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. (We skipped
when doing perturbations).
• Quantization of angular momentum split between
E+R Chaps 7+8


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
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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/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  Vz
 ( x, y )   x ( x) y ( y )
• solve 2 differential equations and get
E  Ex  E y 
 2 2
2 ma 2
(nx2  n y2 )
• 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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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
)
1
 sin

2 M 2r 2
2
d
d
[ E  V (r )] 
( sindd )  constant  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. Sets value on m quantum
number
• 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
a j 2 
j ( j 1)l (l 1)
( j 2)( j 1)
aj
• series ends if a value equals 0
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
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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. They
are related to angular momentum (discuss later)
Ylm  lm m
 spherical harm onics
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