BARTON ZWIEBACH: We were faced last time
with a question of interpretation
of the Schrodinger wave function.
And so to recap the main ideas that we were looking at,
we derive this Schrodinger equation,
basically derived it from simple ideas--
having operators, energy operator, momentum operator,
and exploring how the de Broglie wavelength associated
to a particle would be a wave that would solve the equation.
And the equation was the Schrodinger equation,
a free Schrodinger equation, and then
we added the potential to make it
interacting and that way, we motivated the Schrodinger
equation and took this form of x and t psi of x and t.
And this is a dynamical equation that governs the wave function.
But the interpretation that we've
had for the wave function, we discussed what Born said,
was that it's related to probabilities
and psi squared multiplied by a little dx
would give you the probability to find
the particle in that little dx at some particular time.
So psi of x and t squared dx would be the probability
to find the particle at that interval dx around x.
And if you're describing the physics of your Schrodinger
equation is that of a single particle,
which is the case here--
one coordinate, the coordinate of the particle, this integral,
if you integrate this all over space,
must be 1 for the probability to make sense.
So the total probability of finding the particle must be 1,
must be somewhere.
If it's in one part of another part
or another part, this probabilities-- for this
to be a probability distribution,
it has to be well-normalized, which means 1.
And we said that this equation was interesting but somewhat
worrisome, because if the normalization of the wave
function satisfies, if this holds for t equal t-nought
then the Schrodinger equation, if you know the wave function
all over space for t equal t-nought, which is what you
would need to know in order to check that this is working,
a t equal t-nought, you take the psi of x and t-nought,
integrate it.
But if you know psi of x and t-nought for all x,
then the Schrodinger equation tells you
what the wave function is at a later time.
Because it gives you the time derivative of the wave
function in terms of data about the wave
function all over space.
So automatically, the Schrodinger equation
must make it true that this will hold at later times.
You cannot force the wave function to satisfy this at all
times.
You can force it maybe to satisfy at one time,
but once it satisfies it at this time, then it will evolve,
and it better be that at every time later,
it still satisfies this equation.
So this is a very important constraint.
So we'll basically develop this throughout the lecture today.
We're going to make a big point of this trying to explain why
the conditions that we're going to impose on the wave function
are necessary; what it teaches you about the Hamiltonian,
we'll teach you that it's a Hermitian operator;
what do you learn about probability--
you will learn that there is a probability current;
and all kinds of things will come out of taking seriously
the interpretation of this probability,
the main point being that we can be sure it behaves
as a probability at one time, but then for later times,
the behaviors and probability the Schrodinger equation must
help--
must somehow be part of the reason this works out.
So that's what we're going to try to do.
Now, when we write an equation like this, and more explicitly,
this means integral of psi star of x and t, psi of x and t dx
equal 1.
You can imagine that not all kind of functions
will satisfy it.
In particular, any wave function, for example,
that at infinity approaches a constant will never
satisfy this, because if infinity, you
approach a constant, then the integral
is going to be infinite.
And it's just not going to work out.
So the wave function cannot approach a finite number,
a finite constant as x goes to infinity.
So in order for this to hold--
order to guarantee this can even hold, can conceivably hold,
it will require a bit of boundary conditions.
And we'll say that the limit as x
goes to infinity or minus infinity--
plus/minus infinity of psi of x and t will be equal to 0.
It better be true.
And we'll ask a little more.
Now, you could say, look, certainly
the limit of this function could not
be in number, because it would be non-zero number,
the interval will diverge.
But maybe there is no limit.
The wave function is so crazy that it can be integrated,
but suddenly, it has a little spike
and it just doesn't have a normal limit.
That could conceivably be the case.
Nevertheless, it doesn't seem to happen in any example that
is of relevance.
So we will assume that the situations are not
that crazy that this happened.
So we'll take wave functions that necessarily
go to 0 at infinity.
And that certainly is good.
You cannot prove it's a necessary condition,
but if it holds, it simplifies many, many things,
and essentially, if the wave function is good enough to have
a limit, then the limit must be 0.
The other thing that we will want
is that d psi/dx, the limit as x goes to plus/minus infinity
is bounded.
That is, yes, the limit may exist and it may be a number,
but it's not infinite.
And In every example that I know of--
in fact, when this goes to 0, this goes to 0 as well--
but this is basically all you will ever need in order
to make sense of the wave functions and their integrals
that we're going to be doing.
Now you shouldn't be too surprised
that you need to say something about this wave function
in the analysis that will follow,
because the derivative--
you have the function and its derivative,
because certainly, there are two derivatives here.
So when we manipulate these quantities
inside the integrals, you will see very soon--
single derivatives will show up and we'll have to control them.
So the only thing that I'm saying
is that when you see a wave function that
satisfies this property, you know
that unless the function is extremely crazy,
it's a function that goes to 0 at plus/minus infinity.
And it's the relative pursuant it also goes to 0,
but it will be enough to say that it maybe goes to a number.
Now there's another possibility thing for confusion
here with things that we've been saying before.
We've said before that the physics of a wave function
is not altered by multiplying the wave function by a number.
We said that psi added to psi is the same state; psi
is the same state as square root of 2 psi--
all this is the same physics, but here it
looks a little surprising if you wish, because if I have a psi
and I got this already working out,
if I multiply psi by square root of 2, it will not hold.
So there seems to be a little maybe something with the words
that we're been using.
It's not exactly right and I want
to make sure there is no room for confusion here,
and it's the following fact.
Here, this wave function has been normalized.
So there's two kinds of wave functions that you can have--
wave functions that can be normalized and wave functions
that cannot be normalized.
Suppose somebody comes to you and gives you a psi of x and t.
Or let's assume that--
I'll put x and t.
No problem.
Now suppose you go and start doing this integral--
integral of psi squared dx.
And then you find that it's not equal to 1
but is equal to some value N, which
is different from 1 maybe.
If this happens, we say that psi is normalizable, which
means it can be normalized.
And using this idea that changing the value--
the coefficient of the function--
doesn't change too much, we simply say, use
instead psi prime, which is equal to psi over
square root of N. And look what a nice property this psi prime
has.
If you integrate psi prime squared, it would be equal--
because you have psi prime here is squared,
it would be equal to the integral of PSI squared divided
by the number N--
because there's two of them--
dx, and the number goes out and you have the integral of psi
squared dx, but that integral was exactly N, so that's 1.
So if your wave function has a finite integral
in this sense, a number that is less than infinity,
then psi can be normalized.
And if you're going to work with probabilities,
you should use instead this wave function,
which is the original wave function divided by a number.
So they realize that, in some sense,
you can delay all of this and you can always
work with wave functions that are normalizable,
but only when you're going to calculate your probabilities.
You can take the trouble to actually normalize them
and those are the ones you use in these formulas.
So the idea remains that we work flexibly with wave functions
and multiply them by numbers and nothing changes as long as you
realize that you cannot change the fact that the wave function
is normalizable by multiplying it by any finite number,
it will still be normalized.
And if it's normalizable, it's equivalent to a normalized wave
function.
So those two words sound very similar,
but they're a little different.
One is normalizable, which means it has an integral of psi
squared finite, and normalize is one
that already has been adjusted to do this
and can be used to define a probability distribution.
OK.
So that, in a way of introduction
to the problem that we have to do,
our serious problem is indeed justifying
that the time evolution doesn't mess up the normalization
and how does it do that?