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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So, we'll get started. And as I mentioned,
to some degree this is going to be review on
the setting of our notation and conventions clear. So, our first topic is
the Schrodinger equation. So this Schrodinger
equation is an equation that takes the following form. I h bar partial
derivative of this object called the wave function
that depends on x and t is equal to minus h squared
over 2m second derivative with respect to x plus v
of x and t Psi of x and t. And that's the full equation. That's the Schrodinger equation. Now actually, this is not
the Schrodinger equation in most generality, but it's
the Schrodinger equation for the case that you
have a potential that depends on x and t. For the case that we are doing
non-relativistic physics, because this thing you may
remember is p squared over 2m is the kinetic energy operator. So p squared over 2m
is non-relativistic. That's a non-relativistic
kinetic energy. So this is non-relativistic. Moreover, we have
just one x here. That means it's a
particle in one dimension. So we've done a few things,
but this is generally enough to illustrate our ideas. And the most
important thing that should be said at
this point is that Psi of x and t-- which is
the wave function-- belongs to the complex numbers. It's a complex number. And that's by necessity. If Psi would be real, this
quantity-- the right hand side-- would be real. The potential is a real number. On the left hand side, on the
other hand, if Psi is real, its derivative would be real,
and this would be imaginary. So, it's just impossible to get
the solution of this equation if Psi is real. So, Psi complex is really
the fundamental thing that can be said about
this wave function. Now, you've used complex
numbers in physics all the time, and even in electromagnetism,
you use complex numbers. But you use them really
in an auxiliary way only. You didn't use them in an
absolutely necessary way. So, for example. In E&M, you had an electric
field, for example, for a circularly polarized wave. And you would write it as this. Let me put the z here. Zero. X hat plus y hat--
those are unit vectors. I is a complex number. It's the square root of minus 1. E to the IKZ minus omega t. You typically wrote
things like that, but, in fact, you
always meant real part. An electric field
is a real quantity. And the Maxwell's equations
are real equations. This is a circularly
polarized wave. And this whole
thing-- by the time you take the real part of
this, all these complex numbers play absolutely no role. It's just a neat way of writing
a complicated electric field in which the x component and the
y component are out of phase, and that you have a wave at
the same time propagating in the z direction. So this-- in the
here, E is real, and all i's are auxiliary. This is completely
different from the case of the Schrodinger equation. This i there is fundamental. The Psi is the
dynamical variable, and it has to be complex. So, we make a few remarks
about the Schrodinger equation to get started. First remark is that this
is first order differential equation in time. This has implications. Those two derivatives
are maybe-- for some funny
Hamiltonians, you can have even more than two
derivatives or more complicated things. But definitely there's just
one derivative in time. So, what this means
is that if you know the wave function
all over space, you can calculate
what it's going to be a little time later. Because if you know
it all over space, you can calculate
this right hand side and know what is
the time derivative. And with the time
derivative, you can figure it out what
it's going to be later. A first order differential
equation in time is something that if you know
the quantity at one time, the differential
equation tells you what it's going to be later. So, that's really sufficient. Psi of x-- of all x's-- at some
time t naught determines Psi at all times. Second property,
fundamental property. The equation is linear. So, if you have
two solutions, you can form a third by
superimposing them, and you can superimpose them
with complex coefficients. So, if you have two
solutions, Psi 1 and Psi 2, then a 1 Psi 1 plus A2
Psi 2 is a solution. And here the a's belong
to the complex numbers. So A 1 and A 2 are
complex numbers. As far as complex
numbers are concerned, the first thing you
just need to know is the definition of the
length of a complex number. So, if you have z, a
typical name people use for a complex number,
having two components. A plus ib, where
a and b are real. There's the definition of
the complex conjugate, which is a minus ib, and
there's the definition of the length of the
complex number, which is square root of a
squared plus b squared, which is the square
root of z times z star. So, that's for your
complex number. So, the property that this makes
this into a physical theory and goes beyond math
is what you know is the interpretation of the
wave function as a probability. So, what do we construct? We construct p of x and
t, or sometimes called the row of x and t as a density. And it's defined
as Psi star of x t. Now, here the notation
means this Psi star-- we'd put the star
here-- it really means Psi of x and
t complex conjugate. You complex conjugate
the wave function. And you get that. We'd put the star here,
and typically don't put the parentheses, unless
you have to complex conjugate something that's a
little ambiguous. So, Psi star of x and
t times Psi of x and t. And this is called the
probability density. Probability density. And the interpretation is
that if you take p of x and t and multiply by little dx,
this is the probability to find the particle in the
interval x comma x plus dx at time t. So, this is our
probability density. It's a way to make physics
out of the wave function. It's a postulate. And so the consequence
of this postulate, since we're describing
just one particle, is that we must have the
particle as somewhere. So, if we add the probabilities
that the particle is somewhere all over space, this
is the probability that the particle is in
this little dx we integrated that must be equal to 1. And this must hold
for all times. In terms of things
to notice here, maybe one thing you can
notice is the units of Psi. The units of Psi must be 1
over square root of length, because when we square it,
then we multiply it by length, we get one, which has no units. Key property of the
Schrodinger equation. We will revisit the Schrodinger
equation later and derive it, sort of the way [? De ?]
[? Rack ?] derives it in his textbook. As just a consequence of
unitary time evolution, it would be a very
neat derivation. It will give you a
feeling that you really understand something deep
about quantum mechanics. And it will be
true, that feeling. But here, we're going to
go the other way around. Just simply ask the
question-- suppose you have a wave function
such that the integral of this quantity at some
specific time is equal to one. Will this integral be
equal to one for all times, given that it is one
at some given time? Now, you say, well,
why do you ask that? I ask that because actually
this could be a problem. We've said that if you know the
wave function all over space at one time, it's
determined everywhere. So any time later. Therefore, if I know the wave
function at time equal zero is good-- time equal t zero--
is a good wave function, I might warranty that when I
saw the Schrodinger equation, the wave function will be
normalized, well, later? Yes, you are. And it's a simple or
interesting exercise that we'll call it
the quick calculation that I'll leave
it for you to do. Which is show that d dt of
this integral Psi of x and t squared dx is equal to zero. So, basically what
this is saying. You got one but, think of
this integral-- I'm sorry, I'm missing a dx here-- think
of this integral for all times. Now it could be a
function of time, because you put an
arbitrary time here. The integral might
depend on time. So, it's a good question to
think of that integral that may be a function of time
and take its derivative. If its derivative is
zero for all times, and that sometimes equal to
one, it will be one forever. So, you must show
that this is true. Now, this I think you've done
one way or another several ways maybe in 804. But I ask you to do it again. So this is left for you as a
way to warm up on this object. And you will see actually
that it's a little subtle. It's a little delicate,
because how is it going to go? You're going to go in and take
the derivative of Psi Psi star. You're going to take
the derivative of Psi and you're going to use
the Schrodinger equation. You're going to take the
derivative of Psi star, and you're going to use
the complex conjugate of the Schrodinger equation. It's going to be a little messy. But then you're going to
do integration by parts, and you're going to get zero,
but only if you throw away the terms at infinity. And what gives you the
right to throw them away? You will have to think. And the answer is that
you will throw them away if the wave function
goes to zero at infinity, which must do it. The wave function must
go to zero at infinity, because if it didn't
go to zero at infinity, it went to a
constant at infinity, it would pick up an
un-normalizable thing here. So the wave function definitely
has to go to zero at infinity. But that will also
not be quite enough if you're careful about
what you're doing. You will have to demand that the
derivative of the wave function doesn't blow up. It's not asking too much,
but it's asking something. A function could go
to zero, presumably, and its derivative at
the same time blow up, but it would be a very
pathological function. This will bring us to
something that we said. We're going to
try to be precise, but it's not so
easy to be precise. When you try to be
precise, you can exaggerate and go
precise to a point that you're paralyzed with
fear with every equation. We don't want to get that far. We want you to notice what
happens and just look at it and state what you need. Why can't we be precise? Because at the end of
the day, this equation is extraordinarily
complicated, and maybe crazy. The potential is crazy enough. So, functions-- mathematicians
can invent crazy functions, things like a function that is
one for every rational number and zero for every
rational number. Put that for a potential here,
and who knows what one gets. So, we're going to
take mild functions. We're not going to make
them a very complicated, and we're going to be stating
very soon what we need. So, what you need
for this to work is that the function
goes to zero and the relative goes to zero. Yes. AUDIENCE: The potential
has to be real always? PROFESSOR: The potential
is real at this moment. Yes. For the discussion
that we're doing here, v is also a real number. AUDIENCE: So it
can't be complex? PROFESSOR: Sorry? AUDIENCE: Can it be complex? PROFESSOR: It could be
in certain applications for particles in
electromagnetic fields. You can have
something that looks like a complex Hamiltonian. So we will not discuss that
in this couple of lectures, but maybe later. Yes. AUDIENCE: Are there
any conditions that the potential has
to be time-dependent? PROFESSOR: Well, at this
moment, I put it time dependent. Also, it complicated
potentials, but they're sometimes necessary. And we will discuss
some of them. We will have very simple
time dependencies. Otherwise, it's difficult
to solve this equation. But very soon-- in
about five minutes, I will say-- let's consider
time-independent things to review the things that are a
little more basic and important and that you should
definitely remember well. OK, so that's this part of
the Schrodinger equation. I want to remind you of
another concept called the current--
probability current. Probability current. What is it? It's a j of x and
t-- that you will review in the homework--
is given by h over m, the imaginary part of
Psi star d Psi over dx. So, it's a real quantity. And it's called a
probability current. And it goes together with
this probability density, this probability density
that we wrote over here. So it's the current
associated to that density. Let's think a second
what this means. In electromagnetism, you have
currents and charged densities. So in E&M, you have a current. It's a vector and
a charged density. Now, this current
could also be a vector. If you're working in
more than one dimension, it would be a vector. But if you have
electromagnetism, the most famous thing associated
to electromagnetism currents and charged densities is the
so-called conservation law. This differential
equations satisfied by the current and the density. Divergence of j plus d
Rho dt is equal to zero. That means charge conservation. You may or may
not remember that. If you don't, it's a good
time to review it in E&M and check on that,
discuss it in recitation. Think about it. This means charge conservation
as we usually understand, and the way to do it--
I'm saying just in words-- is you think of
a volume, you can see how much charge
is inside, and you see that the rate of
change of the charge is proportional to the current
that is escaping the volume. Which is to say, charge is
never destroyed or created. It can escape a volume,
because the charges are moving, but if it doesn't escape, well,
the charge remains the same. So, this is charge conservation. And this is the same thing. So the divergence
of j in this case reduces to dj dx plus
d Rho dt equals zero. It has a very similar
interpretation. So, perhaps in
equations, it's easier to think of interpretation. Consider the real line
and the point a and b, with a less than b. And define the
probability pab of t of finding the particle in
this interval between a and b at any time. You should be able to
show-- and it's again another thing to review. This you can review. And this review as well. You will use this differential
equation, things like that, to show that dpab dt-- the
rate at which the probability that you find the particle
in this interval changes depends on what the
current is doing here and what the current
is doing here. So, it's actually given by j of
a and t minus j at b at time t. You can show, and
please try to show it. So, what does that mean? You can have the particle
here at any time. But if you want to know how
the probability changing, you must see how
it's leaking from a or how it's leaking from b. Now j's are defined,
by convention, positive to the right. So, if there's a current--
a bit of current at a, it increases the probability. This particle is sort of
moving into the interval. And here at b, there's a
positive current decreases the probability. Finally, for wave functions,
the last thing we say is that these wave functions
are-- you want them normalized, but we can work with
them and they're physically equivalent if they
differ just by a constant. So Psi 1 and Psi 2 are
said to be equivalent if Psi 1 of x and t is equal to
some complex constant of Psi 2 of x and t. Now, you would say,
well, I don't like that. I like normalized
wave functions, and you could have
a point there. But even if these are
normalized functions, they could differ by a phase. And they would still be
physically equivalent. This part of the definition
of the theory-- the definition of the theory is that these
wave functions are really physically equivalent
and indistinguishable. And that puts a constraint on
the way we define observables. Any observable should
have this property that, whether we used this
wave function or the other, they give you the
same observables. So, if your wave
functions are normalized, this can be complex
constant of length one. Then one normalized implies
the other is normalized. If they're not normalized, you
can say, look, the only reason I'm not normalizing
it because I don't gain all that much by
normalizing it, in fact. I can do almost everything
without normalizing the wave function. So, why should I bother? And we'll explain that
also as well very soon. So, this is something
that this part of the physical interpretation
that we should keep. So, now we've reviewed
the Schrodinger equation. Next thing we want to say is
the most important solutions of the Schrodinger equations
are those energy Eigenstates, stationary states. And let's just go
through that subject and explain what it was. So, I'm going to
start erasing here. So we're going to look at--
whoops-- stationary solutions. Now, I've used this
week wave function with a capital
Psi for a purpose, because I want to distinguish
it from another Psi that we're going to
encounter very soon. So, stationary solutions. And we'll take it-- from now
assume v is time-independent. The case is
sufficiently important that we may as well do it. So, in this case, the
Schrodinger equation is written as I h bar
d Psi dt, and we'll write it with something
called an h hat acting on Psi. And h hat at this
point is nothing else than minus h squared over 2m
second derivative with respect to x plus v of x. We say that h hat is an operator
acting on the wave function Psi on the right. Operator acting on that--
what does that mean? Basically, when we say an
operator acts on some space, we mean that it takes
elements of that space and moves them
around in the space. So, you've got a wave function,
which is a complex number that depends on x and t
ultimately, and then you act with this thing, which
involves taking derivatives, multiplying by v
of x, and you still got some complex
function of x and t. So, this is called the
Hamiltonian operator, and it's written like that. This Hamiltonian operator
is time-independent. So, what is a stationary state? A stationary state-- the way
it's defined is as follows. A stationary state of energy
e-- which is a real number-- is a Psi of x and t
of the following form. It's a simple form. It's a pure exponential in
time times a function that just depends on x. So, it's a pretty simple object. So what is it? We say that this is
a stationary state. e to the minus i Et
over H bar Psi of x. And this Psi is in purpose
different from this Psi. It doesn't have the
bar at the bottom, and that signals
to you that that's the time-independent one. So this also belongs
to the complex numbers, but doesn't depend on time. So, it's called stationary
because, as it turns out, when we will compute expectation
values of any observable on this state, in
this stationary state, it will be time-independent. In particular, you
know, one thing that observable is the
probability density. And when you look at that,
you have Psi star and Psi. Since E is real,
this phase cancels-- this is really a face,
because E is real. Therefore, Psi star Psi, the
e cancels, and all the time dependence cancels
and goes away. Same thing here for the j. The x derivative over here
it doesn't do anything to that phase. Therefore, the phase e
to the i Et over H bar cancels from there two. And the current also
has no time dependence. So, this will be the case
for any operator that is called a
time-independent operator. It will have time-independent
expectation values. So you can ask anything about
some familiar operator-- energy operator, momentum operator,
angular momentum operator-- all the famous operators
of quantum mechanics, and it will have real
expectation values. So, as you, you're
supposed to now plug this into this equation. And it's a famous result. Let's just do it. Plug back into the top equation. So, we have I H bar. The DET will only
act on the phase, because the Psi has
no time-dependence. And on the other hand,
on the right hand side, the H has nothing
to do with time, and therefore it can slide
through the exponential until it hits Psi. So here we have H-- well,
I'll put the exponential in front-- H on little Psi. So, we multiply here,
and what do we get? Well, the H bars cancel. The i at minus i gives you one. You get that E in front. So you get E times
this phase Psi of x. And the phase is
supposed to be here, but I cancel it with
this phase as well. And I get on the
right hand side H Psi. I will put it as
a left hand Psi. And this is the time-independent
Schrodinger equation. So far this is really
a simple matter. We've written a solution
that will represent the stationary state,
but then this energy should be such that you
can solve this equation. And as you've
learned before, it's something not so easy
to solve that equation. So what do we want to
say about this equation? Well, we have a lot to
say, and a few things will be pointed out now
that are very important. So, we have a
differential equation now. This differential equation has
second derivatives with respect to x. Now it has no time derivatives. The time has been factored out. Time is not a problem anymore. This equation, in
fact, looks quite real in that it seems that
Psi could even be real here. And in fact, yes,
there's no problem with this Psi being real. The total Psi just can't
be real in general. But this one can be
a real, and we'll consider those cases as well. So, things that
we want to say is that this is a second
order differential equations in space. So second order differential
equation in space. You could write it here. The H operator has
partial derivatives, but this time time,
you might as well say that this is minus
h squared over 2m. The second Psi vx squared
plus v of x tines Psi of x. Because Psi only depends
on x, might as well write it as complete derivative. So, second order
differential equation. And therefore, the
strategy for this equation is a little out there in
relation to the Schrodinger equation. We said, in the
Schrodinger equation, we know the wave function
everywhere, you know it later. Here, if you know it at one
point-- the wave function-- and you know the derivative
at that one point, you have it everywhere. Why is that? Because that's how you solve
a differential equation. If you know the wave
function and the derivative at the point, you go to
the equation and say, I know the wave function and
I know the first derivative, and I know the
second derivative. So, a little later I can know
what the first derivative is, and if I know what the first
derivative is a little later, I can then know what the wave
function is a little later, and you just integrate
it numerically. So, you just need to know the
wave function Psi of x zero and Psi prime at x zero
suffice for a solution when v is regular. But this v is not too
complicated, or too strange, because you can always
find exceptions. You have the square
well potential, and you say, oh, I know
the wave function is here and its derivative is zero. Does that determine
the solution? No, because it's infinite. There's no space here, really,
and you should work here. So, basically, unless v
is really pathological, Psi and Psi prime are enough
to solve for everything. And that actually means
something very important, that if Psi is equal to zero
at x zero is equal to zero, and Psi prime at x zero
is equal to zero, then under these regular conditions,
Psi of all x is zero. Because you have a
differential equation which the initial value is zero,
the Psi prime is zero. And you go through
the equation, you see that every solution
has to be zero. It's the only possibility here. So what happens now is the
following-- that you have a physical understanding
that your wave function, when it becomes zero-- it may
do it slowly that it's becoming zero, but never quite being
zero-- but if it's zero, it does it with Psi prime
different from zero, so the wave function
is not zero all over. So, this is a pretty important
fact that is useful many times when you try to understand
the nature of solutions. So what else do we have here? Well, we have energy
Eigenstates on the spectrum. So, what is an
energy Eigenstate? Well, it's a solution
of this equation. So a solution Psi--
a solution for Psi is an energy Eigenstate. Then, this set of values
of E is this spectrum. And these two values
of E-- if there's a value of E that has
more than one solution, we say the spectrum
is degenerate. So a degenerate spectrum is
more than one Psi for a given E. So, these are just definitions,
but they're used all the time. So, our energy Eigenstates
are the solutions of this. The funny thing
about this equation is that sometimes
the requirement that Psi be normalized
means that you can't always find the solution
for any value of E. So, only specific
values of Es are allowed-- you know that for
the harmonic oscillator, for example-- and
therefore there's something called the spectrum,
which is the allowed values. And many times you
have degeneracies, and that makes for very
interesting physics. Let's say a couple more things
about the nature of this wave function. So, what kind of
potentials do we allow? We will allow potentials
that can fail to be bounded. What do we allow? We allow failure of continuity. Certainly, we must allow
that in our potentials that we consider,
because you have even the finite square well. The potential is not continuous. You can allow as well
failure to be bounded. So, what is a typical example? The harmonic oscillator,
the x squared potential. It's not bounded. It goes to infinity. So, we can fail
to be continuous, but we can fail at one
point, another point, but we shouldn't fail at
infinitely many points, presumably. So, it's piecewise continuous. It can fail to be bounded, and
it can include delta functions. Which is pretty interesting,
because a lot of physics uses delta functions,
but a delta function is a complicated thing. We'll include delta
functions but not derivatives of them, nor powers. So we won't take anything more
strange than delta functions, collections of delta functions. So, this is really how delicate
your potentials will be. They will not be more
complicated than that. But for that, we
will assume, and it will be completely
consistent to require the following for
the wave functions. So Psi is continuous-- Psi of
x-- is continuous and bounded. And its derivative is bounded. Psi prime is bounded. AUDIENCE: What about Psi's
behavior at infinity? PROFESSOR: Sorry? AUDIENCE: What kind
of extra conditions do we have to impose of
Psi's behavior at infinity? PROFESSOR: Well, I will not
impose any condition that is further than that,
except the condition that they've been normalizable. And even that we will
be a little-- how would I say, not too
demanding on that. Because there will
be wave functions, like momentum Eigenstates
that can't be normalized. So, we'll leave it at that. I think probably this is
what you should really box, because for a
momentum Eigenstate, e to the ipx over h bar. This is a momentum Eigenstate. This is continuous. It's bounded. The derivative is bounded. It is not normalizable,
but it's so useful that we must include in the
list of things that we allow. So, bound states
and non-bound states are things that are
not normalizable. So, I don't put normalization. Now, if you put normalization,
then the wave function will go to zero at infinity. And that's all you
would want to impose. Nothing else. So, really in some
sense, this is it. You don't want more than that. AUDIENCE: Is normalization
sufficient to ensure the derivative also goes
to zero at infinity? PROFESSOR: Sorry? AUDIENCE: Is normalization
sufficient to ensure that the-- PROFESSOR: Not that I know. I don't think so. AUDIENCE: Then why is
integration by price generically valid? PROFESSOR: It's probably
valid for restricted kinds of potentials. So you could not
prove it in general. So, you know,
there may be things that one can generalize and
be a little more general, but I'm trying to
be conservative. I know that for any
decent potential-- and we definitely need
Psi prime bounded. And wave functions that
go to zero, the only ones I know that also have
Psi prime going to zero. But I don't think it's easy
to prove that's generic, unless you make
more assumptions. So, all right. So, this we'll have
for our wave functions, and now I want to say
a couple of things about properties
of the Eigenstates. Now, we will calculate
many of these Eigenstates, but we need to understand
some of the basic properties that they have. And there's really two
types of identities that I want you to be
very aware that they play some sort of dual role-- a
pretty interesting dual role-- that has to do with
these wave functions. So, the Eigenstates
of-- Eigenstates of H hat-- these are the
energy Eigenstates. you can consider them
and make a list of them. So, you have an energy E zero
less than or equal an E 1, E 2. Just goes like that. And you have a Psi zero, Psi 1. All this wave functions. And then H hat Psi N
is equal to E N Psi N. You have a set of solutions. So, this is what will happen
if you have a good problem. A reasonable potential, and
nothing terribly strange going on. There would be a
lot of solutions, and they can be chosen
to be orthonormal. Now at first sight, it's a
funny term to use-- orthonormal. This is a term that
we use for vectors. Two vectors are
orthogonal, and we say they're orthonormal if they
have unit length, and things like that. But what do we mean the two
functions are orthonormal? Well, our function's vectors. Well, that's a little dubious. But the way we will think
in quantum mechanics is that, in some
sense, functions are vectors in an infinite
dimensional space. So, they're just vectors,
but not in three dimensions. Why? Think of it. If you have a function,
you have to give values-- independent values-- at many
points-- Infinitely many. And if you give
all those values, you've got the function. If you have a vector, you
have to give components, and you've got the vector. So, in a sense, to
give a function, I have to give a lot of numbers. And I can say the
first vector is the value along the direction--
the first component is the value around zero. The second unit vector is the
value of about 0.01, 9.02, going on and on. And then list of
them, and you have a vector of infinite dimensions. You say, totally
useless. [LAUGHTER] No, it's not totally useless. Actually, if you
visualize that-- and we'll do it
later more-- you will be able to understand many
formulas as natural extensions. So, what does it mean that these
two functions are orthonormal? Well, a dot product,
or orthonormality, is to say that the
dot product is zero. And the way we dot product
functions Psi m and Psi n of x is we take their values at
the same point with star one, and we integrate. And, if this is
equal to delta mn, we say the functions
are orthonormal. So, ortho, for
orthogonal, which says that if m is different from
n, the Kronecker delta, that symbol is equal to
one if the two labels are the same, or zero otherwise. If they're different,
you get zero. The inner product--
this left hand side is called the
inner product-- is zero. On the other hand, if they are
the same, if m is equal to n, it says that the
Psi squared is one. Kind of like a wave function
that is well normalized. So we say normal
for orthonormal. So these are orthonormal wave
functions, and that's good. This is called orthonormality. But then there is a
more subtle property, which is that this
set of functions is enough to expand any
function in this interval that you're doing your
quantum mechanics. So, if you have any
reasonable function, it can be written as a
superposition of these ones. So, this differential
equation furnishes for you a collection of functions
that are very useful. So this is orthonormality. This is also
completeness, which is to say that any
function can be written as a sum of of this function. So I will write it as this. Psi of x-- an
arbitrary Psi of x-- can be written as
bm Psi n of x n equals zero to infinity,
where the bn's are complex. So, this is an assumption, but
it's a very solid assumption. When you study differential
equations of this type-- Sturm-Liouville
problem-- this is one thing that
mathematicians prove for you, and it's not all that easy. But the collection
of wave functions is good in this sense. It provides you a
complete set of things that any function can be
written in terms of that. I'm not saying this satisfies
any particular equation. You see, this function satisfies
very particular equations-- those equations-- but this
is an arbitrary function. And it can be written
as a sum of this. See, these equations
have different energies for different Psi's. This Psi here satisfies
no obvious equation. But here is a problem
that this is useful for. Suppose you're given a wave
function at, at the given time, you know what it looks like. So, here is your wave function. Psi. And you know that
Psi at x and time equals to zero happens to
be equal to this Psi of x that we wrote above. So, it's equal to bn Psi n of x. Well, if you know that, if you
can calculate this coefficient, the wave function of time
equals zero is known, say, and it was given
by this thing, which is then written in this form. If you can write
it in this form, you've solved the problem
of time evolution, because then Psi
of x at any time is just simply obtained by
evolving each component. Which is bn e to the minus
iEnt over h bar Psi n of x. So this is the important result. Now, look what has happened. We have replaced each term. We added this exponential. Why? Because then each one
of these is a solution of the full
Schrodinger equation. And therefore a superposition
with complex coefficients is still a solution of
the Schrodinger equation. Therefore, this thing
I've put by hand is, you would say it's ad hoc. No, it's not. We've put it by
hand, yes, but we've produced a solution of the
Schrodinger equation, which has another virtue. When t is equal to
zero, it becomes what you know the
wave function is. So, since this solves the
Schrodinger equation-- time equals zero gives
you the right answer. And you remember that
the Schrodinger equation, if you know that
time equals zero, the term is a wave
function everywhere-- this is the solution. It's not just a solution. It's the solution. So, you've solved this equation,
and it's a very nice thing. It all depends, of
course, on having found the coefficients bn. Because typically
at time equals zero, you may know what
the wave function is, but you may not
know how to write it in terms of these
coefficients bn. So, what do you do then? If you don't know
those coefficients, you can calculate them. How do you calculate them? Well, you use orthonormality. So you actually take this and
integrate against another Psi star. So you take a Psi star
sub m and integrate-- multiply and integrate. And then the right hand side
will get the Kronecker delta that will pick out one term. So, I'm just saying in
words a two line calculation that you should do if you
don't see this as obvious. Because it's a
kind of calculation that you do a few times in life. Then it becomes obvious
and you never do it again. It's minus infinity to infinity
dx Psi m star of x Psi of x dx. So, bm is given
by this quantity, or bn is given by this quantity. You obtain it from here
plus orthonormality. So, once you have
this bn, you can do something that may-- if
you look at these things and say, well, I'm
bored, what should I do? I say, well, you have bm. Plug it back. What happens then? You say, why would
I plug it back? I don't need to plug it back. And that's true, but it's
not a crazy thing to do, because it somehow must
lead to some identity. Because you solve an equation
and then plug it back and try to see if
somehow it makes sense. So either it makes sense, or
you learned something new. So, we were supposed
to calculate the bn's. And now we have them, so
I can plug this back here. So what do I Get Psi of x now is
equal to the sum from n equals zero to infinity. bn-- but this bn is the integral
of Psi n star of x prime. I put here Psi of x prime. dx prime. I don't want to confuse
the x's with x prime, so I should put the x
primes all over here. Psi n of x. Well, can I do the integral? No. So, have I gained anything? Well, you've gained something
if you write it in a way that Psi is equal to
something times Psi. That doesn't look
all that simple, but we can at least organize it. Let's assume things
are convergent enough that you can change orders
of sums and integrals. That's an assumption
we always do. I'll write it like this. dx prime. And now I'll put the
sum here equals zero to infinity of Psi
n star of x prime. And I'll put the other
Psi here as well. The Psi n of x over here. I'll put the parentheses, and
finally the Psi of x prime here. So, now it's put in a nice way. And it's a nice way
because it allows you to learn something new
and interesting about this. And what is that? That this must be a
very peculiar function, such that integrated
against Psi gives you Psi. And what could it be? Well, this is of
the form, if you wish-- the x prime--
some function of x and x prime--
times Psi of x prime. So, this k is this thing. Well, you can try to
think what this is. If you put the delta
function here-- which may be a little
bit of a cheat-- you will figure out
the right answer. This must be a function
that sort of picks out the value of the function
at x by integrating. So it only cares
about the value at x. So, it must be a delta function. So, in fact, this
is a delta function, or should be a delta function. And therefore the
claim is that we now have a very curious identity
that looks as follows. It looks like n equal zero
to infinity, Psi n star of x prime Psi n
of x is actually delta of x minus x prime. So, this must be true. If what we said at
the beginning is true, that you can expand any function
in terms of the Eigenfunctions, then, well, that's not
such a trivial assumption. And therefore, it allows you
to prove something fairly surprising, that
this must be true, that this identity must be true. And I want you to realize
and compare and contrast with this identity here. One is completeness. One is orthonormality. There are two kinds
of sums going on here. Here is sum over space, and
you keep labels arbitrary-- label indices arbitrary. So, sum over space. These functions depend
on space and on labels. Sum over space, and
keep the labels, and you get sort of a
unit matrix in this space, in the space of labels. Here, you keep the positions
arbitrary, but sum over labels. And now you get
like a unit matrix in the space of positions. Something is one--
but actually infinite, but you couldn't do better--
when x is equal to x prime. So, if you think
of it as a matrix, this function in x and x prime
is a very strange matrix, with two indices, x and x prime. And when x is different
from x prime, it's zero, but when x is equal
to x prime, it's one. But it has to be
a delta function, because continuous variables. But it's the same idea. So, actually if you think
of these two things, x and m as dual variables, this
is a matrix variable, and then you're sort of
keeping these two indices open and summing over
the other index. Multiplying in one way
you get a unit matrix. Here, you do the
other way around. You have a matrix in m and n. This is a more familiar
matrix, but then you sum over the other things. So, they're dual,
and two properties that look very different in the
way you express them in words. One is that they're orthonormal. The other is that
they're complete. And then suddenly
then the mathematics tells you there's a nice
duality between them. So, the last thing
I want to say today is about expectation
values, which is another concept we
have to review and recall. So let's give those ideas. So, if we have a
time-dependent operator-- no, a time independent-- we'll do
a time-independent operator, I'm sorry. Time-Independent operator. And this operator
will be called A hat. No time dependence
on the operator. So, then we have the
expectation value of this operator on
a normalized state. So what does that mean? The expectation value of this
operator on a state-- on a wave function here. Now, this wave function
is time-dependent. So this expectation
value of this operator is expected to be
a function of time. And how is it defined? It's defined by doing
the following integral. Again, from minus
infinity to infinity, dx Psi star of x and t,
and then the operator A acting on Psi of x and t. And Psi is supposed to
be a normalized state. So, notice the notation here. We put the Psi here because
of the expectation-- whenever somebody asks you
the expectation value for an operator, it
has to be on a given state. So you put the state. Then you realize that this is
a time-dependent wave function typically, so it
could depend on time. Now, we said about
stationary states that if the state is
stationary, there's a single time exponential here. There's just one term, e to
the minus iEt over h bar. And if A, of course, is a
time-independent operator, you won't care about
the exponential. You will cancel this
one, and there will not be a time dependence there. But if this state
is not stationary-- like most states
are not stationary-- remember it's very important. If you have a stationary
state, and you superimpose another stationary state,
the result is not stationary. Stationary is a
single exponential. More than one exponential
is not stationary. So when you have this, you
could have time dependence. So that's why we wrote it. Whenever you have a state
that is not stationary, there is time dependence. Now, you could do
the following thing. So here is a simple but
important calculation that should be done. And it's the expectation
value of H. So what is the expectation value
of the Hamiltonian at time t on this wave function Psi
that we've computed there? So, we would have to
do that whole integral. And in fact, I ask
you that you do it. It's not too hard. In fact, I will say
it's relatively simple. And you have H on
Psi of x and t, and then you must substitute
this Psi equal the sum of bn Psi n. And you have two sums. And the H acting on each
side n-- you know what it is. And then the two sums-- you
can do the integral using orthonormality. It's a relatively
standard calculation. You should be able to do it. If you find it hard, you
will see it, of course, in the notes. But it's the kind of thing
that I want you to review. So, what is the answer here? It's a famous answer. It's bm squared En. So, you get the expected
value of the energy. It's a weighted average over all
of the stationary states that are involved in this state
that you've been building. So your state has a little
bit of Psi zero, Psi 1, Psi 2, Psi 3. And for each one, you square its
component and multiply by En. And this is time-independent. And you say, well,
you told me that only for stationary states,
things are time-independent. Yes, only for stationary
states, all operators are time-independent,
but the Hamiltonian is a very special operator. It's an energy
operator, and this is a time independent system. It's not being
driven by something, so you would expect the
energy to be conserved. And this is pretty
much the statement of conservation of energy,
the time-independence of this thing. My last remark is technical
about normalizations, and it's something
you may find useful. If you have a wave function
that is Psi, which is not normalized, you may say,
OK, let's normalize it. So, what is the
normalized wave function? The normalized wave
function is Psi divided by the square root of
the integral of Psi star Psi dx. You see, this is a number, and
you take the square root of it. And this is the Psi of x and t. If the Psi is not normalized,
this thing is normalized. So, think of doing this here. Suppose you don't
want to work too hard, and you want to normalize
your wave function. So, your Psi is not normalized. Well, then this is
definitely normalized. You should check that. Square it, an integrate
it, and you'll see. You'll get one. But then I can then calculate
the expectation value of A on that state,
and wherever I see a Psi that should be normalized,
I put this whole thing. So what do I end up with? I end up with this
integral from infinity to infinity dx Psi star A A
hat Psi divided by the integral from minus infinity to
infinity of Psi star Psi dx. If you don't want to normalize
a wave function, that's OK. You can still calculate
its expectation value by working with a
not-normalized wave function. So in this definition,
Psi is not normalized, but you still get
the right value. OK, so that's it for today. Next time we'll do properties
of the spectrum in one dimension and begin something new called
the variational problem. All right. [APPLAUSE] Thank you, thank you.
