Lecture 21 – Schrodinger’s equation
Ch 9
pages 446-451; 455-463
Wave Equations - The Classical Wave Equation
In classical mechanics, wave functions are obtained by
solving a differential equation. For example, the onedimensional wave equation for a vibrating string with linear
mass r (units of kg/m) and tension T (units of force) is:
 2  ( x, t ) r  2  ( x, t )

0
T
x 2
t 2
c
T
r
The wave velocity is given by c and the wave function
quantifies the vertical displacement of the string as a function
of x and t
Wave Equations - The Classical Wave Equation
 2  ( x, t ) r  2  ( x, t )

0
T
x 2
t 2
c
T
r
Any function of the form: n ( x, t )  An cos( 2v n t ) sin( 2x / n )
is a solution of the wave equation, where the specific forms
for the wave frequency nn and the wavelength n are
determined by the details of the problem
For example, for a harmonically vibrating string, fixed at x=0
and x=L (i.e. with boundary conditions (0, t )  ( L, t )  0
nn 
nc
2L
and  n 
for n  1,2,3...
2L
n
The frequencies nn are the harmonics of the vibrating string.
Wave Equations - The Classical Wave Equation
 2  ( x, t ) r  2  ( x, t )

0
T
x 2
t 2
c
T
r
n ( x, t )  An cos( 2v n t ) sin( 2x / n )
Any linear combination of wave functions
( x, t )   c n n ( x, t )
n
(where cn are constant) is also a solution to the wave equation.
A wave function that is independent of time is called a
standing wave. The wave equation for a standing wave is:
 2 ( x )
 ( x )  0
2
x
Wave Equations - The Classical Wave Equation
 2  ( x, t ) r  2  ( x, t )

0
T
x 2
t 2
c
n ( x, t )  An cos( 2v n t ) sin( 2x / n )
T
r
 2 ( x )
 ( x )  0
x 2
Where  is a constant. If the boundary conditions are
(0)  ( L)  0
then any function of the form:
is a solution if
 n 
n   
 L 
2
 ( x )  sin
nx
L
Standing waves
Wave functions and experimental observables
Particle wave functions are obtained by solving a quantum
mechanical wave equation, called the Schroedinger equation
In classical mechanics, the solution to the wave equation
(x,t) describes the displacement (e.g. of a string) as a
function of time and place
In quantum mechanics, Schrodinger and Heisenberg
introduced an analogous concept called wave function (x,t)
Schroedinger’s quantum mechanical wave equation
Schrodinger introduced his famous equation to calculate the
value of the wave function for a particle in a potential V(x,t),
the time-dependent Schroedinger equation is:
 2  ( x, t )
ih ( x, t )

V
(
x
,
t
)

(
x
,
t
)

2
t
8 2 m x 2
h2
If the potential V is independent of time, the wave function
has the simple form:
i 2Et / h
( x, t )  Ce
 ( x)
Where C is a constant and f satisfies the time-independent
Schroedinger’s equation which has a simpler form:
 2 ( x )
 2
 V ( x, t ) ( x )  E ( x )
2
8 m x
h2
Schroedinger’s quantum mechanical wave equation
 2  ( x, t )
ih ( x, t )

V
(
x
,
t
)

(
x
,
t
)

2
t
8 2 m x 2
h2
 2 ( x )
 2
 V ( x, t ) ( x )  E ( x )
2
8 m x
h2
Solutions of this equation are independent of time, and are
called stationary or standing particle waves, in analogy to
classical standing waves in a vibrating string.
Schroedinger’s equation for a particle in an infinitely deep
well is:
 2 ( x )
 2
 E ( x )
2
8 m x
h2
It is identical in form to the classical equation for a standing
wave.
Wave functions and experimental observables
Its physical interpretation is not immediate
The square of the wave unction (x,t)2 characterizes the
electron distribution in space and is a measure of the
probability of finding an electron (or any other particle) at a
certain time and place
For example, the probability of finding a particle within a
certain volume in space is given by:

V
2
( x, t )dx
Wave functions and experimental observables
Although the wave function is a mathematical concept, it is of
fundamental importance and can be directly measured (sort
of)
For example, X-ray diffraction experiments directly measure
(apart from a Fourier transform) the square of the electron
distribution of the material
The chemical bond can only be described and understood by
calculating wave functions for the electron in a molecule and
so on.
Electron density from x-ray crystallography
What do we actually measure? Operators
Ĥ  E
The Hamiltonian:
Ĥ  KE  ( PE )
The Wavefunction: 
The Hamiltonian is
an operator
Describes a system in a given
state
E = energy
Operators are associated with observables
Position
x
multiply by x
Momentum
px

ih
x
Kinetic energy
kx


2m x
Potential energy
V(x)
2
multiply by V(x)
Wave functions and experimental observables
We can calculate the energy of a particle from its wave
function and any other property of the system
A fundamental tenet of quantum mechanics is that
observables can be derived once the wave function is known.
However, the duality of matter introduced earlier introduces
a probabilistic nature to measurements, so that we can
calculate observables only in a probabilistic sense
Wave functions and experimental observables
This is done through the expectation value of a variable O,
which can be calculated using the expression:
 O   *Odx
Given a function of a complex variable f=a+ib, the complex
conjugate f*=a-ib
For example, the average position of an electron in a molecule
is given by:
 x   * xdx
time-independent Schroedinger equation: particle in a box
Using the Schroedinger Equation we can obtain the energies
and wave functions for a particle in a box. Particle-in-a-Box
refers to a particle of mass m in a potential defined as:
0 if 0  x  L
V ( x)  

  otherwise 
The wave function has the form
x, t   ei 2tE / h x 
Where (x) is obtained by solving the time-independent
Schroedinger equation:
2
2
  ( x)
 V ( x ) ( x )  E ( x )
2
2
8 m x
h
With the requirement that
 (0)   ( L)  0
to reflect the boundary conditions imposed by the potential V(x).
time-independent Schroedinger equation: particle in a box
0 if 0  x  L
V ( x)  

  otherwise 
x, t   ei 2tE / h x 
Since the particle must remain in the box where V(x)=0, the
Schroedinger equation simplifies to:
 2
 2
 E ( x )
8 m x 2
h2
Which can be rearranged to a familiar form
 2
2


 ( x)  0
2
x
8 2 mE
 
h2
2
The general solution to this equation is:
 ( x )  A cos x  B sin x
time-independent Schroedinger equation: particle in a box
0 if 0  x  L
V ( x)  

  otherwise 
x, t   ei 2tE / h x 
 ( x )  A cos x  B sin x
However, the boundary conditions (the particle cannot leave
the box!)
 ( 0)   ( L )  0
Can only be satisfied if A=0 and

n
L
n=0, 1, 2, 3, …
The analogy with standing waves (vibrating strings) is well worth
noting. Since, by definition: 2 8 2 mE  n  2
 
h
2


 L 
The quantized energy is obtained by substituting the expression
for the wave function into Schrodinger’s equation and solving for
the energy.
h 

2
2
8 2 m x 2
 E ( x )
time-independent Schroedinger equation: particle in a box
 2
 2
 E ( x )
8 m x 2
h2
It is found to be:
 ( x) 
2
 nx 
sin 

L  L 
n2h2
En 
8mL2
The lowest energy level (n=1) is called ground state, the others
are called excited states. These are very important concepts in
spectroscopy.
We can use the result to calculate the probability that the
particle in state n is at position x: P   ( x )  A sin  nx 
2
2
n
2


 L 
To determine the constants An we can recall that all probabilities
must sum to one because the particle must be somewhere in the
box
 nx 
A sin
dx  1
L

0
2
n
2


 L 
time-independent Schroedinger equation: particle in a box
L
A
2
n
0
 nx 
sin 2 
dx  1
 L 
2
 n x 
P   ( x )  An2 sin 2 

 L 
Which can be rearranged to give:
Final answer:
 ( x) 
2
 nx 
sin 

L  L 
1
2  nx 
0 sin  L dx  An2
L
An2 
2
L
Wavefunctions for particle in the box
• What does the energy look like?
2
2
nh
E
8mL2
n = 1, 2, …
Energy is quantized
E

*
Application/Example
• Consider the following dye molecule, the length of
which can be considered the length of the “box” an
electron is limited to:
+
N
L=8Å
N
What wavelength of light corresponds to E from n=1 to
n=2?
2
h2
h
2
2
2
19
E 
n

n

2

1

2.8
x
10
J

initial 
2  final
2 
8mL
8m(8 Å )
  700nm
(experimental: 680 nm)
Solving the Quantum Mechanical Wave Equation
If the potential is independent of time i.e. V=V(x), the
Schroedinger equation can be solved as follows.
1. Assume the wave function is a product of a function
dependent only on x and a function only dependent on t:
( x, t )   ( x ) (t )
2. Substitute that expression into the Schroedinger equation:
 2 ( x ) (t )
ih
 (t )

V
(
x
)

(
x
)

(
t
)


(
x
)
2
t
8 2 m
x 2
h2
3. Divide both sides of the equation by
( x, t )   ( x ) (t )
1  2 ( x )
ih 1  (t )

V
(
x
)

2  (t ) t
8 2 m  ( x ) x 2
h2
Because the left-hand-side of the equation is dependent only on x,
and the right-hand-side is dependent only on t, both sides must
equal a constant. It can be shown to be the energy E.
Solving the Quantum Mechanical Wave Equation
4. The time equation:
Can be re-written as:
ih 1  (t )
E
2  (t ) t
ih  (t )
 E (t )
2 t
This equation has the general solution:  (t )  e i 2Et / h
5. The space-dependent equation:
 2 ( x )
 V ( x ) ( x )  E ( x )
2
2
8 m x
h2
is called the stationary or time independent Schroedinger
equation. The solution (x) depends on the potential V(x) and the
boundary conditions imposed
Solving the Quantum Mechanical Wave Equation
To summarize, the general solution to the Schroedinger
equation (if the potential V(x) is independent of time):
 2  ( x, t )
ih ( x, t )

V
(
x
,
t
)

(
x
,
t
)

2
t
8 2 m x 2
h2
is
( x, t )   ( x )e i 2Et / h
Where (x) is obtained by solving the time-independent
Schroedinger equation:
 2 ( x )
 V ( x ) ( x )  E ( x )
2
2
8 m x
h2
Particle in a 3D box
The Schroedinger equation for a particle in a threedimensional box with dimensions a, b, c is:
h 2   2 ( x, y, z, t )  2 ( x, y, z, t )  2 ( x, y, z, t )  ih ( x, y, z, t )

 


t
8 2 m 
x 2
y 2
z 2
 2
This equation can be solved exactly as for a one-dimensional case
by assuming:
( x, y , z, t )   x ( x ) y ( y ) z ( z ) (t )
As before
 (t )  e i 2Et / h
The time-independent wave equation is:
h 2   2 ( x, y , z , t )  2 ( x, y , z , t )  2 ( x, y , z , t ) 

  E x  E y  E z


8 2 m 
x 2
y 2
z 2

 x (0)   x ( a )   y (0)   y (b)   z (0)   z (c )  0
Particle in a 3D box
h 2   2 ( x, y , z , t )  2 ( x, y , z , t )  2 ( x, y , z , t ) 

  E x  E y  E z


8 2 m 
x 2
y 2
z 2

This equation can be further separated into three identical
equations of the form of the one-dimensional particle-in-abox equation. The result is that the energy is a sum of three
identical terms:
2
2
 2 n2

E  Ex  E y  Ez 
h nx
nz
y




8m  a 2 b 2 c 2 
The wave function is a product of the form:
 ( x, y , z ) 
8
 n x   n y x   n z x 
 sin 
sin  z  sin 

abc  a   b   c 