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00:00:00,970 --> 00:00:03,820
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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00:14:52,930 --> 00:14:54,820
to get confused
with the notation.

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00:14:54,820 --> 00:14:57,540
If I make a little typo
in what I'm writing

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00:14:57,540 --> 00:15:00,480
it can confuse you
for a long time.

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So, so far these are numbers.

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00:15:05,990 --> 00:15:09,100
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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00:15:20,590 --> 00:15:22,660
I want to make one more-- yes?

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00:15:22,660 --> 00:15:25,610
AUDIENCE: Should that say E psi?

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00:15:25,610 --> 00:15:27,550
PROFESSOR: Oh yes,
thank you very much.

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00:15:31,520 --> 00:15:32,020
Thank you.

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00:15:34,530 --> 00:15:43,985
Now, the question is, can I
move this p close to the psi.

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00:15:50,860 --> 00:15:51,360
Opinions?

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00:15:53,960 --> 00:15:54,900
Yes?

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00:15:54,900 --> 00:15:57,260
AUDIENCE: Are you asking
if it's just a constant?

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00:15:57,260 --> 00:15:59,710
PROFESSOR: Correct,
p is a constant.

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00:15:59,710 --> 00:16:01,460
p hat is not a constant.

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00:16:01,460 --> 00:16:02,330
Derivatives are not.

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00:16:02,330 --> 00:16:04,490
But p at this
moment is a number.

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00:16:04,490 --> 00:16:06,750
So it doesn't care
about the derivatives.

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00:16:06,750 --> 00:16:07,610
And it goes in.

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00:16:07,610 --> 00:16:16,180
So I'll write it as
1 over 2m h/i d dx,

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00:16:16,180 --> 00:16:22,480
and here, output p psi,
where is that number.

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00:16:22,480 --> 00:16:28,750
But now, p psi, I can
write it as whatever it is,

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00:16:28,750 --> 00:16:41,550
which is h/i d dx, and p
psi is again, h/i d dx psi.

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00:16:41,550 --> 00:16:44,340
So here we go.

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00:16:44,340 --> 00:16:48,180
We have obtained, and
let me write the equation

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00:16:48,180 --> 00:16:51,240
in slightly reversed form.

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00:16:51,240 --> 00:16:57,780
Minus, because of the two
i's, 1 over 2m, two partials

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00:16:57,780 --> 00:17:03,660
derivatives is a second order
partial derivative on psi,

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00:17:03,660 --> 00:17:09,329
h bar squared over
2m d second dx psi.

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00:17:09,329 --> 00:17:13,904
That's the whole right-hand
side, is equal to E psi.

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00:17:21,200 --> 00:17:26,890
So the number E
times psi is this.

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00:17:26,890 --> 00:17:33,732
So we could call this
thing the energy operator.

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00:17:45,040 --> 00:17:48,090
And this is the energy operator.

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00:17:54,190 --> 00:18:03,800
And it has the property that
the energy operator acting

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00:18:03,800 --> 00:18:06,620
on this wave function
is, in fact, equal

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00:18:06,620 --> 00:18:09,530
to the energy times
the wave function.

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00:18:12,780 --> 00:18:19,970
So this state again is
an energy eigenstate.

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00:18:19,970 --> 00:18:22,710
Energy operator on the
state is the energy

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00:18:22,710 --> 00:18:23,920
times the same state.

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00:18:23,920 --> 00:18:46,110
So psi is an energy eigenstate,
or a state of definite energy,

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00:18:46,110 --> 00:18:48,970
or an energy
eigenstate with energy,

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00:18:48,970 --> 00:18:56,100
E. I can make it clear for
you that, in fact, this energy

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00:18:56,100 --> 00:18:58,890
operator, as you've
noticed, the only thing

244
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.