Download SCPY152 Lecture 20 Quantum Physics in 3 Dimensions

SCPY152
Lecture 20 Quantum Physics in 3 Dimensions
Udom Robkob, Physics-MUSC
March 17, 2015
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Today Topics
Rigid potential cube
Isotropic harmonic oscillator in 3D
Rigid potential sphere
Quantization of angular momentum
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
Let there be an infinite potential cube of size a with
potential function

x, y, z ≤ 0
 ∞ ,
0 , 0 < x, y, z < a
V (x, y, z) =

∞ ,
a, y, z ≥ a
Udom Robkob, Physics-MUSC
(1)
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
There are no particle waves outside the cube.
Inside the cube, there are three independent standing
waves in three directions;
r
n πx 2
1
sin
,
ϕn1 (x) =
a
a
r
n πy 2
2
ϕn2 (y) =
sin
,
a
a
r
n πz 2
3
ϕn3 (z) =
sin
.
a
a
(2)
The corresponding energies of each direction are
En1 =
n22 π 2 ~2
n23 π 2 ~2
n21 π 2 ~2
,
E
=
,
E
=
,
n
n
2
3
2ma2
2ma2
2ma2
(3)
with n1 , n2 , n3 = 1, 2, 3, ... are the three quantum numbers.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
The composite wave function inside the cube come from
mulitplication of the three standing waves;
ϕn1 ,n2 ,n3 (x, y, z) = ϕn1 (x)ϕn2 (y)ϕn3 (z)
3/2
n πx n πy n πz 2
1
2
3
sin
sin
sin
.
=
a
a
a
a
(4)
The corresponding composite energy come from addition of
the three energies;
En1 ,n2 ,n3 = En1 + En2 + En3 = (n21 + n22 + n23 )
Udom Robkob, Physics-MUSC
π 2 ~2
. (5)
2ma2
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
Examples of energy levels:
n1 n2 n3 En1 ,n2 ,n3
1
2
1
1
2
2
1
3
1
1
2
1
1
2
1
2
1
2
1
3
1
2
1
1
1
2
1
2
2
1
1
3
2
Udom Robkob, Physics-MUSC
π 2 ~2
2ma2
3
6
6
6
9
9
9
11
11
11
12
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
”Degeneracy” is the number of degenerated energy levels,
i.e, having different quantum numbers but the same energy
value.
For examples,
E1,1,1 = 3
E2,2,2 = 12

E2,1,1 = E1,2,1 = E1,1,2 = 6 
E2,2,1 = E2,1,2 = E1,2,2 = 9

E3,1,1 = E1,3,1 = E1,1,3 = 11
nondegenerate
deneracy = 3
Degenertaed states come from symmetry, i.e., symmetry of
the potential cube. Look the same under 90 degree
rotations.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
These waves and energies come from Schrodinger equation
inside the cube, which appear as
2
∂
∂2
∂2
+
+
ϕn1 ,n2 ,n3 (x, y, z)
∂x2 ∂y 2 ∂z 2
2m
En1 ,n2 ,n3 ϕn1 ,n2 ,n3 (x, y, z) = 0.
(6)
~2
This equation can be rewritten in the form
1 d2 ϕn1
2mEn1
1 d2 ϕn2
2mEn2
+
+
+
ϕn1 dx2
~2
ϕn2 dy 2
~2
1 d2 ϕn3
2mEn3
+
+
= 0.
(7)
ϕn3 dz 2
~2
+
This is known as a method of separation of variables.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
It contains three Schrodinger equations in three
independent directions.
So we can say that in D dimensions with separable
potential function,
V (x1 , x2 , ..., xD ) = V (x1 ) + V (x2 ) + ... + V (xD ),
the method of separation of independent variables can be
applied.
The resulting resolution will appear in the form
ϕn1 ,n2 ,...,nD (x1 , x2 , ..., xD ) = ϕn1 (x1 )ϕn2 (x2 )...ϕnD (xD )
En1 ,n2 ,...,nD = En1 + En2 + ... + EnD . (8)
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Rigid Potential Cube
Each of the separated part will satisfy the corresponding
Schrodinger’s equation
2
d
2m
+ 2 (Eni − V (xi )) ϕni (xi ) = 0.
(9)
~
dx2i
For i = 1, 2, ..., D.
Note also that we will get D quantum numbers,
n1 , n2 , ..., nD , for a particle living in D-dimenions.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Isotropic Harmonic Oscillator in 3D
Let us consider the isotropic harmonic oscillator in
3-dimensions. Its potential function appear in the form
1
mω 2 (x2 + y 2 + z 2 ).
2
= V (x) + V (y) + V (z)
V (x, y, z) =
(10)
It is a separable function. Its quantized energies will be
En1 ,n2 ,n3
= En1 + En2 + En3
3
= ~ω n1 + n2 + n3 +
,
2
(11)
with n1 , n2 , n3 = 0, 1, 2, ... .
Here its wave functions are not interesting, because they
are not easy to display.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Isotropic Harmonic Oscillator in 3D
Another viewpoint of this problem can be done by writing
in spherical coordinates: (r, θ, φ):
Relations to carthesian coordinate system appear as
x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
2
2
2
x +y +z =r
2
(12)
(13)
Here we have mathematical relations: x = x(r, θ, φ),
y = y(r, θ, φ), and z = z(r, θ).
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Isotropic Harmonic Oscillator in 3D
The isotropic harmonic oscillator potential appear in
spherical coordinates as
1
V (r) = mω 2 r2 .
2
(14)
Now let us rewrite Schrodinger equation in spherical
coordinates.
First of all, we have to determine the variations
dx = dr sin θ cos φ + r cos θ cos φdθ − r sin θ sin φdφ
(15)
dy = dr sin θ sin φ + r cos θ sin φdθ + r sin θ cos φdφ
(16)
dz = dr cos θ − r sin θdθ.
Udom Robkob, Physics-MUSC
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SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Schrodinger Equation in Spherical Coordinates
These variations can be written in matrix form as

 


dx
sin θ cos φ cos θ cos φ − sin φ
dr
 dy  =  sin θ sin φ cos θ sin φ cos φ  

rdθ
dz
cos θ
− sin θ
0
r sin θdφ
(18)
By invsersion, we can get mathematical relations:
r = r(x, y, z), θ = θ(x, y, z), and φ = φ(x, y).
Then inverse relations will appear as

 


dr
sin θ cos φ − sin θ sin φ cos θ
dx

 =  − cos θ cos φ cos θ sin φ sin θ   dy 
rdθ
r sin θdφ
− sin φ
− cos φ
0
dz
(19)
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Schrodinger Equation in Spherical Coordinates
We now change the differentiations from cartesian system
to shperical system by using the chain rule as
∂
∂x
∂
∂y
∂
∂z
=
=
=
∂r ∂
∂θ ∂
∂φ ∂
+
+
∂x ∂r ∂x ∂θ ∂x ∂φ
∂r ∂
∂θ ∂
∂φ ∂
+
+
∂y ∂r ∂y ∂θ
∂y ∂φ
∂r ∂
∂θ ∂
∂φ ∂
+
+
∂z ∂r ∂z ∂θ
∂z ∂φ
(20)
(21)
(22)
Exercise: Give full details expressions of above equations.
Then we calculate
∂2
∂2
∂2
+
+
= ∇2
∂x2 ∂y 2 ∂z 2
in terms of ∂/∂r, ∂/∂θ, and ∂/∂φ.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Schrodinger Equation in Spherical Coordinates
We found that
∇
2
1 ∂
2 ∂
r
=
r2 ∂r
∂r
1
1 ∂
∂
1 ∂2
+ 2
sin θ
+
(23)
r
sin θ ∂θ
∂θ
sin2 θ ∂φ2
Let us define the angular differential operator
1 ∂
∂
1 ∂2
2
2
L = −~
sin θ
+
sin θ ∂θ
∂θ
sin2 θ ∂φ2
(24)
It is just the angular momentum sqaured operator written
in spherical coordinates;
~ = ~r × p~,
L
L2 = L2x + L2y + L2z
(25)
where Lx = ypz − zpy = −i~(yd/dz − zd/dy), etc.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Quantization of Angular Momentum
Exercise: Write Lx , Ly , Lz operators interms of angular
differentiations, d/dθ, d/dφ.
With separation of wave function:
ϕ(x, y, z) → ϕ(r, θ, φ) = R(r)Y (θ, φ)
(26)
The angular function Y (θ, φ) is found to satisfy the
following equations
L2 Yl,m (θ, φ) = l(l + 1)~2 Yl,m (θ, φ)
(27)
Lz Yl,m (θ, φ) = m~Yl,m (θ, φ)
(28)
where
Yl,m (θ, φ) → spherical harmonics
(29)
l
=
0, 1, 2, 3, ...
(30)
m
=
−l, −(l − 1), ..., (l − 1), l
(31)
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Quantization of Angular Momentum
Note that l is known as ”angular momentum quantum
number” and m is known as ”magnetic moment quantum
number”.
Examples of the quantized values of the angular momentum
l = 0, m = 0 → L2 = 0, Lz = 0
l = 1, m = 0, ±1 → L2 = 2~2 , Lz =
(32)
±~
0
(33)

 ±2~
2
2
±~ (34)
l = 2, m = 0, ±1, ±2 → L = 6~ , Lz =

0
Exercise: Give the quantized values of angular momentum
at l = 3, 4.
What is the meaning of these quantized values?
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Quantization of Angular Momentum
Let us look at the picture of angular momentum vector;
L2 is measured of length or magitude of the angular
momentum.
Lz is measured of z-component, or orientation, of the
angular momentum.
Quantization of the angular momentum means having
discrete lengths and discrete orientations, with respect to
z-direction.
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Quantization of Angular Momentum
Exercise: Draw the angular momentum diagrams in
quantum states l = 3, 4 .
We can look at the corresponding state functions, or
spherical harmoincs Yl,m (θ, φ), as in the following figures;
Udom Robkob, Physics-MUSC
SCPY152 Lecture 20 Quantum Physics in 3 Dimensio
Quantization of Angular Momentum
Here are their detail expressions;
r
1 1
Y0,0 =
2 π
r
1 3
cos θ
Y1,0 =
2 π
r
1
2 ±iφ
Y1,±1 = ∓
e
sin θ
2 2π
r
1 5
Y2,0 =
3 cos2 θ − 1
4 π
r
1 15 ±iφ
Y2,±1 = ∓
e
sin θ cos θ
2 2π
r
1 15 ±2iφ 2
Y2,±2 =
e
sin θ
4 2π
Udom Robkob, Physics-MUSC
(35)
(36)
(37)
(38)
(39)
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SCPY152 Lecture 20 Quantum Physics in 3 Dimensio