the separated radial and orbital parts of the schrodinger equation: note that the angular momentum...
Post on 20-Dec-2015
213 views
TRANSCRIPT
( )[ ] )1(sin
1
sin
cos1222
22
2
2
2
2
22
+−=−⎥⎦
⎤Θ+⎢
⎣
⎡ Θ
Θ−=−+⎥
⎦
⎤⎢⎣
⎡+ ll
h l θθθθ
θm
dd
dd
rUEmr
drdR
rdrRd
Rr
( )[ ] )1(22
2
2
2
22
+−=−+⎥⎦
⎤⎢⎣
⎡+ ll
hrUE
mrdrdR
rdrRd
Rr
)1(sin
1
sin
cos12
22
2
+−=−⎥⎦
⎤Θ+⎢
⎣
⎡ ΘΘ
− lll θθθθ
θm
dd
dd
The separated radial and orbital parts of the Schrodinger equation:
Note that the angular momentum equation does not depend on the form of the potential, but it does relate the “magnetic quantum number” to the angular
momentum quantum number.
The radial equation does depend on the form of the potential…it relates the total energy to the angular momentum.
Note that the magnetic quantum number is independent of the energy. This leads to degenerate states.
ErUd
d
rmd
d
d
d
rmdr
dR
rdr
Rd
mR=+
ΦΘ
−⎥⎦
⎤⎢⎣
⎡ Θ+
ΘΘ
−⎥⎦
⎤⎢⎣
⎡+
−)(
sin1
2cot
12
22 2
2
22
2
2
2
2
2
2
22
φθθθ
θhhh
Θ+−=Θ−Θ
+Θ
)1(sin
1
sin
cos2
22
2
lll θθθθ
θm
dd
dd
( )φφ
Φ−=Φ 22
2
lmdd
( ) ( ) ( ) ( ) ( )rERrRrUrRmrdr
dR
rdr
Rd
m=+
++⎥⎦
⎤⎢⎣
⎡+−
2
2
2
22
2
12
2
hllh
tim eYrRtr ωφθφθ −=Ψ ),()(),,,( ll
The Schrodinger equation for a spherically symmetric potential:
where:
The final separated forms:Gives us three quantum numbers…analogous to Ex, Ey, and Ez in the cartesian case…
lKh ll ±±±== ,,2,1,0mmLz
Lz must be less than L, and cannot equal L, otherwise Lx=Ly=0, and all three components of the momentum would be known simultaneously in violation of the uncertainty principle!
)1(cos
+==
lllmLz
Lθ e v
A revolving charge gives rise to a magnetic field.
€
−h2
2m
d2R
dr2+
2
r
dR
dr
⎡
⎣ ⎢
⎤
⎦ ⎥+
l l +1( )h2
2mr2R r( ) +U r( )R r( ) = ER r( )
€
L2
2mr2
€
Korbital =1
2mv 2
L = mvr
∴Korbital =m
2
L
mr
⎛
⎝ ⎜
⎞
⎠ ⎟
2
=L
2
2mr2
energykineticradial
dr
rRd
dr
dR
rdr
Rdr
2
2
2
2 )(2=⎥
⎦
⎤⎢⎣
⎡+
the potentialexpression for kinetic energy
kinetic plus potential energy gives the total energy
kpm
pKE h== ;
2
2kp
m
pKE h== ;
2
2
position x x
momentum p
potential energy U U(x)
kinetic energy K
total energy E
xi ∂∂h
2
22
2 xm ∂∂
−h
ti
∂∂
h
observable
operator
Eigenvalues are a constant of the motion.
Consider the time independent Schrodinger equation…
Ψ=Ψ⎥⎦
⎤⎢⎣
⎡+
−ExU
dx
d
m)(
2 2
22h
When applied to a wavefunction, this expression yields E…energy eigenvalues
For example these are the energy eigenvalues you find when applying this to a simple harmonic oscillator potential
02012
3
0
aZr
ea
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛
( )rRn nll
02
0
23
0
22
02 aZr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
02
0
23
0
232
12 aZr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛
03
2
00
23
0 27
2
3
212
303 a
Zr
ea
Zr
a
Zr
a
Z −
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛
03
00
23
0 61
3
24
313 a
Zr
ea
Zr
a
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
03
2
0
23
0 527
22
323 a
Zr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
,...3,2,12 2
2
0
2
=⎭⎬⎫
⎩⎨⎧
−= nnZ
ake
En
n=1 E=-13.6 eV
n=2 E= -3.4 eV
n=3 E= -1.5 eV
n=4 E= -0.8 eV
allowed transitions
forbidden transition
l=3
l=2
l=1
angular momentum must be conserved
…photons carry angular momentum.
1,2,1,0 −= nKl 1,2,1,0 −= nKl
ll l ≤≤− m ll l ≤≤− m
K,3,2,1=n K,3,2,1=nEnergy: ,...3,2,12 2
2
0
2
=⎭⎬⎫
⎩⎨⎧
−= nnZ
ake
En
Assuming that no more than one electron can occupy each state, there are a total of states.2n
There are a total of orbitals within each subshell.
12 +l
There are a total of n subshells.
Preview: Actually, we will find in Chapter 9 that two electrons can occupy the same orbital if they have different “spin”.
02012
3
0
aZr
ea
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛
( )rRn nll
02
0
23
0
22
02 aZr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
02
0
23
0
232
12 aZr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛
03
2
00
23
0 27
2
3
212
303 a
Zr
ea
Zr
a
Zr
a
Z −
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛
03
00
23
0 61
3
24
313 a
Zr
ea
Zr
a
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
03
2
0
23
0 527
22
323 a
Zr
ea
Zr
a
Z −
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
( ) timn eYrRtr ωφθφθ −=Ψ ,)(),,,( l
ll( ) tim
n eYrRtr ωφθφθ −=Ψ ,)(),,,( lll
22
22
)()(generally
shell symmetricy sphericall afor 4)(
rRrrP
drrrP
=
Ψ= π
drrRrdrrP )()(1 2
00 ∫∫∞∞
==
∫∫∞∞
==0
3
0)()( drrRrdrrrPr
