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Okay. So, in today's lecture, I'm going to
introduce some of the quantities we saw

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last time, a little more formally. So,
we'll talk about observables with respect

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to continuous quantum states. I'll go over
Schrodinger's equation for, for free

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particle on a, on a line more precisely.
And then, then we are going to step on

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and, and talk about particle in a box.
Okay, which will give us a toy model, you

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know, a very simplified model in which we
can actually talk about atomic qubits you

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know, this simple model is surprisingly
accurate and, and it will allow you to

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picture what a qubit might look like.
Okay, so, so, so, let's start, let's you

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know, so, r, what we, what we want to do
is we want to model the particle on a line

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and to begin with, we think of, you know,
we think of the particle which can be in

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one of a discrete number of states on the
line say, say, it's in one of these K

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different locations. And then, of course,
we'd write, write the, state of this

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particle as a superposition sum over, sum
over j of alpha sub j, j where j goes from

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zero to K-1. And we might also wish to
wish to, create an observable which gives

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us the position of the system. So, let's
say, we have a position, observable M. So,

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M is going to be her mission matrix. So, M
= M conjugate transpose. And what, what

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and it, it therefore it has an orthonormal
set of eigenvectors with real eigenvalues.

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S, o what are, what are the eigenvectors
of this observable M going to be? Well, if

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it's a position observable, then the
eigenvectors should, should clearly be

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the, be the states of definite positions
which is j. So, so, its a standard basis

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states and, and M, of course, will be
diagonal in the standard basis and then

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what we want, want the position value to
be. So, well, we wanted to be, of course,

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this matrix, right? So Mj, if you have the
definite state j. And it gives us this.

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It, it diagonally was, is also j. Right,
and, and so, that, that's what that' s

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what it gives us. So, for, for, for
example you know if we happen to be in the

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state, one / square root three, zero +
square root two / 3,1. This is not a state

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which, you know, the, here, the, the
position of the particle is not definite.

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So, if you have, if you have a, a thousand
such, such particles, each prepared in the

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same state and we measure each of these
1,000 particles, then roughly one / three

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of them, you know, roughly 333, will,
will, will end, you know, when we measure

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them, they will be, they will be at
position zero. So, the, so, the outcome

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will be zero and for roughly 666, that
position would be one, right, ? O course

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if you want a definite outcome maybe we
should be in one of the eigenstates, which

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is, which is either zero or one or two,
etc. Okay, so, so that's, that's a quick

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recap of, of, of, of the discrete state of
the events. And now, in the, in the case

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of you know, in the case that we don't
want this discrete states but we want the

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particle to be anywhere on the line then
we replace this. This are, are the state

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vector psi, which are finite dimensional
vector, k-dimensional vector by a wave

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function, psi (x), right? So, think of psi
(x) as the amplitude being at x. So, psi

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(x) is, is a function from the rear line
to the complex numbers, okay. So now, we

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have you know, it's as though we are
replacing this, this k-dimensional vector

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psi, right, often through alpha k - one, b
this infinite dimensional function psi

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(x). So, so, think of k is becoming very,
very large and continuous and that's what,

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that's what this function is. And, of
course, we must have an normalization

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condition and that is given by integral
for minus infinity to infinity, of psi (x)

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magnitude ^two dx = 0.We also have a
notion of inner product, so we have an

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inner product. So, if you have two
different wave functions, psi (x) and phi

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(x), then the inner product, is just an
integral from minus infinity to infinity

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of phi (x) conjugates psi (x) dx. So, the
way, let me point out something that , I

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guess, if alpha is a complex number, a +
ib, then we, we've been using the notation

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alpha star to denote a - ib, it's a, it's
a conjugate of alpha. And, I also

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interchangeably use the notation alpha
bar. So, depend, you know, sometimes I use

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alpha star, sometimes alpha bar, depending
upon which is convenient to write, and so,

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you should interpret them to be exactly
the same thing. Okay, alright, so now,

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let's, let's look at the fact that
observable M was Hermitian, which means

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that M = M conjugate transverse. Another
way of writing that is, this is the same

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thing as saying that, the ij entry of
alright? You know, make, make sure you,

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you understand why this is the ith and jth
entry of M, is equal to, if you look at

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the jth, ith entry of M, it's a conjugate
of that. Now, it's can also, this, this is

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also called self-adjoint. And of course,
this, this was not just for basis vectors

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i and j, but it's actually holds for any
two states, phi and psi. So, so, M is

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self-adjoint to Hermitian, if and only if
this holds for any two states psi and phi.

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Okay? So, convince yourself that this,
this really is trivially equivalent to, to

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the previous, previous assertion. Okay.
So, so, what's, what corresponds to

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Hermitian, to, to a self-adjoint to an
observable in the, in the, in the finite

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dimensional case? Well, it's just, you
know, you know, it's, it's just going to

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be some linear mapping on wave functions,
right? So, an observable is, is a linear

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function, it's self-adjoint Mapping M that
maps wave functions to wave functions So,

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for example, there's, there's a position
observable, x hat, and the way x hat works

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is x hat, when it's applied to a wave
function psi (x), it gives us some wave

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function phi (x), where, phi (x) is just x
phi (x). Okay. So, what, where did this

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come from? So, remember, what, what, what,
we had for the position, you know, if you

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had a position observable in the finite
dimensional case, we wrote it down like

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this. Write it, it was, so, where, where,
this was the position of the particle. It

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was at position zero, one or etc. And when
we, when we applied it to some, to some

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state, alpha zero, so on, to alpha K-1, we
just got the, a new state which was zero

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times alpha zero, one times alpha 1e, K -
one times alpha K - one, right. So, so,

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this corresponds to psi (x), right. So,
at, at x psi (x) is the amplitude for

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being at that point. What does phi (x)
tell us? Well, phi (x) is the observable

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applied to psi (x), and what should it
tell us? Well it should, it should

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multiply that, that amplitude by the
position, which is the eigenvalue of that,

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of, right? So, so, these are the
eigenvectors. So, so they're, so the, the

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position eigenvectors are, are x, itself,
write zero or one , etc, right? So, the

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position eigenvectors are, are, in, in
this case, are, are, of the form j, and

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so, so, j is, so, so, whatever is in this
jf location, the delta function of j, it

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gets multiplied by, by j, alright? So, you
get j alpha jM. And that's exactly what's

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happening with the position. So, and of
course, you can see that this is, you

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know, this is self-adjoint. And, you know,
submission, etc. Okay. So, the other

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interesting operator is the momentum
operator. It is denoted by p hat and it's

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i h bar d / dx. Okay, so, what, what this
means is that when you, when you apply

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this, two of a function psi. So, so, p hat
when applied to psi (x) is i H bar d / dx

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of psi (x). Okay, so, let's try to wrap
our heads around it and see why is this a

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Hermitian operator? So, the easiest thing
to do is to just think about it in terms

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of the discrete analog. Som let's, let's
try to understand the discrete analog. So,

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again, remember we, we are thinking about
it like this, we have a particle on the

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line. It's, you know, its at one of a
discreet, we have a lot, you know, we have

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these evenly spaced points and it's in one
of these evenly spaced points, a finite

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side of them. And so, so, we write our
state vector as, alpha NOT, alpha one

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through alpha K - one. And now, how should
we write the derivative? You know, d / dx.

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Well, d / dx has the following
interpretation. So, to, to get the, the

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derivative at this point j, one way I can
figure out the derivative at, at this

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point, if, if I have some function is at,
at x, I can look at the point to the right

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of it, x + delta x and look at the
function value there, I can't look at the

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point to the left of it, x - delta x and
look at the value there. Take the

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difference and divide by two delta x.
Okay, so, what does this correspond to?

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Well, what it corresponds to, I claim, is,
if you have ones on the off diagonal, zero

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on the diagonal, and -one below the
diagonal. Okay, so, why is that? Well, you

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see because, because if you have, if you
have let's say, alpha j, alpha j - one,

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alpha j + one, and now, now imagine,
what's the, what's it going to look like

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when you multiply by the, what's the jth
row going to give us? Well, you take this

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vector and slide it along the, along the
jth row and , of course, alpha j

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multiplied with zero, alpha j - one with
-one, and alpha j + one with +one. So,

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you'll get (alpha j + one) - (alpha J -
one), which is exactly what we wanted

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here, that's the first derivative. So,
this matrix corresponds to the first

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derivative. But it is Hermitian. Well, no,
because if you take it's conjugate, well,

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it's real sub-conjugate does nothing and
when you take transpose, you get negative

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of what it used to be. Okay, so, so now,
how do we make it ission? Well, it's easy,

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we just multiply by i. So, when, when we
multiply by i, these gets, get changed to

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i's and these gets changed to -i's, okay?
And now, what happens when we take the

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conjugate? Well, when, when we get the
con, take the conjugate, the i gets

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replaced by -i, -i.i. And then, when we
take the transpose, we get back what we

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started from. So, so, indeed, this matrix
is the, is, is self-adjoint to Hermitian.

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And so, similarly, this, this momentum
operator is self-adjoint. It's something

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that you can verify. Okay, so, so now,
having done that, we can start trying to

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piece together what, what changed equation
for, for particle on a l ine should look

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like. Okay. So, so, remember, we had a, a
momentum operator [unknown] which is i h

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bar d / dx and now, we want to, we want to
understand what Schrodinger equation

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should look like for a particle on a line.
So, we have a free particle on a line and

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okay, so, what Schrodinger equation tells
us is this that i h bar d psi / dt = h

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psi, right? Okay, so, to, to understand
what Schrodinger equation says for this

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particle on the line, all we have to do is
understand what H is. Well, what is H? H

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is the Hamiltonian. It's the energy
operator. So, what does the energy of a

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particle on a line? Well, the particle,
you know, classically, the energy would

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be, would be the potential energy plus the
kinetic energy. In our case, the, the

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particle is a free particle so we have a
free particle which means, by definition,

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that there's no potential energy. What
about the kinetic energy? Well,

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classically, the kinetic energy is p^2 /
2M, where M is the mass and p for

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momentum. Okay, so now, this is realistic
way of guessing what the, what the

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Hamiltonian is going to be for in the, in
the quantum case. So, what you do is as

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you do something which is which sounds a
bit ad hoc, which is you, you, you look at

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the classical case and you replace your
momentum by the momentum operator, okay.

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And you know, to make it sound very
official and make it sound correct

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physicists have invented a name for this
procedure. They call it the correspondence

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principle. Okay. So, so, so using this
correspondence principle, which is a nice

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way of guessing what the, what, what, what
this basic equation of motion should be.

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We, we get that i h bar d psi / dt = b hat
squared / 2M psi. So, what's P hat

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squared? Well you just apply P hat twice.
So, it's i h bar d / dx of i h bar d / dx

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00:19:33,004 --> 00:19:38,003
one / 2M psi. So, that's really, if you're
applying, if you take the derivative

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twice, it's the second derivative. I^2 is
-one so it's -h bar square / 2M d^2 / dx^2

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of psi. And this is exactly the form that
we intuitively came up with last time. S

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o, last time we said, let's not worry
about the constants and we, we said,

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00:20:02,097 --> 00:20:09,076
Schrodinger's equation for free particles
should look in form, something like i d

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00:20:09,076 --> 00:20:15,037
psi / dt is d^2 psi / dx^2. And that's
what this other way of looking at

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00:20:15,037 --> 00:20:19,008
Schrodinger's equation also tells us.