Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 5. Quantum mechanics – Schrödinger’s equation •
superposition principle • time-independent Schrödinger equation •
continuity of Ψ and its derivative • probability interpretation of the
wave function Ψ • normalization of Ψ
Sources: Feynman Lectures I, Chap. 37;
Merzbacher (2nd edition) Chap. 4;
Merzbacher (3rd edition) Chap. 3 Sects. 1-3, 5;
6,7 ‫ יחידות‬,‫פרקים בפיסיקה מודרנית‬
Schrödinger’s equation
In 1926, Heisenberg was 25. Schrödinger
was 14 years older, and more reasonable.
In that year, each of them invented a new
mechanics. Schrödinger’s new mechanics
was an equation for a continuous wave.
Heisenberg’s new mechanics was a matrix
algebra – full of “quantum jumps”. Yet
they made exactly the same predictions!
We begin with Schrödinger’s equation. In a few weeks, we will
see how Heisenberg’s matrix algebra jumps out of it.
Schrödinger made a very reasonable analogy:
Fundamentally, there are no “light rays”. The rule that “the
angle of incidence equals the angle of reflection” is only an
approximation. So is Snell’s law. Rules about focal length
are fictions used by lens-makers. All these “light rays” are
actually electromagnetic waves.
Schrödinger made a very reasonable analogy.
Fundamentally, there are no “light rays”. The rule that “the
angle of incidence equals the angle of reflection” is only an
approximation. So is Snell’s law. Rules about focal length
are fictions used by lens-makers. All these “light rays” are
actually electromagnetic waves.
So Schrödinger assumed that paths
of material objects – trajectories and
orbits – are approximations to “matter
waves”. He asked, “What equation
do these matter waves obey?”
Schrödinger first tried to find a relativistic equation for
matter waves, but he failed. (Paul Dirac succeeded two
years later.) Then he tried to find a non-relativistic equation.
Schrödinger first tried to find a relativistic equation for
matter waves, but he failed. (Paul Dirac succeeded two
years later.) Then he tried to find a non-relativistic equation.
Each physical system has its own version of Schrödinger’s
equation!
Schrödinger first tried to find a relativistic equation for
matter waves, but he failed. (Paul Dirac succeeded two
years later.) Then he tried to find a non-relativistic equation.
Each physical system has its own version of Schrödinger’s
equation!
Here is Schrödinger’s equation for a one-dimensional
particle of mass m in a potential V(x):

2 2
i  ( x, t )  
 ( x, t )  V ( x )  ( x, t )
2
t
2m x
where we define   h/2  1.0545716 3  10
34
Js.
We don’t know what this equation means…but, as we shall
soon see, neither did Schrödinger! So let’s explore: Assume
V(x) = 0 (i.e. the particle is free) and solve for Ψ :

 
i  ( x , t )  

(
x
,
t
)
t
2m x 2
2
2
If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields
i d
1 2 d 2
T (t )  
X ( x)
2
T (t ) dt
X ( x) 2m dx
We don’t know what this equation means…but, as we shall
soon see, neither did Schrödinger! So let’s explore: Assume
V(x) = 0 (i.e. the particle is free) and solve for Ψ :

 
i  ( x , t )  

(
x
,
t
)
t
2m x 2
2
2
If Ψ is of the form Ψ(x,t) = X(x) T(t), then substitution yields
i d
1 2 d 2
T (t )  constant  
X ( x)
2
T (t ) dt
X ( x) 2m dx
i d
1 2 d 2
T (t )  constant  
X
(
x
)
T (t ) dt
X ( x) 2m dx 2
Try T (t )  e it and X ( x)  e ikx which together imply
  constant   2 k 2 / 2m .
We recognize the Planck-Einstein law, E  h   ,
and de Broglie’s law, p  h /   k .
 2k 2 p 2

So the constant is E    constant 
.
2m
2m
We found solutions to Schrödinger’s equation

2 2
i  ( x , t )  

(
x
,
t
)
t
2m x 2
ikxit
2



k
/ 2m .

(
x
,
t
)

e
of the form k
, where
These are not all the solutions, but we obtain all the solutions
by “superposing” these solutions:
 ( x, t ) 


A(k )e
ikxit
dk
,
2
where A(k) is any complex function of k and   k / 2m.
Superposition principle
If Ψ1(x,t) and Ψ2(x,t) are solutions of Schrödinger’s equation, then
αΨ1(x,t) + βΨ2(x,t)
is another solution of Schrödinger’s equation, for any complex
numbers α and β.
The superposition principle follows directly from the fact that
Schrödinger’s equation is always linear in Ψ:
 2 2


i ( x, t )  

V
(
x
)

(
x
,
t
)

2
t
2
m

x


Superposition principle
Namely, if
 2 2


i 1 ( x, t )  
 V ( x) 1 ( x, t )
2
t
 2m x

and
 2 2


i 2 ( x, t )  
 V ( x) 2 ( x, t )
2
t
 2m x

then we multiply the first equation by α and the second equation
by β and sum them to get the equation that 1 ( x, t )  2 ( x, t )
satisfies.
Now let’s try to solve Schrödinger’s equation for V(x) ≠ 0 using
the same method of separating out the dependence on t. We
already know that we can write
 ( x, t )  e it ψ( x)
and if we substitute this Ψ(x,t) into Schrödinger’s equation,
 2 2


i ( x, t )  
 V ( x) ( x, t ) ,
2
t
 2m x

we obtain an equation for ψ(x):
 2 d 2

 ψ( x)  
 V ( x) ψ( x) .
2
 2m dx

Since   E we write
 2 d 2

Eψ( x)  
 V ( x) ψ( x)
2
 2m dx

,
which is called the time-independent Schrödinger equation.
Since   E we write
 2 d 2

Eψ( x)  
 V ( x) ψ( x)
2
 2m dx

,
which is called the time-independent Schrödinger equation.
Question: Do solutions of the time-independent Schrödinger
equation obey the superposition principle?
Suppose ψ1(x) is a solution for E1 and ψ2(x) is a solution for E2.
Does it matter if E1= E2?
What about the solutions i ( x, t )  e
iEi t / 
ψi ( x) ?
Time-independent Schrödinger equation
Let’s try to solve the time-independent Schrödinger equation
 2 d 2

Eψ( x)  
 V ( x) ψ( x)
2
 2m dx

,
for a simple finite “square” potential well V(x):
V(x)
V0
−L/2
0
L/2
x
Time-independent Schrödinger equation
Inside:
2 d 2
Eψ( x)  
ψ( x )
2
2m dx
, | x| L/2 ;
the solution is a linear combination of eikx and e−ikx or,
equivalently, of cos kx and sin kx , where k =
2mE /  .
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Time-independent Schrödinger equation
Outside:
2 d 2
( E  V0 ) ψ( x)  
ψ( x )
2
2m dx
, | x| L/2 ;
the solution is a linear combination of eik′x and e−ik′x if E ≥ V0 ,
or a linear combination of ek′x and e−k′x if E ≤ V0 , where
k′ = 2m | E  V0 | /  .
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Continuity of ψ(x) and its derivative
What about at x = ± L/2? The left side of the equation, Eψ(x) or
(E − V0) ψ(x), is finite; but the right side is not finite unless both
ψ(x) and its derivative dψ(x)/dx are both continuous.
Therefore, the solutions of the time-independent Schrödinger
equation must be continuous with continuous first derivative.
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Continuity of ψ(x) and its derivative
Let’s guess that because of the symmetry of V(x), the solutions
ψ(x) are either symmetric or anti-symmetric. (Afterwards, we will
check that this guess is correct.) We’ll save time because we’ll
have to match the inside and outside solutions only at one point
(let’s say, at x = L/2).
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Probability interpretation of ψ(x)
But before we continue, we note that Schrödinger’s idea of a
“matter wave” for electrons turned out to be totally unworkable.
Just think about the electric charge of an electron: if an electron
is really a matter wave, then we should be able to find bits of its
charge in different places. No such bits of charge have ever been
seen! On the contrary, the charge of the electron is, as far as we
know, indivisible.
Consider a two-slit interference
experiment with electrons. If we
look at the charges that hit the
final screen, we always find whole
electrons with charge −e, never
parts of an electron.
Probability interpretation of ψ(x)
But before we continue, we note that Schrödinger’s idea of a
“matter wave” for electrons turned out to be totally unworkable.
Just think about the electric charge of an electron: if an electron
is really a matter wave, then we should be able to find bits of its
charge in different places. No such bits of charge have ever been
seen! On the contrary, the charge of the electron is, as far as we
know, indivisible.
Indeed, a few months after Schrödinger published his equation,
Max Born gave ψ(x) and Ψ(x,t) a new interpretation: they are
probability waves, and |ψ(x)|2 = [ψ(x)]*ψ(x) is the probability
density to find the electron at the point x. Likewise, |Ψ(x,t)|2 =
[Ψ(x,t)]* Ψ(x,t) is the probability density to find the electron at
the point x at time t. This interpretation is still accepted today.
Normalization
As long as electrons don’t simply disappear (conservation of
charge) the probability to find the electron somewhere must
be 1. The sum of all probabilities must be 1. In the case of
the continuous variable x, the sum is an integral, so we have
the mathematical requirement


| ψ( x) | dx  1 
2


| ( x, t ) |2 dx
This mathematical requirement is called normalization.
Let’s return now to the time-independent Schrödinger equation for
the finite well and focus on the case E < V0. For this case we
concluded that the outside solutions must be a linear combination
of ek′x and e−k′x, where k′ = 2m | E  V0 | /  . But normalization
requires that the outside solution can only be e−k′|x| .
V0
outside
inside
−L/2
0
outside
L/2
x
We considered two types of solutions to the time-independent
Schrödinger equation: symmetric and anti-symmetric. For the
symmetric solution, the continuity conditions at x = L/2 are
kL
A cos
 A' e k 'L / 2 ,
2
kL
 Ak sin
  A' k ' e k 'L / 2
2
continuity of ψ at x  L/2 ;
,
continuity of
dψ
at x  L/2 ;
dx
V0
outside
inside
ψ( x)  A' e k ' x
ψ( x)  A cos kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
Therefore, the continuity conditions for the symmetric solution
imply k tan kL/2 = k′, which in turn implies discrete energy levels!
From the definitions of k and k′ we have
kL
2 2m(V0  E ) 2mV0
2
k tan
 (k ' ) 


k
2
2
2
2
2
and we can solve this equation graphically by setting each side
to a variable y2; the solutions are the intersections of the two
2
curves y = k tan kL/2 and y2 + k2 = 2mV0 /  .
V0
outside
inside
ψ( x)  A' e k ' x
ψ( x)  A cos kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
y2 + k2 = 2mV0 / 
yL
2
2
y = k tan kL/2
2
1
1
2
kL/2
V0
outside
inside
ψ( x)  A' e k ' x
ψ( x)  A cos kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
For the anti-symmetric solution, the continuity conditions at
x = L/2 are
kL
A sin
 A' e k 'L / 2 ,
2
kL
Ak cos
  A' k ' e k 'L / 2
2
continuity of ψ at x  L/2 ;
,
continuity of
dψ
at x  L/2 ;
dx
V0
outside
inside
ψ( x)   A' e k ' x
ψ( x)  A sin kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
Therefore, the continuity conditions for the symmetric solution
imply k cot kL/2 = –k′ . From the definitions of k and k′ we have
kL
2 2m(V0  E ) 2mV0
2
k cot
 ( k ' ) 


k
2
2
2
2
2
and we can solve this equation graphically by setting each side
to a variable y2; the solutions are the intersections of the two
2
curves y = k cot kL/2 and y2 + k2 = 2mV0 /  .
V0
outside
inside
ψ( x)   A' e k ' x
ψ( x)  A sin kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
y2 + k2 = 2mV0 /  2
yL
2
y = k cot kL/2
2
1
2
1
kL/2
V0
outside
inside
ψ( x)   A' e k ' x
ψ( x)  A sin kx
−L/2
0
outside
L/2
ψ( x)  A' e k ' x
x
The first three energy levels: the ground state is symmetric, the
next state is anti-symmetric, the next state is symmetric, etc.
First excited
state
Ground state
−L/2
0
L/2
The first three energy levels: the ground state is symmetric, the
next state is anti-symmetric, the next state is symmetric, etc.
First excited
state
Classically
forbidden
region
Ground state
−L/2
0
L/2
The first three energy levels: the ground state is symmetric, the
next state is anti-symmetric, the next state is symmetric, etc.
As long as V0 is finite, there are finitely many energy levels.
For V0 infinite (infinite “square” well), there are infinitely many
levels, and ψ(x) vanishes outside the well (|x| > L/2, classically
forbidden region). The derivative of ψ(x) is discontinuous at
x = L/2 but this exception to the continuity rule is due to infinite
V0. The first three normalized wave functions and energies are
ψ1 ( x) 
2
x
 2 2
cos
, E1 
,
2
L
L
2mL
ψ 2 ( x) 
2
2 x
4 2  2
sin
, E2 
,
2
L
L
2mL
ψ3 ( x) 
2
3 x
9 2  2
cos
, E3 
.
2
L
L
2mL
Conclusions:
1. There is no physical system exactly like an electron in a
one-dimensional square well potential. But it is a model we
can solve, and we learn from it some features of quantum
mechanics and quantum behavior. We can also use it as an
approximate model, e.g. of an electron in a wire segment.
2. We learn that energy levels can be quantized, i.e. energies
can take discrete values.
3. We saw that symmetries of the model (here, parity) show up
in the symmetries of the solutions.
4. We saw a unique quantum effect: quantum electrons can
penetrate regions that are forbidden to classical electrons!