Lecture 5: Momentum operator, Schrödinger equation, and interpretation of the wavefunction.

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L5.1 Momentum operator, energy operator, and a differential equation (20:33)

L5.2 Free Schrödinger equation (09:56)

L5.3 The general Schrödinger equation. x, p commutator (17:58)

L5.4 Commutators, matrices, and 3-dimensional Schrödinger equation (16:12)

L5.5 Interpretation of the wavefunction (08:01)

L5.1 Momentum operator, energy operator, and a differential equation (20:33)

MITOCW | watch?v=ELBh60GU5yE
PROFESSOR: Last time, we talked about the Broglie wavelength. And our conclusion was, at the end of the
day, that we could write the plane wave that corresponded to a matter particle, with some
momentum, p, and some energy, E. So that was our main result last time, the final form for the
wave.
So we had psi of x and t that was e to the i k x minus i omega t. And that was the matter wave
with the relations that p is equal to h bar k. So this represents a particle with momentum, p,
where p is h bar times this number that appears here, the wave number, and with energy, E,
equal to h bar omega, where omega is that number that appears in the [? term ?] exponential.
Nevertheless, we were talking, or we could talk, about non-relativistic particles. And this is our
focus of attention. And in this case, E is equal to p squared over 2m. That formula that
expresses the kinetic energy in terms of the momentum, mv.
So this is the wave function for a free particle. And the task that we have today is to try to use
this insight, this wave function, to figure out what is the equation that governs general wave
functions. So, you see, we’ve been led to this wave function by postulates of the Broglie and
experiments of Davisson, and Germer, and others, that prove that particles like electrons have
wave properties.
But to put this on a solid footing you need to obtain this from some equation, that will say, OK,
if you have a free particle, what are the solutions. And you should find this solution. Perhaps
you will find more solutions. And you will understand the problem better.
And finally, if you understand the problem of free particle, there is a good chance you can
generalize this and write the equation for a particle that moves under the influence of
potentials. So basically, what I’m going to do by trying to figure out how this wave emerges
from an equation, is motivate and eventually give you, by the middle of this lecture, the
Schrodinger equation. So that’s what we’re going to try to do.
And the first thing is to try to understand what kind of equation this wave function satisfies. So
you want to think of differential equations like wave equations. Maybe it’s some kind of wave
equation. We’ll see it’s kind of a variant of that. But one thing we could say, is that you have
this wave function here. And you wish to know, for example, what is the momentum. Well you
should look at k, the number that multiplies the x here, and multiply by h bar. And that would
give you the momentum.
But another way of doing it would be to do the following. To say, well, h bar over i d dx of psi of
x and t, calculate this thing. Now, if I differentiate with respect to x, I get here, i times k going
down. The i cancels this i, and I get h bar k. So, I get h bar k times the exponential. And that is
equal to the value of the momentum times the wave.
So here is this wave actually satisfies a funny equation, not quite the differential equation we’re
looking for yet, but you can act with a differential operator. A derivative is something of a
differential operator. It operates in functions, and takes the derivative. And when it acts on this
wave function, it gives you the momentum times the wave function. And this momentum here
is a number. Here you have an operator. An operator just means something that acts on
functions, and gives you functions. So taking a derivative of a function is still a function. So
that’s an operator.
So we are left here to think of this operator as the operator that reveals for you the momentum
of the free particle, because acting on the wave function, it gives you the momentum times the
wave function. Now it couldn’t be that acting on the wave function just gives you the
momentum, because the exponential doesn’t disappear after the differential operator acts. So
it’s actually the operator acting on the wave function gives you a number times the wave
function. And that number is the momentum.
So we will call this operator, given that it gives us the momentum, the momentum operator, so
momentum operator. And to distinguish it from p, we’ll put a hat, is defined to be h bar over i d
dx. And therefore, for our free particle, you can write what we’ve just derived in a brief way,
writing p hat acting on psi, where this means the operator acting on psi, gives you the
momentum of this state times psi of x and t. And that’s a number. So this is an operator state,
number state.

So we say a few things, this language that we’re going to be using all the time. We call this
wave function, this psi, if this is true, this holds, then we say the psi of x and t is an eigenstate
of the momentum operator. And that language comes from matrix algebra, linear algebra, in
which you have a matrix and a vector. And when the matrix on a vector gives you a number
times the same vector, we say that that vector is an eigenvector of the matrix. Here, we call it
an eigenstate. Probably, nobody would complain if you called it an eigenvector, but eigenstate
would be more appropriate. So it’s an eigenstate of p.
So, in general, if you have an operator, A, under a function, phi, such that A acting on phi is
alpha phi, we say that phi is an eigenstate of the operator, and in fact eigenvalue alpha. So,
here is an eigenstate of p with eigenvalue of p, the number p, because acting on the wave
function gives you the number p times that wave function. Not every wave function will be an
eigenstate. Just like, when you have a matrix acting on most vectors, a matrix will rotate the
vector and move it into something else. But sometimes, a matrix acting in a vector will give you
the same vector up to a constant, and then you’ve got an eigenvector. And here, we have an
eigenstate.
So another way of expressing this, is we say that psi of x and t, this psi of x and t, is a state of
definite momentum. It’s important terminology, definite momentum means that if you
measured it, you would find the momentum p. And the momentum-- there would be no
uncertainty on this measurement. You measure, and you always get p. And that’s what,
intuitively, we have, because we decided that this was the wave function for a free particle with
momentum, p. So as long as we just have that, we have that psi is a state of definite
momentum.
This is an interesting statement that will apply for many things as we go in the course. But now
let’s consider another aspect of this equation. So we succeeded with that. And we can ask if
there is a similar thing that we can do to figure out the energy of the particle.
And indeed we can do the following. We can do i h bar d dt of psi. And if we have that, we’ll
take the derivative. Now, this time, we’ll have i h bar. And when we differentiate that wave
function with respect to time, we get minus i omega times the wave function. So i times minus i
is 1. And you get h bar omega psi. Success, that was the energy of the particle times psi.
And this looks quite interesting already. This is a number, again. And this is a time derivative of
the wave function. But we can put more physics into this, because in a sense, well, this
differential equation tells you how a wave function with energy, E, what the time dependence
of that wave function is.
But that wave function already, in our case, is a wave function of definite momentum. So
somehow, the information that is missing there, is that the energy is p squared over 2m. So we
have that the energy is p squared over 2m. So let’s try to think of the energy as an operator.
And look, you could say the energy, well, this is the energy operator acting on the function
gives you the energy. That this true, but it’s too general, not interesting enough at this point.
What is really interesting is that the energy has a formula. And that’s the physics of the
particle, the formula for the energy depends on the momentum. So we want to capture that.
So let’s look what we’re going to do. We’re going to do a relatively simple thing, which we are
going to walk back this. So I’m going to start with E psi. And I’m going to invent an operator
acting on psi that gives you this energy. So I’m going to invent an O.
So how do we do that? Well, E is equal to p squared over 2m times psi. It’s a number times
psi. But then you say, oh, p, but I remember p. I could write it as an operator. So if I have p
times psi, I could write it as p over 2m h bar over i d dx of psi.
Now please, listen with lots of attention. I’m going to do a simple thing, but it’s very easy to get
confused with the notation. If I make a little typo in what I’m writing it can confuse you for a
long time. So, so far these are numbers. Number, this is a number times psi. But this p times
psi is p hat psi which is that operator, there. So I wrote it this way.
I want to make one more-- yes?
AUDIENCE: Should that say E psi?
PROFESSOR: Oh yes, thank you very much. Thank you. Now, the question is, can I move this p close to the
psi. Opinions? Yes?
AUDIENCE: Are you asking if it’s just a constant?
PROFESSOR: Correct, p is a constant. p hat is not a constant. Derivatives are not. But p at this moment is a
number. So it doesn’t care about the derivatives. And it goes in. So I’ll write it as 1 over 2m h/i
d dx, and here, output p psi, where is that number. But now, p psi, I can write it as whatever it
is, which is h/i d dx, and p psi is again, h/i d dx psi.
So here we go. We have obtained, and let me write the equation in slightly reversed form.
Minus, because of the two i’s, 1 over 2m, two partials derivatives is a second order partial
derivative on psi, h bar squared over 2m d second dx psi. That’s the whole right-hand side, is
equal to E psi.
So the number E times psi is this. So we could call this thing the energy operator. And this is
the energy operator. And it has the property that the energy operator acting on this wave
function is, in fact, equal to the energy times the wave function.
So this state again is an energy eigenstate. Energy operator on the state is the energy times
the same state. So psi is an energy eigenstate, or a state of definite energy, or an energy
eigenstate with energy, E. I can make it clear for you that, in fact, this energy operator, as
you’ve noticed, the only thing that it is is minus h squared over 2m d second dx squared.
But where it came from, it’s clear that it’s nothing else but 1 over 2m p hat squared, because p
hat is indeed h/i d dx. So if you do this computation. How much is this? This is A p hat times p
hat, that’s p hat squared. And that’s h/i d dx h/i d dx. X And that gives you the answer. So the
energy operator is p hat squared over 2m.
All right, so actually, at this moment, we do have a Schrodinger equation, for the first time. If
we combine the top line over there. I h bar d dt of psi is equal to E psi, but E psi I will write it as
minus h squared over 2m d second dx squared psi.

L5.2 Free Schrödinger equation (09:56)

MITOCW | watch?v=7euh_iwzSGo
PROFESSOR: This is a wonderful differential equation, because it carries a lot of information. If you put this psi,
it’s certainly going to be a solution. But more than that, it’s going to tell you the relation between k and omega. So
if you try your-- we seem to have gone around in circles.
But you’ve obtained something very nice. First, we claim that that’s a solution of that equation and has the deep
information about it. So if you try again, psi equal e to the ikx minus i omega t, what do we get?
On the left hand side, we get ih minus i omega psi. And on the right hand side, we get minus h squared over 2m
and two derivatives with respect to x. And that gives you an ik times and other ik. So ik squared times psi.
And the psis cancel from the two sides of the equation. And what do we get here? h bar omega. It’s equal to h
squared k squared over 2m, which is e equal p squared over 2m.
So it does the whole job for you. That differential equation is quite smart. It admits these as solutions.
Then, this will have definite momentum. It will have definite energy. But even more, when you try to see if you
solve it, you find the proper relation between the energy and the momentum that tells you you have a particle.
So this is an infinitely superior version of that claim that that is a plane wave that exists. Because for example,
another thing that you have here is that this equation is linear. Psi appears linearly, so you can form solutions by
superposition.
So the general solution, now, is not just this. This is a free particle Schrodinger equation. And you might say, well,
the most general solution must be that, those plane waves. But linearity means that you can compose those plane
waves and add them. And if you can add plane waves by Fourier theorem, you can create pretty much all the
things you want.
And if you have this equation, you know how to evolve free particles. Now, you can construct a wave packet of a
particle and evolve it with the Schrodinger equation and see how the wave packet moves and does its thing. All
that is now possible, which was not possible by just saying, oh, here is another wave.
You’ve worked back to get an equation. And this is something that happens in physics all the time. And we’ll
emphasize it again in a few minutes. You use little pieces of evidence that lead you-- perhaps not in a perfectly
logical way, but in a reasonable way-- to an equation.
And that equation is a lot smarter than you and all the information that you put in. That equation has all kinds of
physics. Maxwell’s equations were found after doing a few experiments. Maxwell’s equation has everything in it, all
kinds of phenomena that took years and years to find.
So it’s the same with this thing. And the general solution of this equation would be a psi of x and t, which would be
a superposition of those waves. So you would put an e to the ikx minus i omega t. I will put omega of k because
that’s what it is.
Omega is a function of k. And that’s what represents our free particles-- omega of kt. And this is a solution. But so
will be any superposition of those solutions. And the solutions are parametrized by k. You can choose different
momenta and add them.
So I can put a wave with one momentum plus another wave with another momentum, and that’s perfectly OK. But
more generally, we can integrate. And therefore, we’ll write dk maybe from minus infinity to infinity. And we’ll put a
phi of k, which can be anything that’s not part of the differential equation.
Now, this is the general solution. You might probably say, wow, how do we know that? Well, I suggest you try it. If
you come here, the ddt will come in. We’ll ignore the k. Ignore this. And just gives you the omega factor here.
That ddx squared-- we’ll ignore, again, all these things, and give you that. From the relation omega minus k
equals 0, you’ll get the 0. And therefore, this whole thing solves the Schrodinger equation-- solves the Schrodinger
equation.
So this is very general. And for this, applies what we said yesterday, talking about the velocity of the waves. And
this wave, we proved yesterday, that moves with a group velocity, v group, which was equal to d omega dk at
some k0, if this is localized at k0.
Otherwise, you can’t speak of the group velocity this thing will not have a definite group velocity. And the omega
dk-- And you have this relation between omega and k, such a to way that this is the evp, as we said yesterday.
And this is ddp of b squared over 2m, which is p over m, which is what we call the velocity of the particle.
So it moves with the proper velocity, the group velocity. That’s actually a very general solution. We’ll exploit it to
calculate all kinds of things. A few remarks that come from this equation.
Remarks. 1, psi cannot be a real. And you can see that because if psi was real, the right hand side would be real.
This derivative would be real because the relative of a real function is a real function. Here you have an imaginary
number.
So structurally, it is forbidden to have full wave functions that are real. I call these full wave functions because we’ll
talk sometime later about time independent wave functions. But the full wave function cannot be real.
Another remark is that this is not the wave equation of the usual type-- not a usual wave equation. And what a
usual wave equation is something like d second phi dx squared minus 1 over v squared d second phi dt squared
equals zero. That’s a usual wave equation.
And the problem with that wave equation is that it has real solutions. Solutions, phi that go like functions of x minus
is vt, plus minus x over vt. And we cannot have those real solutions.
So we managed to get a wave, but not from a usual wave equation. This, waves also all move with some same
velocity, velocity, v, of the wave. These waves don’t do that. They have a group velocity.
It’s a little bit different situation. And what has happened is that we still kept the second derivative, with respect to
x. But in time, we replaced it by first derivative. And we put an i. And somehow, it did the right job for us.

L5.3 The general Schrödinger equation. x, p commutator (17:58)

MITOCW | watch?v=rwzg8iEOc8s
PROFESSOR: ih bar d psi dt equal E psi where E hat is equal to p squared over 2m, the operator. That is the
Schrodinger equation. The free particle Schrodinger equation-- you should realize it’s the
same thing as this. Because p is h bar over i ddx.
And now Schrodinger did the kind of obvious thing to do. He said, well, suppose I have a
particle moving in a potential, a potential V of x and t-- potential. Then the total energy is
kinetic energy plus potential energy.
So how about if we think of the total energy operator. And here is a guess. We’ll put the just p
squared over 2m, what we had before. That’s the kinetic energy of a particle. But now add
plus V of x and t, the potential.
That is reasonable from your classical intuition. The total energy is the sum of them. But it’s
going to change the Schrodinger equation quite substantially.
Now, most people, instead of calling this the energy operator, which is a good name, have
decided to call this the Hamiltonian. So that’s the most popular name for this thing. This is
called the Hamiltonian H.
And in classical mechanics, the Hamiltonian represents the energy expressed in terms of
position and momenta. That’s what the Hamiltonian is, and that’s roughly what we have here.
The energy is [? in ?] [? terms ?] [? of ?] momenta and position.
And we’re going to soon be getting to the position operator, therefore. So this is going to be
the Hamiltonian. And we’ll put the hat as well.
So Schrodinger’s inspiration is to say, well, this is going to be H hat. And I’m going to say that
ih bar d psi dt is equal to H hat psi. Or equivalently, ih bar ddt of psi of x and t is equal to minus
h squared over 2m, [? v ?] [? second ?] dx squared-- that’s the p squared over 2m-- plus V of
x and t, all multiplying psi.
This is it. This is the full Schrodinger equation. So it’s a very simple departure. You see, when
you discover the show that the equation for a free particle, adding the energy was not that
difficult. Adding the potential energy was OK.
We just have to interpret this. And maybe it sounds to you a little surprising that you multiply
this by psi. But that’s the only way it could be to be a linear equation. It cannot be that psi is
acted by this derivative, but then you add v.
It would not be a linear equation. And we’ve realize that the structure of the Schrodinger
equation is d psi dt is equal to an energy operator times psi.
The whole game of quantum mechanics is inventing energy operators, and then solving these
equations, then see what they are. So in particular, you could invent a potential and find the
equations. And, you see, it looks funny. You’ve made a very simple generalization. And now
you have an equation.
And now you can put the potential for the hydrogen atom and calculate, and see if it works.
And it does. So it’s rather unbelievable how very simple generalizations suddenly produce an
equation that has the full spectrum of the hydrogen atom. It has square wells, barrier
penetration, everything. All kinds of dynamics is in that equation.
So we’re going to say a few more things about this equation now. And I want you to
understand that the V, at this moment, can be thought as an operator. This is an operator,
acts on a wave function to give you a function.
This is a simpler operator. It’s a function of x and t. And multiplying by a function of x and t
gives you a function of x and t. So it is an operator. Multiplying by a given function is an
operator. It changes all the functions.
But it’s a very simple one. And that’s OK, but V of x and t should be thought as an operator.
So, in fact, numbers can be operator. Multiplication by a number is an operator. It adds on
every function and multiplies it by a number, so it’s also an operator.
But x has showed up. So it’s a good time to try to figure out what x has to do with these things.
So that’s what we’re going to do now. Let’s see what’s x have to do with things.
OK, so functions of x, V of x and t multiplied by wave functions, and you think of it as an
operator. So let’s make this formal. Introduce an operator, X hat, which, acting on functions of
x, multiplies them by x.
So the idea is that if you have the operator X hat acting on the function f of x, it gives you
another function, which is the function x times f of x-- multiplies by x. And you say, wow, well,
why do you have to be so careful in writing something so obvious? Well, it’s a good idea to do
that, because otherwise you may not quite realize there’s something very interesting
happening with momentum and position at the same time, as we will discover now.
So we have already found some operators. We have operators P, x, Hamiltonian, which is p
squared over 2m. And now you could put V of x hat t. You know, if here you put V of x hat,
anyway, whatever x hat does is multiplied by x.
So putting V of x hat here-- you may want to do it, but it’s optional. I think we all know what we
mean by this. We’re just multiplying by a function of x.
Now when you have operators, operators act on wave functions and give you things. And we
mentioned that operators are associated or analogs of matrices. And there’s one fundamental
property of matrices. The order in which you multiply them makes a difference.
So we’ve introduced two operators, p and x. And we could ask whether the order of
multiplication matters or not. And this is the way Heisenberg was lead to quantum mechanics.
Schrodinger wrote the wave equation. Heisenberg looked at operators and commutation
relations between them. And it’s another way of thinking of quantum mechanics that we’ll use.
So I want to ask the question, that if you have p and x and you have two operators acting on a
wave function, does the order matter, or it doesn’t matter? We need to know that. This is the
basic relation between p and x.
So what is the question? The question is, if I have-- I’ll show it like that-- x and p acting on a
wave function, phi, minus px acting on a wave function, do I get 0? Do I get the same result or
not?
This is our question. We need to understand these two operators and see how they are
related. So this is a very good question. So let’s do that computation.
It’s, again, one of those computation that is straightforward. But you have to be careful,
because at every stage, you have to know very well what you’re doing. So if you have two
operators like a and b acting on a function, the meaning of this is that you have a acting on
what b acting on phi gives you.
That’s what it means to have two things acting. Your first act with the thing on the right. You
then act on the other one.
So let’s look at this thing-- xp phi minus px phi. So for the first one, you would have x times p
hat on phi minus p hat times x on phi, phi of x and t maybe-- phi of x and t.
OK, now what do we have? We have x hat acting on this. And this thing, we already know what
it is-- h over i ddx of phi of x and t-- minus p hat and x, acting on phi, is little x phi of x and t.
Now this is already a function of x and t. So an x on it will multiply it by x. So this will be h over i
x ddx of phi. It just multiplies it by x at this moment-- minus here we have h bar over i ddx of x
phi.
And now you see that when this derivative acts on phi, you get a term that cancels this. But
when it acts on x, it gives you an extra term. ddx of x is minus h over i phi-- or ih phi.
So the derivative acts on x or an phi. When it acts on phi, gives you this term. When it acts on
x, gives you the thing that is left over.
So actually, let me write this in a more clear way. If you have an operator, a linear operator A
plus B acting on a function phi, that’s A phi plus B phi. You have linear operators like that.
And we have these things here. So this is actually equal to x hat p hat minus p hat x hat on
phi. That’s what it means when you have operators here.
So look what we got, a very surprising thing. xp minus px is an operator. It wants to act on
function. So we put a function here to evaluate it. And that was good.
And when we evaluate it, we got a number times this function. So I could say-- I could forget
about the phi. I’m simply right that xp minus px is equal to ih bar.
And although it looks a little funny, it’s perfectly correct. This is an equality between operators–
equality between operators. On the left-hand side, it’s clear that it’s an operator. On the righthand side, it’s also an operator, because a number acts as an operator on any function it
multiplies by it.
So look what you’ve discovered, this commutator. And that’s a notation that we’re going to use
throughout this semester, the notation of the commutator. Let’s introduce it here.
So if you have two operators, linear operators, we define the commutator to be the product in
the first direction minus the product in the other direction. This is called the commutator of A
and B.
So it’s an operator, again, but it shows you how they are non-trivial, one with respect to the
other. This is the basis, eventually, of the uncertainty principle. x and p having a commutator
of this type leads to the uncertainty principle.
So what did we learn? We learned this rather famous result, that the commutator of x and p in
quantum mechanics is ih bar.

L5.4 Commutators, matrices, and 3-dimensional Schrödinger equation (16:12)

MITOCW | watch?v=m7UT2Hr465o
PROFESSOR: This is very important. This is the beginning of the uncertainty principle, the matrix formulation
of quantum mechanics, and all those things. I want to just tabulate the information of matrices.
We have an analog, so we have operators. And we think of them as matrices.
Then in addition to operators, we have wave functions. And we think of them as vectors. The
operators act on the wave functions or functions, and matrices act on vectors. We have
eigenstate sometimes and eigenvectors.
So matrices do the same thing. They don’t necessarily commute. There are very many
examples of that. I might as well give you a little example that is famous in the theory of spin,
spin 1/2. There is the Pauli matrices. Sigma 1 is equal to 1, 1, 0, 0. Sigma 2 is 0 minus i, i 0,
and sigma 3 is 1 minus 1, 0, 0.
And a preview of things to come-- the spin operator is actually h bar over 2 sigma. And you
have to think of sigma as having three components. That’s where it is. Spins will be like that.
We won’t have to deal with spins this semester. But there it is, that spin 1/2. Somehow these
matrices encode spin 1/2.
And you can do simple things, like sigma 1 times sigma 2. 0, 1, 1, 0 times 0 minus i, i, 0. Let’s
see if I can get this right. i, 0, 0 minus i. And you can do sigma 2 sigma 1 0 minus i, i 0, 0, 1, 1,
0 equals minus i, 0, 0, i, i. So I can go ahead here.
And therefore, sigma 1 commutator with sigma 2 is equal to sigma 1, sigma 2 minus sigma 2,
sigma 1. And you can see that they’re actually the same up to a sign, so you get twice. So you
get 2 times i 0, 0 minus i. And this is 2i times 1 minus 1, 0, 0. And that happens to be the
sigma 3 matrix. So sigma 1 and sigma 2 is equal to 2i sigma 3.
These matrices talk to each other. And you would say, OK, these matrices commute to give
you this matrix. This thing commutes to give you a number so that surely it’s a lot easier. You
couldn’t be more wrong.
This is complicated, extraordinarily complicated to understand what this means. This is very
easy. This is 2 by 2 matrices that you check. In fact, you can write matrices for x and p. This
correspondence is not just an analogy. It’s a concrete fact. You will learn-- not too much in this
course, but in 805-- how to write matrices for any operator. They’re called matrix
representations.
And therefore, you could ask how does the matrix for x look. How does the matrix for p look?
And the problem is these matrices have to be infinite dimensional. It’s impossible to find two
matrices whose commutator gives you a number.
Something you can prove in math is actually not difficult. You will all prove it through thinking a
little bit. There’s no two matrices that commute to give you a number. On the other hand, very
easy to have matrices that commute to give you another matrix.
So this is very strange and profound and interesting, and this is much simpler. Spin 1/2 is
much simpler. That’s why people do quantum computations. They’re working with matrices
and simple stuff, and they go very far. This is very difficult. x and p is really complicated.
But that’s OK. The purpose of this course is getting familiar with those things. So I want to now
generalize this a little bit more to just give you the complete Schrodinger equation in three
dimensions. So how do we work in three dimensions, three-dimensional physics?
There’s two ways of teaching 804-- it’s to just do everything in one dimension, and then one
day, 2/3 of the way through the course-- well, we live in three dimensions, and we’re going to
add these things. But I don’t want to do that. I want to, from the beginning, show you the threedimensional thing and have you play with three-dimensional things and with one-dimensional
things so that you don’t get focused on just one dimension.
The emphasis will be in one dimension for a while, but I don’t want you to get too focused on
that. So what did we have with this thing? Well, we had p equal h bar over i d dx. But in three
dimensions, that should be the momentum along the x direction.
We wrote waves like that with momentum along the x direction. And py should be h bar over i
d dy, and pz should be h bar over i d dz-- momentum in the x, y, and z direction. And this
corresponds to the idea that if you have a wave, a de Broglie wave in three dimensions, you
would write this-- e to the i kx minus omega t, i omega t.
And the momentum would be equal to h bar k vector, because that’s how the plane wave
works. That’s what de Broglie really said. He didn’t say it in one dimension. Now, it may be
easier to write this as p1 equal h bar over i d dx1, p2 h bar over i d dx2, and p3 h bar over i d
dx3 so that you can say that all these three things are Pi equals h bar over i d dxi-- and maybe
I should put pk, because the i and the i could get you confused-- with k running from 1 to 3.
So that’s the momentum. They’re three momenta, they’re three coordinates. In vector
notation, the momentum operator will be h bar over i times the gradient. You know that the
gradient is a vector operator because d dx, d dy, d dz. So there you go. The x component of
the momentum operators, h bar over i d dx, or d dx1, d dx2, d dx3. So this is the momentum
operator.
And if you act on this wave with the momentum operator, you take the gradient, you get this–
so p hat vector. Now here’s a problem. Where do you put the arrow? Before or after the hat? I
don’t know. It just doesn’t look very nice either way. The type of notes I think we’ll use for
vectors is bold symbols so there will be no proliferation of vectors there.
So anyway, if you have this thing being the gradient acting on this wave function, e to the i kx
minus i omega t, that would be h over i, the gradient, acting on a to the i kx vector minus i
omega t. And the gradient acting on this-- this is a vector-- actually gives you a vector. So you
can do component by component, but this gives you i k vector times the same wave function.
So you get hk, which is the vector momentum times the wave function.
So the momentum operator has become the gradient. This is all nice. So what about the
Schrodinger equation and the rest of these things? Well, it’s not too complicated. We’ll say one
more thing.
So the energy operator, or the Hamiltonian, will be equal to p vector hat squared over 2m plus
a potential that depends on all the coordinates x and t, the three coordinates. Even the
potential is radial, like the hydrogen atom, is much simpler. There are conservation laws.
Angular momentum works nice. All kinds of beautiful things happen. If not, you just leave it as
x and p.
And now what is p hat squared? Well, p vector hat squared would be h bar over-- well, I’ll write
this-- p vector hat dotted with p vector hat. And this is h over i gradient dotted with h over i
gradient, which is minus h squared Laplacian.
So your Schrodinger equation will be ih bar d psi dt is equal to the whole Hamiltonian, which
will be h squared over 2m. Now Laplacian plus v of x and t multiplied by psi of x vector and t.
And this is the full three-dimensional Schrodinger equation.
So it’s not a new invention. If you invented the one-dimensional one, you could have invented
the three-dimensional one as well. The only issue was recognizing that the second dx squared
now turns into the full Laplacian, which is a very sensible thing to happen.
Now, the commutation relations that we had here before-- we had x with p is equal to ih bar.
Now, px and x failed to commute, because d dx and x, they interact. But px will commute with
y. y doesn’t care about x derivative.
So the p’s failed to commute. They give you a number with a corresponding coordinate. So
you have the i-th component of the x operator and the j-th component of the p operator–
these are the components-- give you ih bar delta ij, where delta ij is a symbol that gives you 1
if i is equal to j and gives you 0 if i is different from j.
So here you go. X and px is 1 and 1. Delta 1, 1 is 1. So you get ih bar. But if you have x with
py or p2, you would have delta 1, 2, and that’s 0, because the two indices are not the same.
So this is a neat way of writing nine equations. Because in principle, I should give you the
commutator of x with px and py and pc, y with px, py, pc, and z with px, py, pc. You’re seeing
that, in fact, x just talks to px, y talks to py, z talks to pz.
So that’s it for the Schrodinger equation. Our goal is going to be to understand this equation.
So our next step is to try to figure out the interpretation of this psi. We’ve done very nicely by
following these things.
We had a de Broglie wave. We found an equation. Which invented a free Schrodinger
equation. We invented an interacting Schrodinger equation. But we still don’t know what the
wave function means.

L5.5 Interpretation of the wavefunction (08:01)

MITOCW | watch?v=R-5hjmV-bdY
PROFESSOR: interpretation of the wave function.
–pretation–
the wave function.
So you should look at what the inventor said. So what did Schrodinger say? Schrodinger
thought that psi represents particles that disintegrate. You have a wave function.
And the wave function is spread all over space, so the particle has disintegrated completely.
And wherever you find more psi, more of the particle is there. That was his interpretation.
Then came Max Born. He said, that doesn’t look right to me. If I have a particle, but I solve the
Schrodinger equation. Everybody started solving the Schrodinger equation.
So they solved it for a particle that hits a Coulomb potential. And they find that the wave
function falls off like 1 over r.
OK, the wave function falls off like 2 over r. So is the particle disintegrating? And if you
measure, you get a little bit of the particle here? No.
Max Born said, we’ve done this experiment. The particle chooses some way to go. And it goes
one way, and when you measure, you get the full particle. The particle never disintegrates.
So Schrodinger hated what Max Born said. Einstein hated it. But never mind. Max Born was
right. Max Born said, it represents probabilities.
And why did they hate it? Because suddenly you lose determinism. You can just talk about
probability. So that was sort of funny.
And in fact, neither Einstein nor Schrodinger ever reconciled themselves with the probabilistic
interpretation. They never quite liked it. It’s probably said that the whole Schrodinger cat
experiment was a way of Schrodinger to try to say how ridiculous the probability interpretation
was.
Of course, it’s not ridiculous. It’s right. And the important thing is summarized, I think, with one
sentence here. I’ll write it. Psi of x and t does not tell how much of the particle–
is at x at time t.
But rather–
what is the probability–
probability–
–bility–
to find it–
at x at time t. So in one sentence, the first clause is what Schrodinger said, and it’s not that. It’s
not what fraction of the particle you get, how much of the particle you get. It’s the probability of
getting.
But that requires–
a little more precision. Because if a particle can be anywhere, the probability of being at one
point, typically, will be 0. It’s a continuous probability distribution.
So the way we think of this is we say, we have a point x. Around that point x, we construct a
little cube.
d cube x. And the probability-- probability dp, the little probability to find the particle at xt in the
cube, within the cube–
the cube–
is equal to the value of the wave function at that point. Norm squared times the volume d cube
x.
So that’s the probability to find the particle at that little cube. You must find the square of the
wave function and multiply by the little element of volume. So that gives you the probability
distribution.
And that’s, really, what the interpretation means. So it better be, if you have a single particle–
particle, it better be that the integral all over space-- all over space-- of psi squared of x and t
squared must be equal to 1.
Because that particle must be found somewhere. And the sum of the probabilities to be found
everywhere must add up to 1. So it better be that this is true.
And this poses a set of difficulties that we have to explore. Because you wrote the Schrodinger
equation. And this Schrodinger equation tells you how psi evolve in time.
Now, a point I want to emphasize is that the Schrodinger equation says, suppose you know
the wave function all over space. You know it’s here at some time t0. The Schrodinger
equation implies that that determines the wave function for any time.
Why? Because if you know the wave function throughout x, you can calculate the right hand
side of this equation for any x. And then you know how psi changes in time.
And therefore, you can integrate with your computer the differential equation and find the
wave function at a later time all over space, and then at a later time. So knowing the wave
function at one time determines the wave function at all times.
So we could run into a big problem, which is-- suppose your wave function at some time t0
satisfies this at the initial time. Well, you cannot force the wave function to satisfy it at any time.
Because the wave function now is determined by the Schrodinger equation.
So you have the possibility that you normalize the wave function well. It makes sense at some
time. But the Schrodinger equation later, by time evolution, gives you another thing that
doesn’t satisfy this for all times.
So what we will have to understand next time is how the Schrodinger equation does the right
So what we will have to understand next time is how the Schrodinger equation does the right
thing and manages to make this consistent. If it’s a probability at some time, at a later time it
will still be a probability distribution.

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