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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, so let me go
back to what we were doing. The plan for today
is as follows. We're going to look at
this unitary time evolution and calculate this operator
u, given the Hamiltonian. That will be the first
order of business today. Then we will look
at the Heisenberg picture of quantum mechanics. And the Heisenberg picture
of quantum mechanics is one where the operators,
the Schrodinger operators, acquire time dependence. And it's a pretty useful way of
seeing things, a pretty useful way of calculating
things as well, and makes the relation between
classical mechanics and quantum mechanics more obvious. So it's a very important tool. So we'll discuss that. We'll find the Heisenberg
equations of motion and solve them for a
particular case today. All this material is not so
to be covered in the test. The only part-- of course,
the first few things I will say today about solving
for the unitary operator you've done in other ways, and
I will do it again this time. So going back to what we
were saying last time, we postulated unitary
time evolution. We said that psi at t was given
by some operator U of t t0 psi t0. And then we found
that this equation implied the Schrodinger
equation with a Hamiltonian given by the
following expression. ih dU dt of t t0
u dagger of t t0. So that was our derivation
of the Schrodinger equation. We start with the
time evolution. We found that, whenever we
declare that states evolve in time in that way, they
satisfy a first order time differential equation
of the Schrodinger form in which the Hamiltonian
is given in terms of U by this equation. And we talked about
this operator. First we showed that it
doesn't depend really on t0. Then we showed that
it's Hermitian. It has units of energy. And as you may have seen
already in the notes, there is a very
clear correspondence between this
operator and the way the dynamics follows with
the ideas of Poisson brackets that are the precursors
of commutators from classical mechanics. So that's in the notes. I will not go in detail in this. Many of you may have not
heard of Poisson brackets. It's an interesting
thing, and really that will be good enough. So our goal today
is to find U given H, because as we mentioned
last time, for physics it is typically more
easy to invent a quantum system by postulating
a Hamiltonian and then solving it than
postulating a time evolution operator. So our goal in general is to
find U of t t0 given H of t. That's what we're
supposed to do. So the first thing
I'm going to do is multiply this equation by u. By multiplying this equation
by a u from the right, I will write first this
term. ih dU dt of t t0 is equal to H of t U of t t0. So I multiplied this
equation by u from the right. This operator is unitary,
so u dagger u is one. That's why this equation
cleaned up to this. Now there's no confusion
really here with derivatives, so I might this well write
them with normal derivatives. So I'll write this
equation as d dt of U t t0 is equal to H of t U of t t0. You should be able to
look at that equation and say I see the
Schrodinger equation there. How? Imagine that you have a psi
of t0 here, and you put it in. Then the right hand side becomes
h and t acting on psi of t. And on the left hand
side, this psi of t0 can be put inside the derivative
because it doesn't depend on t. Therefore this becomes
ih bar d dt of psi of t. So the Schrodinger
equation is there. OK so now let's solve this. We'll go through three cases. Case one, h is time independent. So we're doing this
sort of quickly. So H of t is really H like that. No explicit time
dependence there. So what do we have? ih bar. Let's write dU dt
is equal H times U. And we tried to write a
solution of the form U use equal to e to the minus
iHt over h bar times U0. Does that work? Well, we can think du dt and ih. So we get ih. When I take dU dt, I have to
differentiate this exponential. And now in this exponential,
this full operator H is there. But we are differentiating
with respect to time. And H doesn't depend
on time, so this is not a very difficult situation. You could imagine the
power series expansion. And H, as far as
this derivative goes, is like if it would
be even a number. It wouldn't make any
difference if it's an operator. So the derivative with
respect to time of this thing is minus iH over h times
the same exponential. Moreover, the position
of this h could be here, or it could be to the right. It cannot be to the
right of U0 though, because this is a matrix, a
constant matrix that we've put in here as a possible thing
for boundary condition. So so far we've taken
this derivative, and then i's cancel, the h
bar cancels, and you get H. But this whole
thing is, again, U. So the equation has been solved. So try this. And it works. So having this solution
we can write, for example, that U of t t0 is going
to be e to the minus iHt over h bar, some
constant matrix. When t is equal to t0, this
matrix becomes the unit matrix. So this is e to the minus
iHt0 over h bar times U0. And therefore from
here, U0 is the inverse of this matrix, which is nothing
else but e to the iHt0 over h bar. So I can substitute back
here what U0 is and finally obtain U of t t0 is e to the
minus iH over h bar t minus t0. And this is for h
time independent. And that's our solution. There's very little
to add to this. We discussed that in
recitation on Thursday. This unitary operator
you've been seeing that from the beginning of
the course in some sense, that you evolve
energy eigenstate. If this acts on any
energy eigenstate, h is an energy-- if you act
here on an energy eigenstate, the energy eigenstate is an
eigenstate precisely for H, you can put just
the number here. That is e to the, say,
alpha h on a state psi n is equal to e to the
alpha en psi n if h on psi n is equal to en on psi n. So the function of an operator
acting on an eigenstate is just the function
evaluated at the eigenvalue. So this is a rule that you've
been using a really long time. OK, so when h is time
independent, that's what it is. How about when h has a
little time dependence? What do I call a
little time dependence? A little time dependence
is an idea, the sign to make it possible for
you to solve the equation, even though it has
some time dependence. So you could have Hamiltonians
that are time dependent, but still have a
simplifying virtue. So H of t is time dependent. But assume that H at t1 and h
at t2 commute for all t1 and t2. So what could that be? For example, you know that the
particle in a magnetic field, the spin in a magnetic field
is minus gamma B dot the spin. And you could have a time
dependent magnetic field, B of t times the spin. I'm not sure this is the
constant gamma that they usually call gamma,
but it may be. Now then if the magnetic field
is time dependent, but imagine its direction is
not time dependent. So if its direction is not time
dependent, then, for example, you would have here minus
gamma Bz of t times Sz. And the Hamiltonian at
different times commute because Sz commutes with itself,
and the fact that it's time independent doesn't
make it fail to commute. So if you have a magnetic field
that is fixed in one direction but change in time, you
can have a situation where your Hamiltonian
is time dependent, but still at different
times it commutes. And you will discuss such
case because it's interesting. But later on as we do
nuclear magnetic resonance, we will have the
more interesting case in which a magnetic field
rotates and therefore it's not that simple. So what happens if you have a
time dependent Hamiltonian that actually commutes? Well, the claim
is that U of t t0 is given by a natural extension
of what we had before. You would want to put
exponential of minus iHt, but the reason this worked
was because the derivative with respect to time brought
down an iH over h bar. So one way to fix this is to
put t t0 H of t prime dt prime. So this is an
answer to try this. Look at this. If the Hamiltonian were
to be time independent, you could take it out. And then you would
get t minus t0. That brings you back to this
case, so this looks reasonable. So let me call this
quantity R of t. And then you notice
that R dot of t, the derivative of this
quantity with respect to time. Well, when you differentiate
an integral the upper argument, you get just the
integrand evaluated at the time represented
by the upper argument of the upper limit
of integration. So this is H of t. And now here comes
a crucial point. You're trying to differentiate. This U is really e
to the R. And you're trying to differentiate to see
if the equation holds dU dt. So what is the dU dt? Would be d dt of 1 plus R
plus RR plus 1 3 factor RRR. And now what happens? You differentiate here, and
the first term is R dot. Here You, would have one
half R dot R plus R R dot. And then 1 over 3 factorial,
but three factors. R dot RR plus R R
dot R plus RR R dot. But here is the
claim R dot commutes with R. Claim R
dot and R commute. Why is that? Well, R dot depends on
H. And R is an integral of H as well, but the H
at different times commute anyway, so this must be true. There's no place where you
can get a contribution, because R dot is like an H,
and here's an integral of H. So since the Hamiltonians
are assumed to commute, R dot commutes with
R. And this becomes like a normal derivative of
an exponential in which you can move the R dot to
the left everywhere. And you're differentiating
the usual thing. So this is R dot and times
the exponential of R. So actually that
means that we've got pretty much our answer,
because R dot is minus i over h bar H of t. And e to the R is U,
so we got dU dt equals this, which is the
same as this equation. The only reason a derivative
with respect to time will not give the usual thing is if
R and R dot fail to commute, and they don't. So you could put the R dot here. You can put R dot
on the other side, because it commutes with
R, but it's better here. And therefore you've got
this very nice solution. So the solution is not that bad. Now finally, I want to discuss
for a second the general case. So that's case-- there was a
1, a 2, a 3 H of t general. What can you do? Well, if H of t is
general, there's not too much you can do. You can write
something that will get you started doing
things, but it's not obviously terribly useful. But it's interesting
anyway that there's a way to write something
that makes sense. So here it is. U of t and t0. I'll write the answer
and explain how it looks, and then you will
see that it's OK. It's interesting. But it probably is not
the most practical way you can solve this problem. So here it is. There's an acronym
for this thing. T it's called the time
ordered exponential. This operator does something
to the exponential function. So it's a definition. So I have to say what
this time ordered exponential is, and
it's the following. You take the exponential
and just begin to expand. So 1 minus i over
h bar-- or I'll put like this, plus minus
i over h bar integral from t0 to t of dt1 H of t1. So far, so good. I've just expanded this. Now if I would
continue expanding, I would get something that
doesn't provide the solution. You see, this thing
is the solution when the Hamiltonian at
different times commute. So it's unlikely to
be the solution when they don't commute. In fact, it's not the solution. So what is the next term here? The next term is you
think of the exponential as you would expand as usual. So you will have here plus one
half of this thing squared. So I will put something and then
erase it, so maybe don't copy. One half minus i
over h bar squared. And you would say, well, t0
to t dt prime H of t prime. t0 to t dt double prime
H of double prime. Well, that would be
just an exponential. So what is a time
ordered exponential? You erase the one half. And then for notation
call this t1 and t1. And then the next integral
do it only up to time t1, and call this t2. So t1 will always
be greater than t2, because t2 is integrated
from t0 to t1. And as you integrate here over
the various t1's, you just integrate up to that value. So you're doing less
of the full integral then you should be
doing, and that's why the factor of one
half has disappeared. This can be continued. I can write the next one would
be minus i over h bar cubed integral t0 to t H of t1
integral t0 to t1 dt2 H of t2. And then they next
integral goes up to t2. So t0 to t2 dt3 H of t3. Anyway, that's a time
ordered exponential. And I leave it to you to
take the time derivative, at least to see that the first
few terms are working exactly the way they should. That is, if you take a
time derivative of this, you will get H times that thing. So since it's a
power series, you will differentiate
the first term, and you will get
the right thing. Then the second term and you
will start getting everything that you need. So it's a funny object. It's reassuring that
something like this success, but in general, you
would want to be able to do all these
integrals and to sum them up. And in general,
it's not that easy. So it's of limited usefulness. It's a nice thing
that you can write it, and you can prove things
about it and manipulate it. But when you have a
practical problem, generally that's not
the way you solve it. In fact, when we will discuss
the rotating magnetic fields for magnetic resonance, we
will not solve it in this way. We will try to figure out
the solution some other way. But in terms of
completeness, it's kind of pretty in
that you go from the exponential to the
time ordered exponential. And I think you'll see
more of this in 806. So that's basically
our solution for H and for the unitary
operator U in terms of H. And what we're
going to do now is turn to the Heisenberg
picture of quantum mechanics. Yes, questions? AUDIENCE: Why does
R dot [INAUDIBLE]? PROFESSOR: Because that's
really a property of integrals. d dx integral up
to x from x0 g of x prime dx prime is
just equal to g of x. This is a constant
here, so you're not varying the integral
over in this limit. So if this limit would
also be x dependent, you would get
another contribution, but we only get the
contribution from here. What's really happening
is you're integrating up to x, then up to x plus
epsilon subtracting, so you pick up the value of the
function of the upper limit. Yes? AUDIENCE: So what happens to
the T that was pre factor? PROFESSOR: What
happens to this T? AUDIENCE: Yeah, what happens? PROFESSOR: That's just a symbol. It says time order the
following exponential. So at this stage,
this is a definition of what t on an
exponential means. AUDIENCE: OK. PROFESSOR: It's not--
let me say T is not an operator in the usual
sense of quantum mechanics or anything like that. It's an instruction. Whenever you have an
exponential of this form, the time ordered
exponential is this series that we've written down. It's just a definition. Yes? AUDIENCE: So when we have
operators in differential equations, do we
still get [INAUDIBLE]? PROFESSOR: If we have what? AUDIENCE: If we have operators
in differential equations do we still get unique
[INAUDIBLE] solutions? PROFESSOR: Yes, pretty much. Because at the end
of the day, this is a first order matrix
differential equation. So it's a collection of first
order differential equations for every element of a matrix. It's pretty much the
same as you have before. If you know the operator
at any time, initial time, with the differential
equation you know the operator at a little bit time later. So the operator is
completely determined if you know it initially and
the differential equation. So I think it's
completely analogous. It's just that it's
harder to solve. Nothing else. One last question. AUDIENCE: So let's say
that we can somehow fly in this unitary operator,
and then we have a differential equation, and we
somehow, let's say, get a wave function out of it. What is the interpretation
of that wave function? PROFESSOR: Well,
it's not that we get the wave
function out of this. What really is
happening is that you have learned how to calculate
this operator given H. And therefore now you're able
to evolve any wave function. So you have solved
the dynamical system. If somebody tells you a time
equals 0, your system is here, you can now calculate where it's
going to be at the later time. So that's really all
you have achieved. You now know the solution. When you're doing
mechanics and they ask you for an orbit problem,
they say at this time the planet is here. What are you supposed to find? x is a function of time. You now know how it's
going to develop. You've solved
equations of motion. Here it's the same. You know the wave
function of time equals. If you know it at any time,
you've solved problem. OK, so Heisenberg picture
of quantum mechanics. Heisenberg picture. So basically the
Heisenberg picture exists thanks to the existence
of the Schrodinger picture. Heisenberg picture
of quantum mechanics is not something
that you necessarily invent from the beginning. The way we think of it is we
assume there is a Schrodinger picture that we've developed in
which we have operators like x, p, spin, Hamiltonians,
and wave functions. And then we are going to
define a new way of thinking about this, which is
called the Heisenberg picture of the
quantum mechanics. So it all begins by considering
a Schrodinger operator As hat, which is s
is for Schrodinger. And the motivation comes
from expectation values. Suppose you have time
dependent states, in fact, matrix elements. One time dependent state alpha
of t, one time dependent state beta of t. Two independent time
dependent states. So you could ask what is
the matrix element of A between these two time dependent
states, a matrix element. But then, armed with
our unitary operator, we know that As is here, and
this state beta at time t is equal to U of t
comma 0 beta at time 0. And alpha t is equal to
alpha at 0 U dagger of t0. So the states have
time dependence. But the time
dependence has already been found, say, in principle,
if you know U dagger. And then you can speak about the
time dependent matrix elements of the operator As
or the matrix element of this time dependent
operator between the time equals 0 states. And this operator is
sufficiently important that this operator is
called the Heisenberg version of the operator s. Has time dependence, and it's
defined by this equation. So whenever you have
Schrodinger operator, whether it be time dependent
or time independent, whatever the
Schrodinger operator is, I have now a
definition of what I will call the
Heisenberg operator. And it is obtained by acting
with a unitary operator, U. And operators always act
on operators from the left and from the right. That's something that operators
act on states from the left. They act on the state. But operators act on
operator from the left and from the right,
as you see them here, is the natural, ideal
thing to happen. If you have an operator that's
on another from the right only or from the left
only, I think you have grounds to be
suspicious that maybe you're not doing things right. So this is the
Heisenberg operator. And as you can
imagine, there's a lot of things to be said
about this operator. So let's begin with a remark. Are there questions about
this Heisenberg operator. Yes? AUDIENCE: Do we know anything
about the Schrodinger operator? PROFESSOR: You have
to speak louder. AUDIENCE: Is the
Schrodinger operator related to the
Hamiltonian [INAUDIBLE]? PROFESSOR: Any
Schrodinger operator, this could be the
Hamiltonian, this could be x hat, it
could be Sz, could be any of the
operators you know. All the operators you know
are Schrodinger operators. So remarks, comments. OK, comments. One, at t equals 0
A Heisenberg becomes identical to A
Schrodinger at t equals 0. So look why. Because when t is equal to 0,
U of t-- of 0 0 is the operator propagates no state, so
it's equal to the identity. So this is a wonderful
relation that tell us you that time equals
0 the two operators are really the same. And another simple remark. If you have the unit operator
in the Schrodinger picture, what is the unit operator
in the Heisenberg picture? Well, it would be U
t 0 dagger 1 U t 0. But 1 doesn't matter. U dagger with U is 1. This is a 1 Schrodinger,
and therefore it's the same operator. So the unit operator
is the same. It just doesn't
change whatsoever. OK, so that's good. But now this is something
interesting also happens. Suppose you have
Schrodinger operator C that is equal to the product
of A with B, two Schrodingers. If I try to figure
out what is CH, I would put U dagger-- avoid
all the letters, the t 0. It's supposed to be t 0. Cs U. But that's equal
U dagger As Bs U. But now, in between
the two operators, you can put a U U dagger,
which is equal to 1. So As U U dagger
Bs U. And then you see why this is really nice. Because what do you get is
that C Heisenberg is just A Heisenberg times B Heisenberg. So if you have C Schrodinger
equals A Schrodinger, B Schrodinger, C Heisenberg is
A Heisenberg B Heisenberg. So there's a nice correspondence
between those operators. Also you can do is
for commutators. So you don't have to
worry about this thing. So for example, if A
Schrodinger with B Schrodinger is equal to C Schrodinger,
then by doing exactly the same things, you see that
A Heisenberg with B Heisenberg would be the commutator
equal to C Heisenberg. Yes? AUDIENCE: That argument
for the identity operators being the same in both pictures. If the Hamiltonian
is time independent, does that work for any
operator that commutes with the Hamiltonian? PROFESSOR: Hamiltonian
is [INAUDIBLE]. AUDIENCE: Because then you
can push the operator just through the exponential
of the Hamiltonian. PROFESSOR: Yeah, we'll
see things like that. We could discuss that
maybe a little later. But there are some cases,
as we will see immediately, in which some operators are
the same in the two pictures. So basically operators that
commute with the Hamiltonian as you say, since U
involves the Hamiltonian, and this is the
Hamiltonian, if the operator commutes with the Hamiltonian
and you can move them across, then they are the same. So I think it's definitely true. So we will have an
interesting question, in fact, whether the
Heisenberg Hamiltonian is equal to the
Schrodinger Hamiltonian, and we'll answer that very soon. So the one example that here I
think you should keep in mind is this one. You know this is true. So what do you knowing
the Heisenberg picture? That X Heisenberg of t times
P Heisenberg of t commutator is equal to the Heisenberg
version of this. But here was the unit operator. And therefore this is just ih
bar times the unit operator again, because
the units operator is the same in all pictures. So these commutation relation
is true for any Heisenberg operator. Whatever commutation relation
you have of Schrodinger, it's true for
Heisenberg as well. OK, so then let's talk
about Hamiltonians. Three, Hamiltonians. So Heisenberg
Hamiltonian by definition would be equal to U dagger
t 0 Schrodinger Hamiltonian times U of t 0. So if the Schrodinger
Hamiltonian-- actually, if Hs at t1 commutes
units with Hs at t2, the Schrodinger Hamiltonian
is such that for all t1 and t2 they commute with each other. Remember, if that is the
case, the unitary operator is any way built
by an exponential. It's this one. And the Schrodinger
Hamiltonians commute. So as was asked in
the question before, this thing commutes
with that, and you get that they are the same. So if this is happening, the
two Hamiltonians are identical. And we'll have the
chance to check this today in a nice example. So I will write in this
as saying the Heisenberg Hamiltonian as a
function of time then is equal to the
Schrodinger Hamiltonian as a function of time. And this goes Hs of t1
and Hs of t2 commute. OK, now I want you
to notice this thing. Suppose the Hs of t is some
Hs of x,p, and t, for example. OK, now you come and
turn it into Heisenberg by putting a U dagger from the
left and a U from the right. What will that do? It will put U dagger from the
left, U dagger on the right. And then it will start
working it's way inside, and any x that it will find
will turn into a Heisenberg x. Any p will turn
into Heisenberg p. Imagine, for example,
any Hamiltonian is some function of x. It has an x squared. Well the U dagger
and U come and turn this into x Heisenberg squared. So what I claim here happens
is that H Heisenberg of t is equal to U dagger H
Schrodinger of x, p, t, U. And therefore this
becomes H Schrodinger of x Heisenberg of t, P
Heisenberg of t, and t. So here is what the
Heisenberg Hamiltonian is. It's the Schrodinger Hamiltonian
where X's, and P's, or spins and everything has
become Heisenberg. So the equality of
the two Hamiltonians is a very funny condition on
the Schrodinger Hamiltonian, because this is supposed to
be equal to the Schrodinger Hamiltonian, which
is of x, p, and t. So you have a function
of x, p, and t. And you put X
Heisenberg P Heisenberg, and somehow the whole
thing is the same. So this is something very
useful and we'll need it. One more comment,
expectation values. So this is three. Comment number four on
expectation values, which is something you've already--
it's sort of the way we began the discussion and
wanted to make sure it's clear. So four, expectation values. So we started with this
with alpha and beta, two arbitrary states,
matrix elements. Take them equal and to
be equal to psi of t. So you would have psi
t As psi t is, in fact, equal to psi 0 A
Heisenberg psi 0. Now that is a key equation. You know you're doing
expectation value at any given time of a Schrodinger operator,
turn it into Heisenberg and work at time equals 0. It simplifies life tremendously. Now this is the key identity. It's the way we motivated
everything in a way. And it's written in
a way that maybe it's a little too schematic,
but we write it this way. We just say the
expectation value of As is equal to the
expectation value of AH. And this, well, we
save time like that, but you have to
know what you mean. When you're computing
the expectation value for a
Schrodinger operator, you're using time
dependent states. When you're computing
the expectation value of the
Heisenberg operator, you're using the time equals
0 version of the states, but they are the same. So we say that the
Schrodinger expectation value is equal to the Heisenberg
expectation value. We right it in the bottom,
but we mean the top equation. And we use it that way. So the Heisenberg
operators, at this moment, are a little mysterious. They're supposed to be
given by this formula, but we've seen that
calculating U can be difficult. So calculating the
Heisenberg operator can be difficult sometimes. So what we try to do
in order to simplify that is find an equation that
is satisfied by the Heisenberg operator, a time
derivative equation. So let's try to find
an equation that is satisfied by the Heisenberg
operator rather than a formula. You'll say, well,
this is better. But the fact is that seldom you
know U. And even if you know U, you have to do this
simplification, which is hard. So finding a differential
equation for the operator is useful. So differential equation
for Heisenberg operators. So what do we want to do? We want to calculate ih bar d
dt of the Heisenberg operator. And so what do we get? Well, we have several things. Remember, the
Schrodinger operator can have a bit of
time dependence. The time dependence would be
an explicit time dependence. So let's take the time
derivative of all this. So you would have three terms. ih bar dU dagger dt
As U plus U dagger As dU dt plus-- with an ih bar--
U dagger ih bar dAs minus dt. dAs dt and U. Well, you have these equations. Those were the
Schrodinger equations we started with today. The derivatives of U, or
the derivatives of U dagger. so what did we have? Well, we have that ih bar dU dt
was HU-- H Schrodinger times U. And therefore ih
bar dU dagger dt. I take the dagger of this. I would get a minus sign. I would put it on
the other side. Is equal to U dagger
Hs with a minus here. And all the U's are
U's of t and t0. I ran out of this thick chalk. So we'll continue
with thin chalk. All right, so we are here. We wrote the time
derivative, and we have three terms to work out. So what are they? Well we have this
thing, ih bar this. So let's write it. ih bar d
d dt of As-- of A Heisenberg, I'm sorry-- Is equal to that
term is minus U dagger Hs A Schrodinger U. The next term plus ih
bar dU dt on the right. So we have plus U dagger
As Hs dU dt, so U. Well, that's not bad. It's actually quite nice. And then the last term, which
I have very little to say, because in general, this
is a derivative of a time dependent operator. Partial with respect to time,
it would be 0 if As depends, just say, on X, on P, on
Sx, or any of those things, has to have a particular t. So I will just leave this as
plus ih bar dAs dt Heisenberg. The Heisenberg version
of this operator using the definition that anything,
any operator that we have a U dagger in front,
a U to the right, is the Heisenberg
version of the operator. So I think I'm doing all
right with this equation. So what did we have? Here it is. ih bar d dt
of A Heisenberg of t. And now comes the
nice thing, of course. This thing, look at it. U dagger U. This
turns everything here into Heisenberg. H Heisenberg, A Heisenberg. Here you have A Heisenberg H
Heisenberg, and what you got is the commutator between them. So this thing is A Heisenberg
commutator with H Heisenberg. That whole thing. And then you have plus
ih bar dAs dt Heisenberg. So that is the Heisenberg
equation of motion. That is how you can calculate
a Heisenberg operator if you want. You tried to solve this
differential equation, and many times
that's the simplest way to calculate the
Heisenberg operator. So there you go. It's a pretty
important equation. So let's consider
particular cases immediately to just get some intuition. So remarks. Suppose As has no
explicit time dependence. So basically, there's
no explicit t, and therefore this
derivative goes away. So the equation becomes
ih bar dAh, of course, dt is equal to Ah
Heisenberg sub h of t. And you know the
Heisenberg operator is supposed to be simpler. Simple. If the Schrodinger operator
is time independent, it's identical to the
Schrodinger Hamiltonian. Even if the Schrodinger
operator has time dependence, but they commute, this
will become the Schrodinger Hamiltonian. But we can leave it like that. It's a nice thing anyway. Time dependence of
expectation value. So let me do a little remark on
time dependence of expectation values. So suppose you have the usual
thing that you want to compute. How does the expectation value
of a Schrodinger operator depend on time? You're faced with that
expectation value of As, and it changes in
time, and you want to know how you
can compute that. Well, you first say,
OK, ih bar d dt. But this thing is nothing but
psi 0 A Heisenberg of t psi 0. Now I can let the
derivative go in. So this becomes psi 0
ih bar dAh dt psi 0. And using this, assuming that
A is still no time dependence, A has no explicit
time dependence, then you can use
just this equation, which give you
psi 0 Ah Hh psi 0. So all in all, what
have you gotten? You've gotten a rather simple
thing, the time derivative of the expectation values. So ih bar d dt. And now I write
the left hand side as just expectation value
of H Heisenberg of t. And on the left hand side has
to the A Schrodinger expectation value, but we call
those expectation values the same thing as a
Heisenberg expectation value. So this thing becomes the right
hand side is the expectation value of A Heisenberg
H Heisenberg like that. And just the way we say
that Heisenberg expectation values are the same as
Schrodinger expectation values, you could as well
write, if you prefer, as d dt of A Schrodinger
is equal to the expectation value of A Schrodinger
with H Schrodinger. It's really the same equation. This equation we derived
a couple of lectures ago. And now we know that
the expectation values of Schrodinger operators are
the same as the expectation value of their
Heisenberg counterparts, except that the states are
taking up time equals 0. So you can use either
form of this equation. The bottom one is one
that you've already seen. The top one now looks almost
obvious from the bottom one, but it really took
quite a bit to get it. One last comment
on these operators. How about conserved operators? What are those things? A time independent As
is set to be conserved if it commutes with a
Schrodinger Hamiltonian. If As commutes with As equals 0. Now you know that if As with
Hs is 0, Ah with Hh is 0, because the map between
Heisenberg and Schrodinger pictures is a commutator that
is valued at the Schrodinger picture is valued in the
Heisenberg picture by putting H's. So what you realize from
this is that this thing, this implies Ah
commutes with Hh. And therefore by
point 1, by 1, you have to dAh dt is equal to 0. And this is nice, actually. The Heisenberg operator is
actually time independent. It just doesn't depend on time. So a Schrodinger operator,
it's a funny operator. It doesn't have time in there. It has X's, P's, spins, and
you don't know in general, if it's time independent
in the sense of conserve of expectation values. But whenever As
commutes with Hs, well, the expectation
values don't change in time. But as you know, this
d dt can be brought in, because the states are
not time dependent. So the fact that this
is 0 means the operator, Heisenberg operator, is
really time independent. Whenever you have a
Schrodinger operator, has no t, the Heisenberg one
can have a lot of t. But if the operator
is conserved, then the Heisenberg operator
will have no t's after all. It will really be conserved. So let's use our last 10
minutes to do an example and illustrate much of this. In the notes, there
will be three examples. I will do just one in lecture. You can do the other ones
in recitation next week. There's no need really
that you study these things at this moment. Just try to get whatever you
can now from the lecture, and next week you'll
go back to this. So the example is the
harmonic oscillator. And it will illustrate the
ideas very nicely, I think. The Schrodinger
Hamiltonian is p squared over 2m plus 1/2 m omega
squared x hat squared. OK, I could put x Schrodinger
and p Schrodinger, but that would be just far
too much. x and p are the operators
you've always known. They are Schrodinger operators. So we leave them like that. Now I have to write the
Heisenberg Hamiltonian. Well, what is the
Heisenberg Hamiltonian? yes? AUDIENCE: It's identical. PROFESSOR: Sorry? AUDIENCE: It's identical. PROFESSOR: Identical, yes. But I will leave that
for a little later. I will just assume,
well, I'm supposed to do U dagger U.
As you said, this is a time independent
Hamiltonian. It better be the
same, but it will be clearer if we now write
what it should be in general. Have a U dagger and
a U from the right. They come here, and they turn
this into P Heisenberg over 2m plus 1/2 m omega
squared x Heisenberg. OK, that's your
Heisenberg Hamiltonian. And we will check, in fact,
that it's time independent. So how about the operators X
Heisenberg and P Heisenberg. What are they? Well, I don't know
how to get them, unless I do this
sort of U thing. That doesn't look too bad
but certainly would be messy. You would have to do
an exponential of e to the minus iHt over
t with the x operator and another exponential. Sounds a little complicated. So let's do it the way the
equations of the Heisenberg operators tell you. Well, X and P are time
independent Schrodinger operators, so that equation
that I boxed holds. So ih dx Heisenberg
dt is nothing else than X Heisenberg commuted
with H Heisenberg. OK, can we do that commutator? Yes. Because X Heisenberg,
as you remember, just commutes with P Heisenberg. So instead of the
Hamiltonian, you can put this. This is X Heisenberg P
Heisenberg squared over 2m. OK well, X Heisenberg
P Heisenberg is like you had X and P.
So what is this commutator? You probably know it by now. You act with this
and these two p. So it acts on one, acts on the
other, gives the same on each. So you get P Heisenberg times
the commutator of X and P, which is ih bar
times a factor of 2. So we could put hats. Better maybe. And then what do we get? The ih there and ih cancels. And we get some
nice equation that says dX Heisenberg dt is
1 over m P Heisenberg. Well, it actually
looks like an equation in classical mechanics. dx dt is P over m. So that's a good thing about the
Heisenberg equations of motion. They look like
ordinary equations for dynamical variables. Well, we've got this one. Let's get P. Well, we didn't
get the operator still, but we got an equation. So how about P dP dt. So ih dP Heisenberg dt would be
P Heisenberg with H Heisenberg. And this time only the
potential term in here matters. So it's P Heisenberg
with 1/2 m omega squared X Heisenberg squared. So what do we? We get 1/2 m omega squared. Then we get again a factor of 2. Then we get one left over Xh. And then a P with Xh,
which is a minus ih bar. So the ih bars cancel, and we
get dPh dt is equal to-- the h bar cancelled-- m
omega squared Xh. Minus m. All right, so these are our
Heisenberg equations of motion. So how do we solve for them now? Well, you sort of have
to try the kind of things that you would do classically. Take a second derivative
of this equation. d second Xh dt squared
would be 1 over m dPh dt. And the dPh dt
would be [INAUDIBLE] 1 over m times minus
m omega squared Xh. So d second Xh dt squared is
equal to minus omega squared Xh, exactly the
equation of motion of a harmonic oscillator. It's really absolutely
nice that you recover those equations
that you had before, except that now you're
talking operators. And it's going to simplify your
life quite dramatically when you try to use these
operators, because, in a sense, solving for the time
dependent Heisenberg operators is the same as finding the
time evolution of all states. This time the operators
change, and you will know what they change like. So you have this. And then you write Xh is equal
to A cosine omega t plus B sine omega t where A
and B are some time independent operators. So Xh of t, well,
that's a solution. How about what is P? Ph of t would be m dX m dx dt. So you get minus m omega
sine omega tA plus m omega cosine omega tB. OK, so that's it. That is the most
general solution. But it still doesn't look like
what you would want, does it? No, because you haven't used
the time equals 0 conditions. At time equals 0, the
Heisenberg operators are identical they to the
Schrodinger operators. So at t equals 0,
Xh of t becomes A, but that must be X hat,
the Schrodinger operator. And at t equals
0, Ph of t becomes equal to this is 0
m omega B. And that must be equal to
the P hat operator. So actually we have already
now A and B. So B from here is P hat over m omega. And therefore Xh
of t is equal to A, which is X hat cosine
omega t plus B, which is P hat over m
omega sine omega t. An Ph of t is here. A is-- Ph of t is m omega
B, which is [INAUDIBLE] P. So it's P hat cosine
omega t minus m omega X hat sine omega t. So let's see. I hope I didn't make mistakes. P hat minus m omega
X hat sine omega t. Yep, this is correct. This is your whole solution
for the Heisenberg operators. So any expectation value
of any power of X and P that you will want to
find its time dependence, just put those
Heisenberg operators, and you will calculate things
with states at time equals 0. It will become very easy. So the last thing
I want to do is complete the promise
that we had about what is the Heisenberg Hamiltonian. Well, we had the Heisenberg
Hamiltonian there. And now we know the
Heisenberg operators in terms of the Schrodinger one. So Hh of t is equal to
Ph-- 1/2m Ph squared. So I have P hat cosine omega t
minus m omega X hat sine omega t squared plus 1/2 m
omega squared Xh squared. So X hat cosine
omega t plus P hat over m omega sine
omega t squared. So that's what the
Heisenberg Hamiltonian is. So let's simplify this. Well, let's square these things. You have 1/2m cosine squared
omega t P hat squared. Let's do the square of this one. You would have
plus 1/2m m squared omega squared sine
squared omega t X squared. And then we have
the cross product, which would be plus--
or actually minus 1/2m. The product of these two things. m omega sine omega
t cosine omega t. And you have Px plus XP. OK, I squared the first terms. Now the second one. Well, let's square
the P squared here. What do we have? 1/2 m omega squared
over m squared omega squared sine squared
of omega t P squared. The x plus 1/2 m
omega squared cosine squared omega t X squared. And the cross term. Plus 1/2 m omega squared over
m omega times cosine omega t sine omega t XP plus PX. A little bit of work,
but what do we get? Well, 1/2 m. And here we must
have 1/2 m, correct. 1/2 m. Sine squared omega t P squared. So this is equal
1/2 m P squared. These one's, hall here you
have 1/2 m omega squared. So it's 1/2 m omega
squared cosine and sine squared is X hat squared. And then here we have
all being over 2. And here omega over 2,
same factors, same factors, opposite signs. Very good. Schrodinger Hamiltonian. So you confirm that this
theoretical expectation is absolutely correct. And what's the meaning? You have the
Heisenberg Hamiltonian written in terms of the
Heisenberg variables. But by the time you substitute
these Heisenberg variables in, it just becomes identical to
the Schrodinger Hamiltonian. All right, so that's
all for today. I hope to see in office
hours in the coming days. Be here Wednesday 12:30,
maybe 12:25 would be better, and we'll see you then. [APPLAUSE]
