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QUANTUM IN : SEPARABLE SYSTEMS
1D Systems 3D Systems
x
r
(
x ˆ y ˆ z ˆ
)
= i x ˆ + j y ˆ + k z ˆ
� d
p ˆ = + j + k
i dx
p
(
p ˆ
x
p ˆ
y
p ˆ
z
)
= i
i ∂ x i ∂ y i ∂ y
⎡ x ˆ, p ˆ ⎤ i � ⎡ ⎤
⎡ ⎤
⎣ ⎦
⎡
⎣
x ˆ, p ˆ
x
⎤
⎦
= i �
⎣
y p ˆ, ˆ
y
⎦
= i �
⎣
z ˆ, p ˆ
z ⎦
= i �
p
2
2
d
2
p
2
2
2
2
2
2
2
T = = T = = + +
2 m 2 m dx
2
2 m 2 m ∂ x
2
2 m ∂ y
2
2 m ∂ y
2
ψ
(
x
)
ψ
(
x y z , ,
)
O
ˆ
=
∫
ψ
(
x
)
O
ˆ
ψ
(
x
)
dx O
ˆ
=
∫
ψ
(
x y z , ,
)
O
ˆ
ψ
(
x y z , ,
)
dx dy dz
By fiat, operators corresponding to different axes commute with one another.
p y
= yp p p
xy yx etc.
p p
z z z x x z
Further, operators in one variable have no effect on functions of another:
xf
(
y )
= f (
y )
p f (
x )
= f (
x )
f
(
z )
p =
( )
x
z
p
z
x
p
x
f z etc.
The Time Independent Schrödinger Equation becomes:
2
2
2
2
2
2
2
+ V
(
x
, y
, z ˆ )
⎥
(
x , y , z )
= E ψ
(
x , y , z )
2 m
∂ x ∂ y ∂ z
2
the Laplacian
2
2
⇒
− ∇ + V
(
x
, y
, z ˆ )
ψ (
x , y , z )
= E ψ (
x , y , z )
2 m
H
2
2
+ V
(
x
)
y z Hamiltonian operator in 3D
2 m
(
, y z , )
(
x , , )
H ψ x E y z 3D Schrödinger equation
(Time Independent)
Separation of variables
IF V
(
x
y
z
)
= V
x
(
x
)
+ V
y
(
y
)
+ V
z
(
z
)
2
2
2
2
2
2
( )
x
( )
y
( )
z
( )
, − x y + − + V
H x , y z = + V
+ − + V
z
then
⎣
2 m ∂ x
2 m ∂ y
2 m ∂ z
= H
+ H
+ H
x y z
⇒ Schrödinger’s Eq. becomes:
H
+ H
+ H
ψ x , y , z = E ψ x , ,
x y z
( ) (
y z )
Then try solution of form ψ (
x , y , z )
=ψ
x
(
x )
ψ
y
(
y )
ψ
z
(
z )
(separation of variables)
Where we assume that the 1D functions satisfy the appropriate 1D TISE:
H
(
x )
= E ψ
(
x )
x x x x
H
y
y
(
y )
= E
y
y
(
y )
H
(
z )
= E ψ
(
z )
z z z z
First term:
H
x
x
( x ) ψ
y
( y ) ψ
z
( z ) =ψ
y
( y ) ψ
z
( z ) H
x
x
( x ) =ψ
y
( y ) ψ
z
( z ) E
x
x
( x )
E ψ
( ) ( )
= ψ z
x x
x ψ
y
y
z
( )
Same for H
and H
⇒
y z
H
ψ = E ψ
H
x
+ H
y
+ H
z
⎦
x
(
x )
y
(
y )
z
(
z )
(
E
x
+ E
y
+ E
z
)
x
(
x )
y
(
y )
z
(
z )
E = E + E + E
x y z
Thus, if the Hamiltonian has this special form, the eigenfunctions of the 3D
Hamiltonian are just products of the eigenfunctions of the 1D Hamiltonian and
the situation is equivalent to doing three separate 1D problems.
Conclusion: Wavefunctions multiply and the energies add if H
is separable into
Harmonic potentials are also important for describing the vibrations of
polyatomic molecules. For example, in HCO x might correspond to the CH
stretch, y to the C O stretch and z to the H C O bend.
In either case, the general potential is given by
V ( x , y , z ) = k x + k y + k z = V x + V y + V z
x
2
y
2
2
x
( )
y
( )
z
( )
z
Now, because the potential is a
sum of an x potential, a y
potential and a z potential, we
can easily write the
Hamiltonian down as a sum:
H
= H
+ H
+ H
x y z
p ˆ
x
2
2
H
= + k x ˆ
x x
2 m 2
p
y
2
2
H = + k y ˆ
y y
2 m 2
H
p ˆ
z
2
k z ˆ
2
z z
2 m 2
where each 1D Hamiltonian describes a particle subject to a Harmonic potential
with the appropriate spring constant (k x
, k
y
or k
z
). Based on the discussion
above, we can immediately write down all the eigenfunctions and eigenvalues:
1
−
α
x
x
2
−
α
y
y
2
α
⎞
4 −
α
z
z
2
n n
x y
n
(
x , , )
= N H
x n
x
(
α
x
1/
x
)
e
2
N H
y n
y
(
α
y
1/
y
)
e
2
z
⎟ z n
z
(
z
1/
)
2
ψ y z N H α z e
z
π
⎠
E
x
n
z
n
x
� ω
x
n
y
� ω
y ⎥
n
x
� ω
n n z
y
(
mk
x
)
1 2
k
x
α = ω = etc.
x x
� m
Notice again that for the 3D problem we have 3 quantum numbers, and the
energy and wavefunction depend on all three simultaneously.
Degeneracies
In 3D, there are a number of interesting things that can happen that we didn’t
see in 1D. One example of this is that, in 3D it is possible for two different
eigenfunctions of the Hamiltonian to have the same energy. When this happens,
these two states are called degenerate. If there are three, four… states with
the same energy, they are said to be threefold, fourfold … degenerate. This
never happened for the Particle in a Box or the Harmonic Oscillator. In fact,
one can show that for bound states in 1D, one never has degeneracy; every state
has its own energy. However, in 3D we can find degeneracy very easily. For
example, if our spring constants are all the same:
k = k = k ≡ k ⇒ ω =ω =ω ≡ω
x y z x y z
⇒ E
n n n
= � ω
n
x
y
z
x y z
The ground state has an energy E
000
=
2
3
� ω. There is only one way I can get this
energy ( n
x
= 0, n
y
= 0, n
z
= 0 ), so it is not degenerate. However, there are three
ways I can get the first excited state energy
5
2
� ω : n
x
= 1, n
y
= 0, n
z
= 0 ,
n
x
= 0, n
y
= 1, n
z
= 0 or n
x
= 0, n
y
= 0, n
z
= 1. So we find that if we choose all the
spring constants to be equal,
E
000
= nondegenerate level
E
100
= E
010
= E
001
= 3-fold degenerate level
etc.
We will typically draw this with a picture like:
E
…
E 110
E 011
E
101
E
200
E
020
E
002
6 fold degenerate
E
100
E
010
E
001
3 fold degenerate
E 000
Non degenerate
Note the wavefunctions are distinct:
100
(
x , , )
010
(
x , y z , )
001
(
x , y , )
y z z
This leads to an interesting effect. Suppose we make up a wavefunction that is
a sum of two degenerate states, say
ψ ( x , y , z ) = a ψ
010
x , , ) + b ψ
001
( y z ( x , y , z )
E
000
z
nondegenerate level
E
100
= E
010
z
2-fold degenerate level
E
001
z
nondegenerate level
Example 2: Particle in 3 D box
x
z
y
b
a
c
V
(
x , y , z )
= V
x
(
x )
+ V
y
(
y )
+ V
z
(
z )
V
( )
x = 0 0 ≤ x ≤ a
x
V
y
( )
y = 0 0 ≤ y ≤ b
V
( )
z = 0 0 ≤ z ≤ c
z
V x , V y , V z = ∞ otherwise
x
( )
y
( )
z
( )
Inside the box: Outside the box:
2
2
2
2
2
2
2
(
x , y , z )
= E ψ
(
x , y , z )
(
x , y , z
)
2 m
∂ x ∂ y ∂ z
We can again apply separation of variables since H
= H
x
+ H
y
+ H
z
where
H
ˆ
x
, H
ˆ
y
, H
ˆ
z
are each 1D particle in a box Hamiltonians. So the solutions to the
3D equation are products of the 1D solutions
⇒
(
x , y , z )
x
(
x )
y
(
y )
z
(
z )
n n n
where from the 1D problem we have the solutions
1
2
n
x
π x
2
n
x
2
n
x
( )
x =
sin
n
x
= 1, 2,3,... E
n
2 x
a
⎠ ⎝
a
⎠
8 m a
1
2 ⎛
n
y
π y
2
n
2
y
n
y
( )
y =
sin
n
y
= 1, 2,3,... E
n
2 y
b
⎠
⎝
b
8 m b
1
n
z
( )
z =
2
sin
n
z
π z
n
z
= 1, 2,3,... E
n
z
2
n
2
z
2
c
⎠ ⎝
c
⎠
8 m c
Where the energy is now a function of all three Quantum numbers:
If the symmetry is “broken”, i.e. a ≠ b ≠ c then the degeneracy is lifted.
h
2
h
2
E
112
2
2
2
≠ E
121
2
2
2
8 m ⎝
a b c ⎠
8 m ⎝
a b c ⎠
Summary
3 D box
1
2
n
x
π x ⎞
n
y
π y
⎞ ⎛
n
z
π z ⎞
n n n
(
x , , )
sin
sin
y z sin
x y z
abc ⎠ ⎝
a
⎠ ⎝
b
c ⎠
h
2
n
2
n
2
n
2
n = 1,2,3,... n = 1,2,3,... n = 1,2,3,...
n n n x y z
E
x y z
x
2
2
y
2
z
8 m
a b c
