The following
content is provided under a Creative
Commons license. Your support will help MIT
OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare ocw.mit.edu. PROFESSOR: Anything lingering
and disturbing or bewildering? No? Nothing? All right. OK, so the story so far is
basically three postulates. The first is that the
configuration of a particle is given by, or described
by, a wave function psi of x. Yeah? So in particular, just to
flesh this out a little more, if we were in 3D, for
example-- which we're not. We're currently in
our one dimensional tripped out tricycles. In 3D, the wave function
would be a function of all three
positions x, y and z. If we had two particles,
our wave function would be a function of the
position of each particle. x1, x2, and so on. So we'll go through lots of
details and examples later on. But for the most
part, we're going to be sticking with
single particle in one dimension for the
next few weeks. Now again, I want to emphasize
this is our first pass through our definition
of quantum mechanics. Once we use the language and
the machinery a little bit, we're going to develop a more
general, more coherent set of rules or definition
of quantum mechanics. But this is our first pass. Two, the meaning of
the wave function is that the norm squared psi of
x, norm squared, it's complex, dx is the probability of
finding the particle- There's an n in their. Finding the particle-- in the
region between x and x plus dx. So psi squared
itself, norm squared, is the probability density. OK? And third, the
superposition principle. If there are two
possible configurations the system can be in,
which in quantum mechanics means two different
wave functions that could describe the system
given psi 1 and psi 2, two wave functions
that could describe two different configurations
of the system. For example, the particles here
or the particles over here. It's also possible to find
the system in a superposition of those two psi is
equal to some arbitrary linear combination alpha
psi 1 plus beta psi 2 of x. OK? So some things to note--
so questions about those before we move on? No questions? Nothing? Really? You're going to make he
threaten you with something. I know there are questions. This is not trivial stuff. OK. So some things to note. The first is we
want to normalize. We will generally
normalize and require that the integral over
all possible positions of the probability density
psi of x norm squared is equal to 1. This is just saying that the
total probability that we find the particle somewhere
had better be one. This is like saying if
I know a particle is in one of two
boxes, because I've put a particle in
one of the boxes. I just don't remember which one. Then the probability that
it's in the first box plus probability that
it's in the second box must be 100% or one. If it's less, then the particle
has simply disappeared. And basic rule, things
don't just disappear. So probability
should be normalized. And this is our prescription. So a second thing to note is
that all reasonable, or non stupid, functions
psi of x are equally reasonable as wave functions. OK? So this is a very
reasonable function. It's nice and smooth. It converges to 0 infinity. It's got all the nice
properties you might want. This is also a
reasonable function. It's a little annoying,
but there it is. And they're both perfectly
reasonable as wave functions. This on the other
hand, not so much. So for two reasons. First off, it's discontinuous. And as you're going to
show in your problem set, discontinuities are very
bad for wave functions. So we need our wave
functions to be continuous. The second is over some
domain it's multi valued. There are two different
values of the function. That's also bad, because
what's the probability? It's the norm squared,
but if it two values, two values for the probability,
that doesn't make any sense. What's the probability
that I'm going to fall over in 10 seconds? Well, it's small, but it's not
actually equal to 1% or 3%. It's one of those. Hopefully is much
lower than that. So all reasonable
functions are equally reasonable as wave functions. And in particular, what
that means is all states corresponding to
reasonable wave functions are equally reasonable
as physical states. There's no primacy in wave
functions or in states. However, with that said,
some wave functions are more equal than others. OK? And this is important,
and coming up with a good
definition of this is going to be an important
challenge for us in the next couple of lectures. So in particular,
this wave function has a nice simple
interpretation. If I tell you this
is psi of x, then what can you tell me about the
particle whose wave function is the psi of x? What can you tell me about it? What do you know? AUDIENCE: [INAUDIBLE]. PROFESSOR: It's here, right? It's not over here. Probability is basically 0. Probability is large. It's pretty much here with
this great confidence. What about this guy? Less informative, right? It's less obvious what this
wave function is telling me. So some wave functions are
more equal in the sense that they have-- i.e. they have simple
interpretations. So for example,
this wave function continuing on infinitely, this
wave function doesn't tell me where the particle is,
but what does it tell me? AUDIENCE: Momentum. PROFESSOR: The
momentum, exactly. So this is giving me
information about the momentum of the particle because it
has a well defined wavelength. So this one, I would also say
is more equal than this one. They're both perfectly
physical, but this one has a simple interpretation. And that's going to
be important for us. Related to that is that any
reasonable function psi of x can be expressed as a
superposition of more equal wave functions,
or more precisely easily interpretable wave functions. We saw this last time
in the Fourier theorem. The Fourier theorem said look,
take any wave function-- take any function, but I'm
going to interpret in the language of
quantum mechanics. Take any wave function which
is given by some complex valued function, and it
can be expressed as a superposition
of plane waves. 1 over 2pi in our normalization
integral dk psi tilde of k, but this is a set
of coefficients. e to the ikx. So what are we doing here? We're saying pick a value of k. There's a number
associated with it, which is going to be an
a magnitude and a phase. And that's the
magnitude and phase of a plane wave, e to the ikx. Now remember that
e to the ikx is equal to cos kx plus i sin kx. Which you should all know,
but just to remind you. This is a periodic function. These are periodic functions. So this is a plane wave
with a definite wavelength, 2pi upon k. So this is a more equal
wave function in the sense that it has a
definite wavelength. We know what its momentum is. Its momentum is h bar k. Any function, we're saying, can
be expressed as a superposition by summing over all
possible values of k, all possible
different wavelengths. Any function can be expressed
as a superposition of wave functions with a
definite momentum. That make sense? Fourier didn't think
about it that way, but from quantum
mechanics, this is the way we want
to think about it. It's just a true statement. It's a mathematical fact. Questions about that? Similarly, I claim that I can
expand the very same function, psi of x, as a
superposition of states, not with definite
momentum, but of states with definite position. So what's a state with
a definite position? AUDIENCE: Delta. PROFESSOR: A delta
function, exactly. So I claim that any
function psi of x can be expanded a
sum over all states with a definite position. So delta of-- well, what's a
state with a definite position? x0. Delta of x minus x0. OK? This goes bing when
x0 is equal to x. But I want a sum over all
possible delta functions. That means all
possible positions. That means all possible
values of x0, dx0. And I need some
coefficient function here. Well, the coefficient function
I'm going to call psi of x0. So is this true? Is it true that I
can take any function and expand it in a superposition
of delta functions? Absolutely. Because look at what
this equation does. Remember, delta
function is your friend. It's a map from integrals
to numbers or functions. So this integral, is
an integral over x0. Here we have a
delta of x minus x0. So this basically says
the value of this integral is what you get by
taking the integrand and replacing x by x0. Set x equals x0, that's
when delta equals 0. So this is equal to the argument
evaluated at x0 is equal to x. That's your psi of x. OK? Any arbitrarily ugly function
can be expressed either as a superposition of states
with definite momentum or a superposition of states
with definite position. OK? And this is going to be true. We're going to find this
is a general statement that any state can be expressed
as a superposition of states with a well defined
observable quantity for any observable
quantity you want. So let me give you just a
quick little bit of intuition. In 2D, this is a perfectly
good vector, right? Now here's a question
I want to ask you. Is that a superposition? Yeah. I mean every vector
can be written as the sum of other
vectors, right? And it can be done in an
infinite number of ways, right? So there's no such
thing as a state which is not a superposition. Every vector is a
superposition of other vectors. It's a sum of other vector. So in particular we often
find it useful to pick a basis and say look, I know what
I mean by the vector y, y hat is a unit vector
in this direction. I know what I mean
by the vector x hat. It's a unit vector
in this direction. And now I can ask, given that
these are my natural guys, the guys I want to attend
to, is this a superposition of x and y? Or is it just x or y? Well, that's a superposition. Whereas x hat itself is not. So this somehow is about finding
convenient choice of basis. But any given vector
can be expressed as a superposition of
some pair of basis vectors or a different pair
of basis vectors. There's nothing hallowed
about your choice of basis. There's no God given
basis for the universe. We look out in the universe
in the Hubble deep field, and you don't see somewhere
in the Hubble deep field an arrow going x, right? So there's no natural
choice of basis, but it's sometimes
convenient to pick a basis. This is the direction of
the surface of the earth. This is the direction
perpendicular to it. So sometimes
particular basis sets have particular meanings to us. That's true in vectors. This is along the earth. This is perpendicular to it. This would be slightly strange. Maybe if you're leaning. And similarly, this
is an expansion of a function as a
sum, as a superposition of other functions. And you could have done this
in any good space of functions. We'll talk about that more. These are particularly
natural ones. They're more equal. These are ones with different
definite values of position, different definite
values of momentum. Everyone cool? Quickly what's the momentum
associated to the plane wave e to the ikx? AUDIENCE: [INAUDIBLE]. PROFESSOR: h bar k. Good. So now I want to
just quickly run over some concept questions for you. So whip out your clickers. OK, we'll do this verbally. All right, let's try this again. So how would you interpret
this wave function? AUDIENCE: e. PROFESSOR: Solid. How do you know
whether the particle is big or small by looking
at the wave function? AUDIENCE: [INAUDIBLE]. PROFESSOR: All right. Two particles described by
a plane wave of the form e to the ikx. Particle one is a smaller
wavelength than particle two. Which particle has
a larger momentum? Think about it, but
don't say it out loud. And this sort of defeats the
purpose of the clicker thing, because now I'm supposed to be
able to know without you guys saying anything. So instead of
saying it out loud, here's what I'd like you to do. Talk to the person
next to you and discuss which one has the larger AUDIENCE: [CHATTER]. All right. Cool, so which one has
the larger momentum? AUDIENCE: A. PROFESSOR: How come? [INTERPOSING VOICES] PROFESSOR: RIght,
smaller wavelength. P equals h bar k. k equals 2pi over lambda. Solid? Smaller wavelength,
higher momentum. If it has higher
momentum, what do you just intuitively expect to
know about its energy? It's probably higher. Are you positive about that? No, you need to know how the
energy depends on the momentum, but it's probably higher. So this is an
important little lesson that you probably all
know from optics and maybe from core mechanics. Shorter wavelength
thing, higher energy. Higher momentum for sure. Usually higher energy as well. Very useful rule of
thumb to keep in mind. Indeed, it's particle one. OK next one. Compared to the wave function
psi of x, it's Fourier transform, psi tilde of x
contains more information, or less, or the
same, or something. Don't say it out loud. OK, so how many people
know the answer? Awesome. And how many people
are not sure. OK, good. So talk to the person next to
you and convince them briefly. All right. So let's vote. A, more information. B, less information. C, same. OK, good you got it. So these are not hard ones. This function, which is
a sine wave of length l, 0 outside of that region. Which is closer to true? f has a single well defined
wavelength for the most part? It's closer to true. This doesn't have to be exact. f has a single well
defined wavelengths. Or f is made up of a wide
range of wavelengths? Think it to yourself. Ponder that one for a minute. OK, now before we
get talking about it. Hold on, hold on, hold on. Since we don't have
clickers, but I want to pull off
the same effect, and we can do this,
because it's binary here. I want everyone close your eyes. Just close your eyes,
just for a moment. Yeah. Or close the eyes of
the person next to you. That's fine. And now and I want you to vote. A is f has a single
well defined wavelength. B is f has a wide
range of wavelengths. So how many people think
A, a single wavelength? OK. Lower your hands, good. And how many people think B,
a wide range of wavelengths? Awesome. So this is exactly what happens
when we actually use clickers. It's 50/50. So now you guys need to talk
to the person next to you and convince each
other of the truth. AUDIENCE: [CHATTER]. All right, so the volume sort of
tones down as people, I think, come to resolution. Close your eyes again. Once more into the
breach, my friends. So close your eyes, and
now let's vote again. f of x has a single,
well defined wavelength. And now f of x is made up
of a range of wavelengths? OK. There's a dramatic
shift in the field to B, it has a wide range
of wavelengths, not a single wavelength. And that is, in fact,
the correct answer. OK, so learning happens. That was an empirical test. So does anyone want
to defend this view that f is made of a wide
range of wavelengths? Sure, bring it. AUDIENCE: So, the sine
wave is an infinite, and it cancels out
past minus l over 2 and positive l over
2, which means you need to add a bunch
of wavelengths to actually cancel it out there. PROFESSOR: Awesome, exactly. Exactly. If you only had the thing
of a single wavelength, it would continue with a single
wavelength all the way out. In fact, there's a
nice way to say this. When you have a sine wave, what
can you say about it's-- we know that a sine
wave is continuous, and it's continuous
everywhere, right? It's also differentiable
everywhere. Its derivative is continuous
and differentiable everywhere, because it's a cosine, right? So if yo you take a
superposition of sines and cosines, do you ever
get a discontinuity? No. Do you ever get something whose
derivative is discontinuous? No. So how would you ever
reproduce a thing with a discontinuity
using sines and cosines? Well, you'd need some infinite
sum of sines and cosines where there's some technicality
about the infinite limit being singular, because
you can't do it a finite number of
sines and cosines. That function is continuous,
but its derivative is discontinuous. Yeah? So it's going to take an
infinite number of sines and cosines to reproduce
that little kink at the edge. Yeah? AUDIENCE: So a finite
number of sines and cosines doesn't mean finding-- or
an infinite number of sines and cosines doesn't mean
infinite [? regular ?] sines and cosines, right? Because over a finite
region [INAUDIBLE]. PROFESSOR: That's true, but
you need arbitrarily-- so let's talk about that. That's an excellent question. That's a very good question. The question here
is look, there's two different things you
can be talking about. One is arbitrarily large and
arbitrarily short wavelengths, so an arbitrary
range of wavelengths. And the other is
an infinite number. But an infinite number
is silly, because there's a continuous variable here k. You got an infinite
number of wavelengths between one and 1.2, right? It's continuous. So which one do you mean? So let's go back
to this connection that we got a minute
ago from short distance and high momentum. That thing looks like it has
one particular wavelength. But I claim, in
order to reproduce that as a superposition of
states with definite momentum, I need arbitrarily
high wavelength. And why do I need arbitrarily
high wavelength modes? Why do we need to arbitrarily
high momentum modes? Well, it's because of this. We have a kink. And this feature, what's the
length scale of that feature? It's infinitesimally
small, which means I'm going to have to--
in order to reproduce that, in order to probe
it, I'm going to need a momentum that's
arbitrarily large. So it's really about the
range, not just the number. But you need arbitrarily
large momentum. To construct or detect an
arbitrarily small feature you need arbitrarily
large momentum modes. Yeah? AUDIENCE: Why do
you [INAUDIBLE]? Why don't you just
say, oh you need an arbitrary small wavelength? Why wouldn't you just
phrase that [INAUDIBLE]? PROFESSOR: I chose
to phrase it that way because I want an emphasize
and encourage-- I emphasize you to think and
encourage you to conflate short distance and
large momentum. I want the connection between
momentum and the length scale to be something that
becomes intuitive to you. So when I talk about
something with short features, I'm going to talk about it as
something with large momentum. And that's because in a
quantum mechanical system, something with
short wavelength is something that carries
large momentum. That cool? Great. Good question. AUDIENCE: So earlier you
said that any reasonable wave function, a possible
wave function, does that mean
they're not supposed to be Fourier transformable? PROFESSOR: That's
usually a condition. Yeah, exactly. We don't quite
phrase it that way. And in fact, there's a
problem on your problem set that will walk you
through what we will mean. What should be
true of the Fourier transform in order for this
to reasonably function. And among other things--
and your intuition here is exactly right--
among other things, being able to have a Fourier
transform where you don't have arbitrarily high
momentum modes is going to be an
important condition. That's going to turn to be
related to the derivative being continuous. That's a very good question. So that's the optional
problem 8 on problem set 2. Other questions? PROFESSOR: Cool, so that's
it for the clicker questions. Sorry for the technology fail. So I'm just going to
turn this off in disgust. That's really irritating. So today what I want
to start on is pick up on the discussion of the
uncertainty principle that we sort of
outlined previously. The fact that when we have a
wave function with reasonably well defined position
corresponding to a particle with reasonably
well defined position, it didn't have a reasonably
well defined momentum and vice versa. The certainty of
the momentum seems to imply lack of knowledge about
the position and vice versa. So in order to do that, we
need to define uncertainty. So I need to define for
you delta x and delta p. So first I just want
to run through what should be totally
remedial probability, but it's always useful
to just remember how these basic things work. So consider a set
of people in a room, and I want to plot the number
of people with a particular age as a function of the
age of possible ages. So let's say we have 16
people, and at 14 we have one, and at 15 we have 1,
and at 16 we have 3. And that's 16. And at 20 we have 2. And at 21 we have 4. And at 22 we have 5. And that's it. OK. So 1, 1, 3, 2, 4, 5. OK, so what's the
probability that any given person in this group of
16 has a particular age? I'll call it a. So how do we compute
the probability that they have age a? Well this is easy. It's the number that have
age a over the total number. So note an important thing,
an important side note, which is that the sum
over all possible ages of the probability that you
have age a is equal to 1, because it's just going to
be the sum of the number with a particular age over the
total number, which is just the sum of the number
with any given age. So here's some questions. So what's the most likely age? If you grabbed one
of these people from the room with
a giant Erector set, and pull out a person,
and let them dangle, and ask them what
their age is, what's the most likely they'll have? AUDIENCE: 22. PROFESSOR: 22. On the other hand,
what's the average age? Well, just by eyeball roughly
what do you think it is? So around 19 or 20. It turns out to
be 19.2 for this. OK. But if everyone had a little
sticker on their lapel that says I'm 14, 15, 16, 20,
21 or 22, how many people have the age 19.2? None, right? So a useful thing is
that the average need not be an observable value. This is going to come
back to haunt us. Oops, 19.4. That's what I got. So in particular how
did I get the average? I'm going to define
some notation. This notation is
going to stick with us for the rest of
quantum mechanics. The average age,
how do I compute it? So we all know this, but let
me just be explicit about it. It's the sum over
all possible ages of the number of
the number of people with that age times
the age divided by the total number of people. OK? So in this case, I'd go 14,14,
16, 16, 16, 20, 20, 21, 21, 21 21, 22, 22, 22, 22, 22. And so that's all
I've written here. But notice that I can
write this in a nice way. This is equal to the sum
over all possible ages of a times the ratio of Na to N
with a ratio of Na to n total. That's just the probability
that any given person has a probability a. a times probability of a. So the expected value is the
sum over all possible values of the value times the
probability to get that value. Yeah? This is the same equation,
but I'm going to box it. It's a very useful relation. And so, again, does the
average have to be measurable? No, it certainly doesn't. And it usually isn't. So let's ask the same thing
for the square of ages. What is the average
of a squared? Square the ages. You might say, well, why
would I ever care about that? But let's just be
explicit about it. So following the
same logic here, the average of a squared,
the average value of the square of the
ages is, well, I'm going to do exactly
the same thing. It's just a squared, right? 14 squared, 15 squared, 16
square, 16 squared, 16 squared. So this is going to give me
exactly the same expression. So over a of a squared
probability of measuring a. And more generally, the expected
value, or the average value of some function of
a is equal-- and this is something you
don't usually do-- is equal to the sum over a of
f of a, the value of f given a particular value of a,
times the probability that you measure that value of
a in the first place. It's exactly the same
logic as averages. Right, cool. So here's a quick question. Is a squared equal to the
expected value of a squared? AUDIENCE: No. PROFESSOR: Right, in
general no, not necessarily. So for example,
the average value-- suppose we have a Gaussian
centered at the origin. So here's a. Now a isn't age, but it's
something-- I don't know. You include infants or whatever. It's not age. Its happiness on a given day. So what's the average value? Meh. Right? Sort of vaguely neutral, right? But on the other hand,
if you take a squared, very few people have
a squared as zero. Most people have a
squared as not a 0 value. And most people are
sort of in the middle. Most people are sort of
hazy on what the day is. So in this case,
the expected value of a, or the average
value of a is 0. The average value of a
squared is not equal to 0. Yeah? And that's because the squared
has everything positive. So how do we characterize--
this gives us a useful tool for characterizing
the width of a distribution. So here we have a distribution
where its average value is 0, but its width is non-zero. And then the expectation
value of a squared, the expected value of
a squared, is non-zero. So how do we define the
width of a distribution? This is going to be
like our uncertainty. How happy are you today? Well, I'm not sure. How unsure are you? Well, that should give
us a precise measure. So let me define three things. First the deviation. So the deviation is going to be
a minus the average value of a. So this is just take
the actual value of a and subtract off the
average value of a. So we always get something
that's centered at 0. I'm going to write it like this. Note, by the way, just a
convenient thing to note. The average value of a
minus it's average value. Well, what's the
average value of 7? AUDIENCE: 7. PROFESSOR: OK, good. So that first term is
the average value of a. And that second term
is the average value of this number, which is
just this number minus a. So this is 0. Yeah? The average value
of a number is 0. The average value
of this variable is the average value of
that variable, but that's 0. So deviation is not a terribly
good thing on average, because on average the
deviation is always 0. That's what it means to
say this is the average. So the derivation
is saying how far is any particular
instance from the average. And if you average
those deviations, they always give you 0. So this is not a
very good measure of the actual width
of the system. But we can get a nice measure by
getting the deviation squared. And let's take the mean
of the derivation squared. So the mean of the derivation
squared, mean of a minus the average value of a squared. This is what I'm going to
call the standard deviation. Which is a little odd,
because really you'd want to call it the
standard deviation squared. But whatever. We use funny words. So now what does it mean if
the average value of a is 0? It means it's centered
at 0, but what does it mean if the standard
deviation of a is 0? So if the standard
deviation is 0, one then the distribution
has no width, right? Because if there was
any amplitude away from the average
value, then that would give a non-zero
strictly positive contribution to this average expectation,
and this wouldn't be 0 anymore. So standard deviation
is 0, as long as there's no width, which is
why the standard deviation is a good useful measure
of width or uncertainty. And just as a note, taking
this seriously and taking the square, so standard
deviation squared, this is equal to the
average value of a squared minus twice a times the average
value of a plus average value of a quantity squared. But if you do this
out, this is going to be equal to a squared
minus 2 average value of a average value of a. That's just minus
twice the average value of a quantity squared. And then plus average
value of a squared. So this is an alternate way of
writing the standard deviation. OK? So we can either write it in
this fashion or this fashion. And the notation for
this is delta a squared. OK? So when I talk about
an uncertainty, what I mean is, given
my distribution, I compute the
standard deviation. And the uncertainty
is going to be the square root of the
standard deviations squared. OK? So delta a, the words
I'm going to use for this is the uncertainty in a given
some probability distribution. Different probability
distributions are going to give me different delta a's. So one thing that's
sort of annoying is that when you
write delta a, there's nothing in the notation
that says which distribution you
were talking about. When you have multiple
distributions, or multiple possible probability
distributions, sometimes it's useful to just put given
the probability distribution p of a. This is not very often used,
but sometimes it's very helpful when you're doing calculations
just to keep track. Everyone cool with that? Yeah, questions? AUDIENCE: [INAUDIBLE]
delta a squared, right? PROFESSOR: Yeah, exactly. Of delta a squared. Yeah. Other questions? Yeah? AUDIENCE: So really it should
be parentheses [INAUDIBLE]. PROFESSOR: Yeah, it's just
this is notation that's used typically, so I didn't
put the parentheses around precisely to alert you to the
stupidities of this notation. So any other questions? Good. OK, so let's just do the same
thing for continuous variables. Now for continuous variables. I'm just going to
write the expressions and just get them
out of the way. So the average value of
some x, given a probability distribution on x where x
is a continuous variable, is going to be equal
to the integral. Let's just say x is
defined from minus infinity to infinity, which is pretty
useful, or pretty typical. dx probability
distribution of x times x. I shouldn't use curvy. I should just use x. And similarly for
x squared, or more generally, for f of x,
the average value of f of x, or the expected value of
f of x given this probability distribution, is going to
be equal to the integral dx minus infinity to infinity. The probability distribution
of x times f of x. In direct analogy to
what we had before. So this is all just mathematics. And we define the
uncertainty in x is equal to the expectation
value of x squared minus the expected value
of x quantity squared. And this is delta x squared. If you see me dropping an
exponent or a factor of 2, please, please, please tell me. So thank you for that. All of that is just straight up
classical probability theory. And I just want to write this
in the notation of quantum mechanics. Given that the
system is in a state described by the wave function
psi of x, the average value, the expected value of x, the
typical value if you just observe the particle
at some moment, is equal to the integral over
all possible values of x. The probability distribution,
psi of x norm squared x. And similarly, for
any function of x, the expected value is going to
be equal to the integral dx. The probability distribution,
which is given by the norm squared of the wave function
times f of x minus infinity to infinity. And same definition
for uncertainty. And again, this notation
is really dangerous, because the expected value of
x depends on the probability distribution. In a physical system,
the expected value of x depends on what the
state of the system is, what the wave function
is, and this notation doesn't indicate that. So there are a couple of ways
to improve this notation. One of which is-- so this is,
again, a sort of side note. One way to improve
this notation x is to write the expected
value of x in the state psi, so you write psi as a subscript. Another notation that
will come back-- you'll see why this is a
useful notation later in the semester-- is
this notation, psi. And we will give meaning
to this notation later, but I just want to
alert you that it's used throughout books, and
it means the same thing as what we're talking about
the expected value of x given a particular state psi. OK? Yeah? AUDIENCE: To calculate the
expected value of momentum do you need to transform the-- PROFESSOR: Excellent question. Excellent, excellent question. OK, so the question
is, how do we do the same thing for momentum? If you want to compute the
expected value of momentum, what do you have to do? Do you have to do some Fourier
transform to the wave function? So this is a
question that you're going to answer
on the problem set and that we made a
guess for last time. But quickly, let's
just think about what it's going to be
purely formally. Formally, if we want to know the
likely value of the momentum, the likely value the momentum,
it's a continuous variable. Just like any other
observable variable, we can write as the integral
over all possible values of momentum from,
let's say, it could be minus infinity to infinity. The probability of having that
momentum times momentum, right? Everyone cool with that? This is a tautology, right? This is what you
mean by probability. But we need to know if we have
a quantum mechanical system described by state
psi of x, how do we can get the probability
that you measure p? Do I want to do this now? Yeah, OK I do. And we need a guess. Question mark. We made a guess at the
end of last lecture that, in quantum
mechanics, this should be dp minus infinity to infinity
of the Fourier transform. Psi tilde of p up
to an h bar factor. Psi tilde of p, the Fourier
transform p norm squared. OK, so we're guessing that the
Fourier transform norm squared is equal to the probability
of measuring the associated momentum. So that's a guess. That's a guess. And so on your problem set
you're going to prove it. OK? So exactly the same
logic goes through. It's a very good
question, thanks. Other questions? Yeah? AUDIENCE: Is that p
the momentum itself? Or is that the probability? PROFESSOR: So this
is the probability of measuring momentum p. And that's the value p. We're summing over all p's. This is the probability,
and that's actually p. So the Fourier
transform is a function of the momentum in the same
way that the wave function is a function of
the position, right? So this is a function
of the momentum. It's norm squared
defines the probability. And then the p on
the right is this p, because we're computing
the expected value of p, or the average value of p. That make sense? Cool. Yeah? AUDIENCE: Are we then
multiplying by p squared if we're doing all p's? Because we have the dp times
p for each [INAUDIBLE]. PROFESSOR: No. So that's a very good question. So let's go back. Very good question. Let me phrase it in
terms of position, because the same
question comes up. Thank you for asking that. Look at this. This is weird. I'm going to phrase this as a
dimensional analysis question. Tell me if this is the same
question as you're asking. This is a thing with
dimensions of what? Length, right? But over on the
right hand side, we have a length and a probability,
which is a number, and then another length. That looks like
x squared, right? So why are we getting something
with dimensions of length, not something with
dimensions of length squared? And the answer is this
is not a probability. It is a probability density. So it's got units of
probability per unit length. So this has dimensions
of one over length. So this quantity,
p of x dx, tells me the probability, which is a
pure number, no dimensions. The probability to find the
particle between x and x plus dx. Cool? So that was our
second postulate. Psi of x dx squared
is the probability of finding it in this domain. And so what we're doing is we're
summing over all such domains the probability times the value. Cool? So this is the difference
between discrete, where we didn't have these
probability densities, we just had numbers, pure
numbers and pure probabilities. Now we have probability
densities per unit whatever. Yeah? AUDIENCE: How do you
pronounce the last notation that you wrote? PROFESSOR: How do you pronounce? Good, that's a good question. The question is, how do
we pronounce these things. So this is called
the expected value of x, or the average value of
x, or most typically in quantum mechanics, the
expectation value of x. So you can call it
anything you want. This is the same thing. The psi is just to denote
that this is in the state psi. And it can be
pronounced in two ways. You can either say
the expectation value of x, or the expectation
of x in the state psi. And this would be
pronounced one of two ways. The expectation value of x in
the state psi, or psi x psi. Yeah. That's a very good question. But they mean the same thing. Now, I should emphasize that you
can have two ways of describing something that mean
the same thing, but they carry different
connotations, right? Like have a friend
who's a really nice guy. He's a mensch. He's a good guy. And so I could see
he's a nice guy, I could say he's
[? carinoso ?], and they mean different things
in different languages. It's the same idea, but they
have different flavors, right? So whatever your
native language is, you've got some analog of this. This means something in
a particular mathematical language for talking
about quantum mechanics. And this has a different flavor. It carries different
implications, and we'll see what
that is later. We haven't got there yet. Yeah? AUDIENCE: Why is there a
double notation of psi? PROFESSOR: Why is there
a double notation of psi? Yeah, we'll see later. Roughly speaking, it's because
in computing this expectation value, there's a psi squared. And so this is to
remind you of that. Other questions? Terminology is one of the
most annoying features of quantum mechanics. Yeah? AUDIENCE: So it seems like
this [INAUDIBLE] variance is a really convenient
way of doing it. How is it the
Heisenberg uncertainty works exactly as it does for
this definition of variance. PROFESSOR: That's a
very good question. In order to answer
that question, we need to actually work out
the Heisenberg uncertainty relation. So the question is, look, this
is some choice of uncertainty. You could have chosen some
other definition of uncertainly. We could have considered
the expectation value of x to the fourth
minus x to the fourth and taken the
fourth root of that. So why this one? And one answer is, indeed,
the uncertainty relation works out quite nicely. But then I think
important to say here is that there are many ways
you could construct quantities. This is a convenient
one, and we will discover that it has nice
properties that we like. There is no God given reason why
this had to be the right thing. I can say more, but I don't
want to take the time to do it, so ask in office hours. OK, good. The second part of
your question was why does the Heisenberg
relation work out nicely in terms of
these guys, and we will study that in
extraordinary detail. We'll see that. So we're going to derive it
twice soon and then later. The later version is better. So let me work
out some examples. Or actually, I'm going
to skip the examples in the interest of time. They're in the
notes, and so they'll be posted on the web page. By the way, the first 18
lectures of notes are posted. I had a busy night last night. So let's come back to
computing expectation values for momentum. So I want to go
back to this and ask a silly-- I want to make some
progress towards deriving this relation. So I want to start over on
the definition of the expected value of momentum. And I'd like to do it directly
in terms of the wave function. So how would we do this? So one way of saying this is
what's the average value of p. Well, I can phrase this
in terms of the wave function the following way. I'm going to sum over
all positions dx. Expectation value of x
squared from minus infinity to infinity. And then the momentum
associated to the value x. So it's tempting to write
something like this down to think maybe
there's some p of x. This is a tempting
thing to write down. Can we? Are we ever in a position
to say intelligently that a particle--
that an electron is both hard and white? AUDIENCE: No. PROFESSOR: No,
because being hard is a superposition of being
black and white, right? Are we ever in a position
to say that our particle has a definite position
x and correspondingly a definite momentum p. It's not that we don't get too. It's that it doesn't
make sense to do so. In general, being in
a definite position means being in a
superposition of having different values for momentum. And if you want a sharp
way of saying this, look at these relations. They claim that any
function can be expressed as a superposition of states
with definite momentum, right? Well, among other things a
state with definite position, x0, can be written
as a superposition, 1 over 2pi integral dk. I'll call this delta tilde of k. e to the ikx. If you haven't played with
delta functions before and you haven't seen this,
then you will on the problem set, because we
have a problem that works through a
great many details. But in particular, it's
clear that this is not-- this quantity can't be
a delta function of k, because, if it were, this
would be just e to the ikx. And that's definitely
not a delta function. Meanwhile, what can you say
about the continuity structure of a delta function. Is it continuous? No. Its derivative isn't continuous. Its second derivative. None of its derivatives
are in any way continuous. They're all absolutely
horrible, OK? So how many momentum
modes am I going to need to superimpose in order
to reproduce a function that has this sort of structure? An infinite number. And it turns out
it's going to be an infinite number with the
same amplitude, slightly different phase, OK? So you can never say
that you're in a state with definite position
and definite momentum. Being in a state with
definite position means being in a superposition
of being in a superposition. In fact, I'm just going
right down the answer here. e to the ikx0. Being in a state with
definite position means being in a
superposition of states with arbitrary momentum
and vice versa. You cannot be in a state
with definite position, definite momentum. So this doesn't work. So what we want is we
want some good definition. So this does not work. We want some good
definition of p given that we're working
with a wave function which is a function of x. What is that good
definition of the momentum? We have a couple of hints. So hint the first. So this is what we're after. Hint the first is
that a wave-- we know that given a wave with wave
number k, which is equal 2pi over lambda, is associated,
according to de Broglie and according to
Davisson-Germer experiments, to a particle-- so
having a particle-- a wave, with wave number k or
wavelength lambda associated particle with momentum
p is equal to h bar k. Yeah? But in particular, what is a
plane with wavelength lambda or wave number k look like? That's e to the iks. And if I have a
wave, a plane wave e to the iks, how do I
get h bar k out of it? Note the following, the
derivative with respect to x. Actually let me
do this down here. Note that the derivative with
respect to x of e to the ikx is equal to ik e to the ikx. There's nothing up my sleeves. So in particular, if
I want to get h bar k, I can multiply by h
bar and divide by i. Multiply by h bar, divide by
i, derivative with respect to x e to the ikx. And this is equal to
h bar k e to the ikx. That's suggestive. And I can write this
as p e to the ikx. So let's quickly
check the units. So first off, what are
the units of h bar? Here's the super
easy to remember the units of-- or
dimensions of h bar are. Delta x delta p is h bar. OK? If you're ever in doubt,
if you just remember, h bar has units of
momentum times length. It's just the easiest
way to remember it. You'll never forget it that way. So if h bar has units of
momentum times length, what are the units of k? 1 over length. So does this
dimensionally make sense? Yeah. Momentum times length divided
by length number momentum. Good. So dimensionally we
haven't lied yet. So this makes it tempting
to say something like, well, hell h bar upon i
derivative with respect to x is equal in some-- question
mark, quotation mark-- p. Right? So at this point it's just
tempting to say, look, trust me, p is h bar upon idx. But I don't know
about you, but I find that deeply,
deeply unsatisfying. So let me ask the question
slightly differently. We've followed the
de Broglie relations, and we've been led
to the idea that using wave functions
that there's some relationship
between the momentum, the observable quantity that
you measure with sticks, and meters, and stuff, and this
operator, this differential operator, h bar upon on i
derivative with respect to x. By the way, my notation for
dx is the partial derivative with respect to x. Just notation. So if this is supposed
to be true in some sense, what is momentum have
to do with a derivative? Momentum is about
velocities, which is like derivatives with
respect to time, right? Times mass. Mass times derivative with
respect to time, velocity. So what does it have to do with
the derivative with respect to position? And this ties into the
most beautiful theorem in classical mechanics, which
is the Noether's theorem, named after the mathematician who
discovered it, Emmy Noether. And just out of
curiosity, how many people have seen Noether's
theorem in class. Oh that's so sad. That's a sin. OK, so here's a statement
of Noether's theorem, and it underlies an enormous
amount of classical mechanics, but also of quantum mechanics. Noether, incidentally,
was a mathematician. There's a whole wonderful
story about Emmy Noether. Ville went to her
and was like, look, I'm trying to understand
the notion of energy. And this guy down the
hall, Einstein, he has a theory called general
relativity about curved space times and how that has
something to do with gravity. But it doesn't
make a lot of sense to me, because I don't even
know how to define the energy. So how do you define
momentum and energy in this guy's crazy theory? And so Noether, who
was a mathematician, did all sorts of beautiful
stuff in algebra, looked at the problem
and was like I don't even know what it means in
classical mechanics. So what is a mean in
classical mechanics? So she went back to
classical mechanics and, from first
principles, came up with a good definition of
momentum, which turns out to underlie the modern idea
of conserved quantities and symmetries. And it's had enormous
far reaching impact, and say her name would praise. So Noether tells us the
following statement, to every symmetry-- and I
should say continuous symmetry-- to every symmetry is associated
a conserved quantity. OK? So in particular, what
do I mean by symmetry? Well, for example, translations. x goes to x plus some length l. This could be done for
arbitrary length l. So for example, translation
by this much or translation by that much. These are translations. To every symmetry is associated
a conserved quantity. What symmetry is
associated to translations? Conservation of momentum, p dot. Time translations, t
goes to t plus capital T. What's a conserved
quantity associated with time
translational symmetry? Energy, which is
time independent. And rotations. Rotational symmetries. x, as a vector, goes to
some rotation times x. What's conserved by virtue
of rotational symmetry? AUDIENCE: Angular momentum. PROFESSOR: Angular momentum. Rock on. OK So quickly, I'm not going to
prove to you Noether's theorem. It's one of the most beautiful
and important theorems in physics, and you
should all study it. But let me just
convince you quickly that it's true in
classical mechanics. And this was observed long
before Noether pointed out why it was true in general. What does it mean to have
transitional symmetry? It means that, if I
do an experiment here and I do it here, I get
exactly the same results. I translate the system
and nothing changes. Cool? That's what I mean by
saying I have a symmetry. You do this thing,
and nothing changes. OK, so imagine I have a
particle, a classical particle, and it's moving
in some potential. This is u of x, right? And we know what the
equations of motion are in classical
mechanics from f equals ma p dot is equal to
the force, which is minus the gradient of u. Minus the gradient of u. Right? That's f equals ma in
terms of the potential. Now is the gradient of u 0? No. In this case, there's a force. So if I do an experiment
here, do I get the same thing as doing my experiment here? AUDIENCE: No. PROFESSOR: Certainly not. The [? system ?] is not
translationally invariant. The potential breaks that
translational symmetry. What potential has
translational symmetry? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, constant. The only potential that has full
translational symmetry in one dimension is translation
invariant, i.e. constant. OK? What's the force? AUDIENCE: 0. PROFESSOR: 0. 0 gradient. So what's p dot? Yep. Noether's theorem. Solid. OK. Less trivial is
conservation of energy. I claim and she claims--
and she's right-- that if the system has
the same dynamics at one moment and a few moments
later and, indeed, any amount of time later,
if the laws of physics don't change in
time, then there must be a conserved
quantity called energy. There must be a
conserved quantity. And that's Noether's theorem. So this is the first
step, but this still doesn't tell us what
momentum exactly has to do with a derivative
with respect to space. We see that there's
a relationship between translations and
momentum conservation, but what's the relationship? So let's do this. I'm going to define an
operation called translate by L. And what translate
by L does is it takes f of x and it maps
it to f of x minus L. So this is a thing that
affects the translation. And why do I say that's
a translation by L rather than minus L. Well,
the point-- if you have some function like this,
and it has a peak at 0, then after the translation, the
peak is when x is equal to L. OK? So just to get the
signs straight. So define this operation,
which takes a function of x and translates it by L, but
leaves it otherwise identical. So let's consider
how translations behave on functions. And this is really cute. f of x minus L can be
written as a Taylor expansion around the point x-- around
the point L equals 0. So let's do Taylor
expansion for small L. So this is equal to f of x
minus L derivative with respect to x of f of x plus L squared
over 2 derivative squared, two derivatives of x, f
of x plus dot, dot, dot. Right? I'm just Taylor expanding. Nothing sneaky. Let's add the next
term, actually. Let me do this on
a whole new board. All right, so we
have translate by L on f of x is equal to f of x
minus L is equal to f of x. Now Taylor expanding
minus L derivative with respect to x of f
plus L squared over 2-- I'm not giving
myself enough space. I'm sorry. f of x minus L is equal to
f of x minus L with respect to x of f of x plus L
squared over 2 to derivatives of x f of x minus
L cubed over 6-- we're just Taylor expanding--
cubed with respect to x of f of x and so on. Yeah? But I'm going to write this in
the following suggestive way. This is equal to 1 times
f of x minus L derivative with respect to x f
of x plus L squared over 2 derivative with
respect to x squared times f of x minus L cubed over 6
derivative cubed with respect to x plus dot, dot, dot. Everybody good with that? But this is a series that
you should recognize, a particular Taylor series
for a particular function. It's a Taylor expansion for the AUDIENCE: Exponential. PROFESSOR: Exponential. e to the minus L derivative
with respect to x f of x. Which is kind of awesome. So let's just check to make
sure that this makes sense from dimensional grounds. So that's a derivative with
respect to x as units of 1 over length. That's a length, so
this is dimensionless, so we can exponentiate it. Now you might look at me and
say, look, this is silly. You've taken an
operation like derivative and exponentiated it. What does that mean? And that is what it means? [LAUGHTER] OK? So we're going to do this all
the time in quantum mechanics. We're going to do things
like exponentiate operations. We'll talk about
it in more detail, but we're always
going to define it in this fashion as a
formal power series. Questions? AUDIENCE: Can you
transform operators from one space to another? PROFESSOR: Oh, you totally can. But we'll come back to that. We're going to talk about
operators next time. OK, so here's where we are. So from this what
is a derivative with respect to x mean? What does a derivative
with respect to x do? Well a derivative
with respect to x is something that generates
translations with respect to x through a Taylor expansion. If we have L be
arbitrarily small, right? L is arbitrarily small. What is the translation by
an arbitrarily small amount of f of x? Well, if L is
arbitrarily small, we can drop all the
higher order terms, and the change is just Ldx. So the derivative
with respect to x is telling us about
infinitesimal translations. Cool? The derivative with
respect to a position is something that
tells you, or controls, or generates infinitesimal
translations. And if you exponentiate
it, you do it many, many, many times in
a particular way, you get a macroscopic
finite translation. Cool? So this gives us three things. Translations in x are generated
by derivative with respect to x. But through Noether's
theorem translations, in x are associated to
conservation of momentum. So you shouldn't be so
shocked-- it's really not totally shocking-- that in
quantum mechanics, where we're very interested in the action
of things on functions, not just in positions, but
on functions of position, it shouldn't be totally shocking
that in quantum mechanics, the derivative with
respect to x is related to the momentum in
some particular way. Similarly, translations
in t are going to be generated
by what operation? Derivative with respect to time. So derivative with respect to
time from Noether's theorem is associated with
conservation of energy. That seems plausible. Derivative with respect
to, I don't know, an angle, a rotation. That's going to be
associated with what? Angular momentum? But angular momentum
around the axis for whom this is
the angle, so I'll call that z for the moment. And we're going to see these
pop up over and over again. But here's the thing. We started out with these
three principles today, and we've let ourselves to
some sort of association between the momentum and
the derivative like this. OK? And I've given you
some reason to believe that this isn't totally insane. Translations are
deeply connected with conservation of momentum. Transitional symmetry
is deeply connected with conservation momentum. And an infinitesimal
translation is nothing but a derivative
with respect to position. Those are deeply
linked concepts. But I didn't derive anything. I gave you no
derivation whatsoever of the relationship between
d dx and the momentum. Instead, I'm simply
going to declare it. I'm going to declare that,
in quantum mechanics-- you cannot stop me--
in quantum mechanics, p is represented by an
operator, it's represented by the specific operator h bar
upon I derivative with respect to x. And this is a declaration. OK? It is simply a fact. And when they say it's a fact,
I mean two things by that. The first is it is a fact
that, in quantum mechanics, momentum is represented
by derivative with respect to x times h bar upon i. Secondly, it is a fact that,
if you take this expression and you work with the rest
of the postulates of quantum mechanics, including
what's coming next lecture about operators
and time evolution, you reproduce the physics
of the real world. You reproduce it beautifully. You reproduce it so well that
no other models have even ever vaguely come close to the
explanatory power of quantum mechanics. OK? It is a fact. It is not true in
some epistemic sense. You can't sit back
and say, ah a priori starting with the integers we
derive that p is equal to-- no, it's a model. But that's what physics does. Physics doesn't tell
you what's true. Physics doesn't tell
you what a priori did the world have to look like. Physics tells you
this is a good model, and it works really well,
and it fits the data. And to the degree that
it doesn't fit the data, it's wrong. OK? This isn't something we derive. This is something we declare. We call it our model, and then
we use it to calculate stuff, and we see if it
fits the real world. Out, please, please leave. Thank you. [LAUGHTER] I love MIT. I really do. So let me close
off at this point with the following observation. [LAUGHTER] We live in a world
governed by probabilities. There's a finite probability
that, at any given moment, that two pirates might
walk into a room, OK? [LAUGHTER] You just never know. [APPLAUSE] But those probabilities can be
computed in quantum mechanics. And they're computed
in the following ways. They're computed
the following ways as we'll study in great detail. If I take a state, psi of x,
which is equal to e to the ikx, this is a state that has
definite momentum h bar k. Right? We claimed this. This was de Broglie
and Davisson-Germer. Note the following,
take this operator and act on this wave
function with this operator. What do you get? Well, we already
know, because we constructed it to
have this property. P hat on psi of
x-- and I'm going to call this psi sub
k of x, because it has a definite k-- is equal
to h bar k psi k of x. A state with a definite
momentum has the property that, when you hit it with
the operation associated with momentum, you get back the
same function times a constant, and that constant is
exactly the momentum we ascribe to that plane wave. Is that cool? Yeah? AUDIENCE: Question. Just with notation,
what does the hat above the p [INAUDIBLE]? PROFESSOR: Good. Excellent. So the hat above the
P is to remind you that P is on a number. It's an operation. It's a rule for
acting on functions. We'll talk about that in
great detail next time. But here's what I
want to emphasize. This is a state which is equal
to all others in the sense that it's a perfectly
reasonable wave function, but it's more equal because it
has a simple interpretation. Right? The probability that I measure
the momentum to be h bar k is one, and the
probability that I measure it to be anything
else is 0, correct? But I can always consider a
state which is a superposition. Psi is equal to alpha, let's
just do 1 over 2 e to the ikx. k1 x plus 1 over root
2 e to the minus ikx. Is this state a state
with definite momentum? If I act on this
state-- I'll call this i sub s-- if I act on this state
with the momentum operator, do I get back this
state times a constant? No. That's interesting. And so it seems to
be that if we have a state with definite momentum
and we act on it with momentum operator, we get
back its momentum. If we have a state
that's a superposition of different momentum and we act
on it with a momentum operator, this gives us h bar k 1,
this gives us h bar k2. So it changes
which superposition we're talking about. We don't get back
our same state. So the action of this
operator on a state is going to tell us something
about whether the state has definite value of the momentum. And these coefficients
are going to turn out to contain all the information
about the probability of the system. This is the
probability when norm squared that will measure the
system to have momentum k1. And this coefficient
norm squared is going to tell us
the probability that we have momentum k2. So I think the
current wave function is something like a
superposition of 1/10 psi pirates plus 1 minus
is 1/100 square root. To normalize it
properly psi no pirates. And I'll leave you with
pondering this probability. See you guys next time. [APPLAUSE] CHRISTOPHER SMITH:
We've come for Prof. Allan Adams. PROFESSOR: It is I. CHRISTOPHER SMITH: When in
the chronicles of wasted time, I see descriptions
of fairest rights, and I see lovely
shows of lovely dames. And descriptions of ladies
dead and lovely nights. Then in the bosom of
fair loves depths. Of eyes, of foot,
of eye, of brow. I see the antique pens
do but express the beauty that you master now. So are all their praises but
prophecies of this, our time. All you prefiguring. But though they had
but diving eyes-- PROFESSOR: I was wrong
about the probabilities. [LAUGHTER] CHRISTOPHER SMITH:
But though they had but diving eyes,
they had not skill enough you're worth to sing. For we which now behold
these present days have eyes to behold. [LAUGHTER] But not tongues to praise. [APPLAUSE] It's not over. You wait. ARSHIA SURTI: Not marbled with
gilded monuments of princes shall outlive this
powerful rhyme. But you shall shine more
bright in these contents that unswept stone
besmear its sluttish tide. When wasteful war shall
statues overturn and broils root out the work of masonry. Nor Mars his sword. Nor war's quick fire shall
burn the living record of your memory. Gainst death and all oblivious
enmity shall you pace forth. Your praise shall
still find room, even in the eyes
of all posterity. So no judgment arise till
you yourself judgment arise. You live in this and
dwell in lover's eyes. [APPLAUSE] CHRISTOPHER SMITH: Verily
happy Valentine's day upon you. May your day be filled
with love and poetry. Whatever state you're in,
we will always love you. [LAUGHTER] [APPLAUSE] Signed, Jack Florian,
James [INAUDIBLE]. [LAUGHTER] PROFESSOR: Thank you, sir. Thank you. CHRISTOPHER SMITH: Now we go. [APPLAUSE]
