Download Schroedinger’s equation in 3-dimensions Rohlf 7.6 P213-216

Schroedinger’s equation
in 3-dimensions
Rohlf 7.6 P213-216
Schroedinger’s Equation


∂Ψ( r ,t)



2
−
∇ Ψ( r ,t) + V ( r )Ψ( r ,t) = i
2m
∂t
2
Separation of variables:
Time equation:
Space equation:


Ψ( r ,t) = ψ ( r )T (t)
∂T iα
+ T =0
∂t 
2m

∇ ψ + 2 α − V (r ) ψ = 0

2
(
)
Solution to the t dependent equation:
∂T iα
+ T =0
∂t 
Assume an exponential solution:
Insert into the diff. eq. above:
iα bt
bAebt +
Ae = 0

iα
b+
=0

The solution is then
T = Aebt
A∂ebt iα bt
+ Ae = 0
∂t

iα
b=−

T = Ae
− iω t
T = Ae
ω =
α

−
iα
t

Solution to time equation.
α=E
∂T iE
+ T =0
∂t 
E = hf =  ω
T = Ae−iω t
T=
E
−i t
Ae 
Solution to time equation.
T=
E
−i t
Ae 
= Ae−iω t
Schroedinger space equation in Cartesian coordinates
∂2ψ ∂2ψ ∂2ψ 2m
+ 2 + 2 + 2 E − V (x, y, z) ψ = 0
2
∂x
∂y
∂z

(
)
2m
2m
2m p 2 p 2
2
E
−
V
=
KE
=
=
=
k
2
2
 2 2m  2
(
)


note: p and k depend on x, y and z.
∂2ψ ∂2ψ ∂2ψ
2
+
+
+
k
ψ =0
2
2
2
∂x
∂y
∂z
Specialize to 2-dimensions.!
Solve for 2-dimensional infinite wall box of dimensions Lx = Ly = L
Assume the solution can be factorized. !
ψ (x, y) = X (x)Y ( y)
Inside the box the particle is free
k2
p2
2mE
=
= 2m
E
−V
=
( )
2m
2
2
Outside the box, and on the boundries of the box ψ = 0.
Outside the box, and on the boundries of the box ψ=0.
Inside the box free particle with constant |k|
Insert ψ (x, y) = X (x)Y ( y) in Schroedinger's equation:
∂2 ( XY ) ∂2 ( XY )
Y∂2 X X ∂2Y
2
2
+
+
k
XY
=
+
+
k
XY = 0
2
2
2
2
∂x
∂y
∂x
∂y
Divide by XY
Y ∂2 X
X ∂ 2Y
1 ∂ 2 X 1 ∂ 2Y
2
2
+
+
k
=
+
+
k
=0
XY ∂x 2 XY ∂y 2
X ∂x 2 Y ∂y 2
1 ∂2 X
1 ∂ 2Y
2
2
=
−
−
k
=
−
γ
X ∂x 2
Y ∂y 2
(
)
∂ 2Y
+ k 2 − γ 2 Y = 0,
2
∂y
∂2 X
+γ 2X = 0
2
∂x
Exercise
∂2 X
∂x 2
∂2Y
+γ X = 0
2
∂y 2
+ β 2Y = 0
Show the following are solutions:
X (x) ∝ e±iγ x
(
β2 ≡ k2 − γ 2
Y ( y) ∝ e±iβ y
)
Assume an exponential solution:!
∂2 X
∂x 2
+γ 2X = 0
X = Aeax
a 2 eax + γ 2 eax = 0
∂ 2Y
∂y
2
(
)
+ k 2 − γ 2 Y = 0,
b2 eby + k 2 eby − γ 2 eby = 0
a = ±iγ
Y = Beby
b = ±i k 2 − γ 2
≡ ± iβ
∂2 X
∂x 2
∂2Y
+γ X = 0
2
∂y 2
+ β 2Y = 0
(
β2 ≡ k2 − γ 2
The following are solutions:
(
)
=
Ax sin(γ x) + Bx cos(γ x)
(
)
=
Ay sin( β y) + By cos( β y)
X (x) ∝ eiγ x ± e−iγ x
Y ( y) ∝ eiβ y ± e−iβ y
)
Solution for a free particle.
(
ψ (x, y) = X (x)Y ( y) = C eiγ x ± e−iγ x
)(
(
eiβ y ± e−iβ y
)
)(
ψ (x, y) = X (x)Y ( y) = Ax sin(γ x) + Bx cos(γ x) Ay sin( β y) + By cos( β y)
Apply boundary condition.
at x=0 and x = L
ψ = 0 ∴ Bx = 0, γ L = nπ
at y=0 and y = L
ψ = 0 ∴ By = 0,
β L = mπ
nπ
γ = kx =
L
β = ky =
mπ
L
)
(
)
(
)
X (x) = A eiγ x ± e−iγ x
Y ( y) = B eiβ y ± e−iβ y
knm =
Enm =
2mEnm

nπ
(
2mL
2 2
2
2 2
+m π
)
=
= k xn
mπ
= k ym
L
2
k 2 − k xn
Enm =
2
2
nπ
L
=
β =
k2 − γ 2 =
β = k ym =
2
γ
⇒
2
 2 knm
2m
2
(
2 2
=
k xn + k 2ym
2m
n
(
2mL
π 22
2
2
k 2 = k xn
+ k 2ym = knm
2
+ m2
)
)
Homework
Apply the method of separation of variables
to obtain the energy levels of a rigid wall
cubical box having sides Lx,Ly,Lz=L
π 22 2
2
2
E
=
n
+
m
+
l
nml 2mL2
(
)
where n, m and l are integer quantum numbers
(note, Rohlf uses n1,n2 and n3 instead of n,m and l)
Solution:
Assume ψ (x, y, z) = X (x)Y ( y)Z(z) . Insert in Schroedinger's equation. Divide by XYZ:
YZ ∂2 X
XZ ∂2 Y
XY ∂2 Z
1 ∂2 X 1 ∂2Y 1 ∂2 Z
2
+
+
+k =
+
+
+ k2 = 0 = 0
2
2
2
2
2
2
XYZ ∂x
XYZ ∂y
XYZ ∂z
X ∂x
Y ∂y
Z ∂z
(
1 ∂2 X
1 ∂2Y 1 ∂2 Z
=−
−
− k 2 = −γ 2
2
2
2
X ∂x
Y ∂y
Z ∂z
(
)
1 ∂2Y
1 ∂2 Z
=−
− k 2 − γ 2 = −β 2
2
2
Y ∂y
Z ∂z
∂2 X
∂x
2
+γ X = 0
2
∂2Y
∂y
2
)
1 ∂2Y 1 ∂2 Z
+
+ k 2 − γ 2 = 0,
2
2
Y ∂y
Z ∂z
∂2Y
∂y
∂2 Z
2
+β Y =0
∂z
2
∂2 Z
2
+β Y =0
2
(
+ k −γ −β
2
2
∂z
2
2
∂x
(
2
+γ 2X = 0
)
+ k2 − γ 2 − β2 Z = 0
) Z = ∂z
∂2 Z
2
+ ηZ = 0
)(
(
∂2 X
)(
ψ (x, y, z) = X (x)Y ( y)Z(z) = Ax sin(γ x) ± Bx cos(γ x) Ay sin( β y) ± Bz cos( β y) Az sin(η z) ± Bz cos(η z)
Ψ(x, y, z) = C sin(γ x) sin( β y) sin(η z)
Apply boundary condition at x, y, x each at 0, L
nπ
mπ
lπ
γ =
,β=
,η=
L
L
L
π2 2
with k = γ + β + η = 2 n + m2 + l 2
L
2
2
2
2
(
)
)
Exercise
Suppose an Oxygen atom (Z=8) can be approximated by a cubical potential
of side L=5 A (=0.5 nm).
What are the quantum energies, in eV, of each electron assuming at most only 2 electrons may occupy each quantum state. Density of states and Fermi energy.
Rohlf Ch. 12 Sec.12.6
Quantized momentum k
kx =
nπ
L
ky =
mπ
L
kz =
lπ
L
k space or momentum space

k

ky

kz
k 2 = k x2 + k 2y + k z2

kx
2 2
p2  k
2 2
E=
=
=
k x + k 2y + k z2
2m
2m
2m
Surfaces of constant k2 have constant energy
(
)
All states at a radius k have the same energy.
ky
dk
kx
kz
π π π π3
π3
Volume per state = Δk xn Δk ym Δk zl =
= 3 =
LLL L
V
This is the same for all shapes of volumes V for high quantum numbers.
Volume in spherical sector =
14 3
1
π k = π k3
83
6
Number of states up to k (or Ek.) !
Rohlf sec. 14.2, p375-377
!
1 4π 3
π3
Volume in k space up to k vk =
k . Volume per state vs =
.
8 3
V
Number of states up to k :
Number of states up to E :
π 3
k
vk
V 3
N =
= 6
=
k .
2
3
vs
6π
π
V
2
k =
2m

2
V ⎛ 2m ⎞
N=
⎜
⎟
6π 2 ⎝  2 ⎠
⎛ 2m ⎞
k =⎜
E
⎝  2 ⎟⎠
3
E
3/2
E 3/2 .
3
2
Number of states up to E :
V ⎛ 2m ⎞
N=
⎜
⎟
6π 2 ⎝  2 ⎠
3/2
E 3/2 .
Number of states up to E per unit volume :
N
1 ⎛ 2m ⎞
=
V 6π 2 ⎜⎝  2 ⎟⎠
3/2
E 3/2
N
1 ⎛ 2m ⎞
Number of states up to E per unit volume :
=
V 6π 2 ⎜⎝  2 ⎟⎠
3/2
At T =0, electrons fill all states, 2 per state, to the Fermi energy:
Ne
1 ⎛ 2m ⎞
= ne = 2 ⎜ 2 ⎟
V
3π ⎝  ⎠
EF =
3/2
3/2
EF
⎛ 2m ⎞
Fermi energy : ne =
⎜⎝ 2 ⎟⎠ EF
2/3

3π 2
2/3
2
h ⎛ 3ne ⎞
8m ⎜⎝ π ⎟⎠
1
( )
2/3
where h = 2π 
E 3/2