>> Welcome back to
Chemistry 131A.
Today we're going to talk
about Model 1D Quantum Systems,
in particular the particle
in a box, as it's called.
The box. What's the box?
Well, to make things
mathematically simple,
when we were normalizing
the wave function
and doing various things, we
took the limits on integrals
to be plus or minus infinity
because it made the math easy.
What it really did is it made
certain terms to away to zero.
And now we're going
to do a similar thing.
We're going to trap a particle
inside a region of space
by saying that the
potential energy,
if the particle gets
outside this region,
is going to go to plus infinity.
And since a particle
cannot have infinite energy,
that traps the particle in
the interior of the box.
The reason why we do
this is just, again,
mathematical simplicity.
This is the limit to make
it mathematically simple
of what could be quite
a realistic thing
where the energy gets high if
a particle gets too far away
from where it should be.
So our condition
on the potential
energy, v of x, is this.
It's zero if the particle is
in the box, which we'll say is
between x equals 0 and x equals
L. And it's infinity otherwise,
as the particle is outside
that region of the box.
And as I said, the reason is
just mathematical simplicity.
And we will see that what that
amounts to in this case is
that the particle can't
penetrate the edge
of the box at all.
Even though it's a
wave, it cannot get in.
So it's reflected off.
Well, if suppose we
said well why is that?
Why does the wave function
have to vanish outside the box?
The answer is we know that
psi star psi tells us the
probability that the
particle's there.
And if that's non-zero and we
multiply by v equals infinity,
we get an infinite energy.
But if we specify the energy of
the particle as not infinite,
then we can't have any
probability outside the box.
The expectation value,
in other words,
of the energy would be far too
big if we let even a little bit
of the wave function
slip outside the box.
In addition to the wave function
being zero outside the box,
the wave function has to be zero
right at the edges of the box.
And the reason for that is
that we want the wave
function to be continuous.
We want the wave function to
be a well-defined function
that we can take the
derivative of, and so forth.
So we want it to be zero
right at the edge of the box.
And that's pretty much
like a guitar string.
If you put your finger
on a fret,
you're holding the
string down there.
You can pluck the string with
different amounts of force
to get different
amounts of energy.
And the other side is held
and you get a certain
standing wave pattern.
And that's pretty much
exactly what the wave function
for the particle is going
to be doing inside
this spatial region.
So, as I said, the continuity
of psi is the reason why we
can say it's zero at the edge.
And that means that the kinds
of functions that we can have
to have a zero at x equals 0
and another zero at x equals L.
But inside the box,
there's nothing.
There's no potential
energy at all
until the particle
encounters the edge of the box,
it pretty much thinks
it's a free particle.
And we already solved
in a previous lecture
what the wave function is
for a free particle.
And so what we can
do is just piggyback
on that use those solutions.
And then we have a
couple of extra criteria
to normalize the wave function
and to make sure that it's zero
at the edges of the box.
Inside the box we have
the time-dependent --
time-independent, excuse
me -- Schrodinger equation,
which I've written here
as minus h-bar squared
over 2m second derivative
of psi.
And there's no v, so I've left
that out, is equal to E psi.
And that will give us
the energy eigenfunctions
for the wave function
inside the box.
We already know the
general solution.
It's Ae to the ipx
upon h-bar plus Be
to the minus ipx upon
h-bar, where p squared
over 2m is the kinetic
energy of the particle.
But now we have two
additional equations.
We have that the value, psi, the
probability amplitude at zero,
the edge of the box
should be zero.
And that is equal to A plus B,
because e to the minus i 0 is
still e to the 0, and that's 1.
So that goes away.
We just have A plus
B is equal to 0.
And then at the other
edge of the box,
psi at L is also equal to 0.
And here we have the
same kind of thing
but we have two additional
things.
We have e to the plus
ipL upon h-bar and e
to the minus ipL upon h-bar.
Well, the first equation
just means that A is equal
to minus B, or B is equal
to minus A, same thing.
And that means that we can write
the second equation as using A
for both sides as Ae to
the ipL upon h-bar minus Ae
to the minus ipL upon h-bar.
And so we have the two
exponential functions have
to be equal.
We don't want A to be 0,
or then everything is 0.
And the wave function vanishes.
And therefore, there's
a condition
that these two counter rotating
things be adding up to 0
at the edge of the box when
x is equal to L. And what
that condition is
going to mean is
that when these two
corkscrews have to add up to 0,
they can't just be
pointing like that,
they have to be pointing
opposite each other,
it's going to mean that
there's a restriction
on the wavelength of
the wave function.
And that means that only
certain energies are going
to be allowed.
A very useful relationship
here is Euler's formula,
which I've written out for
you: e to the i theta is equal
to cos theta plus i sine theta.
If you haven't had a
course in complex analysis,
you may not have encountered
this kind of formula,
but it is in fact by far one
of the most useful relations
you can ever imagine.
And I'll just take
a little aside here
and show you how you can use
this formula to derive all kinds
of other formulas which some
people try to commit to memory.
Or they even write them on
bits of paper and stick them
on their desk so they
won't forget them
if they're doing a lot of
trigonometric problems.
But let's do a practice problem.
Let's use the above
relationship,
e to the i theta is equal to
cos theta plus i sine theta
to derive the double angle
formula for the cosine
and the sine functions.
So if we have cos 2 theta, we
want to express that in terms
of cosine theta and sine theta.
And if we have sine 2 theta we
want to express that in terms
of cosine theta and sine theta.
And rather than memorizing
some kind of formula
that doesn't make much
sense, we're just going
to use the properties
of the exponentials,
which are very easy to
remember and do a little bit
of algebra, which is quick.
About all we have to remember is
that i squared is
equal to minus 1.
And then we're in the clear.
So here is the answer.
Rather than memorizing
the things,
here's what we're going to do.
We're going to set a
relationship and say e
to the 2i theta is equal to
e to the i theta squared,
because that's how
exponentials work
when you have something
in the exponent.
And e to the i theta
squared is e
to the i theta times
e to the i theta.
On the left hand side, then,
we have cosine 2 theta plus i
sine 2 theta, because that's e
to the 2i theta, or e to
the i 2 theta, if you like.
On the right hand side, where
we have e to the i theta times e
to the i theta, we have cosine
plus i sine times cosine plus
i sine.
And if we work that out
on the right hand side,
that's cosine squared theta plus
2i cosine theta sine theta plus
i squared sine theta.
And at this point, all
we have to remember is
that i squared is
equal to minus 1.
And therefore we end up with
cosine squared theta plus 2i
sine theta cos theta
minus sine squared theta.
So what? We have cosine 2 theta
plus i sine 2 theta on one side
and we have this thing with
three terms on the other side.
Well, these are two
complex numbers.
And we're saying these two
complex numbers are equal,
and therefore the key thing is
that that means the
real parts are equal
and the imaginary
parts are equal.
Because the way we think
of complex numbers is the
real parts along the x
and the imaginary parts along y.
And if the two numbers
are equal,
they have to have the same
value of x and the same value
of y. Well, the real part
on the left hand side
is cosine 2 theta.
The real part on the right hand
side is cos squared theta minus
sine squared theta.
That's the cosine
double angle formula.
The imaginary part on the left
hand side is sine 2 theta.
The imaginary part on the
right hand side is 2 cos theta
sine theta.
And therefore that is the
relationship between the sine
of 2 theta and the sine
and cosine of theta.
And it's very quick and
easy to do stuff like this.
You don't have to
remember anything.
You can work out the
triple angle formula.
You can work out this and that
and the other very
easy and quickly.
And it's much, much, much
better than memorizing formulas
that you don't quite
fully understand.
If you rely on your memory,
you might get it wrong.
And for sure, as you get older,
you'll start making
more and more mistakes.
If you just re-derive it every
time, it's very easy to do,
and much more reliable.
So let me just close by saying
any trigonometric identity
that you've seen can be derived
by using these complex
exponentials
and just relating the
real and imaginary part
without any particular
other talent.
Okay, let's get back to
our particle in a box.
Now, we know that the
wave function vanishes
at the right hand side of
the box, big L. And we put
in those numbers and we
had Ae to the plus ipH --
ipL upon h-bar minus Ae to
the minus i. and therefore,
we just put in cosine
plus i sine
and we get A times cosine pL
upon h-bar plus i sine minus
cosine plus i sine.
And therefore at the
end, we get 2iL --
excuse me 2iA sine
pL upon h-bar.
That's the condition that
that thing equal to 0.
A can't be 0, i is
definitely not 0,
it's the square root of minus 1.
2 is not 0, and therefore the
sine function has to be 0.
The sine function is 0 when the
argument, theta, which I've set
to be equal to pL over h-bar,
is an integer multiple of pi.
In math we work in radians,
we don't work in degrees.
We always put things in
radians, dimensionless.
And the integer values
that are allowed
for theta are n equals
1, 2, 3 and so forth.
And therefore, pL over
h-bar should be n times pi,
where n is some positive
integer.
Again, we can't have n equal 0.
If we have n equals 0,
then the sine function
vanishes everywhere.
And that means the
wave function vanishes.
Therefore, what we've found
here in this little bit
of mathematics is
that the condition
that the particle be
localized, that it be restricted
to some region in space
and not just allowed
to just go wherever it wants,
results right away in the result
that the momentum and the energy
of the particle are quantized.
They can only occur
in set amounts.
We can take our result
a step further now
by doing the following.
Let's solve for this
quantized momentum.
Here's what we find.
The momentum is equal
to n pi h-bar over L.
And since h-bar is h over 2 pi,
then the momentum
is just mh over 2L.
And that's just using
the definition of h-bar.
That means the energy
is quantized,
because the particle inside the
box only has kinetic energy p
squared over 2m.
And if the momentum's quantized,
that means the energy's
quantized.
And the energy of
the nth allowed state
that can be occupied
is n squared h squared
over 8mL squared.
Just by squaring the momentum
and dividing it by 2m.
This is an important formula
for a couple of reasons.
First, it shows you how the
quantization is occurring.
Second, it shows you that
the ladder of states is going
up like the squares, not evenly,
not getting closer
and closer together.
And what we'll find in the
future, in a future lecture,
is that that has to do a lot
with the shape of the potential.
So if we imagine that
the potential is written
as a power series, we
could write v of x is equal
to something times x squared.
And then we could say
it's x to the 4th.
Or we could make it
higher and higher.
And if we made it x to the
infinity, it would be 0
when x is less than 1,
just say inside the box.
And then it would suddenly blow
up to infinity the second
you got higher than that.
And that fact that the power
of the potential is very high is
the reason why these states get
farther and farther apart.
If the power is lower,
they may be evenly spaced
or they may get closer
and closer together.
Putting everything together now,
in what we've got, we've figured
out that everything's quantized.
We know the form of
the wave function.
And therefore we can write psi n
of x is 2i capital A
sine px upon h-bar.
And then we can express
that in terms
of our quantization condition
using the quantized momentum n
pi x over L. And so
now h-bar is out of it
and we just have this
nice wave in the box.
As the quantum number
increases, the number of nodes
in the wave is going
to increase.
And that is the condition
that the energy increased.
And that's of course
exactly what we expect.
But it's nice that
it just comes out
and that it makes perfect sense.
In fact, what we're going
to find in the future is
that the so-called nodes,
the places where the
wave function crosses 0,
are ever so important because
they give us a really strong
clue as to what kind of
behavior we're going to see.
Where you have a node
in the wave function,
that means that the particle
is never found there.
Because psi is 0, so
psi star psi is 0.
And the particle's
never found there.
And in fact, if it crosses
through 0, when you square it ,
it gets very small because it's
small and negative on one side
and it's small and
positive on the other side.
So the probability
smoothly goes to 0.
And that means, although the
box is uniform, there are places
in the box where we do not ever
expect to find the particle
if we try to measure
its position.
And that certainly
seems different
than any classical
situation that we might think
about where we have
a ball inside a box.
And the ball could
be anywhere in there.
Of course, the quantum state
of such a system has huge n,
and we could never see this
kind of very subtle behavior
because the de Broglie
wavelength
of the ball is far too short.
Well, we still need to
determine capital A. B went away
because it was related to A.
And the only way we
can determine A is
by normalizing the
wave function.
Recall that normalization
just means that we insist,
if this wave function is going
to be measuring probability,
that the probability that
the particle be somewhere,
in this case somewhere
inside the box,
is 100 percent, or unity.
We can't have it
be less than that.
We can't have it
be more than that.
So let's go ahead and
normalize the particle
in a box energy eigenfunctions.
We'll do the ground state, but
it's the same for the others.
And this will be
practice problem 9.
This is a simple
calculus problem.
However, before you do
any kind of integration,
what you should check is
whether you can do anything
to the integrand to make
the integration easier.
And we certainly can,
because the easiest function
to integrate or differentiate
is the exponential function.
And if we can cast our
problem in terms of that,
then we're out of the woods.
If we use some other
function where we have to play
around a lot or we don't
know the antiderivative,
then that could be problematic.
Now, here's what
we need to have.
The integral from minus infinity
to infinity of psi
star psi is 1.
But the wave function
is 0 outside the box.
And so in this case what we
can do is simplify the integral
to the integral from 0 to L,
because that's the only --
it's 0 otherwise, psi
star psi is equal to 1.
And in terms of our
un-normalized psi,
what we have is the integral.
Remember, if we have an i we
have to put it to minus i.
So we have minus 2iA sine n pi
x upon L times 2iA n pi x upon L
minus i times i as
plus 1, of course.
That's why we always take the
complex conjugate to make sure
that we get a real
positive probability.
A doesn't depend on x
and therefore we can pull
it out of the integral.
And we get 4A squared times
the integral from 0 to L
of sine squared n pi x upon
L. And we want that to be 1.
And once we find out what
the value of the integral is,
we can work out what A is.
And then we can go ahead
and put A into our formula
for the particle in
a box wave function.
And then we're done with
the whole nine yards.
Now, you could look up
the integral in a book.
That's what I used to do
when I was a student, often,
if I couldn't do the integral.
You can use software like
Mathematica or Math Cat
or something else, Maple.
Or you could look it
up in integrals.com.
But just for an exercise here,
I want to do the problem,
just do it right through.
And I want to do it by
changing the sign back
to e to the i theta.
Because whenever I'm going to
do a derivative or an integral,
I think maybe if I turn it
back to the exponential,
it's going to be much
easier for me to do it.
And the only formula I need
here is that the integral of e
to the ax dx is equal
to 1 over a times e
to the ax plus some constant.
But if we're doing
a definite integral,
the constant is going
to go away.
It subtracts out when we take
the two limits, of course.
And that's all I need to know.
And the only thing
I need to know
about complex numbers is complex
numbers behave the same way
as real numbers.
If we have e to the some
funny complex thing,
it's the funny complex thing
in the denominator and e
to the funny complex
thing again.
And you just close your
eyes and keep going.
Let's have a look.
We can write our normalization
condition like this.
Rather than the 2i sine,
we use e to the plus i,
e to the minus i. And then the
same thing, e to the plus i,
e to the minus i
times the integral.
And that's equal to 1.
And that simplifies
to this, then.
We end up with A squared
times the integral from 0 to L
of 1 minus e to the i,
minus e to the i plus 1.
And that should be equal to 1.
Now, the complex
exponential, since it's equal
to cos theta plus i sine
theta, if you square it
and you have the square
in the formula there,
then e to the 2i
pi is equal to 1.
And therefore that part goes
away when we evaluate it at 0
and L, both conditions,
it goes away
because of the way
the wavelength
of the wave function is set.
And so we end up with
three terms here,
which I've written out.
But it should be
equal to 1A squared
and then 2x is the
anti-derivative
of 2 evaluated at L and at 0.
That part has to work out.
And then the other two parts
that have a 1 over minus 2ia,
because that was in the
complex exponential here
where I've let a just be
the collection of constants,
p and h-bar and so forth.
And then another term.
And the ones with the 2i in the
dominator, fortunately for us,
we don't have to worry about
because when we evaluate it,
it's 1 at L and it's 1 at 0.
And so when we subtract
those limits,
that part just vanishes
from the equation.
And we're only left
with the first term,
which is 2L minus
0 or 2 times 0.
And therefore, a squared times
that should be equal to 1.
So finally, at the
end of the day,
we find out that capital A is
equal to 1 over -- 2 times --
the square root of 2L.
So here's our final
wave function.
Psi n of x is 2iA.
And that brings the 2 from
the bottom to the top.
And so times sine n pi x
over L. And that's equal
to i times the square root of
2 upon L sine n pi x over L.
That would be our final result
except we still have a bias.
We don't like the i.
It still bugs people.
We've got an imaginary
wave function.
It doesn't change the
probability distribution.
It does not change the energy.
It does not change anything
to multiply a wave function
by a number that's complex
that has unit length.
What you should think about
if you're multiplying a wave
function by a complex number
with unit length is something
like you're changing
the color of it.
But you aren't changing
the shape of it.
And since when you take the
complex conjugate you get the
opposite color, when
you crash them together,
you always get the same thing.
And for that reason, at this
point, we can get rid of the i.
And that's conventional.
And that's called picking the
phase of the wave function.
We're free to choose the
phase to be whatever we like.
We don't have to have
an i. We can have a 1.
We can have a minus
i. It doesn't matter.
But just to make it a little
bit easier to understand
if you're first encountering
it, the conventional choice is
to choose the phase
to just be 1.
And when you pick that phase you
get the final solution here psi
n of x is equal to
the square root of 2
over L times sine n pi x over
L. The lowest energy state
for the particle is n equals 1.
Because remember, we
can't have n equal 0
or the wave function
vanishes everywhere.
And there's nothing anywhere.
There's no particle.
And there's a node at each side
of the box, but nowhere else.
So here's a plot of what the
ground state wave function
looks like.
It's just half a sine wave.
It comes up, reaches an apex
in the center of the box,
which is apparently the
most likely place to be.
And then goes down toward
the edge of the box.
And what we see is that even
though there's no potential
at all in the box, the fact
that there is a potential
setting limits really profoundly
changes what the
wave function does.
It makes it far less
likely that it's going
to be near the edge of the box.
It's as if the wave
function has a sixth sense
and knows hey I don't want
to get too close to the box
because the energy outside
the box is infinite.
Sort of like putting your
hand on a wall and detecting
that there's fire in
the next apartment over.
And then you decide to
stay away from the wall.
And that's pretty much what
the wave function's doing here.
And remember, what we're
plotting here is the
probability amplitude.
We have to take this thing that
we've got and square it in order
to figure out the
probability density.
And when we do that, we find
that the edges are
really very low.
Because they were going up like
x and now they're x squared.
That's even much smaller.
It comes up and then it sort of
peaked in the middle and comes
down again toward
the other edge.
So let me now take
this wave function.
We have an explicit
value for the --
we have an explicit form
for the wave function.
Let's figure out, then,
the expectation value
for the position and the
momentum of the particle
in this ground energy
eigenstate.
We know it's not in
a position eigenstate
because a position eigenstate
means there's one position.
And this has a range
of positions.
It is in a momentum eigenstate,
but it's a super position
of two possible values of
P. So, it's not actually
in a momentum eigenstate.
It's in a superposition
of two equal
and opposite momentum
eigenstates.
Let's, then, go ahead and
take this and just work
through as an exercise
in expectation values.
We're going to have to do
some challenging integrals
because we always have
to do the integral
to do the expectation value.
And in addition, we're going
to have to do some derivatives.
So it seems like we're
going forward and backward
because remember, the momentum
operator is the derivative
times ih-bar.
And therefore we're
going to have to do both.
So recall, here's our formula
for the expectation
value of any operator.
The expectation value
of some operator, omega,
is equal to the integral of psi
omega-hat operating on psi dx.
For our finite box, and
for the position operator,
we then have the
following thing.
The expectation value of
position, the average value
after a large number
of measurements,
is equal to the integral
from 0 to L of square root
of 2 upon L sine pi x upon
L times x, because remember,
the x-hat operator
just multiplies
by x. The number, the variable.
So the hat goes away.
We just have an x. Then
we have the other one.
Since they're both real, psi
and psi star are the same.
And we can do that
integral easily.
And the expectation
value comes out to L
over 2, which is perfect.
The most -- the average of
all the possible measurements
of the box averages
to the middle.
Well, it would have to.
Because the box has symmetry.
How it could it be
anywhere else?
But at least it's interesting
that the math just tells
us what we already knew.
How about the momentum operator?
This is a little bit harder.
So let's work through this.
The expectation value of
p is equal to the integral
of psi star p-hat psi dx.
And now we have the
derivative function
in here and an integral.
And if we clock through it,
we have to take the derivative
of sine pi x upon L. And
that give us a cosine.
And then that also brings out
1 over L. So, we end up with,
in the end, minus 2i h-bar,
times the integral from 0 to L
of sine times some extra
stuff times cosine.
And this isn't so easy to
figure out, but it turns
out that we could rearrange it.
Keep in mind, however, that
the integral has a sine
and a cosine in it.
It doesn't have anything else.
It doesn't have any
imaginary parts.
And we know, because the
momentum operator is a hermitian
operator, that the expectation
value has to be real.
And we have a big
i out in front.
And the only way i times
something can be real is
if the something is 0.
None of the other
things, L and so forth,
have any imaginary part.
And therefore we could just
use that argument to say
that the expectation value
of the momentum is 0.
And that would make
perfect sense
because the particle is not,
on average, going anywhere.
It's trapped in a box.
So how could it have anything
except the expectation value
of the momentum be 0?
The -- now as a further
illustration, let's go forward
and calculate the values
of x squared and p squared.
You'll see in a minute why
we're bothering to do that.
The expectation value of x
squared is the same thing.
We have the integral, again.
And now we have the
integral from 0 to L
of x squared sine
squared pi x upon L dx.
And we have a 2 over
L out in front.
And again, you can
either do this by parts.
You can do this by software.
You can look it up.
And what you find is that the
expectation value of x squared,
the square of the position,
comes out to be L squared
over 3 minus L squared
over 2 times pi squared.
That's what the integral
works out.
You can probably see that
the first part, the L squared
over 3 comes from
integrating x squared times1.
And the other part comes
from integrating the trig
functions a couple of times.
So that tells you pretty
much how that's coming about.
That's the expectation value
of the square of the position.
That's not in the middle,
and it's certainly not 0.
For the expectation
value of p squared,
we have to take two derivatives
and then do the integral.
We have minus ih-bar,
the derivative,
minus ih-bar the derivative.
So we take the derivative once,
the sine goes to the cosine.
We take the derivative again.
The cosine goes to minus the
sine, conveniently cancelling
out the minus i times
minus i, which is minus,
because i times minus
i is plus 1.
So minus i times
minus i is minus 1.
That all goes away.
And of course, we expect
the expectation value
of something squared
to be positive.
If it came out to be
negative or something else,
we would have an indication
that we've made a
mistake in the algebra.
And therefore the expectation
value p squared turns
out to be h squared n
squared over L squared.
And we can combine
these two then,
in some interesting
little exercise.
Why would this be interesting?
And the answer is, we had
the uncertainty principle.
And the uncertainty principle
applies to any wave function.
If we measure the position
and momentum, the uncertainty
in them has to satisfy bigger
than or equal to h-bar over 2.
And now we can figure
out what that is.
And in doing so, we
can really expand
on what this delta
x delta p means.
Let's think of the variance,
which is the deviation
from the mean squared
of the measurement.
The expectation value
of the variance is
the expectation value
of x minus the expectation
value of x all squared.
So the expectation
value of that.
Well, if I write out x minus
the expectation value of x,
keep in mind, x is
a variable here.
And expectation value
of x is just a number.
And when we work it
out, then, if we work it
out we get the expectation value
of x squared minus 2x
times the expectation value
of x plus the expectation
value of x squared.
The expectation value of 2x
times the expectation value
of x is just 2 times the
expectation value of x,
because the expectation value
of the expectation value
is just what you expect.
It's just the expectation
value of x. And therefore,
the second term subtracts
from the last term.
And we get that the variance is
equal to the expectation value
of x squared minus the
expectation value of x squared.
And the same thing
applies for the momentum.
The variance in the momentum
is the expectation value
of the momentum squared,
minus the expectation value
of the momentum squared.
If the distribution
is very narrow,
if in fact they all
have the same momentum,
then the expectation value
of the square is equal
to the square of the
expectation value.
And the variance is 0.
And same thing for position.
If there is no spread
in position,
then the variance is 0.
That just means that
it's a very, very, very,
very narrow distribution.
You could have an exam where
every student gets the mean.
It would have no
variance at all.
And that would be
very hard, then,
to make a curve for that class.
The variance in position, well,
we knew the expectation
value was L over 2.
So we're going to subtract
that and square it.
And we worked out the
expectation value of x squared
by doing that somewhat
nasty integral.
That was L squared over 3 minus
L squared over 2 pi squared.
And that works out with
a little bit of algebra
to be L squared times 1/12
minus 1 over 2 pi squared.
This doesn't look
too encouraging
because usually things are a
little bit neater than that.
But, that's another good
reason to do this problem,
because things aren't always
so neat as they might be
in some setup problem.
The variance in the momentum.
Well, the expectation value
of the momentum was 0.
So that's easy.
So the variance in the momentum
is just the expectation value
of p squared.
And we got that worked
out, A squared pi squared
over L squared, as you can see.
And therefore, if we associate
the uncertainty in position
and momentum with the
square root of the variance.
So the variance has the unit
of square length
and square momentum.
If we take the square root
of that, we get a measure
of the width of the
distribution.
And for the uncertainty in
position, that then becomes L,
has to do with the box.
The bigger the box is,
the bigger delta x is.
Well, that makes sense
because the particle can spread
out if the box is big.
It goes linearly in L, times the
square root of our 1/12 minus 1
over 2 pi squared,
whatever that is.
And for the momentum,
we take the square root
and we get the formula shown.
Now, we had a relationship,
the uncertainty principle
that said delta x,
delta p is greater than
or equal to h-bar over 2.
Let's have a look.
If we take delta h, delta p,
we get the following formula.
And the key here is that when we
simplify it and pull out the pi
to get rid of the pi, that
we get the following thing.
We have pi squared
minus 6 in there,
times 1/3 in the square root.
So we got our h-bar over 2,
and then we've got square root
of 1/3 times pi squared minus 6.
And pi squared is -- I don't
know what it is, but it's bigger
than 3 squared because I do
know that pi is bigger than 3.
So that thing, there, whatever
it is, is bigger than h-bar
over 2 times 1/3 times
3 squared minus 6 rather
than pi squared minus 6.
And that is just h-bar over 2.
And so what we've shown is
that for a particle in a box,
the uncertainty is bigger
than the minimum uncertainty.
Now, Heisenberg did never
say that the uncertainty
and position times
momentum can be --
should be equal to the minimum.
In fact, that's rare.
Usually you can't even
do as well as that.
But, what he did say is
that you can never be less
than the minimum.
And thank goodness, when we
work it out, it doesn't depend
on the length of the box,
the uncertainty principle,
because as the box gets longer,
then the momentum
becomes narrower
as the position gets wider.
And therefore what
we've shown here in kind
of this simple calculation,
which did take a
little bit of time.
But nevertheless, it's not --
it's fairly straightforward,
is that for the ground
state particle in a box,
that we satisfy the
uncertainty principle.
In fact, we're about at least
10 percent too big compared
to the absolute minimum
that we could have.
And if we did the same
calculation for the n equals 2,
n equals 3, whatever state, it
would be the same conclusion,
exactly, is that we would
satisfy the uncertainty
principle for all those states.
And if you want a
challenge, go ahead and try
for something other
than the ground state,
to calculate delta p, delta
x the same way we did,
but just substitute
in your new formula.
And convince yourself that
it always satisfies the
uncertainty principle.
The energy of the particle
increases like n squared.
And the wave function has
more nodes for higher energy.
And what you often see in books
is sort of a composite plot.
And these composite
plots can be confusing
when you first encounter
them, because in actual fact,
what we're doing is
plotting two things at once.
We usually plot a flat
line like a ladder
to indicate this energy; the
particle can be at this energy.
And then we would have these
things going up like n squared.
And then, to give you an idea
of what the wave function
of the particle looks like,
we plot the wave function
of the particle on the
line as if the line were 0,
even though we're moving
the line up in energy.
And you have to get used to
these plots because if you think
of them in terms of energy
or I'm plotting something wavy
here that's going up and down
in energy, you're
missing the whole point.
The energy is given by the line.
And then separately, we're
just plotting the wave function
to see what it looks like.
It doesn't have units of energy.
And we just plot it on the
same graph because we're lazy
and somebody did that early on.
And everybody liked it
once they got used to it.
But when you first
encounter these kind of plots,
they can be very confusing.
You just don't -- finally you
don't what's being plotted.
Here's a typical plot, then,
that you might see in a book.
I've just taken the
energy scaled as n squared.
The first energy level
is 1 squared is 1,
2 squared is 4, 3 squared is 9.
And then, on those energy
levels, treating them basically
as the 0, I've plotted
the first wave function,
which I've called psi 1.
That's the ground state.
And then the next one, which
has one node in the middle.
So that one avoids
being in the middle.
The bottom one likes
being in the middle.
The next one up avoids
being in the middle.
And then the third one,
psi 3, which has two nodes
in the interior of the box,
and that one likes being
in the middle again,
and likes being closer
to the edges of the box.
And if you take very, very,
very high levels of n,
then you'll find that
you just get these little
ripples everywhere.
And basically the probability
of being anywhere is
basically the same.
There's little things like an
egg carton, but those things are
so tiny, tinier than the
width of a nucleon in an atom,
that there's no way we can do
any experiment to try to uncover
that kind of corrugation
in the wave function.
Okay. I want to close there
and ask the following question,
which we'll get to
in the next lecture.
And that has to do with a
phenomenon called tunneling.
We made the potential go to
infinity at the edge of the box.
And the reason we did that is
you're going to see what happens
if we don't do that, which
is that we get set a lot
of math problems to do.
And not all of them are easy.
But what would happen if
instead of being infinite,
the edges just rose up to
some finite value instead?
What would happen
if we had a plot
where we have the
potential be higher
than the energy of the particle?
So classically, the particle
is trapped and has to remain
within the wall forever.
But it doesn't' go to infinity.
And therefore we
can't use the argument
that psi should necessarily
vanish completely,
because if a small part
of psi leaks through,
and that adds a high energy from
the potential being so high,
that doesn't necessarily
mean that all bets are off
because by spreading out more,
psi has had not so
much curvature.
So, it could be that it kind
of relaxes out and pushes out,
slips out a little bit and
that's a better situation
than having it cut off just
because of the edge of the box.
And in fact, what we're going to
find next time is that because
of this wave nature, is that
the wave function always sneaks
in to some forbidden region.
It's sort of like
somebody at 4 a.m. rolling
through the stop sign when
nobody else is around.
And so it's very
interesting, in fact,
that there's a big difference
between classical mechanics
and quantum mechanics because in
quantum mechanics there's always
some chance of escaping
out of jail no matter
how tight the bars are.
And eventually, in things like
radioactive decay, for example,
that's exactly what happens.
So we'll pick up quantum
tunneling in the next lecture.
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