SCPY152 Lecture 20 Quantum Physics in 3 Dimensions

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SCPY152
Lecture 20 Quantum Physics in 3 Dimensions
Udom Robkob, Physics-MUSC
March 17, 2015
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
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
(1)
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.
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 )
π 2 ~2
. (5)
2ma2
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
π 2 ~2
2ma2
3
6
6
6
9
9
9
11
11
11
12
”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.
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.
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)
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.
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.
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).
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.
(17)
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)
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.
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.
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)
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?
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.
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;
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
(35)
(36)
(37)
(38)
(39)
(40)

SCPY152 Lecture 20 Quantum Physics in 3 Dimensions

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