Show me, then.

Okay. Here's how it works.

Imagine that we make some small change to the correct wavefunction (x).

How will the energy change? Here's our equation for the energy:

E

=

dx

(x)

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

Notice we've factored out a (x) on the right-hand side of the integrand, just for clarity in what follows.

Now we make the substitution  ->  +  in this equation. We expand all the terms, and keep only those proportional to . Why? Because we are assuming is small, so ()², ()³, etc. are real small, and can be neglected.

This is what we get:

E + dE	=		dx	(x)	[ [ [	1 --- 2	d2 ---- dx2	+ V(x)	] ] ]	(x)
			+	(x)	[ [ [	1 --- 2	d2 ---- dx2	+ V(x)	] ] ]	(x)
			+	(x)	[ [ [	1 --- 2	d2 ---- dx2	+ V(x)	] ] ]	(x)

The first term is just the original energy E itself. Also, the third term is equal to the second. (You can prove this by integrating by parts. You will need to realize that the wavefunction (x) must go to zero at the limits of integration, because the wavefunction has to end somewhere, that is, the particle can't have a nonzero probability of being everywhere in the universe.)

Thus we can write the change in energy dE with respect to a change in the wavefunction (x) as:

dE

=  2

dx

(x)

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

[1]

Now here comes the important conceptual step: Remember we said that the correct wavefunction (x) is that which minimizes the energy E. This implies that if we drew a sketch of how E depends on ``trial'' forms of (x) in the neighborhood of the actual, correct form, our sketch would look something like the sketch at right.

The key mathematical point about this sketch is that the curve of E versus trial (x) is flat near the correct (x).

Another way of saying this is that any small variation (x) you make in the correct wavefunction must produce zero change in E. (This fact is closely related to the fact in conventional calculus that the minimum of a function is where the first derivative is zero. The only complication here is that E is a function of a function, or what is called a functional. But the theorem still holds, in the sense that the minimum value of a functional is where the first variation -- the equivalent of a first derivative -- is zero.)

So to ensure that the correct (x) minimizes E we must ensure that every conceivable small variation (x) in the correct (x) produces dE = 0 when inserted into Equation [1] above.

How can this be done? Well, one possibility is to require

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

= 0

That would certainly work. But this turns out not to be the only possible trick. To see why, we need to say something about the allowable variations in (x).

Now we can't just make any old change we like to (x), because we need to make sure that the new wavefunction + stays normalized, that is, that the probability of finding the particle summed up (integrated) over all space is 1. Mathematically we must require:

dx

[ (x) + (x) ] [ (x) + (x) ]

= 1

If we expand this and keep only terms first order in in (x)we have:

dx

(x) (x)

+

2 (x) (x)

= 1

Of course the actual correct (x) must be correctly normalized, that is, we must have:

dx

(x) (x)

= 1

[2]

which means that we must require of our variations that:

dx

(x)

(x)

= 0

[3]

OK, let's get back to Equation [1] for dE, which is:

dE

=  2

dx

(x)

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

[1]

Suppose we now require:

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

= (constant) × (x)

[3]

That would turn Equation [1] into:

dE

=  (constant) × 2

dx

(x)

(x)

And Equation [3] now tells us that the integral on the right-hand side will always equal zero.

Marvelous! We've now shown that requiring Equation [3] to hold for the correct (x) ensures that dE = 0 for all conceivable small variations in it, which in turn means that this wavefunction will be the one that minimizes the energy E.

One question remains -- what is this ``constant'' thingy? Looking back at the equation for E itself, which is:

E

=

dx

(x)

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

If we use our marvelous new Equation [3] in this equation we find:

E

= (constant)

dx

(x)

(x)

And using the fact of the normalization of (x) (Equation [2] above), we know the integral on the right-hand side must be equal to just 1. That means the constant must be equal to the energy E itself.

Thus we proved that in order for the correct wavefunction (x) to minimize the energy E, it must satisfy this equation:

[ [ [

1 --- 2

d2 ---- dx2

+ V(x)

] ] ]

(x)

= E (x)

which is the Schrödinger equation.