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