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Transcript
the wave equation
Schrödinger’s Equation
linearity and superposition
expectation values
“I think that I can safely say that nobody understands quantum
mechanics.”—Richard Feynman (Nobel Prize, 1965)
News Flash!
Intel unveils silicon laser.
News Flash!
UMR Physics Professor uses laser to
write quantum dots on the lightestweight material known to man.
5.2 The Wave Equation
I have been claiming for several chapters now that particles
have wave properties, and we have seen quite a few examples
of experimental data that support this claim.
Eventually, I am going to have to face the challenge, and come
up with some serious mathematical theory to support this
claim.
If particles have wave properties and can be described by a
wave function, there must be a wave equation for particles.
I’m going to set out to find it.
But one last time, let’s review the wave equation for waves.
Here’s a “sophisticated” form of the wave equation:
2 y 1 2 y
= 2 2 .
2
x
v t
The solution y(x,t) is a wave traveling with a speed v through
space (one dimension) and time. (The above 1-dimensional
equation can easily be generalized to 3 dimensions.)
What are these  things?
They represent partial derivatives. If F is a function of (xyz),
then when you take F/x, you treat y and z as constants:
F
If F(xyz) = 9xy + x yz , then
= 9y 2 + 2xyz .
x
2
2
Solutions to the wave equation have the form
x

y = Ft ±  .
v

The - sign represents waves traveling in the +x direction, and
the + sign represents waves traveling in the -x direction.
An equivalent functional form is y = F ( k x  t ) .
An example of a solution of the wave equation is the wave
equivalent to a free particle:
y = A e j ( kx - ωt ) = A cos ( kx - ωt ) + j A sin ( kx - ωt ) .
(A free particle is one which is not acted on by external forces,
including those giving rise to a potential.)
If we take the real part of y, we have the displacement of
waves in a stretched string. (Beiser's equations are slightly
different, but equivalent.)
What have we accomplished in this section?
Not much! We wrote down an “improved” form of the wave
equation. We reminded ourselves that if objects are
represented by waves, their wave functions must satisfy some
form of a wave equation.
5.3 Schrödinger's Equation: Time-Dependent Form
In this section we introduce Schrödinger's equation, which you
can think of as a quantum mechanical statement of the
conservation of energy, and which is probably the most
important equation of quantum mechanics.
Let's "derive" Schrödinger's equation.
“Where did we get that from? It's not possible to derive it from anything
you know. It came out of the mind of Schrödinger.”—Richard Feynman
We begin with conservation of energy:
2
p
E =K +U=
+U,
2m
where the potential U represents the effect of the rest of the
universe on the particle (or system) we are studying.*
U can represent the effect of applied electric fields, charged
particles, gravity, springs, etc.
Consider a particle (or system) represented by a wave function
(x,t).
*Do you see any relativity here? No, this is all nonrelativistic. There is a
relativistic form for anybody who needs it.
Multiply the energy conservation equation by (x,t) from the
right:
1

(x,t)
E
=
P2 (x,t) + U(x, t) (x,t) .
2m
If this is to be a wave equation, it must have solutions of the
form
(x,t) = A e
j (kx - ωt)
=Ae
(j/ ) (Px - Et)
.
The second part of the equation comes from E = ħ and P =
ħk.
I made the equation a bit bigger than normal so you could see the quantities in the exponents.
(x,t) = A e j (kx - ωt) = A e(j/
) (Px - Et)
.
Now, “just for kicks,” take 2/x2

 A e(j/
x
) (Px - Et)
 =
jP
 A e(j/
) (Px - Et)

The result is -P2/ ħ2 :
2

(j/
A e
2 
x
2
 jP 
) (Px - Et)
(j/
 =    A e
 

) (Px - Et)

 = -
P2
2
(x, t) .
2
P2
= - 2 (xt) .
2
x
Solving* for P2/2m, we get
2
P2 
2
=.
2
2m
2m x
*Why? Because it appears in our “energy conservation” expression.
Remember that
(x,t) = A e(j/
) (Px - Et)
Once again, “just for kicks,” take /t.

 A e(j/
t
) (Px - Et)
 =
-jE
The result is –jE / ħ:

-jE

=
 

t
 A e(j/
) (Px - Et)

Solving* for E, we get E = j

.
t
Plugging what we have found for E and P2/2m back into
our “conservation of energy” equation gives
2

2 
j
=+U .
2
t
2m x
This is the one-dimensional form of the time-dependent
Schrödinger equation; it is easy to generalize to 3 dimensions.
Although this equation represents the foundation of quantum
mechanics, I won’t make it “official.” I’ll save that for a later
version.
*Why? Because it appears in our “energy conservation” expression.
“Thus the partial differential equation entered theoretical physics as a handmaid, but
has gradually become mistress.”—A. Einstein
Our original equation
1 2
E (x, t) =
p (x, t) + U(x, t) (x, t)
2m
was deceptive, because p really represents the operator /x
(remember, in Newtonian mechanics momentum is related to
velocity, which is the first derivative of position) and E
represents the operator /t.
Our simple “conservation of energy” equation was really a
linear differential equation.
We have “justified” Schrödinger's equation, but not derived it.
That’s OK—we never derive Newton’s laws either. We justify
them, show that they work, and use them. We believe them
because they describe reality.
The same holds for Schrödinger's equation. It is a postulated
first principle, arrived at by observation of physical reality, and
believed in because it successfully describes the universe.
In other words, we believe it because it works.*
2
Ψ
2 
j
=+U .
2
t
2m x
Schrödinger's equation is a linear differential equation for the
wavefunction .
The potential U(x,t), may simply be zero or a numerical
constant, or it may be a complicated operator. U(x,t)
represents the effects of the universe on the particle.
*That’s a sign of strength, not weakness. Schrödinger’s equation has been used to
explain the previously unexplainable and predict the previously unthought. If you
want to replace Schrödinger’s equation with something else, the “something else”
must do everything Schrödinger’s equation does, and more.
2
Ψ
2 
j
=+U .
2
t
2m x
Once we know some boundary conditions for the particle and
the potential U(x,t), in principle we can solve for the wave
function at any time t and position x.
Once we have the wave function, our problem is “solved.”
In this chapter we will solve Schrödinger's equation for some
simple potentials. Let’s get to work…
But before that, just to whet your appetite, this is what we are
leading up to…
from http://www.nearingzero.net
What’s the simplest potential you can think of?
That’s right: U=0, which is the potential for a free particle.
We could take Schrödinger's equation, plug in U=0, and solve
for .
Or we could be “efficient*” and take our free-particle , plug it
into Schrödinger's equation, and see if we come up with an
identity. Let’s try that.
*Lazy?
The free-particle wave function is
 = e-(j/
) (Et - Px)
.
The free-particle Schrödinger equation (U = 0) is
2
Ψ
2Ψ
j
=.
2
t
2m x
Left-hand side:
j
  e-(j/
) (Et - Px)
t
 =- j
jE
e-(j/
) (Et - Px)
=E  .
Right-hand side:
-
2
2m
2  e-(j/
) (Et - Px)
x 2
 =-
2
 -jP 


2m 

2
-(j/
e

2
P
) (Et - Px)
 = 2m .
P2
Setting LHS = RHS: E  =
.
2m
P2
E =
.
2m
Remember—I told you this is
the nonrelativistic version.
Duh. We already knew that.
Of course! But we needed to check for consistency in the
simplest case before spending any more time on Schrödinger's
equation.
What can we do that’s useful?
Patience! There are still a few ideas to introduce.
5.4 Linearity and Superposition
In quantum mechanics, the physics and the math seem to be
forever entangled. That means we can often gain insight by
looking at the math, independent of a particular physical
system.
It also means that wave functions “behave well.”
2

2
j
=+U 
2
t
2m x
Schrödinger's equation is linear in . In other words, it has no
terms independent of , and no terms involving higher powers
of  or its derivatives.
As a consequence of this linearity, if 1 and 2 are solutions to
Schrödinger's equation, then so is the linear combination
 = a11 + a2 2 ,
where a1 and a2 are constants.
A further consequence of this is that wave functions obey
superposition and exhibit interference. (Duh, they
represent waves, so they had better do these things,
and all other things wavelike.)
If a system is represented by a wave function  = a11 + a22,
how do we calculate the probability density for ?
You can’t just add probabilities! You can’t write P = a1P1 + a2P2,
where P1 = 1* 1 and P2 = 2*2!
Instead,
P =  *  =  a 1 1 + a2  2   a 1 1 + a2  2 
*

 a  + a  
 a   + a   a   + a   a   + a   a  
P = a1* 1* + a*2  *2

P = a1* 1*
1
1
*
1
1

1
*
1

2
2
2
*
2
2

*
2
1
1
*
2
*
2
2

P = P1 +P1 + a*2  *2  a 1 1  + a*2  *2  a2  2 
Interference terms!
Waves interfere!
Beiser uses this result to show why electrons “shot” through a
double slit exhibit interference effects (unlike “pure particles”
but like waves).
Be sure to read section 5.4!
2
5.5 Expectation Values
Once we solve Schrödinger's equation for , we know all
about the particle that is knowable within the limits imposed
by the uncertainty principle.
I've already told you how to calculate the probability of finding
the particle in some volume element centered on the
coordinates (x,t) in one dimension or (x,y,z,t) in three
dimensions.
P(x1  x  x 2 ) = 
x2
x1
r2
 *  dx
P(r1  r  r2 ) =   *  dV .
r1
Sometimes we want to calculate the average value of some
measurable quantity. Just as quantum mechanics has its own
way of calculating probabilities, quantum mechanics has its
own special way of calculating averages.
Let's begin with an example.
Suppose we want to find the average of the set of numbers
1,1,1,2,2,3,3,3,3,4,4,4,4,4. How do you calculate the average?
Add the numbers up and divide by the number of
numbers? That works, but what if we have zillions
of numbers. Is there a way to be clever?
The average is
 3×1 +  2 ×2  +  4 ×3 + 5 ×4  .
3 + 2 + 4 +5
In general, the average of Ni numbers having values xi is
N x
x=
N
i
i
i
i
.
i
If the variable x is continuous, you replace the sums by
integrals.
In quantum mechanics, the probability Pi of finding a particle
in an interval dx at xi is
2
Pi dx =  i dx .
Think of * as being “like” N (“how much probability” 
“how many”).
To get QM
average <x>…
N x
x=
N
i
i
i
i
.
i
…replace this
by *…
…replace this
by * with an
extra x in it…
…and replace these by
integrals (because our
variables are continuous).
In QM (quantum mechanics) because we are dealing with
probabilities, we use the term “expectation value” rather than
“average value.”
The expectation value of x is

x

=

x *  dx
-

-
*
  dx
“like” N
,
where we have replaced a discrete variable by a continuous
one and let the sums become integrals.
The expectation value is the quantum mechanical equivalent of
an average value.
If  is normalized, the integral in the denominator equals one,
and

x =  x *  dx .
-
This is not “official” yet, because there is still a problem.
Position, momentum, energy, kinetic energy, etc. are actually
operators, and the order in which we take them is important.
Remember, momentum is related to /x and
energy is related to /t.
The correct approach is to make a “* sandwich” to get an
expectation value.

x =  * x  dx
-
In general, the expectation value of any quantity, including
operators, is

G  x  =  *G  x   dx .
-
Using the operator expression for momentum also prevents us
from using
the “hat” reminds us

pˆ =   pˆ  dx
*
momentum is an operator
-
to claim we’ve found a way to violate the uncertainty principle.
Let's use our wave function (x) = 3 x for 0  x  1
as an example.
Refreshing your memory…
The expectation value of x is like the average probability of
finding a particle at coordinate x. It is the point where you
could “balance” the * plot on your fingertips.
I am going to move the laser
pointer along the x-axis. You
say “stop” when you think I
reached the point where the
red shaded area would
balance on my fingertip.
Now let’s see if the math agrees with you.

x =   x  dx = 
-
*
1
0

 
3x x

4 1
x
3 x dx = 3 x dx = 3
0
4
1
3
3
= .
4
0