )  rR( r )
2
 4u dr
• note L(L+1) term. Angular momentum. Acts like
repulsive potential and goes to infinity at r=0 (ala
classical mechanics)
• energy eigenvalues typically depend on 2 quantum
numbers (n and L). Only 1/r potentials depend only
on n (and true for hydrogen atom only in first order.
After adding perturbations due to spin and
relativity, depends on n and j=L+s).
P460 - 3D S.E.
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Particle in spherical box
• Good first model for nuclei
V (r)  0 r  a
V (r)   r  a
• plug into radial equation. Can guess solutions
 2 d 2u
2 l (l  1)

 (V 
)u  Eu u( r )  rR( r )
2
2
2 dr
2 r
• look first at l=0
d 2u
dr 2
  k 2u

with
k
2 ME

u  A sin(kr)  B cos(kr)
• boundary conditions. R=u/r and must be finite at
r=0. Gives B=0. For continuity, must have R=u=0
at r=a. gives sin(ka)=0 and
En 0 
n 2 2  2
2 Ma 2
 n 00 
n  1,2,3....
sin( nr / a )
1
r
2a
• note plane wave solution. Supplement 8-B
 General
discusses scattering, phase shifts.
terms are

 ik r
e
R( r ) 
r
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Particle in spherical box
• ForLl>0 solutions are Bessel functions. Often
arises in scattering off spherically symmetric
potentials (like nuclei…..). Can guess shape (also
can guess finite well)
• energy will depend on both quantum numbers
Enl  E10 E11 E12 E20 E21 E22 .....
• and so 1s 1p 1d 2s 2p 2d 3s 3d …………….and
ordering (except higher E for higher n,l) depending
on details
• gives what nuclei (what Z or N) have filled
(sub)shells being different than what atoms have
filled electronic shells. In atoms:
Z  2 4 10 ( He  Be  Ne)
1S 2S 2P
• in nuclei (with j subshells)
Z  2 6 8 14 16 ( He  C  O  Si  S )
1s 1 p 3 1 p 1 1d 5 2s 1
2
2
2
2
P460 - 3D S.E.
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H Atom Radial Function
• For V =a/r get (use reduced mass)

1 d  r 2dR  2m  Ze2
l (l  1)

  2  
 E  R 
R
2
r dr  dr    4 0r
r

• Laguerre equation. Solutions are Laguerre
polynomials. Solve using series solution (after
pulling out an exponential factor), get recursion
relation, get eigenvalues by having the series
end……n is any integer > 0 and L<n. Energy
doesn’t depend on L quantum number.
En 
 MZ e
( 4 0 ) 2 2  2 n 2
2 4

 me c 2 2 Z 2
2n2

13.6 eVZ 2
n2
• Where fine structure constant alpha = 1/137 used.
Same as Bohr model energy
• eigenfunctions depend on both n,L quantum
numbers. First few:
4 

 Zr / a
R10  e
0
a0 
2
0
me e
2

C

 0.5 A
R20  (2  Zra0 )e  Zr / 2 a0
R21  Zra0 e  Zr / 2 a0
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H Atom Wave Functions
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H Atom Degeneracy
• As energy only depends on n, more than one state
with same energy for n>1
• ignore spin for now
Energy
n
l
m
D
-13.6 eV
1
0(S)
0
1
-3.4 eV
2
0
0
1
1(P)
-1,0,1
3
0
0
-1.5 eV
3
1
2(D)
-1,0,1
-2,-1,0,1,2
1 Ground State
4 First excited states
1
3
5
Dn
2
9 second excited states
P460 - 3D S.E.
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Probability Density
|  |2  probability
2
|

|
 dVolum e 1  norm alization


2
0
0
0

1
2
1
0
 
or 
0
2
2
|

|
r
sin  d d dr


2
2
|

|
r
d d cos dr

P(r )  r 2 | Rnl |2
• P is radial probability density
• small r naturally suppressed by phase space (no
volume)
• can get average, most probable radius, and width
(in r) from P(r). (Supplement 8-A)
m ost probable
dP
dr
0
average  r   r width  r 
P460 - 3D S.E.
r2  r 2
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Most probable radius
• For 1S state
P ( r )  Ar 2 | R |2  Ar 2 e 2 r / a0
dP
dr
 0  2re
r  a0
 2 r / a0

2r 2
a0
e  2 r / a0
(" peak" )

r   rP ( r ) dr 
3
2
a0
0
(
n 2 a0
Z
[1  12 (1  l (nl21) )]in general)
r 2   r 2 Ar 2 e  2 r / a0 dr  3a02
r 
3a02  94 a02  0.87a0
•
Bohr radius (scaled for different levels) is a good
approximation of the average or most probable
value---depends on n and L
• but electron probability “spread out” with width
about the same size
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Radial Probability Density
P460 - 3D S.E.
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Radial Probability Density
note #
nodes
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Angular Probabilities
P( ,  ) | ( ) |2 | ( ) |2 sin( )
 m  eim |  |2  1
•
no phi dependence. If (arbitrarily) have phi be
angle around z-axis, this means no x,y dependence
to wave function. We’ll see in angular momentum
quantization
00 "1" S states
10  A cos 11 
A
2
sin 
P states
2
2
 10
 12, 1  11
"1"
• L=0 states are spherically symmetric. For L>0,
individual states are “squished” but in arbitrary
direction (unless broken by an external field)
• Add up probabilities for all m subshells for a given
L get a spherically symmetric probability
distribution
P460 - 3D S.E.
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Orthogonality
nlm nl m 
  2

*
2


r
 nlm n 'l 'm ' sin  drd d
0 0 0
  nn ' ll ' mm ' with   Rn lm m
• each individual eigenfunction is also orthogonal.
• Many relationships between spherical harmonics.
Important in, e.g., matrix element calculations. Or
use raising and lowering operators
• example

E  const ant in zˆ
V | E | r cos note
cos is Legendre polynom ial10
 nlm | r cos | n' l ' m'  
 mm ' l (l '1) f (r )  0 m  m' l  l '1
P460 - 3D S.E.
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Wave functions
• build up wavefunctions from eigenfunctions.
• example
 ( r,  ,  , t ) 
1
( 100eiE1t /   2 211eiE2t /   211eiE2t /  )
6
• what are the expectation values for the energy and
the total and z-components of the angular
momentum?

E   H |     * H dvol    *i  dvol
t
• have wavefunction in eigenfunction components
1
1
3
7
E  ( E1  2 E2  E2 )  ( E1  E1 ) 
E1
6
6
4
24
1
2
L  (l0 (l0  1)  2  l1 (l1  1)  l1 (l1  1))
6
1
 (0(0  1)  2 1(1  1)  1(1  1))  1
6
1
1
1
Lz  ( Lz 0  2 Lz1  Lz 1 )  (0  2  1) 
6
6
6
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