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PROFESSOR: Last time, we talked
about the Broglie wavelength.
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And our conclusion was,
at the end of the day,
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that we could write the
plane wave that corresponded
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to a matter particle, with some
momentum, p, and some energy,
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E. So that was our
main result last time,
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the final form for the wave.
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So we had psi of
x and t that was
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e to the i k x minus i omega t.
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And that was the matter
wave with the relations
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that p is equal to h bar k.
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So this represents a
particle with momentum,
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p, where p is h bar times
this number that appears here,
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the wave number,
and with energy,
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E, equal to h bar
omega, where omega
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is that number that appears
in the [? term ?] exponential.
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Nevertheless, we
were talking, or we
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could talk, about
non-relativistic particles.
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And this is our
focus of attention.
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And in this case, E is
equal to p squared over 2m.
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That formula that expresses
the kinetic energy
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in terms of the momentum, mv.
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So this is the wave function
for a free particle.
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And the task that
we have today is
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to try to use this insight,
this wave function,
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to figure out what
is the equation that
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governs general wave functions.
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So, you see, we've been
led to this wave function
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by postulates of the Broglie
and experiments of Davisson,
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and Germer, and
others, that prove
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that particles like electrons
have wave properties.
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But to put this
on a solid footing
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you need to obtain this
from some equation, that
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will say, OK, if you have
a free particle, what
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are the solutions.
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And you should
find this solution.
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Perhaps you will
find more solutions.
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And you will understand
the problem better.
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And finally, if you understand
the problem of free particle,
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there is a good chance you
can generalize this and write
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the equation for a
particle that moves
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under the influence
of potentials.
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So basically, what
I'm going to do
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by trying to figure out how this
wave emerges from an equation,
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is motivate and
eventually give you,
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by the middle of this lecture,
the Schrodinger equation.
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So that's what we're
going to try to do.
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And the first thing is
to try to understand
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what kind of equation this
wave function satisfies.
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So you want to think of
differential equations
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like wave equations.
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Maybe it's some kind
of wave equation.
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We'll see it's kind
of a variant of that.
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But one thing we
could say, is that you
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have this wave function here.
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And you wish to know, for
example, what is the momentum.
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Well you should look at k, the
number that multiplies the x
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here, and multiply by h bar.
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And that would give
you the momentum.
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But another way of doing it
would be to do the following.
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To say, well, h bar over
i d dx of psi of x and t,
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calculate this thing.
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Now, if I differentiate
with respect to x,
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I get here, i
times k going down.
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The i cancels this
i, and I get h bar k.
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So, I get h bar k
times the exponential.
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And that is equal to the value
of the momentum times the wave.
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So here is this wave actually
satisfies a funny equation,
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not quite the differential
equation we're looking for yet,
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but you can act with a
differential operator.
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A derivative is something
of a differential operator.
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It operates in functions,
and takes the derivative.
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And when it acts on
this wave function,
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it gives you the momentum
times the wave function.
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And this momentum
here is a number.
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Here you have an operator.
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An operator just means something
that acts on functions,
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and gives you functions.
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So taking a derivative of a
function is still a function.
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So that's an operator.
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So we are left here to
think of this operator
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as the operator that
reveals for you the momentum
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of the free particle, because
acting on the wave function,
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it gives you the momentum
times the wave function.
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Now it couldn't be that
acting on the wave function
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just gives you the momentum,
because the exponential doesn't
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disappear after the
differential operator acts.
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So it's actually
the operator acting
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on the wave function gives you a
number times the wave function.
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And that number is the momentum.
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So we will call
this operator, given
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that it gives us the momentum,
the momentum operator, so
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momentum operator.
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And to distinguish it
from p, we'll put a hat,
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is defined to be
h bar over i d dx.
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And therefore, for
our free particle,
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you can write what we've
just derived in a brief way,
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writing p hat
acting on psi, where
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this means the
operator acting on psi,
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gives you the momentum of this
state times psi of x and t.
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And that's a number.
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So this is an operator
state, number state.
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So we say a few things,
this language that we're
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going to be using all the time.
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We call this wave function,
this psi, if this is true,
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this holds, then we
say the psi of x and t
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is an eigenstate of
the momentum operator.
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And that language comes from
matrix algebra, linear algebra,
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in which you have a
matrix and a vector.
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And when the matrix
on a vector gives you
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a number times the
same vector, we
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say that that vector is an
eigenvector of the matrix.
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Here, we call it an eigenstate.
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Probably, nobody
would complain if you
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called it an eigenvector,
but eigenstate
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would be more appropriate.
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So it's an eigenstate of p.
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So, in general, if you have an
operator, A, under a function,
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phi, such that A acting
on phi is alpha phi,
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we say that phi is an
eigenstate of the operator,
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and in fact eigenvalue alpha.
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So, here is an eigenstate
of p with eigenvalue
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of p, the number p, because
acting on the wave function
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gives you the number p
times that wave function.
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Not every wave function
will be an eigenstate.
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Just like, when you have a
matrix acting on most vectors,
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a matrix will rotate
the vector and move it
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into something else.
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But sometimes, a matrix
acting in a vector
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will give you the same
vector up to a constant,
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and then you've
got an eigenvector.
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And here, we have an eigenstate.
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So another way of
expressing this,
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is we say that psi of x
and t, this psi of x and t,
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is a state of definite momentum.
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It's important terminology,
definite momentum means
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that if you measured it, you
would find the momentum p.
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And the momentum-- there
would be no uncertainty
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on this measurement.
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You measure, and
you always get p.
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And that's what,
intuitively, we have,
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because we decided
that this was the wave
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function for a free
particle with momentum, p.
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So as long as we
just have that, we
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have that psi is a state
of definite momentum.
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This is an interesting statement
that will apply for many things
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as we go in the course.
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But now let's consider another
aspect of this equation.
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So we succeeded with that.
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And we can ask if there
is a similar thing
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that we can do to figure out
the energy of the particle.
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And indeed we can
do the following.
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We can do i h bar d dt of psi.
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And if we have that,
we'll take the derivative.
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Now, this time,
we'll have i h bar.
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And when we differentiate that
wave function with respect
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to time, we get minus i omega
times the wave function.
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So i times minus i is 1.
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And you get h bar omega psi.
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Success, that was the energy
of the particle times psi.
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And this looks quite
interesting already.
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This is a number, again.
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And this is a time derivative
of the wave function.
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But we can put more physics into
this, because in a sense, well,
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this differential
equation tells you
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how a wave function
with energy, E,
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what the time dependence
of that wave function is.
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But that wave function
already, in our case,
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is a wave function
of definite momentum.
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So somehow, the information
that is missing there,
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is that the energy
is p squared over 2m.
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So we have that the energy
is p squared over 2m.
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So let's try to think of
the energy as an operator.
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And look, you could
say the energy,
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well, this is the energy
operator acting on the function
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gives you the energy.
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That this true, but it's too
general, not interesting enough
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at this point.
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What is really interesting is
that the energy has a formula.
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And that's the physics
of the particle,
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the formula for the energy
depends on the momentum.
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So we want to capture that.
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So let's look what
we're going to do.
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We're going to do a
relatively simple thing, which
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we are going to walk back this.
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So I'm going to
start with E psi.
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And I'm going to invent
an operator acting on psi
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that gives you this energy.
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So I'm going to invent an O.
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So how do we do that?
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Well, E is equal to p
squared over 2m times psi.
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It's a number times psi.
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But then you say, oh,
p, but I remember p.
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I could write it as an operator.
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So if I have p times
psi, I could write it
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as p over 2m h bar
over i d dx of psi.
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Now please, listen
with lots of attention.
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I'm going to do a simple
thing, but it's very easy
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to get confused
with the notation.
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If I make a little typo
in what I'm writing
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it can confuse you
for a long time.
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So, so far these are numbers.
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Number, this is a
number times psi.
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But this p times
psi is p hat psi
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which is that operator, there.
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So I wrote it this way.
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I want to make one more-- yes?
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AUDIENCE: Should that say E psi?
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PROFESSOR: Oh yes,
thank you very much.
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Thank you.
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Now, the question is, can I
move this p close to the psi.
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Opinions?
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Yes?
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AUDIENCE: Are you asking
if it's just a constant?
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PROFESSOR: Correct,
p is a constant.
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p hat is not a constant.
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Derivatives are not.
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But p at this
moment is a number.
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So it doesn't care
about the derivatives.
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And it goes in.
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So I'll write it as
1 over 2m h/i d dx,
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and here, output p psi,
where is that number.
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But now, p psi, I can
write it as whatever it is,
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which is h/i d dx, and p
psi is again, h/i d dx psi.
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So here we go.
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We have obtained, and
let me write the equation
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in slightly reversed form.
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Minus, because of the two
i's, 1 over 2m, two partials
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derivatives is a second order
partial derivative on psi,
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h bar squared over
2m d second dx psi.
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That's the whole right-hand
side, is equal to E psi.
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So the number E
times psi is this.
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So we could call this
thing the energy operator.
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And this is the energy operator.
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And it has the property that
the energy operator acting
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on this wave function
is, in fact, equal
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to the energy times
the wave function.
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So this state again is
an energy eigenstate.
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Energy operator on the
state is the energy
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times the same state.
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So psi is an energy eigenstate,
or a state of definite energy,
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or an energy
eigenstate with energy,
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E. I can make it clear for
you that, in fact, this energy
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operator, as you've
noticed, the only thing
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00:18:58,890 --> 00:19:07,870
that it is is minus h squared
over 2m d second dx squared.
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00:19:07,870 --> 00:19:09,850
But where it came
from, it's clear
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00:19:09,850 --> 00:19:17,420
that it's nothing else but
1 over 2m p hat squared,
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00:19:17,420 --> 00:19:23,210
because p hat is
indeed h/i d dx.
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00:19:23,210 --> 00:19:25,430
So if you do this computation.
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00:19:25,430 --> 00:19:26,510
How much is this?
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00:19:26,510 --> 00:19:32,000
This is A p hat times p
hat, that's p hat squared.
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00:19:32,000 --> 00:19:38,440
And that's h/i d dx h/i d dx.
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00:19:38,440 --> 00:19:42,470
X And that gives you the answer.
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00:19:42,470 --> 00:19:49,650
So the energy operator
is p hat squared over 2m.
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00:19:52,540 --> 00:19:56,110
All right, so actually,
at this moment,
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00:19:56,110 --> 00:20:00,970
we do have a Schrodinger
equation, for the first time.
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00:20:00,970 --> 00:20:06,050
If we combine the
top line over there.
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00:20:06,050 --> 00:20:17,050
I h bar d dt of psi
is equal to E psi,
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00:20:17,050 --> 00:20:23,900
but E psi I will write it
as minus h squared over 2m d
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00:20:23,900 --> 00:20:28,530
second dx squared psi.