Hi, everybody.
Moving forward in this course, we'll now start to actually ...
solve the Schrodinger equation for different chemical environments.
In the next three videos, we'll be looking specifically at ...
an environment called the "particle in a box".
I'd first like to give you a big picture of what the next ...
several lessons are going to look like.
We've already seen the time-independent Schrodinger ...
equation, which is written on this slide in ...
both one dimension and three dimensions.
Recall that it consists of a kinetic energy operator ...
applied to wavefunction, plus a potential energy ...
operator applied to that same wavefunction,
which is equal to a constant, the measured energy,
times the wavefunction.
So far, we've just written the potential ...
energy operator abstractly, as this function V.
But really, this function V is different in ...
every chemical environment, and that's what makes it ...
different to solve the Schrodinger equation from ...
one system to another.
In the next several lessons, we're going to start thinking ...
about what this potential energy operator V looks like,
and how to solve the Schrodinger equation in ...
different environments, for different chemically ...
relevant systems.
The next three videos are going to focus on the particle in a box.
For the particle in a box, the potential energy operator ...
V is a constant, which for convenience we ...
define as zero, inside the box.
And the potential energy operator is infinity outside the box.
So basically, these conditions ensure that ...
the object can be inside the box,
but it cannot be outside the box.
This sounds like sort of an oversimplified model system,
but really it is chemically relevant.
Specifically, the particle in a box does a ...
reasonable job predicting the energy levels of pi orbitals in ...
conjugated molecules.
Next, we'll spend some time ...
thinking about the system that's called the harmonic ...
oscillator, in which the potential energy ...
operator is parabolic. It's defined as kx^2/2,
which you may recognize the potential energy of a Hooke's law spring.
If the potential energy is like a spring,
you might guess that the harmonic oscillator is relevant ...
to the vibrations of diatomic molecules.
And after that, we'll look at a particle on a ...
ring, or a particle on a spherical shell,
in which the potential energy operator is again zero,
but the object is constrained to a particular curve or a ...
particular surface.
This model turns out to be relevant to,
first of all, the angular part of atomic ...
orbitals, and also the rotations of ...
diatomic molecules.
In the next several lessons, we'll see a lot of math,
and it's easy to get lost in that math.
But I want to emphasize as much as I can that these ...
models are not just gratuitous math.
Each of these models, in addition to improving our ...
ability to solve the Schrodinger equation,
actually does capture chemically meaningful phenomena.
Before we look at the particle in a box,
I want to look at a slightly simpler situation,
the movement of an object in one-dimensional free space.
This is the situation where the potential energy operator V ...
equals zero everywhere.
So if we look at the one-dimensional Schrodinger ...
equation, and we set V equal to zero,
we can actually ignore this second term.
So we can rewrite our equation as the kinetic energy ...
operator applied to psi equals E psi.
If we rearrange these terms to put all the constants together,
we're left with the second derivative of the wavefunction ...
psi equals a negative group of constants times psi.
I realize we haven't all taken a course in differential ...
equations, but we do know something ...
about derivatives. So the question I want to ask is:
What do you think the energy eigenfunctions psi(x) are in this case?
In other words, can you think of any functions ...
where the second derivative is equal to a negative constant ...
times the function itself? I'll give you a few seconds to ...
think about that.
Hopefully you have some thoughts.
One type of function were the second derivative equals a ...
negative constant times the function itself is a sine or a cosine of kx.
If you take the second derivative of cos(kx),
you have -k^2 cos(kx).
And same thing with sine.
The second derivative of sin(kx)
is -k^2 sin(kx).
Alternatively, you might have thought of ...
complex exponentials, for example, e^ikx.
The second derivative of e^ikx is i^2 k^2 e^ikx.
And because i^2 = -1,
the second derivative is -k^2 e^ikx,
and it's something similar for e^-ikx.
So any of these functions are valid energy eigenfunctions ...
for a particle in one-dimensional free space.
And actually, sines and cosines are ...
completely equivalent to complex exponentials.
You may remember from Euler's formula that e^(i phi) = ...
cos(phi) + i sin(phi). So actually,
any combination of sines and cosines can alternatively be ...
written as a combination of complex exponentials.
So any of these functions, or sums and differences of them,
is a valid energy eigenfunction.
So let's consider one of these functions:
psi(x) = e^ikx.
What happens if you take measurements of an object in ...
this wavefunction?
In other words, what can we say about its ...
energy, and its momentum,
and its position?
Let's look first of the energy of this wavefunction e^ikx.
To find the energy of a wavefunction,
we have to apply the Hamiltonian operator,
the energy operator, to the wavefunction and find ...
the corresponding eigenvalue.
In this case, the Hamiltonian operator is ...
just the free-space Hamiltonian operator,
-hbar^2 / 2m times the second ...
derivative of the wavefunction.
Then, we can plug in our ...
wavefunction, and we know that the second ...
derivative is equal to -k^2 e^ikx.
We can combine those constants,
and find that we're left with -hbar^2 k^2 /
2m times the original wavefunction itself.
That means that collection of constants,
hbar^2 k^2 / 2m,
is the energy eigenvalue. It's the energy that we would measure.
So, for any value of k,
the measured energy of the wavefunction e^ikx is hbar^2 ...
k^2 / 2m.
Next, let's look at the momentum of ...
this same wavefunction e^ikx.
Remember, the one-dimensional ...
momentum operator is hbar / i times the first derivative.
You may have seen this written as -i hbar times the ...
first derivative.
Because 1/i is the same as -i, these two forms are equivalent.
To find the measured momentum,
we again just have to apply to momentum operator and find ...
the corresponding eigenvalue.
The momentum operator applied to this wavefunction is ...
hbar / i times the first derivative of ...
the wavefunction.
If we plug in our wavefunction, we find that the first derivative is ik e^ikx.
The i's cancel,
and we're left with hbar k e^ikx,
or hbar k times the original wavefunction itself.
So again, the constants in front of the ...
wavefunction are the momentum eigenvalue,
the momentum that we would measure.
So, the measured momentum of ...
the wavefunction e^ikx
is hbar k.
Finally, let's think about the position of ...
this wavefunction.
Remember, the position operator in one ...
dimension is just x. We multiply x by our wavefunction.
So the question is: Is this wavefunction e^ikx a ...
position eigenfunction?
To find that out, we apply our position operator ...
to this wavefunction,
and we get x times e^ikx.
You'll notice that x times e^ikx is not a scalar multiple of ...
e^ikx. Multiplying by x is not the ...
same as multiplying by a constant.
So no, this wavefunction is not a ...
position eigenfunction, and we don't know with ...
certainty where this object is.
In fact, in this case,
we really don't know anything about where this object is.
Remember, the probability of finding an ...
object in a given location is proportional to the complex ...
conjugate of the wavefunction times the wavefunction.
In this case, that's e^-ikx times e^ikx, or 1.
This value is constant at every point in space,
meaning we're just as likely to find the object at any point in ...
space as we are at any other point.
We know nothing about its position.
Let's summarize what we learned about the movement ...
of an object in wavefunction e^ikx in free space.
We found that e^ikx is an energy eigenfunction,
and it corresponds to the energy eigenvalue hbar^2 k^2 ...
/ 2m.
In free space, any value of k,
and therefore any energy, is allowed.
These same wavefunctions e^ikx are also momentum ...
eigenfunctions corresponding to the measured momentum hbar k.
But these wavefunctions e^ikx are not position ...
eigenfunctions. In fact,
we don't know anything about the position of these objects,
which had to be true based on the Heisenberg uncertainty principle.
Because we know the momentum exactly,
we can't know anything about the position.
That brings us back to our first important model system,
the particle in a one-dimensional box of length L.
In this system, the potential energy operator ...
is equal to zero inside the box,
just like free space, and it's equal to infinity ...
outside the box.
This system is important for a few reasons.
For one, it's our first opportunity to see ...
the major differences between classical mechanics and ...
quantum mechanics.
Also, unlike a lot of problems that ...
we'll see in this course, the particle in a box can be ...
solved exactly, which is useful for us to see.
And third, the particle in a box is ...
physically relevant. Specifically,
it allows us to predict the pi orbitals in conjugated ...
molecules and their energies.
We'll spend the next video working through the math of ...
the particle in a box.
But for now, I want to think just a little bit ...
about what we expect the eigenfunctions to look like.
Outside the box, the potential energy operator ...
equals infinity,
which means if the wavefunction was non-zero ...
outside the box, then the wavefunction would ...
have infinite energy, which isn't physically possible.
This means that outside the box,
the value of the wave function has to be zero.
Inside the box, the potential energy operator ...
is zero, just like free space.
So the wavefunction has to look like a free-space ...
wavefunction, a sine or a cosine or a ...
complex exponential.
And the third idea that's going to allow us to find the ...
wavefunctions of a particle in a box is the fact that we know ...
from the first postulate of quantum mechanics that ...
wavefunctions have to be continuous.
So in this case, we know that the ...
wavefunction is zero outside the box,
it's oscillating inside the box as a sine or a cosine or a ...
complex exponential,
and the wavefunction has to meet itself at the two ends of ...
the box, x=0 and x=L.
In the next video, we'll put all of these ideas ...
together and work through the math,
to find the wavefunctions and the energies of a particle in a ...
one-dimensional box. That's all for now.
