QM Lecture 2: The Schrödinger Equation

**By: Tao Steven Zheng (郑涛)**

*This article was originally posted on Brilliant.org by the same author.*

## **From Wave to Particle**

Louis de Broglie taught us that all particles in the universe possess a wavelength. We will consider microscopic objects because they closely follow the wave-particle duality more than macroscopic objects. Consider a particle travelling in space with no potential to hold it down: we will call this the free particle. Since microscopic particles also behave like waves, we will model the free particle as a sinusoidal plane wave enveloped in a corresponding **wave-packet**. The sinusoidal plane wave propagates in space and evolves in time; thus, we can express it as a complex exponential function

Ψ
(
x
,
t
# )

e

i
(
k
x
−
ω
t
)

{\displaystyle \Psi (x,t) = {e}^{i(kx-\omega t)} }

and wave-packet (Figure 1)

Ψ
(
x
,
t
# )

1

2
π

∫

−
∞

∞

ϕ
(
k
)

e

i
(
k
x
−
ω
t
)

d
k
.

{\displaystyle \Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int\_{-\infty}^{\infty}{\phi(k)}{e}^{i(kx-\omega t) }dk. }

Using de Broglie’s equations of momentum

# k

h
ν

{\displaystyle p=\hbar k=\frac{h}{\nu} }

 and energy

# ω

p

2

2
m

{\displaystyle E= \hbar \omega = \frac{{p}^{2}}{2m} }

, we can rewrite the wave-packet equation as

Ψ
(
x
,
t
# )

1

2
π
ℏ

∫

−
∞

∞

ϕ
(
p
)

e

i
(
p
x
−
E
t
)

/

ℏ

d
p
.

{\displaystyle \Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int\_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}dp. }

What can we do with this integral so that we can describe a physical system? Why not take derivatives until two sides of an equation match? Let’s begin by taking the time derivative and two derivatives in space:

∂
Ψ

∂
t

=

1

2
π
ℏ

∫

−
∞

∞

ϕ
(
p
)

(

−
i
E

ℏ

)

e

i
(
p
x
−
E
t
)

/

ℏ

d
p

{\displaystyle \frac { \partial \Psi }{ \partial t } =\frac { 1 }{ \sqrt { 2\pi \hbar } } \int \_{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { -iE }{ \hbar } \right) e }^{ i(px-Et)/\hbar }dp } }

∂
Ψ

∂
x

=

1

2
π
ℏ

∫

−
∞

∞

ϕ
(
p
)

(

i
p

ℏ

)

e

i
(
p
x
−
E
t
)

/

ℏ

d
p

{\displaystyle \frac { \partial \Psi }{ \partial x } =\frac { 1 }{ \sqrt { 2\pi \hbar } } \int \_{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { ip }{ \hbar } \right) e }^{ i(px-Et)/\hbar }dp } }

∂

2

Ψ

∂

x

2

=

1

2
π
ℏ

∫

−
∞

∞

ϕ
(
p
)

(

−

p

2

ℏ

2

)

e

i
(
p
x
−
E
t
)

/

ℏ

d
p

.

{\displaystyle \frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } } =\frac { 1 }{ \sqrt { 2\pi \hbar } } \int \_{ -\infty }^{ \infty }{ \phi (p){ \left( \frac { -{ p }^{ 2 } }{ { \hbar }^{ 2 } } \right) e }^{ i(px-Et)/\hbar }dp }. }

Now there is one final step to do: balancing. If we multiply the time derivative by

i
ℏ

{\displaystyle i\hbar }

 and the second spatial derivative by

−

ℏ

2

2
m

{\displaystyle \frac{-{\hbar}^{2}}{2m} }

, we get

i
ℏ

∂
Ψ

∂
t

=

−

ℏ

2

2
m

∂

2

Ψ

∂

x

2

.

{\displaystyle i\hbar \frac { \partial \Psi }{ \partial t } = \frac{-{\hbar}^{2}}{2m}\frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } }. }

It is customary in physics to describe physical systems in terms of total energy, which is the sum of the kinetic and potential energies. So all we have to do to this free particle is add

V
Ψ

{\displaystyle V\Psi }

 to the right-hand side. Hence,

i
ℏ

∂
Ψ

∂
t

=

−

ℏ

2

2
m

∂

2

Ψ

∂

x

2

V
Ψ
.

{\displaystyle i\hbar \frac { \partial \Psi }{ \partial t } = \frac{-{\hbar}^{2}}{2m}\frac { { \partial }^{ 2 }\Psi }{ \partial { x }^{ 2 } }+ V\Psi. }

The equation uncovered above is the Schrödinger equation for a non-relativistic particle in one spatial dimension!

## **Operators**

Linear algebra, more specifically linear operator algebra, is the language of quantum mechanics. Thus, it is most appropriate to write the Schrödinger equation in operator form. An operator is an instruction that acts on a function. An operator

A
^

{\displaystyle \hat{A}}

 is linear if it satisfies the identity

A
^

[
a
f
(
x
)
b
g
(
x
)
# ]
a

A
^

f
(
x
)
b

A
^

g
(
x
)

{\displaystyle \hat{A}[af(x)+bg(x)] = a\hat{A}f(x)+b\hat{A}g(x) }

where

a
,
b

{\displaystyle a, b}

 are scalars. This is where quantum mechanics starts to get abstract, so I will introduce some concrete examples. Consider the operators

p
^

=
−
i
ℏ

∂

∂
x

{\displaystyle \hat{p} =-i\hbar\frac{\partial}{\partial x} }

 and

x
^

=
x

{\displaystyle \hat{x} = x }

. If we act

p
^

{\displaystyle \hat p}

 on some position quantity

x

{\displaystyle x}

, we get the differential

−
i
ℏ

∂
x

∂
x

=
−
i
ℏ

{\displaystyle -i\hbar \frac{\partial x}{\partial x} = -i\hbar }

. Hence, momentum is a differential operator. Looks straightforward enough! But consider the convenient things we can do to operators, for example:

p
^

2

=

(

−
i
ℏ

∂

∂
x

)

2

{\displaystyle {\hat{p}}^{2} = {\left(-i\hbar \frac{\partial}{\partial x}\right)}^{2} }

 becomes

p
^

2

=
−

ℏ

2

∂

2

∂

x

2

{\displaystyle {\hat{p}}^{2}= -{\hbar}^{2} \frac{{\partial}^{2}}{\partial {x}^{2}} }

.

Now let us look at the Schrödinger equation in operator form:

i
ℏ

∂
Ψ

∂
t

=

[

p
^

2

2
m

V
^

]

Ψ
.

{\displaystyle i\hbar \frac { \partial \Psi }{ \partial t } =\left[\frac{{\hat{p}}^{2}}{2m}+ \hat{V} \right] \Psi. }

For the time-independent case , we replace the time derivative term with total energy, and if we want to be even more compact, we may denote the operator on the right-hand side with the Hamiltonian operator:

H
^

=

T
^

V
^

{\displaystyle \hat{H} = \hat{T} +\hat{V} }

E
# Ψ

H
^

Ψ
.

{\displaystyle E\Psi=\hat{H} \Psi. }

Notice how this is shorter? Well, that's good and all, but not the point. The point is that we are acting a Hamiltonian operator on the **wavefunction**

Ψ

{\displaystyle \Psi}

, which is convenient for reasons unclear at the moment. You see, linear operators can describe many things, but all linear operators follow the same algebra. We have entered the realm where abstract algebra infiltrates all of physics for the convenience of problem solving. So from now on, we will consider operators to act on physical quantities and objects.

## **Determinate States**

When you solve the time-independent Schrödinger equation, you will encounter values of energy that satisfy

E
# Ψ

H
^

Ψ
.

{\displaystyle E\Psi=\hat{H} \Psi. }

These are called **determinate states**, which are the states for which the measurement of a given physical observable consistently returns the same outcome. Mathematically speaking, determinate states are **eigenfunctions** (also called **eigenstates**) of the linear operator

H
^

{\displaystyle \hat{H}}

 with corresponding **energy eigenvalues**

E

{\displaystyle E}

. When a linear operator possesses several eigenvalues, we call the set of eigenvalues its **spectrum**. If two or more linearly independent eigenfunctions share the same eigenvalue, we call those spectra **degenerate**. You will later learn that degeneracies play an important role in describing the atom, and hence, much of quantum chemistry and solid-state physics.

## **Problems**

### **Problem 1: Infinite Square Well**

Consider the potential that is 0 within the region

0
<
x
<
L

{\displaystyle 0 < x < L }

 and infinite elsewhere. This potential is called the **infinite square well**.

**Part 1:** Solve the time-independent Schrödinger equation for this potential with boundary conditions:

Ψ
(
0
# )
Ψ
(
L
# )
0

{\displaystyle \Psi(0) = \Psi(L) = 0 }

. Make sure to normalize the wavefunction

Ψ
(
x
)

{\displaystyle \Psi(x)}

.

**Part 2:** Now that you have found the solution to the particle in a box, find the set of permissible eigenenergies (energy levels). Remember that the wavenumber of a matter-wave is given by

# k

n
π

L

{\displaystyle k = \frac{n \pi}{L}}

 where

# n
1
,
2
,
3
,
.
.
.

{\displaystyle n=1,2,3,...}

. This will help you simplify

E

n

{\displaystyle {E}\_{n} }

.

**Part 3:** Determine the time-evolution of the stationary states in the form

Ψ
(
x
,
t
# )
Ψ
(
x
,
0
)

e

−
i

E
n

t

/

ℏ

{\displaystyle \Psi(x,t) = \Psi(x,0) {e}^{-iE\_n t/\hbar} }

.

Solution