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PROFESSOR: So
today, before we get
into the meat of
today's lecture,
Matt has very kindly--
Professor Evans
has very kindly agreed
to do an experiment.
Yeah, so for those
of you all who
are in recitations
both he and Barton
talked about polarization
in recitation last week.
And Matt will pick
it up from there.
MATTHEW EVANS: So back
to the ancient past--
this was a week ago.
We had our
hyper-intelligent monkeys
that were sorting things.
It all seemed very theoretical.
And in recitation, I said
things about polarizers.
And I said, look, if
we use polarizers,
we can do exactly the same
thing as these monkeys.
We just need to set up a
little polarization experiment
and the results are identical.
You can use the one
figure out the other.
But I didn't have this or
a nice polarizer ready then
to give a demo, so here we go.
What I'm going to show
you is that, if we start
with something polarized here
with all white-- and right
now I have all vertical
polarization here--
and if I just put
on this other box
there, which is going to be
another polarizer, if I put it
the same way, this is all
of our white electrons
coming through all white.
See?
It doesn't really do much.
And if I look at the
black output over here
of the second Keller
sorting box, that's
the same as turning my
polarizer 90 degrees,
so nothing comes out black.
So if we remove this
guy from the middle,
you have just exactly
what you'd expect.
You sort here, you have white,
and you get all white out.
Great.
Everyone thought that was easy.
We all had that figured out.
This box got thrown
in the center here
and it became sort of confusing,
because you thought, well,
they were white--
I'm going to throw
my box in the middle here--
that's this guy at 45 degrees.
And then if I throw this guy
on the end again, the idea was,
well, they were all white
here, so maybe this guy
identified the soft ones
from the white ones.
And now we have white and soft.
And it should still
all be white, right?
So I put this guy on up here.
They should all come out
but they sort of don't.
And if you say,
well, are they black?
Well, no, they're not
really black, either.
They're some sort of strange
combination of the two.
All right, so that's this
experiment done in polarizers.
But let me just play the
polarizer trick a little bit,
because it's fun.
So this is if I say,
vertical polarization
and how many of them
come out horizontal?
So here I'm saying, white,
and how many of them
will come out black?
That's the analogy.
The answer is none of them.
And strangely, if I
take this thing, which
seems to just attenuate--
this is our middle box here--
and I just stuff
it in between them,
I can get something to come
out even though I still
have crossed
polarizers on the side.
So you can see the middle
region is now brighter
and you can still see
the dark corners there
of the crossed polarizers.
And as I turn this guy around,
I can make that better or worse.
The maximum is
somewhere right there,
and then it goes off again.
So this is a way
of understanding
our electron-sorting,
hyper-intelligent monkeys
in terms of polarizations.
And here it's just
a vector projected
on another vector projected
on another vector-- something
everybody knows how to do.
So here's the
polarization analogy
of the Stern-Gerlach experiment.
PROFESSOR: Awesome.
So the polarization analogy
for interference effects
in quantum mechanics
is a canonical one
in the texts of
quantum mechanics.
So you'll find lots of
books talking about this.
It's a very useful
analogy, and I
encourage you to
read more about it.
We won't talk about
it a whole lot more,
but it's a useful one.
All right, before
I get going, any
questions from last lecture?
Last lecture was pretty
much self-contained.
It was experimental results.
No, nothing?
All right.
The one thing that I
want to add to the last
lecture-- one last
experimental observation.
I glossed over
something that's kind
of important, which
is the following.
So we started off
by saying, look,
we know that if I
have a ray of light,
it's an electromagnetic
wave, and it
has some wavelength lambda.
And yet the photoelectric
effect tells us
that, in addition to having the
wavelength lambda, the energy--
it has a frequency, as
well, a frequency in time.
And the photoelectric
effect suggested
that the energy is
proportional to the frequency.
And we write this as h nu and
h bar is equal to h upon 2 pi
and omega is equal to 2 pi nu.
So this is just the angular
frequency, rather than
the number-per-time frequency.
And h bar is the
reduced Planck constant.
So I'll typically write h
bar omega rather than h nu,
because these two pi's will
just cause us endless pain if we
don't use the bar.
Anyway, so to an
electromagnetic wave,
we have a wavelength
and a frequency
and the photoelectric
effect led us
to predict that the
energy is linearly
proportional to the frequency,
with the linear proportionality
coefficient h bar-- Planck
constant-- and the momentum
is equal to h upon
lambda, also known
as-- I'm going to write
this as h bar k, which
is equal to h upon
lambda, where here, again,
h bar is h upon 2 pi.
And so k is equal
to 2 pi upon lambda.
So k is called the
wave number and you
should have seen this in 8.03.
So these are our basic
relations for light.
We know that light, as
an electromagnetic wave,
has a frequency
and a wavelength--
or a wave number, an
inverse wavelength.
And the claim of the
photoelectric effect
is that the energy and
the momenta of that light
are thus quantized, that
light comes in chunks.
So it has a wave-like
aspect and it also
has properties that are more
familiar from particles.
Now, early on shortly after
Einstein proposed this,
a young French physicist named
de Broglie said, well, look,
OK, this is true of light.
Light has both wave-like and
particle-like properties.
Why is it just light?
The world would be
much more parsimonious
if this relation were true
not just of light, but also
of all particles.
I am thus conjecturing,
with no evidence whatsoever,
that, in fact,
this relation holds
not just for light,
but for any object.
Any object with momentum p has
associated to it a wavelength
or a wave number,
which is p upon h bar.
Every object that has energy
E has associated with it
a wave with frequency omega.
To those electrons that we send
through the Davisson-Germer
experiment apparatus, which are
sent in with definite energy,
there must be a frequency
associated with it, omega
and a wavelength lambda
associated with it.
And what we saw from the
Davisson-Germer experiment
was experimental confirmation
of that prediction--
that electrons have both
particulate and wave-like
features simultaneously.
So these relations are called
the de Broglie relations or "de
BROG-lee"-- I leave it up to
you to decide how to pronounce
that.
And those relations are going
to play an important role for us
in the next few lectures.
I just wanted to give them
a name and a little context.
This is a good example of
parsimony and elegance--
the theoretical
elegance leading you
to an idea that turns out
to be true of the world.
Now, that's a dangerous
strategy for finding truth.
Boy, wouldn't it be nice if--?
Wouldn't it be nice if we
didn't have to pay taxes
but we also had Medicare?
So it's not a terribly
useful guide all the time,
but sometimes it
really does lead you
in the right direction.
And this is a great example
of physical intuition,
wildly divorced from
experiment, pushing you
in the right direction.
I'm making it sound a little
more shocking than-- well,
it was shocking.
It was just shocking.
OK, so with that said,
let me introduce the moves
for the next few lectures.
For the next several lectures,
here's what we're going to do.
I am not going to give you
experimental motivation.
I've given you
experimental motivation.
I'm going to give you a set
of rules, a set of postulates.
These are going to be the
rules of quantum mechanics.
And what quantum
mechanics is for us
is a set of rules to allow
us to make predictions
about the world.
And these rules will be awesome
if their predictions are good.
And if their predictions are
bad, these rules will suck.
We will avoid bad rules
to the degree possible.
I'm going to give you what
we've learned over the past 100
years-- wow-- of developing
quantum mechanics.
That is amazing.
Wow.
OK, yeah, over
the past 100 years
of developing quantum mechanics.
And I'm going to give them to
you as a series of postulates
and then we're going to work
through the consequences,
and then we're going to spend
the rest of the semester
studying examples to develop
an understanding for what
the rules of quantum
mechanics are giving you.
So we're just going to
scrap classical mechanics
and start over from scratch.
So let me do that.
And to begin, let me start with
the definition of a system.
And to understand
that definition,
I want to start with classical
mechanics as a guide.
So in classical mechanics--
let's think about the easiest
classical system you can--
just a single particle
sitting somewhere.
In classical mechanics
of a single particle,
how do you specify the
configuration, or the state--
just different words
for the same thing--
how do you specify
the configuration
or state of the system?
AUDIENCE: By position
and momentum.
PROFESSOR: Specify the
position and momentum, exactly.
So in classical mechanics, if
you want to completely specify
the configuration of a
system, all you have to do
is give me x and
p for my particle.
And if you tell me
this, I know everything.
If you know these numbers,
you know everything.
In particular, what
I mean by saying
you know everything is that,
if there's anything else
you want to measure--
the energy, for example.
The energy is just some function
of the position and momentum.
And if you know the
position and momentum,
you can unambiguously
calculate the energy.
Similarly, the angular
momentum, which is a vector,
you can calculate
it if you know x
and p-- which is just r cross p.
So this gives you complete
knowledge of the system.
There's nothing more to
know if you know that data.
Now, there are certainly
still questions
that you can't answer
given knowledge of x and p.
For example, are there
14 invisible monkeys
standing behind me?
I'm here.
I'm not moving.
Are there 14 invisible
monkeys standing behind me?
You can't answer that.
It's a stupid question, right?
OK, let me give you
another example.
The electron is x
and p, some position.
Is it happy?
Right, so there are still
questions you can't answer.
The point is, complete
knowledge of the system
to answer any physically
observable question--
any question that could
be meaningfully turned
into an experiment,
the answer is
contained in knowing
the state of the system.
But this can't possibly be
true in quantum mechanics,
because, as you saw
in the problem set
and as we've
discussed previously,
there's an uncertainty
relation which
says that your knowledge--
or your uncertainty, rather,
in the position of
a particle, quantum
mechanically--
I'm not even going
to say quantum mechanically.
I'm just going to
say the real world.
So in the real world,
our uncertainty
in the position of
our point-like object
and our uncertainty
in the momentum
is always greater than or
roughly equal to something
that's proportional
to Planck's constant.
You can't be arbitrarily
confident of the position
and of the momentum
simultaneously.
You worked through a
good example of this
on the problem set.
We saw this in the
two-slit experiment
and the interference
of electrons.
This is something we're
going to have to deal with.
So as a consequence,
you can't possibly
specify the position
and the momentum
with confidence of a system.
You can't do it.
This was a myth.
It was a good approximation--
turned out to be false.
So the first thing
we need is to specify
the state, the
configuration of a system.
So what specifies the
configuration of a system?
And so this brings us
to the first postulate.
The configuration, or
state, of a system--
and here again,
just for simplicity,
I'm going to talk about a single
object-- of a quantum object
is completely specified by
a single function, a wave
function, which I
will denote generally
psi of x, which is
a complex function.
The state of the quantum
object is completely
specified once you know the wave
function of the system, which
is a function of position.
Let me emphasize that this is
a first pass at the postulates.
What we're going to do is go
through the basic postulates
of quantum mechanics, then we'll
go through them again and give
them a little more generality.
And then we'll go through
them again and give them
full generality.
That last pass is 8.05.
So let me give
you some examples.
Let me just draw some
characteristic wave functions.
And these are going to turn
out to be useful for us.
So for example, consider
the following function.
So here is 0 and we're
plotting as a function of x.
And then plotting the
real part of psi of x.
So first consider a very
narrowly supported function.
It's basically 0
everywhere, except it
has some particular spot
at what I'll call x1.
Here's another
wave function-- 0.
It's basically 0 except for
some special spot at x2.
And again, I'm plotting
the real part of psi.
And I'm plotting
the real part of psi
because A, psi is a complex
function-- at every point it
specifies a complex number.
And B, I can't draw
complex numbers.
So to keep my head
from exploding,
I'm just plotting the real
part of the wave function.
But you should never forget that
the wave function is complex.
So for the moment,
I'm going to assume
that the imaginary part is 0.
I'm just going to
draw the real parts.
So let me draw a
couple more examples.
What else could be a
good wave function?
Well, those are fine.
What about-- again, we want
a function of x and I'm
going to draw the real part.
And another one.
So this is going to be a
perfectly good wave function.
And let me draw two more.
So what else could be a
reasonable wave function?
Well-- this is harder
than you'd think.
Oh, God.
OK, so that could be the
wave function, I don't know.
That is actually my signature.
My wife calls it a
little [INAUDIBLE].
OK, so here's the deal.
Psi is a complex function.
Psi also needs to not
be a stupid function.
OK so you have to ask me--
look, could it be any function?
Any arbitrary function?
So this is going
to be a job for us.
We're going to
define what it means
to be not-stupid function.
Well, this is a completely
reasonable function--
it's fine.
This is a reasonable function.
Another reasonable function.
Reasonable.
That's a little weird,
but it's not horrible.
That's stupid.
So we're going to
have to come up
with a good definition
of what not stupid means.
So fine, these
are all functions.
One of them is multivalued and
that looks a little worrying,
but they're all functions.
So here's the problem.
What does it mean?
So postulate 2-- The
meaning of the wave function
is that the probability
that upon measurement
the object is found
at the position x
is equal to the norm
squared of psi of x.
If you know the system is
ascribed to the wave function
psi, and you want
to look at point x,
you want to know with
what probability will I
find the particle there,
the answer is psi squared.
Notice that this is a complex
number, but absolute value
squared, or norm squared,
of a complex number
is always a real,
non-negative number.
And that's important because
we want our probabilities
to be real,
non-negative numbers.
Could be 0, right?
Could be 0 chance of something.
Can't be negative 7 chance.
Incidentally, there also
can't be probability 2.
So that means that the
total probability had better
be normalized.
So let me just say this
in words, though, first.
So P, which is the
norm squared of psi,
determines the probability--
and, in particular,
the probability density--
that the object in state psi,
in the state given by the
wave function psi of x,
will be found at x.
So there's the second postulate.
So in particular, when
I say it's a probability
density, what I mean is the
probability that it is found
between the position
x and x plus dx
is equal to P of x dx, which is
equal to psi of x squared dx.
Does that make sense?
So the probability
that it's found
in this infinitesimal
interval is
equal to this density
times dx or psi squared dx.
Now again, it's crucial
that the wave function
is in fact properly normalized.
Because if I say, look,
something could either be here
or it could be
here, what's the sum
of the probability
that it's here
plus the probability
that it's here?
It had better be 1, or there's
some other possibility.
So probabilities
have to sum to 1.
Total probability that you find
something somewhere must be 1.
So what that tells you is
that total probability, which
is equal to the integral over
all possible values of x-- so
if I sum over all possible
values of P of x--
all values-- should
be equal to 1.
And we can write
this as integral dx
over all values of x.
And I write "all" here rather
than putting minus infinity
to infinity because some
systems will be defined
from 1 to minus 1, some systems
will be defined from minus
infinity to infinity-- all
just means integrate over
all possible values-- hold
on one sec-- of psi squared.
AUDIENCE: Are you going
to use different notation
for probability density
than probability?
PROFESSOR: I'm not going to.
Probability density is going
to have just one argument,
and total probability
is going to have
an interval as an argument.
So they're distinct and this
is just the notation I like.
Other questions?
Just as a side note,
what are the dimensions
of the wave function?
So everyone think about
this one for second.
What are the dimensions?
AUDIENCE: Is it 1 over
square root length
PROFESSOR: Awesome.
Yes.
It's 1 over root length.
The dimensions of psi
are 1 over root length.
And the way to see that is
that this should be equal to 1.
It's a total probability.
This is an infinitesimal
length, so this
has dimensions of length.
This has no dimension, so
this must have dimensions of 1
over length.
And so psi itself of x most have
dimensions of 1 over length.
Now, something I
want to emphasize,
I'm going to emphasize,
over and over in this class
is dimensional analysis.
You need to become comfortable
with dimensional analysis.
It's absolutely essential.
It's essential for two reasons.
First off, it's
essential because I'm
going to be merciless
in taking off points
if you do write down a
dimensionally false thing.
If you write down something on
a problem set or an exam that's
like, a length is equal to
a velocity-- ooh, not good.
But the second thing
is, forget the fact
that I'm going to
take off points.
Dimensional analysis is an
incredibly powerful tool
for you.
You can check something
that you've just calculated
and, better yet,
sometimes you can just
avoid a calculation entirely
by doing a dimensional analysis
and seeing that there's only one
possible way to build something
of dimensions length
in your system.
So we'll do that
over and over again.
But this is a question
I want you guys to start
asking yourselves at
every step along the way
of a calculation-- what
are the dimensions of all
the objects in my system?
Something smells like smoke.
So with that said, if that's the
meaning of the wave function,
what physically can we take
away from knowing these wave
functions?
Well, if this is
the wave function,
let's draw the
probability distribution.
What's the probability
distribution?
P of x.
And the probability distribution
here is really very simple.
It's again 0 squared is
still 0 so it's still
just a big spike at x1 and
this one is a big spike at x2.
Everyone cool with that?
So what do you know
when I tell you
that this is the wave function
describing your system?
You know that with
great confidence,
you will find the particle to
be sitting at x1 if you look.
So what this is telling you
is you expect x is roughly x1
and our uncertainty
in x is small.
Everyone cool with that?
Similarly, here you see that
the position is likely to be x2,
and your uncertainty
in your measurement--
your confidence
in your prediction
is another way to say
it-- is quite good,
so your uncertainty is small.
Now what about these guys?
Well, now it's norm squared.
I need to tell you what
the wave function is.
Here, the wave
function that I want--
so here is 0-- is
e to the i k1 x.
And here the wave function
is equal to e to the i k2 x.
And remember, I'm
drawing the real part
because of practical
limitations.
So the real part is just
a sinusoid-- or, in fact,
the cosine-- and similarly,
here, the real part
is a cosine.
And I really should put 0
in the appropriate place,
but-- that worked out well.
So now the question is, what's
the probability distribution,
P of x, associated to
these wave functions?
So what's the norm squared
of minus e to the i k1 x?
If I have a complex number
of phase e to the i alpha,
and I take its norm
squared, what do I get?
1.
Right?
But remember complex numbers.
If we have a complex
number alpha-- or sorry,
if we have a
complex number beta,
then beta squared
is by definition
beta complex
conjugate times beta.
So e to the i alpha, if
the complex conjugate
is e to the minus i alpha,
e to the i alpha times
e to the minus i alpha,
they cancel out-- that's 1.
So if this is the
wave function, what's
the probability distribution?
Well, it's 1.
It's independent of x.
So from this we've learned
two important things.
The first is, this is
not properly normalized.
That's not so key.
But the most important thing is,
if this is our wave function,
and we subsequently measure the
position of the particle-- we
look at it, we say ah,
there's the particle--
where are we likely to find it?
Yeah, it could be anywhere.
So what's the value of x
you expect-- typical x?
I have no idea, no
information whatsoever.
None.
But and correspondingly,
what is our uncertainty
in the position of x
that we'll measure?
It's very large, exactly.
Now, in order to tell you
it's actually infinite,
I need to stretch this off and
tell you that it's actually
constant off to infinity,
and my arms aren't that long,
so I'll just say large.
Similarly here, if
our wave function
is e to the i k2 x--
here k2 is larger,
the wavelength is
shorter-- what's
the probability distribution?
It's, again, constant.
So-- this is 0, 0.
So again, x-- we have no idea,
and our uncertainty in the x
is large.
And in fact it's very large.
Questions?
What about these guys?
OK, this is the real challenge.
OK, so if this is
our wave function,
and let's just say that
it's real-- hard as it
is to believe that-- then what's
our probably distribution?
Well, something
like-- I don't know,
something-- you get the point.
OK, so if this is our
probability distribution,
where are we likely
to find the particle?
Well, now it's a little
more difficult, right?
Because we're unlikely
to find it here,
while it's reasonably likely
to find here, unlikely here,
reasonably likely, unlikely,
like-- you know, it's a mess.
So where is this?
I'm not really sure.
What's our uncertainty?
Well, our uncertainty
is not infinite
because-- OK, my name
ends at some point.
So this is going to go to 0.
So whatever else we know,
we know it's in this region.
So it's not infinite,
it's not small, we'll say.
But it's not arbitrarily
small-- it's not tiny.
Or sorry, it's not
gigantic is what I meant.
Our uncertainty is not gigantic.
But it's still
pretty nontrivial,
because I can say
with some confidence
that it's more likely
to be here than here,
but I really don't know
which of those peaks
it's going to be found.
OK, now what about this guy?
What's the probability
distribution well now
you see why this is a
stupid wave function,
because it's multiply valued.
It has multiple different
values at every value of x.
So what's the probability?
Well, it might be root 2,
maybe it's 1 over root 3.
I'm really not sure.
So this tells us an important
lesson-- this is stupid.
And what I mean by stupid
is, it is multiply valued.
So the wave function--
we just learned
a lesson-- should
be single valued.
And we will explore some more
on your problem set, which
will be posted
immediately after lecture.
There are problems that
walk you through a variety
of other potential pathologies
of the wave function
and guide you to
some more intuition.
For example, the
wave function really
needs to be continuous as well.
You'll see why.
All right.
Questions at this point?
No?
OK.
So these look like pretty
useful wave functions,
because they corresponded
to the particle
being at some definite spot.
And I, for example, am at
a reasonably definite spot.
These two wave
functions, though,
look pretty much useless,
because they give us
no information whatsoever
about what the position is.
Everyone agree with that?
Except-- remember the
de Broglie relations.
The de Broglie relations say
that associated to a particle
is also some wave.
And the momentum
of that particle
is determined by the wavelength.
It's inversely related
to the wavelength.
It's proportional
to the wave number.
Any energy is proportional
to the frequency.
Now, look at those
wave functions.
Those wave functions give
us no position information
whatsoever, but they have
very definite wavelengths.
Those are periodic functions
with definite wavelengths.
In particular, this guy has
a wavelength of from here
to here.
It has a wave number k1.
So that tells us
that if we measure
the momentum of
this particle, we
can be pretty
confident, because it
has a reasonably
well-defined wavelength
corresponding to some
wave number k-- 2 pi
upon the wavelength.
It has some momentum,
and if we measure it,
we should be pretty confident
that the momentum will
be h-bar k1.
Everybody agree with that?
Looks like a sine wave.
And de Broglie
tells us that if you
have a wave of
wavelength lambda,
that corresponds to a
particle having momentum p.
Now, how confident can
we be in that estimation
of the momentum?
Well, if I tell you it's e
to the i k x, that's exactly
a periodic function with
wavelength lambda 2 pi upon k.
So how confident are we?
Pretty confident.
So our uncertainty in
the momentum is tiny.
Everyone agree?
Similarly, for
this wave, again we
have a wavelength-- it's
a periodic function,
but the wavelength
is much shorter.
If the wavelength
is much shorter,
then k is much larger-- the
momentum is much larger.
So the momentum we
expect to measure,
which is roughly h-bar k2,
is going to be much larger.
What about our uncertainty?
Again, it's a perfect
periodic function
so our uncertainty in
the momentum is small.
Everyone cool with that?
And that comes, again, from
the de Broglie relations.
So questions at this point?
You guys are real quiet today.
Questions?
AUDIENCE: So delta
P is 0, basically?
PROFESSOR: It's pretty small.
Now, again, I haven't
drawn this off to infinity,
but if it's exactly
the i k x, then yeah,
it turns out to be 0.
Now, an important thing, so
let me rephrase your question
slightly.
So the question
was, is delta P 0?
Is it really 0?
So here's a problem
for us right now.
We don't have a
definition for delta P.
So what is the
definition of delta P?
I haven't given you one.
So here, when I said
delta P is small,
what I mean is, intuitively,
just by eyeball,
our confidence in that
momentum is pretty good,
using the de Broglie relations.
I have not given
you a definition,
and that will be part of
my job over the next couple
of lectures.
Very good question.
Yeah.
AUDIENCE: How do you code
noise in that function?
PROFESSOR: Awesome.
AUDIENCE: Do you just
have different wavelengths
PROFESSOR: Yeah
AUDIENCE: As you go along?
PROFESSOR: Awesome.
So for example, this-- does
it have a definite wavelength?
Not so much.
So hold that question and wait
until you see the next examples
that I put on this board,
and if that doesn't answer
your question, ask
it again, becayse
it's a very important question.
OK.
AUDIENCE: When you
talk about a photon,
you always say a photon
has a certain frequency.
Doesn't that mean that it
must be a wave because you
have to fix the wave number k?
PROFESSOR: Awesome question.
Does every wave packet of
light that hits your eye,
does it always have a
single, unique frequency?
No, you can take multiple
frequency sources
and superpose them.
An interesting choice
of words I used there.
All right, so the
question is, since light
has some wavelength,
does every chunk of light
have a definite-- this
is the question, roughly.
Yeah, so and the
answer is, light
doesn't always
have a single-- You
can have light
coming at you that
has many different wavelengths
and put it in a prism
and break it up into
its various components.
So you can have a superposition
of different frequencies
of light.
We'll see the same
effect happening for us.
OK, so again, de Broglie
made this conjecture
that E is h-bar omega
and P is h-bar k.
This was verified in the
Davisson-Germer experiment
that we ran.
But here, one of
the things that's
sort of latent in
this is, what he means
is, look, associated to
every particle with energy N
and momentum P is a plane
wave of the form e to the i kx
minus omega t.
And this, properly, in three
dimensions should be k dot x.
But at this point, this is
an important simplification.
For the rest of 8.04,
until otherwise specified,
we are going to be doing
one-dimensional quantum
mechanics.
So I'm going to remove arrow
marks and dot products.
There's going to be one
spatial dimension and one time
dimension.
We're always going to have
just one time dimension,
but sometimes we'll have
more spatial dimensions.
But it's going to be a
while until we get there.
So for now, we're
just going to have kx.
So this is a general plane wave.
And what de Broglie
really was saying
is that, somehow, associated
to the particle with energy E
and momentum P should be
some wave, a plane wave,
with wave number k
and frequency omega.
And that's the wave
function associated to it.
The thing is, not every wave
function is a plane wave.
Some wave functions
are well localized.
Some of them are just
complicated morasses.
Some of them are just a mess.
So now is the most important
postulate in quantum mechanics.
I remember vividly,
vividly, when
I took the analog of this class.
It was called Physics
143A at Harvard.
And the professor
at this point said--
I know him well now,
he's a friend-- he said,
this is what quantum
mechanics is all about.
And I was so psyched.
And then he told me And it
was like, that's ridiculous.
Seriously?
That's what quantum
mechanics is all about?
So I always felt like this
is some weird thing, where
old physicists go crazy.
But it turns out I'm going to
say exactly the same thing.
This is the most important thing
in all of quantum mechanics.
It is all contained in
the following proposition.
Everything-- the two
slit experiments, the box
experiments, all the cool
stuff in quantum mechanics,
all the strange and
counter intuitive stuff
comes directly from
the next postulate.
So here it is.
I love this.
Three-- put a star on it.
Given two possible wave
functions or states--
I'll say configurations--
of a quantum system--
I wish there was "Ride
of the Valkyries"
playing in the background--
corresponding to two
distinct wave functions--
f with an upper ns
is going to be my
notation for functions
because I have to write
it a lot-- psi1 and psi2--
and I'll say, of x--
the system-- is down--
can also be in a superposition.
of psi1 and psi2, where alpha
and beta are complex numbers.
Given any two possible
configurations of the system,
there is also an allowed
configuration of the system
corresponding to
being in an arbitrary
superposition of them.
If an electron can be
hard and it can be soft,
it can also be in an
arbitrary superposition
of being hard and soft.
And what I mean by that
is that hard corresponds
to some particular
wave function,
soft will correspond to some
particular wave function,
and the superposition
corresponds to a different wave
function which is a linear
combination of them.
AUDIENCE: [INAUDIBLE]
combination also
have to be normalized?
PROFESSOR: Yeah, OK, that's
a very good question.
So and alpha and beta
are some complex numbers
subject to the
normalization condition.
So indeed, this wave function
should be properly normalized.
Now, let me step
back for second.
There's an alternate way
to phrase the probability
distribution here,
which goes like this,
and I'm going to put it here.
The alternate statement of
the probability distribution
is that the probability
density at x
is equal to psi of x
norm squared divided
by the integral over
all x dx of psi squared.
So notice that, if we properly
normalize the wave function,
this denominator is equal to 1--
and so it's not there, right,
and then it's equivalent.
But if we haven't
properly normalized it,
then this probability
distribution
is automatically
properly normalized.
Because this is a constant, when
we integrate the top, that's
equal to the bottom,
it integrates to 1.
So I prefer,
personally, in thinking
about this for the first pass
to just require that we always
be careful to choose
some normalization.
That won't always be easy,
and so sometimes it's
useful to forget
about normalizing
and just define the probability
distribution that way.
Is that cool?
OK.
This is the beating soul
of quantum mechanics.
Everything in quantum
mechanics is in here.
Everything in quantum
mechanics is forced on us
from these few principles
and a couple of requirements
of matching to reality.
AUDIENCE: When
you do this-- some
of linear, some of
two wave functions,
can you get interference?
PROFESSOR: Yes.
Excellent.
So the question is, when
you have a sum of two wave
functions, can you get some
sort of interference effect?
And the answer is, absolutely.
And that's exactly
we're going to do next.
So in particular, let me
look at a particular pair
of superpositions.
So let's swap
these boards around
so the parallelism is
a little more obvious.
So let's scrap these
rather silly wave functions
and come up with
something that's
a little more interesting.
So instead of using those as
characteristic wave functions,
I want to build superpositions.
So in particular,
I want to start
by taking an arbitrary--
both of these wave functions
have a simple interpretation.
This corresponds to a
particle being here.
This corresponds to a
particle being here.
I want to take a
superposition of them.
So here's my superposition.
Oops, let's try that again.
And my superposition-- so here
is 0 and here is x1 and here is
x2-- is going to be some
amount times the first one
plus some amount
times the second one.
There's a superposition.
Similarly, I could have taken
a superposition of the two
functions on the
second chalkboard.
And again I'm taking
a superposition
of the complex e to the
i k1 x and e to the i
k2 x and then taking
the real part.
So that's a particular
superposition,
a particular linear combination.
So now let's go back to this.
This was a particle
that was here.
This is a particle
that was there.
When we take the
superposition, what
is the probability distribution?
Where is this particle?
Well, there's some
amplitude that it's here,
and there's some
amplitude that it's here.
And there's rather more
amplitude that it's over here,
but there's still
some probability
that it's over here.
Where am I going to
find the particle?
I'm not so sure anymore.
It's either going to be here
or here, but I'm not positive.
It's more likely
to be here than it
is to be here, but
not a whole lot more.
So where am I going
to find the particle?
Well, now we have
to define this--
where am I going to
find the particle?
Look, if I did this experiment
a whole bunch of times,
it'd be over here more
than it would be over here.
So the average will be
somewhere around here--
it'll be in between the two.
So x is somewhere in between.
That's where we expect
to find it, on average.
What's our uncertainty
in the position?
Well, it's not
that small anymore.
It's now of order x1 minus x2.
Everyone agree with that?
Now, what about this guy?
Well, does this thing
have a single wavelength?
No.
This is like light that
comes at you from the sun.
It has many wavelengths.
In this case, it has just two--
I've added those two together.
So this is a plane wave which
is psi is e to the i k1 x plus
e to the i k2 x.
So in fact, it has two
wavelengths associated with it.
lambda1 lambda2.
And so the probability
distribution now,
if we take the norm squared
of this-- the probability
distribution is the norm
squared of this guy--
is no longer constant, but
there's an interference term.
And let's just see
how that works out.
Let me be very
explicit about this.
Note that the probability
in our superposition of psi1
plus psi2, which I'll call e to
the i k1 x plus e to the i k2
x, is equal to the norm squared
of the wave function, which
is the superposition psi1
plus beta psi2, which
is equal to alpha squared
psi1 squared plus beta squared
psi2 squared plus alpha
star psi1 star-- actually,
let me write this over
here-- beta psi2 plus
alpha psi1 beta star
psi2 star, where star
means complex conjugation.
But notice that this is
equal to-- that first term
is alpha squared times
the first probability,
or the probability of
this thing, of alpha psi1,
is equal to probability 1.
This term, beta
squared psi2 squared,
is the probability that
the second thing happens.
But these terms
can't be understood
in terms of the
probabilities of psi1
or the probability
of psi2 alone.
They're interference terms.
So the superposition
principle, together
with the interpretation of
the probability as the norm
squared of the wave function,
gives us a correction
to the classical addition
of probabilities,
which is these
interference terms.
Everyone happy with that?
Now, here's something very
important to keep in mind.
These things are norms
squared of complex numbers.
That means they're real,
but in particular, they're
non-negative.
So these two are both
real and non-negative.
But what about this?
This is not the norm
squared of anything.
However, this is its
complex conjugate.
When you take something
and its complex conjugate
and you add them
together, you get
something that's
necessarily real.
But it's not
necessarily positive.
So this is a funny thing.
The probability that something
happens if we add together
our two configurations, we
superpose two configurations,
has a positive probability term.
But it's also got
terms that don't
have a definite sign,
that could be negative.
It's always real.
And you can check but
this quantity is always
greater than or equal to 0.
It's never negative,
the total quantity.
So remember Bell's inequality
that we talked about?
Bell's inequality
said, look, if we
have the probability of one
thing happening being P1,
and the probability of
the other thing happening
being P2, the probability
of both things happening
is just P1 plus P2.
And here we see that,
in quantum mechanics,
probabilities
don't add that way.
The wave functions add--
and the probability
is the norm squared
of the wave function.
The wave functions add,
not the probabilities.
And that is what underlies all
of the interference effects
we've seen.
And it's going to be the
heart of the rest of quantum
mechanics.
So you're probably all going,
in your head, more or less
like I was when I took
Intro Quantum, like-- yeah,
but I mean, it's just, you know,
you're adding complex numbers.
But trust me on this one.
This is where it's all starting.
OK so let's go back to this.
Similarly, let's
look at this example.
We've taken the norm squared.
And now we have an
interference effect.
And now, our probability
distribution,
instead of being totally trivial
and containing no information,
our probability distribution
now contains some information
about the position
of the object.
It's likely to be here.
It is unlikely to be
here, likely and unlikely.
We now have some
position information.
We don't have enough
to say where it is.
But x is-- you have
some information.
Now, our uncertainty
still gigantic.
Delta x is still huge.
But OK, we just added
together two plane waves.
Yeah?
AUDIENCE: Why is
the probability not
big, small, small,
big, small, small?
PROFESSOR: Excellent.
This was the real part
of the wave function.
And the wave function
is a complex quantity.
When you take e to
the i k1, and let
me do this on the chalkboard.
When we take e to the i
k1 x plus e to the i k2
x-- Let me write this
slightly differently--
e to the i a plus e to the i
b and take its norm squared.
So this is equal to--
I'm going to write
this in a slightly
more suggestive way--
the norm squared of e to the
i a times 1 plus e to the i b
minus a parentheses
norm squared.
So first off, the norm
squared of a product of things
is the product of
the norm squareds.
So I can do that.
And this overall phase, the norm
squared of a phase is just 1,
so that's just 1.
So now we have the norm squared
of 1 plus a complex number.
And so the norm squared of
1 is going to give me 1.
The norm squared of the complex
number is going to give me 1.
And the cross terms are going
to give me the real part--
twice the real part--
of e to the i b minus
a, which is going to be
equal to cosine of b minus a.
And so what you see
here is that you
have a single frequency
in the superposition.
So good, our
uncertainty is large.
So let's look at
this second example
in a little more detail.
By superimposing two states with
wavelength lambda1 and lambda2
or k1 and k2 we get
something that, OK,
it's still not
well localized-- we
don't know where the
particle is going
to be-- but it's better
localized than it was before,
right?
What happens if we superpose
with three wavelengths,
or four, or more?
So for that, I want to pull
out a Mathematica package.
You guys should all have
seen Fourier analysis
in 18.03, but just in case,
I'm putting on the web page,
on the Stellar page, a
notebook that walks you
through the basics of Fourier
analysis in Mathematica.
You should all be
fluent in Mathematica.
If you're not, you
should probably
come up to speed on it.
That's not what we wanted.
Let's try that again.
There we go.
Oh, that's awesome-- where
by awesome, I mean not.
It's coming.
OK, good.
I'm not even going to mess with
the screens after last time.
So I'm not going to go through
the details of this package,
but what this does is walk
you through the superposition
of wave packets.
So here I'm looking at the
probability distribution coming
from summing up a
bunch of plane waves
with some definite frequency.
So here it's just one.
That's one wave,
so first we have--
let me make this bigger-- yes,
stupid Mathematica tricks.
So here we have
the wave function
and here we have the probability
distribution, the norm squared.
And it's sort of
badly normalized here.
So that's for a single wave.
And as you see, the probability
distribution is constant.
And that's not 0,
that's 0.15, it's
just that I arbitrarily
normalized this.
So let's add two plane waves.
And now what you see is the
same effect as we had here.
You see a slightly more
localized wave function.
Now you have a little bit of
structure in the probability
distribution.
So there's the structure in
the probability distribution.
We have a little
more information
about where the particle
is more likely to be here
than it is to be here.
Let's add one more.
And as we keep adding
more and more plane waves
to our superposition, the wave
function and the probability
distribution associated
with it become
more and more
well-localized until, as we
go to very high numbers of plane
waves that we're superposing,
we get an extremely narrow
probability distribution--
and wave function, for that
matter-- extremely narrow
corresponding to a particle
that's very likely to be here
and unlikely to
be anywhere else.
Everyone cool with that?
What's the expense?
Want have we lost
in the process?
Well we know with
great confidence
now that the particle will be
found here upon observation.
But what will its momentum be?
Yeah, now it's the
superposition of a whole bunch
of different momenta.
So if it's a superposition
of a whole bunch
of different
momenta, here this is
like superposition of a whole
bunch of different positions--
likely to be here,
likely to be here,
likely to be here,
likely to be here.
What's our knowledge
of its position?
It's not very good.
Similarly, now that
we have superposed
many different momenta
with comparable strength.
In fact, here they were
all with unit strength.
We now have no information about
what the momentum is anymore.
It could be anything
in that superposition.
So now we're seeing
quite sharply
the uncertainty relation.
And here it is.
So the uncertainty
relation is now
pretty clear from these guys.
So that didn't work?
And I'm going to leave it alone.
This is enough for
the Fourier analysis,
but that Fourier
package is available
with extensive commentary
on the Stellar web page.
AUDIENCE: Now is
that sharp definition
in the position caused by
the interference between all
those waves and all that--
PROFESSOR: That's
exactly what it is.
Precisely.
It's precisely the interference
between the different momentum
nodes that leads to
certainty in the position.
That's exactly right.
Yeah.
AUDIENCE: So as we're
certain of the position,
we will not be certain
of the momentum.
PROFESSOR: Exactly.
And here we are.
So in this example, we have
no idea what the position is,
but we're quite confident
of the momentum.
Here we have no idea
what the position is,
but we have great
confidence in the momentum.
Similarly here, we have
less perfect confidence
of the position, and here we
have less perfect confidence
in the momentum.
It would be nice to
be able to estimate
what our uncertainty
is in the momentum here
and what our uncertainty
is in the position here.
So we're going to
have to do that.
That's going to be
one of our next tasks.
Other questions?
Yeah.
AUDIENCE: In this half
of the blackboard,
you said, obviously, if
we do it a bunch of times
it'll have more in
the x2 than in the x1.
PROFESSOR: Yes.
AUDIENCE: The average, it will
never physically be at that--
PROFESSOR: Yeah, that's right,
so, because it's a probability
distribution, it won't
be exactly at that point.
But it'll be nearby.
OK, so in order to be more
precise-- And so for example
for this what we do
here's a quick question.
How well do you know the
position of this particle?
Pretty well, right?
But how well do you
know its momentum?
Well, we'd all like to say
not very, but tell me why.
Why is your uncertainty in the
momentum of the particle large?
AUDIENCE: Heisenberg's
uncertainty principle.
PROFESSOR: Yeah,
but that's a cheat
because we haven't actually
proved Heisenberg's uncertainty
principle.
It's just something
we're inheriting.
AUDIENCE: I believe it.
PROFESSOR: I believe it, too.
But I want a better
argument because I
believe all sorts
of crazy stuff.
So-- I really do.
Black holes, fluids, I mean
look, don't get me started.
Yeah.
AUDIENCE: You can take the
Fourier transform of it.
PROFESSOR: Yeah, excellent.
OK, we'll get to
that in just one sec.
So before taking the
Fourier transform, which
is an excellent--
so the answer was,
just take a Fourier
transform, that's
going to give you
some information.
We're going to do
that in just a moment.
But before we do
Fourier transform,
just intuitively, why would
de Broglie look at this
and say, no, that doesn't
have a definite momentum.
AUDIENCE: There's
no clear wavelength.
PROFESSOR: Yeah, there's
no wavelength, right?
It's not periodic by any
stretch of the imagination.
It doesn't look like a thing
with a definite wavelength.
And de Broglie said, look, if
you have a definite wavelength
then you have a
definite momentum.
And if you have a
definite momentum,
you have a definite wavelength.
This is not a wave with
a definite wavelength,
so it is not corresponding
to the wave function
for a particle with
a definite momentum.
So our momentum is
unknown-- so this is large.
And similarly, here, our
uncertainty in the momentum
is large.
So to do better
than this, we need
to introduce the
Fourier transform,
and I want to do that now.
So you should all have seen
Fourier series in 8.03.
Now we're going to do
the Fourier transform.
And I'm going to
introduce this to you
in 8.04 conventions
in the following way.
And the theorem
says the following--
we're not going to prove it by
any stretch of the imagination,
but the theorem
says-- any function
f of x that is sufficiently
well-behaved-- it shouldn't be
discontinuous, it
shouldn't be singular--
any reasonably well-behaved,
non-stupid function f
of x can be built by superposing
enough plane waves of the form
e to the ikx.
Enough may be infinite.
So any function f of x can be
expressed as 1 over root 2 pi,
and this root 2 pi is a choice
of normalization-- everyone
has their own
conventions, and these
are the ones we'll be
using in 8.04 throughout--
minus infinity to infinity
dk f tilde of k e to the ikx.
So here, what we're
doing is, we're
summing over plane waves
of the form e to the ikx.
These are modes with a definite
wavelength 2 pi upon k.
f tilde of k is telling us
the amplitude of the wave
with wavelengths lambda
or wave number k.
And we sum over all
possible values.
And the claim is,
any function can
be expressed as a superposition
of plane waves in this form.
Cool?
And this is for
functions which are
non-periodic on the real line,
rather than periodic functions
on the interval, which is what
you should've seen in 8.03.
Now, conveniently, if
you know f tilde of k,
you can compute f of
x by doing the sum.
But suppose you
know f of x and you
want to determine what
the coefficients are,
the expansion coefficients.
That's the inverse
Fourier transform.
And the statement for
that is that f tilde of k
is equal to 1 over root 2 pi
integral from minus infinity
to infinity dx f of
x e to the minus ikx.
OK, that's sometimes referred
to as the inverse Fourier
transform.
And here's something
absolutely essential.
f tilde of k, the Fourier
transform coefficients
of f of x, are
completely equivalent.
If you know f of x, you
can determine f tilde of k.
And if you know
f tilde of k, you
can determine f of x
by just doing a sum,
by just adding them up.
So now here's the physical
version of this-- oh,
I can't slide that out--
I'm now going to put here.
Oh.
No, I'm not.
I'm going to put that down here.
So the physical
version of this is
that any wave function
psi of x can be expressed
as the superposition
in the form psi of x
is equal to 1 over root 2 pi
integral from minus infinity
to infinity dk psi tilde of k
e to the ikx of states, or wave
functions, with a definite
momentum p is equal to h bar k.
And so now, it's useful
to sketch the Fourier
transforms of each
of these functions.
In fact, we want this up here.
So here we have the function and
its probability distribution.
Now I want to draw the Fourier
transforms of these guys.
So here's psi tilde
of k, a function
of a different
variable than of x,
but nonetheless,
it's illuminating
to draw them next to each other.
And again, I'm
drawing the real part.
And here, x2-- had I had
my druthers about me,
I would have put x2
at a larger value.
Good, so it's further
off to the right there.
I'm so loathe to erase the
superposition principle.
But fortunately,
I'm not there yet.
Let's look at the Fourier
transform of these guys.
The Fourier transform
of this guy-- this is k.
Psi tilde of k well,
that's something
with a definite value of k.
And it's Fourier transform--
this is 0-- there's k1.
And for this guy--
there's 0-- k2.
And now if we look at
the Fourier transforms
of these guys, see,
this way I don't
have to erase the superposition
principle-- and the Fourier
transform of this guy,
so note that there's
a sort of pleasing
symmetry here.
If your wave function
is well localized,
corresponding to a reasonably
well-defined position,
then your Fourier transform
is not well localized,
corresponding to not
having a definite momentum.
On the other hand, if you
have definite momentum,
your position is
not well defined,
but the Fourier transform
has a single peak
at the value of k corresponding
to the momentum of your wave
function.
Everyone cool with that?
So here's a question-- sorry,
there was a raised hand.
Yeah?
AUDIENCE: Are we going
to learn in this class
how to determine the
Fourier transforms
of these non-stupid functions?
PROFESSOR: Yes, that
will be your homework.
On your homework is
an extensive list
of functions for you to
compute Fourier transforms of.
And that will be the job of
problem sets and recitation.
So Fourier series
and computing-- yeah,
you know what's
coming-- Fourier series
are assumed to have been covered
for everyone in 8.03 and 18.03
in some linear
combination thereof.
And Fourier transforms--
[LAUGHTER AND GROANS]
I couldn't help it.
So Fourier transforms
are a slight embiggening
of the space of Fourier
series, because we're not
looking at periodic functions.
AUDIENCE: So when we're
doing the Fourier transforms
of a wave function, we're
basically writing it
as a continuous set
of different waves.
Can we write it
as a discrete set?
So as a Fourier series?
PROFESSOR: Absolutely,
so, however,
what is true about
Fourier series?
When you use a discrete
set of momenta,
which are linear, which are--
It must be a periodic function,
exactly.
So here what we've
done is, we've said,
look, we're writing
our wave function,
our arbitrary wave function,
as a continuous superposition
of a continuous value
of possible momenta.
This is absolutely correct.
This is exactly
what we're doing.
However, that's
kind of annoying,
because maybe you
just want one momentum
and two momenta
and three momenta.
What if you want
a discrete series?
So discrete is fine.
But if you make
that discrete series
integer-related to
each other, which
is what you do with
Fourier series,
you force the function
f of x to be periodic.
And we don't want
that, in general,
because life isn't periodic.
Thank goodness, right?
I mean, there's like one film
in which it-- but so-- it's
a good movie.
So that's the
essential difference
between Fourier series
and Fourier transforms.
Fourier transforms
are continuous in k
and do not assume
periodicity of the function.
Other questions?
Yeah.
AUDIENCE: So
basically, the Fourier
transform associates
an amplitude
and a phase for each of
the individual momenta.
PROFESSOR: Precisely.
Precisely correct.
So let me say that again.
So the question was--
so a Fourier transform
effectively
associates a magnitude
and a phase for each
possible wave vector.
And that's exactly right.
So here there's some
amplitude and phase--
this is a complex
number, because this
is a complex function-- there's
some complex number which
is an amplitude and
a phase associated
to every possible momentum
going into the superposition.
That amplitude may be 0.
There may be no contribution
for a large number of momenta,
or maybe insignificantly small.
But it is indeed
doing precisely that.
It is associating an amplitude
and a phase for every plane
wave, with every different
value of momentum.
And you can compute,
before panicking,
precisely what that
amplitude and phase
is by using the inverse
Fourier transform.
So there's no magic here.
You just calculate.
You can use your calculator,
literally-- I hate that word.
OK so now, here's
a natural question.
So if this is the Fourier
transform of our wave function,
we already knew that this
wave function corresponded
to having a definite--
from de Broglie,
we know that it has
a definite momentum.
We also see that its Fourier
transform looks like this.
So that leads to a
reasonable guess.
What do you think the
probability distribution P of k
is-- the probability
density to find the momentum
to have wave vector h-bar k?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, that's a
pretty reasonable guess.
So we're totally
pulling this out
of the dark-- psi
of k norm squared.
OK, well let's
see if that works.
So psy of k norm
squared for this
is going to give us a nice,
well localized function.
And so that makes
a lot of sense.
That's exactly what
we expected, right?
Definite value of P with
very small uncertainty.
Similarly here.
Definite value of P, with
a very small uncertainty.
Rock on.
However, let's look at this guy.
What is the expected value
of P if this is the Fourier
transform?
Well remember, we have to take
the norm squared, and psi of k
was e to the i k x1--
the Fourier transform.
You will do much
practice on taking
Fourier transforms
on the problem set.
Where did my eraser go?
There it is.
Farewell, principle one.
So what does norm squared
of psi tilde look like?
Well, just like before,
the norm squared
is constant, because the norm
squared of a phase is constant.
And again, the
norm squared-- this
is psi tilde of k norm
squared-- we believe,
we're conjecturing
this is P of k.
You will prove this relation
on your problem set.
You'll prove that it follow
from what we said before.
And similarly, this is
constant-- e to the i k x2.
So now we have no
knowledge of the momenta.
So that also fits.
The momenta is, we have no idea.
And uncertainty is large.
And the momenta is,
we have no idea.
And the uncertainty is large.
So in all these
cases, we see that we
satisfy quite nicely the
uncertainty relation--
small position momentum,
large momentum uncertainty.
Large position
uncertainty, we're
allowed to have small
momentum uncertainty.
And here, it's a little
more complicated.
We have a little bit of
knowledge of position,
and we have a little bit of
knowledge of the momenta.
We have a little bit of
knowledge of position,
and we have a little bit
of knowledge of momenta.
So we'll walk through
examples with superposition
like this on the problem set.
Last questions
before we get going?
OK so I have two things
to do before we're done.
The first is, after lecture
ends, I have clickers.
And anyone who wants
to borrow clickers,
you're welcome to come
down and pick them up
on a first come
first served basis.
I will start using the
clickers in the next lecture.
So if you don't already
have one, get one now.
But the second thing is--
don't get started yet.
I have a demo to do.
And last time I told
you-- this is awesome.
It's like I'm an
experimentalist for a day.
Last time I told you that one of
the experimental facts of lice
is-- of lice.
One of the experimental
facts of lice.
One of the experimental
facts of life
is that there is
uncertainty in the world
and that there is probability.
There are unlikely
events that happen
with some probability,
some finite probability.
And a good example of the
randomness of the real world
involves radiation.
So hopefully you can hear this.
Apparently, I'm not
very radioactive.
You'd be surprised at the
things that are radioactive.
Ah, got a little tick.
Shh.
This is a plate sold at
an Amish county fair.
It's called vaseline ware
and it's made of local clays.
[GEIGER COUNTER CLICKS]
It's got uranium in it.
But
I want to emphasize-- exactly
when something goes click,
it sounds pretty random.
And it's actually a better
random number generator
than anything you'll find in
Mathematica or C. In fact,
for some purposes, the decay
of radioactive isotopes
is used as the perfect
random number generator.
Because it really is totally
random, as far as anyone
can tell.
But here's my favorite.
People used to eat off these.
[MUCH LOUDER, DENSER CLICKS]
See you next time.
