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Okay. So now, we are going to look at a
very simple model of an electron in an

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atom. And this simple model is going to
be, well, it's going to be on the one

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hand, simple enough to, that we are going
to be able to analyze it completely, and

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understand exactly what states are
allowed, and so on. But on the other hand,

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it turns out to be a surprisingly good
model for, they are allowable states of an

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electron in an atom. And therefore, it'll
help us understand, very concretely, what

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an atomic qubit might look like. Okay, so,
so, here, here's what, what that model is,

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its called a particle in a box. And its,
you can think of it as an toy model for a

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hydrogen atom where the electron is
confined in the vicinity of, of the, of

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the proton by Coulomb attraction. So, you
can think of, to a first approximation, to

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even think of this electron as trapped
within a box. It's a, it's a pretty tiny

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box, it's about one angstrom in length or,
or roughly ten^-10 meters. And so of

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course, Coulomb attraction it's, you know,
it's, it's more complicated because it,

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it, it falls off with the, with the square
of the distance. But, but to first

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approximation, we'll, we'll think of the
electron is just confined in this, in a

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box with, with hard, you know, in this
spherical box with hard boundaries. And

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so, we, you know, we can think of this
even though its three-dimensional picture,

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as a one-dimensional picture, because the
only variable is how far you are from,

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from the, from the center. And so, so,
this is our model for, for the, for the

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electron. Its, its confined to a box of,
of length l. And then, it's a free

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particle within this, as long as it stays
in the box. But then, it's not allowed to

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step outside the box because, because of
an infinite potential outside which, which

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makes its effective energy infinite as
soon as it steps outside the box. And, of

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course, this little electron just doesn't
have that kind of energy. So, so, it's,

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it's disallowed from being, being anywhere
outside. Okay ay, so, how are we going to

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figure out what the state of this electron
is? Well, we'll solve Schrödinger

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equation. What's Schrödinger equation? Ih
power d psi / dt = H psi. But, but what is

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H? What's the Hamiltonian in this case?
Well, since we have a free particle, as

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long as we are outside of this as long as
we stay in the box, it's, well, it's p^2 /

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2m psi + vx, where vx is the potential
energy. And so you know, for, for a free,

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free particle this p^2 / 2m, of course,
ends up being -h bar^2 / 2m, d^2 / dx^2.

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And the fact that psi x, you know, we have
a non-zero potential. We can just

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translate into boundary conditions which
say that psi (zero) =psi (l) = zero. Okay,

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so, so now, what, what, what they're left
with is the question of, now that they

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know what the Hamiltonian is we are left
with the, with the problem of solving

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Schrödinger equation for this given
Hamiltonian H. Now remember, how we, how

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we solved Schrödinger equation. Well, you
know, it looked, it looked daunting to ask

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and then we said, the way to solve it is
to first solve for the eigenstates of H

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because those behave very nicely under
Schrödinger equation. So, what, what we

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first do, so their, so here's an outline
of what we do. We first solve H five =

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lambda times five. And once we find such
an eigenstate, then we know that if, if,

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if, if psi at time zero = phi. Then, psi
at time t is e^-i lambda t by h bar phi.

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Okay, so, all we need to do is, is figure
out all the eigenvectors of H and all the

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eigenvalues, the corresponding eigenvalues
and then, as long as we take the initial

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state psi and decompose it into, into
eigenstates of H, we, we understand, we

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can understand how that state will evolve
because, essentially, each eigenstate is

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almost invariant. It, it just, you know,
it, it, it just marches to its, its, its,

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its own drummer which is, which is its
eigenvalue. And it, it, it does so by, by

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this phase precession. E^-1 lambda t / h
bar. Okay, so, so what are the eigenstates

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of, of H? Well, let's guess. That the
eigenstate of this, this operator, we, we

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have H that, that H equal to -h bar^2 / 2m
d^2 / dx^2 has, has an eigenstate of the

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form, phi (x) = e^ikx. Okay. So, so now,
if you apply h to phi (x) is -h bar^ two /

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2m d^2 / dx^2 and e^ikx which is h bar^2 /
2m k^2 e^ikx, right? Because, because you

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got an ik whole squared when you take the
second derivative and i^2 is -1so that

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[unknown] -one. Okay. So, so, this is an
eigenstate with eigenvalue e = h bar^2 k^2

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/ 2m, okay? And of course, we have the
same eigenvalues for the e^ikx as e^-ikx.

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And therefore, we have an eigenspace and
so, our solution is going to be of the

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form phi sub e (x) = (a x e^ikx) + (b x
e^-ikx). This is, all of these for every

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nb, this is an eigenstate of this, this
operator H with eigenvalue e sub k = h

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00:08:00,056 --> 00:08:10,099
bar^2 k^2 / 2m. Now, now, of course, we're
not yet done because we have to impose the

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boundary condition that, that phi (zero)
must be equal to zero and phi (l) must be

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equal to zero. So, it turns out that it's,
it's convenient for us if you, if you

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remember e^ikx is cosine kx + i sine kx.
And similarly to the -ikx is cosine kx - i

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sine kx. And so, instead of writing this
in terms of e^ikx and e^-ikx, you could

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also write this down as a linear
combination of sine kx + d cosine kx.

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Okay. So, let's see. So, so, where, where
are we now? What this leaves us with is we

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have phi sub e (x) is c sine kx + d cosine
kx. And you know, this has energy, e sub k

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= h bar^2 k^2 / 2m. Now, we have to impose
two conditions. We have phi (zero) must be

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equal to zero. Psi (zero) is sine kx =
zero and this is equal to c0 + d1. So,

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that, that already tells us that d = zero.
Okay, and then we can impose the second

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00:10:08,041 --> 00:10:22,006
condition that 5l = one. So, if I said e
(l) equal to, sorry, equal to zero which

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00:10:22,006 --> 00:10:40,067
means that C sine K l = zero So, this
tells us that KL = n pi, where n is an

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00:10:40,067 --> 00:10:57,001
integer. And therefore, we get K which is
a function of n is n pi / L. And so, this

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gives us a quantize function. So, the
energy eigenfunctions are e sub nx is, is

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00:11:05,009 --> 00:11:14,093
h power ^two k sub n^2 / 2m which is h
power^2 n^2 i^2 / 2m, okay. So, n just

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00:11:14,093 --> 00:11:24,051
takes on n is, takes on integral values
1,2,3 and so on. And, what do

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00:11:24,051 --> 00:11:40,058
eigenfunction looks like? Well, phi sub n
(x) is just C sine Kx which is a case of

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00:11:40,058 --> 00:11:56,007
an nx which is n pi / Lx Okay, so, are we
quite done? Well, not quite because we

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must normalize. We must normalize to make
sure that the integral of this from zero

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00:12:06,062 --> 00:12:14,063
to l is one. There's the squared so, so
let's set this out. So, finally for, for

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00:12:14,063 --> 00:12:27,035
we must have integral from zero to l (c)^2
sine^2 n pi / Lx. The x equal to one. Now,

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00:12:27,035 --> 00:12:36,068
of course, if you, if you add in cosine^2
to, to, to this, you would you would get,

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00:12:36,068 --> 00:12:50,048
you would get the integral is C^2 L. And
that's, that's two. So, so, C^2 must be

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00:12:50,048 --> 00:13:02,036
two / L that implies c^2 = two / L or C =
square root two / L. Alright, so, it's,

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00:13:02,036 --> 00:13:09,063
this, you can replace the C by square root
two / L. So, so the wave function is

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00:13:09,063 --> 00:13:15,048
square root two / L times this. And that's
the energy. So, here's a, here's a

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00:13:15,048 --> 00:13:21,092
solution then. So, the solution is, that's
the energy. So, so, so solution is

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00:13:21,092 --> 00:13:31,056
quantized for entity of values, that's the
energy. And that's the corresponding wave

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00:13:31,056 --> 00:13:42,021
function, right, for n = one, two, and so
on. And, if you plot it out, then, when

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you have n = one, you have sine of pi x /
L. Which, which goes like this. Starts at

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00:13:53,072 --> 00:14:03,077
zero, etc. When n =two, you get a full
cycle of the sine, sine wave. When n =

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00:14:03,077 --> 00:14:12,013
three, you got three halves of the cycle
and so on. And so, so, a wave function is

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00:14:12,013 --> 00:14:18,035
quantized, right? These are eigenstates.
Right, these are, these are the

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00:14:18,035 --> 00:14:24,008
eigenstates of H. These are the, meaning,
these are the, these are the wave

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00:14:24,008 --> 00:14:30,006
functions with definite energy. And what's
the energy? Well, that's the corresponding

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00:14:30,006 --> 00:14:35,005
energy, right? So, so, so these wave
functions have an increasing energy,

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00:14:35,005 --> 00:14:42,068
increasing like proportional to n^2. So,
one, four, nine, etc. And these particular

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00:14:42,068 --> 00:14:53,022
eigenstates are, are extending waves,
right. Okay, and of course, if you're

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00:14:53,022 --> 00:15:03,028
given, given any particular arbitrary psi
of x, you would decompose it, you would

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00:15:03,028 --> 00:15:16,025
write it as alpha sub m, psi sub n (x).
And so, if this was, if this was psi sub x

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00:15:16,025 --> 00:15:34,014
at time0, then psi (x) at time t would
just be alpha sub n, e^i e sub n t / h bar

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00:15:34,014 --> 00:15:36,003
psi (x).