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Dynamics of Microscopic
Systems
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The Schrodinger equation

➢ A 2nd order differential equation used to calculate the wavefunction (𝜓, psi) of
a system.
➢ 𝜓 contains all dynamic information of a given system (position, speed,
momentum, energy, etc.).
The time-independent Schrödinger equation for one-dimensional system

or ෡ = 𝐸𝜓
𝐻𝜓

- The first term is the kinetic energy of the particle.

- V(x) is the potential energy of the particle at the point x.
- m is the mass of the particle.
- E is the total energy of the particle.
- ℏ = h/2 = 1.055 x 10-34 J s (modified Planck’s constant).
-𝐻෡ = - (ℏ𝟐/2m) 2+ V (the Hamiltonian operator).
- 2 “del squared” is Laplacian operator; 2 = 2/x2 .
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A simple 𝜓 that satisfies Schrodinger equation

The time-independent
Schrödinger equation
for one-dimensional
system:
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De Broglie relation from Schrodinger equation

The time-independent
Schrödinger equation
for one-dimensional
system
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Born interpretation of the wavefunction

➢ Max Born: the square of

the wavefunction (2) is a
measure of the probability
of finding a particle in a
specific region.
➢ 2 is regarded as a Two advantages:
probability density. 1- The significance of a negative value of ψ
Normally expressed as is ignored.
2- Complex values in ψ will be eliminated.
the square modulus, |ψ|2 |ψ|2 = ψ*ψ, if ψ is complex.

𝑥2
Probability of locating a particle P= න 𝜓(𝑥) 2 d𝑥 One-dimensional region
𝑥1
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Born interpretation of the wavefunction

One-dimensional system Three-dimensional system (probability
of finding e- near a nucleus)

The distance from nucleus = r

Probability of finding a particle Probability of finding the

in the region dx  |ψ(x)|2dx. particle in an infinitesimal
volume d = dxdydz 
|ψ|2 d.
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Born interpretation: Hydrogen atom

The wavefunction of an electron in
the lowest energy state of a P = |ψ|2 d (𝜓2 = e-2r/aº)
hydrogen atom is proportional to e-
r/a
º a. At the nucleus r = 0
(a0 is Bohr radius = 52.9 (constant), r is → P = e0 x 1.0 pm3 = 1
the distance from the nucleus).
➢ Calculate the relative b. At the distance a0
probabilities of finding the → P = e-2 x 1.0 pm3 = 0.14
electron inside a region of volume
dV = 1.0 pm3 located at: The ratio of probabilities
a) the nucleus = 1/0.14 = 7.1
b) a distance a0 from the
nucleus. It is more probable by a factor of 7 to
find the electron at the nucleus than
at a distance of Bohr radius a0 = 52.9
pm (justify??)
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Normalization

The sum over all space of the probabilities (P) must be 1.

If P  1, then  has to be normalized, i.e. multiplied by a constant (N)
where N is the normalization constant.

and

It is here assumed that ψ already includes a normalization factor that

ensures that:
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Normalizing a wavefunction
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Spherical polar coordinates

In three dimensions, normalized
ψ is given by:

The spherical polar

coordinates used for
discussing systems
but d = dxdydz with spherical
symmetry

For systems with spherical

symmetry, it is best to work in The surface of a
sphere is covered by
spherical polar coordinates allowing θ to range
r, θ, and . from 0 to π, and then
sweeping that arc
around a complete
circle by allowing  to
range from 0 to 2π
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Normalizing a wavefunction using spherical coordinates

Normalize the wavefunction
used for the hydrogen atom.
𝜓 = e-r/aº

N2
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Constrains on the wavefunctions

 must: Due to such severe restrictions  acceptable
solutions of Schrödinger eq. (e.g. E) can’t be
(a) be continuous (no any values  a particle may possess only
certain energies  energy is quantized.
disconnection).

(b) have a continuous

slope (no kinks, and first
derivative  0).

(c) have only one value at

each point of space
(single-valued).

(d) be finite everywhere,

must not rise to infinity in a
finite region
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The information in a wavefunction

➢ The Schrodinger equation with zero potential energy

–(ℏ2/2m)(d2𝝍/dx2) = E𝝍
The solution for this equation:
 = Aeikx + Be-ikx E = ℏ𝟐k2 /2m
A and B: constants
➢ To verify that  is a solution:

–ℏ2 d2 –ℏ2
–(ℏ2/2m)(d2𝜓/dx2) = 2𝑚 2(Ae + Be ) = 2𝑚 [A(ik)2eikx + B(-ik)2e-ikx]
ikx -ikx
dx
ℏ2k2
= 2𝑚 [Aeikx + Be-ikx] = E
෡ ψ = Eψ)
➢ Remember (H
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A wavefunction of ex form can also be used

➢ The Schrodinger equation with zero potential energy
–(ℏ2/2m)(d2𝝍/dx2) = E𝝍
The solution for this equation ( is normalized):
 = Aeikx + Be-ikx E = ℏ𝟐k2 /2m
A and B are constants.

To verify that  is a solution:

–ℏ2 d2
–(ℏ2/2m)(d2𝜓/dx2) = 2𝑚 2 (Aeikx + Be-ikx)
dx
–ℏ 2
= 2𝑚 [A(ik)2eikx + B(-ik)2e-ikx]
ℏ2k2
(i2 = -1) = [Aeikx + Be-ikx] = E
2𝑚

෡ = 𝐸𝜓)
Remember (𝐻𝜓

Let’s determine the probability of finding a particle

using the wavefunction  = Aeikx + Be-ikx .
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Probability density |𝝍|𝟐

To find it out, let’s set B = 0 →  = Aeikx

||2 = (Aeikx)*(Aeikx) = (A*e-ikx)(Aeikx) = |A|2 (* means a complex conjugate)
The probability density is independent of x  there is equal probability of finding
the particle along x-axis  position of the particle can’t be predicted.

If we set A = 0 → again ||2 = |B|2  same problem.

Let’s then set A = B   = A( eikx + e-ikx ) = 2A cos(kx) ½ (eikx + e-ikx) = cos(kx)

Hence,
||2 = 4|A|2cos2(kx)

So, the probability density varies

between 0 and 4|A|2

Node, wavefunction = 0