Quantum Mechanics for Engineers 

© Leon van Dommelen 

A.2 An example of variational calculus
The problem to solve in addendum {A.22.1} provides a simple
example of variational calculus.
The problem can be summarized as follows. Given is the following
expression for the net energy of a system:
Here the operator is defined as
The integrals are over all space, or over some other given region.
Further is assumed to be a given positive constant and
is a given function of
the position . The function
will be called the potential and is not given. Obviously the energy
depends on what this potential is. Mathematicians would say that
is a “functional,” a number that depends on what a function is.
The energy will be minimal for some specific potential
. The objective is now to find an equation
for this potential using variational calculus.
To do so, the basic idea is the following: imagine that you start at
and then make an infinitesimally small change
to it. In that case there should be no change
in energy. After all, if there was an negative change in ,
then would decrease. That would contradict the given fact that
produces the lowest energy of all. If there was
an positive infinitesimal change in , then a change in
potential of opposite sign would give a negative change in .
Again that produces a contradiction to what is given.
The typical physicist would now work out the details as follows. The
slightly perturbed potential is written as
Note that the in has been renotated as
. That is because everyone does so in variational
calculus. The symbol does not make a difference, the idea remains the
same. Note also that is a function of position; the
change away from is normally different at
different locations. You are in fact allowed to choose anything you
like for the function , as long as it is
sufficiently small and it is zero at the limits of integration.
Now just take differentials like you typically do it in calculus or
physics. If in calculus you had some expression like , you
would say . (For example, if is
a function of a variable , then
. But physicists usually do not bother with
the ; then they do not have to worry what exactly is
a function of.) Similarly
where
so
For a change starting from :
(Note that by itself gets approximated as
, but is the completely
arbitrary change that can be anything.) Also,
because is a given constant at every position.
Total you get for the change in energy that must be zero
A conscientious mathematician would shudder at the above
manipulations. And for good reason. Small changes are not good
mathematical concepts. There is no such thing as
small
in mathematics. There are just limits where
things go to zero. What a mathematician would do instead is write the
change in potential as a some multiple of a chosen function
. So the changed potential is written as
The chosen function can still be anything that you
want that vanishes at the limits of integration. But it is not
assumed to be small.
So now no mathematical nonsense
is written. The energy for this changed potential is
Now this energy is a function of the multiple . And
that is a simple numerical variable. The energy must be smallest at
= 0, because gives the minimum energy.
So the above function of must have a minimum at
0. That means that it must have a zero derivative at
0. So just differentiate the expression with respect to
. (You can differentiate as is, or simplify first and
bring outside the integrals.) Set this derivative to zero
at 0. That gives the same result (2) as derived by
physicists, except that takes the place of
. The result is the same, but the derivation is
nowhere fishy.
This derivation will return to the notations of physicists. The next
step is to get rid of the derivatives on . Note
that
The way to get rid of the derivatives on is by
integration by parts. Integration by parts pushes a derivative from
one factor on another. Here you see the real reason why the changes
in potential must vanish at the limits of integration. If they did
not, integrations by parts would bring in contributions from the
limits of integration. That would be a mess.
Integrations by parts of the three terms in the integral in the
, , and directions respectively produce
In vector notation, that becomes
Substituting that in the change of energy (2) gives
The final step is to say that this can only be true for whatever
change you take if the parenthetical expression is
zero. That gives the final lookedfor equation for
:
To justify the above final step, call the parenthetical expression
for short. Then the variational statement above is of the form
where can be arbitrarily chosen as long as it is zero
at the limits of integration. It is now to be shown that this implies
that is everywhere zero inside the region of integration.
(Note here that whatever function is, it should not contain
. And there should not be any derivatives of
anywhere at all. Otherwise the above statement is not
valid.)
The best way to see that must be zero everywhere is first assume
the opposite. Assume that is nonzero at some point P. In that
case select a function that is zero everywhere except
in a small vicinity of P, where it is positive. (Make sure the
vicinity is small enough that does not change sign in it.) Then
the integral above is nonzero; in particular, it will have the same
sign as at P. But that is a contradiction, since the integral
must be zero. So the function cannot be nonzero at a point P; it
must be zero everywhere.
(There are more sophisticated ways to do this. You could take
as a positive multiple of that fades away to zero
away from point P. In that case the integral will be positive unless
is everywhere zero. And sign changes in are no longer a
problem.)