Graphical interpretation of the
Schrödinger equation
Christophe De Beule
Outline
•
Schrödinger equation
Relation between the kinetic energy and wave function curvature.
Classically allowed and forbidden regions.
•
Examples
Linear potential.
Finite well.
Hybridization.
Electron in a 1D lattice.
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Schrödinger equation
d
dx
dψ
dx
=−
2m
(E − V (x)) ψ(x)
~2
What is the meaning of the Schrödinger equation?
curvature of ψ at x ∝ −kinetic energy at x × ψ at x
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Allowed region
In classically allowed regions, the kinetic energy is positive.
sign curvature of ψ at x = −sign kinetic energy at x × ψ at x
= −sign ψ at x
The curvature and ψ have opposite sign, so that ψ is concave
towards the axis, therefore oscillating around zero.
ψ(x) > 0
ψ(x) < 0
: ψ(x) is concave
: ψ(x) is convex
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Allowed region
ψ(x) > 0
ψ(x) < 0
: ψ(x) is concave
: ψ(x) is convex
ψ(x)
x
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Forbidden region
In classically forbidden regions, the kinetic energy is negative.
sign curvature of ψ at x = −sign kinetic energy at x × ψ at x
= sign ψ at x
The curvature and ψ have the same sign sign, so that ψ is convex
towards the axis, curving away from zero.
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Summary
In regions of small (large) positive kinetic energy, the wave function
oscillates slowly (rapidly) with large (small) amplitude.
ψ(x)
x
In regions of negative kinetic energy, the wave function curves away
from zero.
ψ(x)
x
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Example 1: Linear potential
Consider a free particle (E > 0) in a 1D linear potential,
V (x) = ax,
a > 0,
with kinetic energy T (x) = E − V (x) = E − ax.
T (x)
E
E/a
x
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Linear potential
In the classically allowed region x < E/a, the kinetic energy,
T (x) = E − ax > 0.
ψ oscillates around zero, faster as x becomes more negative.
ψ(x)
E/a
x
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Linear potential
In the classically forbidden region x > E/a, the kinetic energy,
T (x) = E − ax < 0.
A physical ψ has to curve away from zero exponentially slow.
ψ(x)
E/a
x
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Example 2: Finite well
Consider the bound states (E < 0) of a finite potential well, with
V (x) = −V0 θ(−a < x < a),
2m
d dψ
= 2 (|E| + V ) ψ(x).
dx dx
~
V0
x = −a
x=a
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Finite well
When x < −a, the kinetic energy (−|E|) is negative, and ψ
increases exponentially.
ψ(x)
V0
x = −a
x=a
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Finite well
At x = −a, the kinetic energy (V0 − |E|) becomes positive, and
since ψ is positive, the curvature becomes negative.
ψ curves back towards zero, at a rate dependent on V0 − |E|.
ψ(x)
V0
x = −a
x=a
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Finite well
If V0 − |E| is large enough, the slope will be negative at x = a, and
ψ begins curving upwards again.
ψ diverges unless |E| is such, that the slope at x = a is just right,
so that it increases at an exponentially slow rate.
ψ(x)
V0
x = −a
x=a
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Example 3: Hybridization
Consider a two state system, where |Ai and |Bi are two s like states
that hybridize, e.g. the hydrogen molecule.
√1
2
(|Ai − |Bi)
|Ai
|Bi
∆E ≈ 2 hA| Ĥ |Bi
√1
2
(|Ai + |Bi)
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Hybridization
Symmetric combination
• The wave function is always positive and convex in the center,
• Classically forbidden region with negative kinetic energy.
sign curvature of ψ at x = −sign kinetic energy at x × ψ at x
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Hybridization
Antisymmetric combination
• The wave function has a single node in the center,
• Classically allowed region with positive kinetic energy.
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Hybridization
The kinetic energy is given by,
T (x) =
e2
e2
+
− |E|.
|x + b| |x − b|
T (x)
TS = 0
TA = 0
−b
b
x
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Hybridization
The total kinetic energy is given by,
Z
p̂2
hψ|
|ψi = dx |ψ(x)|2 T (x).
2m
The ground state is given by the combination with the fewest nodes,
which is the symmetric combination for s like states.
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Example 4: Electron in a 1D lattice
The wave function of the electron can be written in a basis of
localized (atom-like) states |ni,
X
|ψi =
|ni hn |ψi .
n
|n − 1i
|ni
|n + 1i
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Stationary states
The stationary states are given by,
X
|ψk i =
|ni eikna .
n
All solutions are contained in the Brillouin zone k ∈ (−π/a, π/a].
The wave function is a superposition of localized orbitals modulated
by the phase factor eikx .
a
xn − a
xn
xn + a
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s like states: wave function
Consider the case where the basis states |ni are s like.
X
|φk=0 i =
|n, si .
n
+
+
+
+
+
X
φk=π/a =
(−1)n |n, si .
n
+
−
+
−
+
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s like states: energy band
For s like basis states, the wave function has,
• no nodes at k = 0,
• one node for each unit cell at k = π/a.
The symmetric combination (k = 0) has the fewest nodes.
E
−π/a
π/a
k
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p like states: wave function
Consider the case where the basis states |ni are p like.
X
|φk=0 i =
|n, pi .
n
−
+
−
+
−
+
−
+
−
+
−
+
X
φk=π/a =
(−1)n |n, pi .
n
−
+
+
−
−
+
+
−
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p like states: energy band
For p like basis states, the wave function has,
• one node for each unit cell at k = 0,
• no nodes at k = π/a.
The antisymmetric combination (k = π/a) has the fewest nodes.
E
−π/a
π/a
k
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Question: dimer lattice
Now we have a lattice with a two atom basis. Take one s or p like
basis state for each atom.
What does the band structure look like, qualitatively?
Hint: first look at a single unit cell.
a
|(n − 1)Bi
|nAi
|nBi
|(n + 1)Ai |(n + 1)Bi
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