Okay. So now, we are going to look at a very simple model of an electron in an
atom. And this simple model is going to be, well, it's going to be on the one
hand, simple enough to, that we are going to be able to analyze it completely, and
understand exactly what states are allowed, and so on. But on the other hand,
it turns out to be a surprisingly good model for, they are allowable states of an
electron in an atom. And therefore, it'll help us understand, very concretely, what
an atomic qubit might look like. Okay, so, so, here, here's what, what that model is,
its called a particle in a box. And its, you can think of it as an toy model for a
hydrogen atom where the electron is confined in the vicinity of, of the, of
the proton by Coulomb attraction. So, you can think of, to a first approximation, to
even think of this electron as trapped within a box. It's a, it's a pretty tiny
box, it's about one angstrom in length or, or roughly ten^-10 meters. And so of
course, Coulomb attraction it's, you know, it's, it's more complicated because it,
it, it falls off with the, with the square of the distance. But, but to first
approximation, we'll, we'll think of the electron is just confined in this, in a
box with, with hard, you know, in this spherical box with hard boundaries. And
so, we, you know, we can think of this even though its three-dimensional picture,
as a one-dimensional picture, because the only variable is how far you are from,
from the, from the center. And so, so, this is our model for, for the, for the
electron. Its, its confined to a box of, of length l. And then, it's a free
particle within this, as long as it stays in the box. But then, it's not allowed to
step outside the box because, because of an infinite potential outside which, which
makes its effective energy infinite as soon as it steps outside the box. And, of
course, this little electron just doesn't have that kind of energy. So, so, it's,
it's disallowed from being, being anywhere outside. Okay ay, so, how are we going to
figure out what the state of this electron is? Well, we'll solve Schrödinger
equation. What's Schrödinger equation? Ih power d psi / dt = H psi. But, but what is
H? What's the Hamiltonian in this case? Well, since we have a free particle, as
long as we are outside of this as long as we stay in the box, it's, well, it's p^2 /
2m psi + vx, where vx is the potential energy. And so you know, for, for a free,
free particle this p^2 / 2m, of course, ends up being -h bar^2 / 2m, d^2 / dx^2.
And the fact that psi x, you know, we have a non-zero potential. We can just
translate into boundary conditions which say that psi (zero) =psi (l) = zero. Okay,
so, so now, what, what, what they're left with is the question of, now that they
know what the Hamiltonian is we are left with the, with the problem of solving
Schrödinger equation for this given Hamiltonian H. Now remember, how we, how
we solved Schrödinger equation. Well, you know, it looked, it looked daunting to ask
and then we said, the way to solve it is to first solve for the eigenstates of H
because those behave very nicely under Schrödinger equation. So, what, what we
first do, so their, so here's an outline of what we do. We first solve H five =
lambda times five. And once we find such an eigenstate, then we know that if, if,
if, if psi at time zero = phi. Then, psi at time t is e^-i lambda t by h bar phi.
Okay, so, all we need to do is, is figure out all the eigenvectors of H and all the
eigenvalues, the corresponding eigenvalues and then, as long as we take the initial
state psi and decompose it into, into eigenstates of H, we, we understand, we
can understand how that state will evolve because, essentially, each eigenstate is
almost invariant. It, it just, you know, it, it, it just marches to its, its, its,
its own drummer which is, which is its eigenvalue. And it, it, it does so by, by
this phase precession. E^-1 lambda t / h bar. Okay, so, so what are the eigenstates
of, of H? Well, let's guess. That the eigenstate of this, this operator, we, we
have H that, that H equal to -h bar^2 / 2m d^2 / dx^2 has, has an eigenstate of the
form, phi (x) = e^ikx. Okay. So, so now, if you apply h to phi (x) is -h bar^ two /
2m d^2 / dx^2 and e^ikx which is h bar^2 / 2m k^2 e^ikx, right? Because, because you
got an ik whole squared when you take the second derivative and i^2 is -1so that
[unknown] -one. Okay. So, so, this is an eigenstate with eigenvalue e = h bar^2 k^2
/ 2m, okay? And of course, we have the same eigenvalues for the e^ikx as e^-ikx.
And therefore, we have an eigenspace and so, our solution is going to be of the
form phi sub e (x) = (a x e^ikx) + (b x e^-ikx). This is, all of these for every
nb, this is an eigenstate of this, this operator H with eigenvalue e sub k = h
bar^2 k^2 / 2m. Now, now, of course, we're not yet done because we have to impose the
boundary condition that, that phi (zero) must be equal to zero and phi (l) must be
equal to zero. So, it turns out that it's, it's convenient for us if you, if you
remember e^ikx is cosine kx + i sine kx. And similarly to the -ikx is cosine kx - i
sine kx. And so, instead of writing this in terms of e^ikx and e^-ikx, you could
also write this down as a linear combination of sine kx + d cosine kx.
Okay. So, let's see. So, so, where, where are we now? What this leaves us with is we
have phi sub e (x) is c sine kx + d cosine kx. And you know, this has energy, e sub k
= h bar^2 k^2 / 2m. Now, we have to impose two conditions. We have phi (zero) must be
equal to zero. Psi (zero) is sine kx = zero and this is equal to c0 + d1. So,
that, that already tells us that d = zero. Okay, and then we can impose the second
condition that 5l = one. So, if I said e (l) equal to, sorry, equal to zero which
means that C sine K l = zero So, this tells us that KL = n pi, where n is an
integer. And therefore, we get K which is a function of n is n pi / L. And so, this
gives us a quantize function. So, the energy eigenfunctions are e sub nx is, is
h power ^two k sub n^2 / 2m which is h power^2 n^2 i^2 / 2m, okay. So, n just
takes on n is, takes on integral values 1,2,3 and so on. And, what do
eigenfunction looks like? Well, phi sub n (x) is just C sine Kx which is a case of
an nx which is n pi / Lx Okay, so, are we quite done? Well, not quite because we
must normalize. We must normalize to make sure that the integral of this from zero
to l is one. There's the squared so, so let's set this out. So, finally for, for
we must have integral from zero to l (c)^2 sine^2 n pi / Lx. The x equal to one. Now,
of course, if you, if you add in cosine^2 to, to, to this, you would you would get,
you would get the integral is C^2 L. And that's, that's two. So, so, C^2 must be
two / L that implies c^2 = two / L or C = square root two / L. Alright, so, it's,
this, you can replace the C by square root two / L. So, so the wave function is
square root two / L times this. And that's the energy. So, here's a, here's a
solution then. So, the solution is, that's the energy. So, so, so solution is
quantized for entity of values, that's the energy. And that's the corresponding wave
function, right, for n = one, two, and so on. And, if you plot it out, then, when
you have n = one, you have sine of pi x / L. Which, which goes like this. Starts at
zero, etc. When n =two, you get a full cycle of the sine, sine wave. When n =
three, you got three halves of the cycle and so on. And so, so, a wave function is
quantized, right? These are eigenstates. Right, these are, these are the
eigenstates of H. These are the, meaning, these are the, these are the wave
functions with definite energy. And what's the energy? Well, that's the corresponding
energy, right? So, so, so these wave functions have an increasing energy,
increasing like proportional to n^2. So, one, four, nine, etc. And these particular
eigenstates are, are extending waves, right. Okay, and of course, if you're
given, given any particular arbitrary psi of x, you would decompose it, you would
write it as alpha sub m, psi sub n (x). And so, if this was, if this was psi sub x
at time0, then psi (x) at time t would just be alpha sub n, e^i e sub n t / h bar
psi (x).