Prof: So the rules of
quantum mechanics are going to
be stated for one last time.
 
Now we're using the fact that
you've heard them over and over,
it's going to be more compact
than before.
So it's really the form of many
postulates, because it's a
theory which is based on
experiment;
it's not deduced by mathematics.
 
You make up some postulates
that condense all the
experimental facts.
 
There are new mathematical
results you will need,
but they are separate.
 
They are deducible
mathematically.
They are not part of the
physics postulates.
The first postulate--by the
way, this is the postulate for
Physics 201.
 
If you know more mathematics
and so on, there's another way
to write the postulate,
but we're not interested in
that.
 
We just want the rules for this
class,
and I don't even care how much
of this you carry in your head,
but I want you to know that in
principle,
every problem I gave you,
every question I asked,
can be answered just by turning
to these postulates,
so you should know what they
are.
So the first postulate is that
every particle--
this is for 1 dimension,
1 particle only--
is given by a certain wave
function,
Y(x),
that contains all the possible
information on that particle.
 
This is the analog of x
and p in classical
mechanics.
 
And we always choose Y
so that Y^(2)dx is
normalized to 1 in all of space.
 
That is a convention;
it's not required.
If you take a Y and you
multiply it by a number,
it stands for the same physical
state.
So you pick the number so that
the squared integral of Y
is 1.
 
The second thing is,
a state of definite momentum.
In other words,
is there a wave function that
describes a particle guaranteed
to have a definite momentum when
I measure it?
 
The answer is yes.
 
That function is denoted by
Y _p(x).
It is not an arbitrary function.
 
This Y is whatever you
like.
This is a particular Y
which assures you that if you
measure my momentum when I'm in
this state,
you will get the value
p, and it looks like this -
e^(ipx/ℏ)
divided by square root of
the length of that universe in
which you're living.
That's the second way to define
this function.
Either I can say this is the
answer, or I can give you a
little way to find this function
in an equivalent way.
The equivalent way says states
of definite momentum obey the
following equation.
 
If you solve this equation,
the function you get is in fact
Y_p.
 
So Y_p
can be defined either way,
either by saying,
solve this equation and then
the function you get is
Y_p,
but you can easily verify that
this function in fact is a
solution to that equation.
 
Now there's a new piece of
mathematics that's not connected
with this.
 
This is a postulate.
 
Mathematics says,
if you live on a ring,
you go around in a circle,
and the x goes to x
 L,
you've got to come back to the
same function.
 
That quantizes p, I have
the value as (2ph/L)
times m,
where m is an integral.
Let me call it n.
 
 m is the particle mass.
 
That is a mathematical
deduction.
You don't have to make that a
postulate.
That's the postulate;
that's a deduction.
Second postulate says--or third
postulate--a state of definite
position, x_0.
 
Namely, if you want to describe
a particle which is guaranteed
to be at x_0,
what function describes that.
I'm going to give the function
the name Y_
X0.
 
You have to be very clear on
what this x_0
is and what x is,
okay?
This p is a label that
says something about the wave
function.
 
This is a function of x,
so x varies from - to
infinity.
 
This is a label for the state.
 
What's special about this state?
 
It's got a momentum p.
What's special about this state?
It's got a definite location
x_0.
What is the function?
 
I think you should know by now
what the function looks like.
Anybody know?
 
State of definite
x_0?
Any guess?
 
Yes?
 
Student: 
>
Prof: Yes,
so I'm going to make it a
little simpler.
 
It's the function,
I'm just going to call it big
spike at x_0.
 
Now big spike at
x_0 is a little
vague and deliberately left so,
because this spike has a little
bit of non 0 value away from
x_0.
And you can make it narrower
and narrower and taller and
taller and that's what has a
technical name,
but I don't want to go there.
 
Let's imagine we can only
measure position to one part in
10 to the 97 meters.
 
Yes?
 
Student:  Is it the
Dirac delta function?
Prof: This one is not,
but eventually it will become
that.
 
But I just want to keep it like
this.
But instead of saying it's a
spike, I can also say it obeys
the following equation.
 
 x Y_x0
(x) =
x_0 times same
function(ie
Y_x0
(x)).
So let me take a minute to
discuss both these equations.
They are very special
mathematical forms,
and this is just for your
own--you don't have to know this
part of it.
 
This will do for this course.
 
If you want to know the whole
story--
see, normally,
when you take a function
Y(x),
and you differentiate it,
you will get a new function,
right?
Take a sine,
you'll get a cosine.
If you take x^(2),
you will get 2x.
So the act of taking
derivatives usually gives you a
brand new function,
a different function.
This one says,
except for those constants here
and here, if you take the
derivative, it should look like
a multiple of the same function.
 
Obviously, that's a very
special function.
You're saying upon
differentiating,
it's going to look the same.
 
Well, not every function has a
property, but we all know one
function does,
the exponential function.
Similarly, if you took a
function, any function Y
(x) and multiplied it by
x, you will get a new
function, right?
 
Cos x will become x
times cos x.
Sine x will become x
times sine x.
e^(x )will become x
times e^(x).
It becomes a completely new
function.
But I am demanding that for
this magical function,
when I multiply by x,
it is essentially a constant
times the same function.
 
You've got to ask yourself,
how can I take a function,
f(x), multiply it by x
everywhere and it looks the
same except for multiplicative
factor.
It looks impossible.
 
But the spike is in fact such a
function.
Let us see why.
 
If we take a random function,
if you multiply it by x,
which looks like this,
it will get taller and taller
as you go to the right.
 
But suppose the function in
question is a spike at
x_0.
 
If you multiply it by x,
you're basically multiplying
only by x_0,
because it has a life only of
x_0.
 
It doesn't exist anywhere else.
 
The only place it's even
non-zero is at the point
x_0.
 
So there's no difference
between multiplying it by
x, and multiplying it by
x_0.
Multiplying by x all
over here is a waste,
because the function is 0
there.
That's why for that spike
function, this is actually true.
You see that?
 
It's the second way that
characterizes it.
So again it says multiplying
by x is generally an
operation that changes the
nature of the function,
but there are special functions
for whom the effect is to simply
rescale the function by a
number.
Find me that function.
 
That's the state of definite
x_0,
and the only answer to that is
a spike.
So we have the option of simply
accepting the left hand side as
postulate 2 and postulate 3,
but there are equivalent ways
to say this.
 
And I will tell you in a minute
why I'm saying this,
because you can ask the
following question - what is the
state of definite energy?
 
The state of definite energy,
I'm going to write it this way
in the right hand side,
is also a solution to some
equation.
 
That equation is −(
â„ ^(2)
/2m)(d^(2)
Y_E/dx ^(2))
V(x)Y_E
(x) =
EY_E
(x).
But I'm trying to tell you that
I'm not going to elevate this to
a postulate,
because I think one of you
noticed already something very
familiar about this equation.
Because, you see,
in classical mechanics,
energy is given by
(p^(2)/2m)
V(x).
 
Do you see that on the left
hand side,
except for the function
Y, I have what here looks
like ½m
x(-iℏd
/dx)^(2)
VY.
That's the left-hand side.
 
So that once you know states of
definite p and states of
definite x obey these
equations,
states of definite anything
else, any function of x
and p, obeys a
similar equation,
but every p is replaced
by −iâ„ 
d/dx,
and every x is just
simply left as an x.
 
So this energy formula does not
require a new axiom,
but if you didn't follow this
logic,
you can simply say I have one
more postulate,
which is that if I have states
of definite energy,
I must solve that equation.
 
But then you will need more and
more postulates,
because you've got zillions of
possible variables you're
interested in.
 
What if somebody wants to know,
what is x^(2)
p^(2)?
 
That's a variable in classical
mechanics.
But I'm not going to write
another postulate for that,
because if you wanted
x^(2)  p^(2),
states of definite x^(2)
p^(2),
it will obey the equation
((-iℏ)^(2)
d^(2)/dx^(2)) x Y
= some constant,
whatever it is, times Y.
 
Namely, p is going to be
replaced by the square of this
derivative.
 
Square of a derivative means do
it twice.
So there's a universal rule for
finding states of definite
anything.
 
Let g be a
general variable in
classical mechanics.
 
In quantum theory,
states of definite g
satisfy the equation where you
first write the classical
formula for g.
 
Maybe g is x^(2)
p^(2).
Replace x simply by
x;
replace p by
-iℏd/dx
and let that multiply as a
function or differentiate some
function,
and that will be a state of
definite g.
 
So I'm just giving you two ways
to do this postulate.
Either you can say states of
definite energy are solutions to
the following equation,
that's HY= E
Y,
and that's what I've been doing
so far,
because I didn't mind in this
class to introduce it as one
more postulate.
If you're looking further
ahead, you should realize that
it is really not an independent
postulate,
that once you know what happens
to x and once you know
happens to p, you can
predict what the equation will
be for any function of x
and p. So if you see
that analogy and it's helpful to
you,
you're welcome to the right
hand side of the board.
If you find it complicated,
add on to your list of things
to remember one more thing.
 
If you want a state of definite
energy, you've got to solve this
equation.
 
Okay, now we come to another
postulate, postulate 4.
Is that right?
 
Postulate 4.
 
If Y(x) is
written as a Σ_g
A_g
Y_g(x)
where g could be
p,
g could be E,
g could be position,
whatever you like,
then the probability that you
will get a value g is
simply
|A_g|^(2),
and A_g is
given by ∫Y_g
'(x)Y
(x)dx.
 
Perhaps you can see now,
if I put g = p,
this is what I call
|A_p|^(2),
and this will be e^(ipx
)etc.
If I put g = energy,
this will be
A_E,
this will be Y'E.
Whatever variable you want,
you do these integrals and
you get the
probabilities.
Then there's another postulate,
postulate 5.
Postulate 5 says if g is
measured and you get one value,
say g_10--this
is a postulate,
you don't have to write
g_10.
You can say one particular
value--then the
Σ_g
A_g
Y_g
collapses to one term,
just
Y_g0.
g_0 is a
particular value of g,
namely the third value,
the fourth value or the ninth
value.
 
This says the state collapses.
 
So originally,
you have a state,
a general one,
which is potentially capable of
giving any possible answer
that's contained in this sum,
with the probability
proportional to the square of
the coefficient.
 
But once you made a measurement
and you got a particular
answer g, call it
g_0,
all the terms of the sum
disappear except the one term,
corresponding to the one answer
you got.
That's a new property of
quantum mechanics,
that the act of measurement
changes the state from being
able to do many things to being
able to do one thing.
The one thing could be
g, if you measure g.
Namely,
it can be momentum,
it can be energy,
it can be position,
but you've got to be very
careful that if you measure
momentum and got a state of
definite momentum,
then you go and measure
position and got a state of
definite position,
and you go back and measure
momentum,
you won't get the same answer.
States of definite momentum
don't know what the position is,
and states of definite position
don't know what the momentum is.
You will never be able to get
them both.
You've got to pick one or the
other.
By the way, I want to point out
to you one postulate seems to be
somewhat missing,
the oldest thing we ever
learned in this class,
right?
Why didn't I mention that?
 
That's actually contained here.
 
It's contained here because if
I say what's the probability
that x has a value
x_0,
I will take integral
Y(x_0
)'Y(x)
dx.
Look at what this is.
 
Y(x_0)'
is a spike at
x_0.
 
Y(x) is some
function.
But the only place where the
products survives is near
x_0.
 
Everywhere else is 0,
so the only place that integral
receives a contribution is
when x is at
x_0.
 
So up to some constant
proportional to area of this
spike, the answer is
proportional to the wave
function at
x_0.
Therefore--I'm sorry,
I mean to say square this,
and that is what I've been
saying is the rule for position.
So again, for position you can
write a separate postulate,
but it's not.
 
The postulate that I gave you
here, if you put g =
position, will give you the same
answer, namely this one.
Again, if you don't want to get
into too many details,
I would say the following is
what you should know.
You want a stated definite
momentum,
or the probability for definite
momentum,
take the integral with
Y_p
. For energy,
take Y_E.
For definite position,
forget all integrals.
Just take Y at the
point x and then square
it.
 
What I'm telling you is taking
Y at the point x
and squaring it is what you
will do if you took the integral
of Y with a spike,
which will filter out,
if you like,
Y at only the one point
x_0.
 
And if you square it,
you will get Y
(x_0).
 
And the last postulate,
all-important postulate 6,
says that the state changes
with time according to the
Schrödinger equation.
 
I'm just going to call it
HY,
but from now on,
it will be -d^(2)
Y/dx^(2)
V(x)Y(x).
If I get tired of writing it,
I will call it HY.
That's of course a postulate,
because you can never derive
Schrödinger's equation any
more than you can derive
Newton's laws.
 
This is, as I said,
one of the most powerful
equations in modern physics,
because it contains all of
non-relativistic quantum
mechanics and all of classical
mechanics,
and all of solid state physics
and super fluids and
superconductors.
Everything comes from this
equation.
A very powerful equation.
 
Okay, now what did we learn
from this equation?
I will mention one thing that's
very important.
It's a consequence of the
equation,
which is, if you start the
system out at time t = 0,
in a state of definite energy
and you see what happens if I
wait some time,
well, it becomes the following
function at time t.
 
It's essentially the same
function, multiplied by
[e^(i)]Et/ℏ.
 
You realize,
generally a wave function,
when you let it evolve,
will flop and wiggle and change
its shape in a complicated way.
 
But if you release it in this
initial condition,
all that happens to it is that.
 
These are called stationary
states, because despite the time
dependence I showed you,
the probability for measuring
anything is time independent.
 
Because when you take the
probability, you find the
absolute value of some integral,
and the absolute value of this
factor just goes away.
 
So it's time independent.
 
That's why we're very
interested.
If you want an atom and the
atom has been sitting for a long
time, it will be in one of these
states of definite energy.
So that's the case of starting
in a state of definite energy,
but then I said,
"Maybe I don't want to
start in a state of definite
energy;
I want to start in an arbitrary
state."
At time 0 I'm going to give you
a Y (x ,0) and I'm
going to ask you,
"What is the fate of this
state,
if I wait some time
p?"
 
The simple answer is,
take Y(x ,0),
calculate
A_E=∫
Y_E'
Y(x,0)dx and then
Y(x,t) =
A_E
Y_E
(x)
e^(−iEt/â„ ).
 
In other words,
if you take your initial state
and expand it--
this is called expanding it--as
a sum over states of definite
energy with some coefficients
you compute at t = 0,
then at any future time,
the coefficients are given
simply by the initial
coefficients,
times this factor.
You can check that this will
satisfy the Schrödinger
equation, because every term in
it is a solution to the
equation.
 
And at t = 0,
when you drop this guy,
it agrees with the initial
state and that's all you want.
Because in this problem,
because it's first order
equation in time,
if my initial state matches
your initial state and satisfies
Schrödinger equation,
it is the answer.
 
There are no two answers.
 
So I want to give you
one--first let me tell you
something that's going to make
you relax.
Everything that I say after
this moment is not in your exam.
I don't want you to worry about
that, because it's an
interesting thing and I want to
tell you all the things you can
do with quantum mechanics.
 
I don't want you to worry about
anything.
Just try to follow this and ask
questions, and get a glimpse of
what you could possibly do with
this theory.
First thing you can do is you
can say, "How do I
understand atoms?"
 
You know the simplest atom in
the world is a hydrogen atom.
It's got a proton and it's got
an electron.
One has charge e,
one has charge -e.
And the proton can be taken to
be so heavy that it is fixed.
Just like when the earth goes
round the sun,
in principle,
the sun is also moving,
but we ignore that,
so we're going to ignore the
motion of the proton.
 
Then the electron,
in classical mechanics,
it has an energy which is
(p_x^(2)
p_y^(2)
p_z^(2))/2m.
That's the kinetic energy,
and the potential energy
-ze^(2)/r, and r
is √(x^(2) y^(2),
z^(2)). Do you understand
that?
That's the classical formula
for energy.
Then our recipe tells us that
in quantum theory,
states of definite energy will
obey the following equation -
-ℏ^(2)/2m x
(d^(2)
Y/dx^(2)
d^(2)Y
/dy^(2)
d^(2)Y
/dz^(2) ) -
(ze^(2)/√(x^(2) y^(2)
z^(2)))Y=
EY.
You have to solve that equation.
 
Now don't worry about how you
solve it.
In fact, even Schrödinger
did not know how to solve it.
He had to ask a mathematician
friend of his how to solve it.
A mathematician owed him a
favor, because at that time,
the mathematician was having an
affair with Schrödinger's
wife and you might say,
"Who owed the favor to
whom?"
 
Actually, Schrödinger owed
a favor to this guy,
because meanwhile,
Schrödinger was trying to
seduce a pair of underage twins.
 
These are all very interesting
things you don't know about the
lives of famous people,
but if you read his biography,
you will find that
Schrödinger's wave function
was all over the classically
forbidden region.
Anyway, he's a very interesting
person.
But what I want you to know is
that even he couldn't solve the
equations.
 
I don't care if you solve it or
not, because he had already done
the great thing.
 
So you go to this equation and
you solve it.
So this math guy helped him
solve it.
Nowadays, undergraduates,
graduates, everybody knows how
to solve it, but the first time
it came, it was quite an
unfamiliar equation.
 
If you solve this thing,
what you find is that the
energy can take only certain
values.
Those values have some number
in front of them a 13.6 electron
volts.
 
That 13.6 electron volts is
some combination of the electric
charge of Planck's constant and
mass of the electron.
Forget all that.
 
Some number,
divided by n^(2) where
n is an integer of 1,2,
3, etc.
That's it.
 
These are the only allowed
energy levels of hydrogen.
If you solve this equation in 3
dimensions and you demand the
functions vanish at infinity,
rather than blow up at
infinity, you will find you can
get solutions only at certain
special energies and these are
the energies you get.
So let's plot these energies.
 
Remember this is energy 0 and
this is n = 1.
n = 1 is at -13.6eV.
 
n = 2 will be -13.6/4eV.
 
n = 3 will be that thing
divided by 9 and so on.
These are the allowed energies.
 
See, that's a great result,
because you are able to find
out for the first time what
energies are available to this
atom,
and furthermore,
you also learn that if this
atom were able to absorb light,
it can do that only say by
going from here to here or going
from here to here,
or going from one allowed level
to another allowed level.
 
And the difference in energy is
E_f -
E_i will be
ℏw of the
photon.
 
That will allow it to go up.
 
Or it can also come down,
and if it comes down from some
level,
n = 4 to n = 2,
it may,
let's see, 1,2,
3,4, it can go from 5 to 2,
it will emit some light.
In fact, this is the only way
we know anything about the
hydrogen atom.
 
No one can see the hydrogen
atom.
You cannot actually see it
because the act of observation
will destroy everything.
 
But we know it's there;
we know it's doing its thing by
shining light on it and seeing
what it absorbs and what it
emits.
 
The understanding of the
quantum world is very different
from classical world.
 
For example,
if Newton says,
"I know the orbits or
ellipses,"
you can go see them.
 
That's what Kepler did,
looked at them for 40 years,
you can see they go in
elliptical orbits.
For hydrogen,
there are no orbits.
There are energy levels and
there are these corresponding
wave functions,
Y (x,
y ,z).
 
And you can plot them and you
know if you start out with any
one particular state,
it will stay that way forever.
And the probability density,
I told you, is also fixed in
time.
 
They are the clouds one finds
in textbook as shape of various
atomic orbits.
 
Okay, yes, there's only one
other thing I should mention,
which is that even though these
are the only allowed energies,
there's more than one state at
a given energy.
You may remember when I did a
particle on a ring,
at a given energy,
there are two states of
momentum.
 
One has got momentum = square
root of 2mE,
other has got momentum square
root of -2mE,
because when you take the
square and divide by 2m,
you get the same energy.
 
So a given energy can have two
states.
Here a given energy can have
many states.
There's only one here and here
you can have 4,
and here you can more.
 
There are ways to calculate
them.
Then it turns out that even
this 1 is actually 2,
because it's another variable
which does not enter our
physics, called the spin of the
electron.
I don't want to even go there,
but it can do one of two
things.
 
So everything I do here,
you've got to double.
So really, there are two states
of energy, lowest energy and
there are 8 here and so on.
 
These magic numbers come out of
doing the Schrödinger
equation.
 
All right, so that's an example
of what you can do with quantum
mechanics.
 
You can solve for the spectrum
of atoms and you can see what
light they will emit,
what light they will absorb.
You can even go beyond that.
 
You can even ask,
what's the rate at which it
will absorb light?
 
What's the rate at which it
will emit light?
Everything can be computed,
because you see,
you not only know the energies,
you also know the corresponding
functions and they are very
important in calculating the
rate of absorption and the rate
of emission.
Okay, that's one thing you can
do.
Now I'm going to tell you about
another uncertainty principle
which takes the following form.
 
It says DE
Dt >
=ℏ.
 
I've got to tell you what it
means.
It looks a lot like this one,
DpDx
> = ℏ,
but it's very different in
nature, because x is the
position of something.
You can try to measure it.
 
 p is the momentum of
something, you can try to
measure it.
 
t is not the time of
something.
t is just time.
 
It's not time of this or that,
and it can be measured
arbitrarily accurately.
 
That is not the problem.
 
So I will have to tell you the
meaning of this uncertainty
relationship.
 
It is different from the others.
 
There are many such things in
quantum mechanics,
but this is unique in the sense
that unlike x and p
which are physical
variables,
time is not a variable
describing a particle.
It's an independent parameter
along which everything happens.
But I'll tell you what this
means,
but before that,
I want to go back to the
uncertainty principle,
which you notice I never
mention, because it is not a
postulate,
it's a consequence.
 
And I told you all about trying
to look at something with a
microscope and have you shine
photons.
And a lot of you tried to find
ways to get around that,
and say, "Maybe if you did
this,
maybe if you did that,
you can beat the uncertainty
principle."
 
So I'm going to put you out of
your misery.
You cannot beat the uncertainty
principle for the following
reason.
 
Forget about quantum mechanics.
 
Let's ask the following
question - what is the
wavelength of this signal?
 
This is 1 meter long.
 
You might say it's 1 meter,
but that's not a correct
answer.
 
This signal doesn't have a 1
meter wavelength.
You know who has a 1 meter
wavelength is this guy who goes
on forever.
 
That is a wave of 1 meter
wavelength.
This was 1 meter for a while,
then completely dead on either
side.
 
So this does not have a
definite wavelength,
even though you think it does.
 
A wavelength is a repetitive
phenomenon in space.
It's got to repeat itself
indefinitely to have a unique
wavelength.
 
So let's take a wave train that
looks like this.
You somehow chop it off nicely
near the end.
It does some number of
oscillations and then it stops.
What is the wavelength
associated with this?
You can associate an
approximate wavelength as
follows.
 
First let me remind you that we
want to think in terms of
k which is 2p/l.
 
It's a reciprocal of a
wavelength, and if you remember,
functions look like kx -
wt when we studied waves.
k is that.
 
What's the meaning of k?
 
k is the rate of change
of phase of the wave,
because if you go a distance
x, the phase changes by
kx.
 
You watch the wave change.
 
Every cycle is worth 2p and it
changes by some amount over some
distance.
 
k is the ratio of how
much phase change you had
divided by the distance you
travel.
Do you understand that?
 
That's k.
 
So let me take this wave and
the k for this will be 2p times
the number of cycles here
divided by the length of that
wave train.
 
But there is no unique n
you can associate with it.
These are all full cycles,
but once you come near the end
there's a little bit of
confusion as to what to do at
the end.
 
How do you close it off?
 
So there's an error or
uncertainty of about 1 cycle
coming from the 2 ends.
 
So the uncertainty in k
is 2 p/L times the
uncertainty in n which is
1 or 2, but that's all it is.
You understand?
 
A finite wave train,
when you round out the edges,
it may not have a full number
of cycles.
You taper it off in some
fashion, but here you've got a
well defined number of cycles.
 
So it's or - 1 coming from the
ends.
So Dk is that.
 
But this is a particle which is
now confined to a region of
length L because the wave
function is 0 beyond that.
So it's a particle whose
position is known to lie inside
this interval of length
L.
So the uncertainty in position
is L.
Again, uncertainty is a
technical definition,
but this is our qualitative
explanation.
So L is really
Dx.
You can write Dx.
 
Dk is 2p.
 
This has nothing to do with
quantum mechanics.
Where did I mention quantum
mechanics?
I'm saying something
intuitively very clear.
If you're defining a
wavelength, you have to let it
run through many cycles.
 
For example,
if you say this guy goes to New
York every week,
and you observed him for only
one week, that's got no meaning.
 
God knows what will happen next
week.
But if you studied a person for
50 weeks and found 50 times the
person went to New York,
it's more credible when you say
this person goes to New York
once a week.
So periodic phenomenon,
either in time or in space,
are defined only after many,
many, many periods.
Therefore a wavelength cannot
be defined for an arbitrarily
short interval.
 
In fact, the longer the
interval, the more wavelengths
you can fit in to say that's my
wavelength.
So wavelength gets more and
more defined as the train gets
longer and therefore the
incompatibility between
wavelength and the length of the
train,
the fact that if it's too short
in one,
it's too big in the other one,
is a classical result.
Quantum mechanics comes in with
a new relation that the momentum
of a particle is connected to
this wave number by
ℏk.
 
Multiply both sides by
ℏ,
ℏ there and
h here,
then it becomes
DxDp
is roughly about an ℏ.
 
So once you concede that a
particle has definite momentum
only if it has a definite
wavelength in its wave function,
that wave function has to be
very extended to have a definite
wavelength and therefore a
definite momentum.
So there is no way you can make
it--there's no way you can take
a wave arbitrarily short in its
extent with an arbitrarily well
defined wavelength.
 
You've got to let it do its
thing many times.
Mathematically what happens is,
there's a mathematical theorem
that says any function you write
down can be built out of waves
of all possible wavelengths with
suitably chosen coefficients.
If your wave has an almost well
defined wavelength,
then the coefficients,
as the function of wavelength,
will have a very sharp peak at
the wavelength corresponding to
this.
 
And as the wave repeats more
and more and more times,
the coefficients will become
sharper and sharper and sharper.
As the wave becomes very small,
if you say "What's the
wavelength of this guy?"
 
it will be very broad.
 
It's nothing to do with quantum
mechanics.
It's just the incompatibility
between two qualities,
you understand?
 
Wavelength needs some time to
express itself.
You cannot do it in a tiny
region.
So if you want to localize the
particle, how can it tell you in
the tiny region what its
frequency of oscillation is in
space?
 
So once you concede that,
once you concede that
wavelength is connected to
momentum, you cannot escape the
uncertainty principle.
 
So now I'm going to come back
to energy and time,
and I come back in the
following way.
Remember, a state of definite
energy E behaves as
follows in time.
 
It looks like Y(x)
times e^(−iEt/
ℏ).
 
So state of definite energy,
you know the particle has
definite energy if you observe
the wave function and it
oscillates in time.
 
And you can write this as
e^(-i)^(w)
^(t),
where w is
E/ℏ.
 
So your particle that has been
in that state forever has a well
defined energy.
 
But if a particle was just put
in that state right now,
the wave function really does
this only from t = 0,
before that it was something
else, then this function has not
oscillated enough times for you
to define its time period.
So if you're watching an
oscillation,
let's say it's an oscillation
in time,
and you say,
"What's the time period of
oscillation?"
 
again, if it extends over a
certain time T,
the frequency w is just the
phase change per unit time.
So it's 2p times the number of
cycles, but there's an
uncertainty in N of order
1 that's 2p/T.
Therefore if the system has
been in existence for a time
T and I call that time as
Dt--
that's the meaning of
Dt--
Dt Dw is 2p,
then quantum mechanics comes in
when you put an
ℏ here,
you put an ℏ
there and you call that the
uncertainty in energy.
 
In other words,
if someone tells me,
"Go see if that clock is a
regular clock.
Tell me if it's running
rhythmically and
periodically."
 
Well, if you give me enough
time, I observe it for 100
cycles, I say it's a good clock.
 
But if you don't give me time
even to finish one cycle,
how can I tell you anything
about its period yet?
It hasn't had time to establish
a period.
Periodic phenomena have to
repeat themselves,
so it takes some time.
 
Therefore for a state to have a
well-defined period,
that is to say,
a well defined energy,
it has to be in existence for
some time.
Then it has a well-defined
energy.
Okay, therefore you can say
that if you've had time to
measure energy only for a finite
time,
then energy will not be defined
to better than this accuracy,
DE
Dt is of order
ℏ.
 
Let me give you an example that
is from classical mechanics,
nothing to do with quantum
mechanics.
You take a bunch of reeds,
you attach them to the wall.
Each one has a different length
maybe.
They have many frequencies of
vibration, natural frequencies
of vibration,
depending on the length and
whatnot.
 
Now if you connect this whole
thing,
apparatus, to some vibrating
gadget,
start shaking the whole thing,
suppose you shake it at a
certain frequency,
omega 0 that matches maybe this
guy.
 
Namely, it resonates with this
guy, so your expectation is,
this one will oscillate like
crazy and the others will not.
So let me look at this rod from
the edge, okay?
I look at them end on,
they all look like this.
They're coming out of the
blackboard.
So I expect this one to be
resonating wildly up and down
and the others to be ignoring
the signal because they are not
at the resonant frequency.
 
But what will happen in real
life is that you will take the
system and you start shaking it,
your intention is to shake it
at a very definite frequency
that matches this guy,
but the system does not know
your plans.
After quarter of a cycle,
it knows you've exerted that
force.
 
It doesn't know you plan to
keep doing this.
So at that time,
what it will do is it will try
to write this as a sum of many
waves of definite frequency.
They will contain many,
many frequencies,
so what you will find is,
this guy also oscillates a
little bit,
that guy also oscillates a
little bit,
and as you wait longer and
longer, so the system caught on
to the fact that you are really
serious and sending a signal of
definite frequency.
And when it has stabilized,
you will find these guys don't
move, these guys don't move;
the guy at the resonant
frequency moves.
 
So you've got to understand,
it takes some time for the
system to know what frequency
you're sending.
If you have not sent many
cycles, if you have sent only
this much,
it does not consider that a
periodic function,
because to it,
the function would really do
that or it could do something
else.
 
It goes with the information it
has.
It goes to the mathematical
tables and finds the Fourier
series for this and sees what
frequencies are there.
And anything that's not 0 can
excite all these things.
And after a while,
when the frequency is well
defined, one reed moves.
 
Same thing happens with atoms.
 
I don't know,
somewhere I drew a picture of
atoms absorbing energy.So take
this atom here in the ground
state of hydrogen.
 
Send light of the frequency
just correct to go to the first
excited state,
to n = 2.
You take a laser or something
of that frequency and you hit
the atom.
 
Take a collection of atoms and
you will think they will all
jump from the ground state to
the first state but nowhere
else.
 
But you will find the minute
you turn on the laser,
even though the laser dials say
this is my frequency,
atoms don't know that.
 
So initially,
it will jump like crazy to all
kinds of energy states and only
after sufficient time,
only after many cycles have
happened,
it will realize, "Hey,
this is the frequency this
guy's sending me."
 
Then it will start going
preferentially to the one state.
Okay, now the final topic I
want to talk about is really the
beginning of a long topic,
but I want to give an
introduction to it because
you're going to need this.
And the question is,
what if I have not just 1
electron but 2,
not 1 particle but 2.
After all, in real life,
there are lots of particles.
What does the quantum mechanics
of more than 1 particle look
like?
 
So if there are 2 particles,
let's say an electron and a
proton,
you will not be surprised that
you will now have a wave
function of 2 variables,
and if you squared that wave
function,
you will get
Y(x_1)
Y(x_2)
--
Y(x_1),
x_2^(2),
will give you the probability
that the proton is at
x_1 and the
electron is at
x_2.
 
You've got 2 things,
you've got to give 2
probabilities,
but once again,
probabilities come from
squaring a function.
This has to be a function of 2
variables, because each guy has
his own position.
 
So you can expect that in
quantum theory of 2 particles,
you'll get a Y that
depends on 2 coordinates,
with 3 particles,
Y that depends on 3.
But let me just stop with 2,
because you can learn some very
profound things in about 5
minutes just by starting here.
Now let's consider,
so let's take a concrete
example.
 
These guys are both in a box,
let's say.
Let's take a simple case,
Yof x1 x2 = Ya of
x1 Yb of x2,
where a and b are wave
functions of energy in a box
with 1 particle.
For example,
this could be the state a and
this could be the state b.
 
The probability to find the
proton at some place and the
electron at some place is not
the same as the probability of
finding them with exchanged
positions.
For example,
the chance of finding the
electron here and the proton in
the middle of the box is 0.
The proton's function vanishes
in the middle of the box.
But the probability of finding
the electron in the middle of
the box and the proton here is
not 0.
That's perfectly okay,
because there are two different
outcomes of the experiment.
 
I look for the particles.
 
I find the electron here and
the proton there.
I say I found this guy here and
that there, and I do it many,
many times and you give me the
odds.
But something very dramatic
happens if the 2 particles are
identical.
 
Identical particles in quantum
mechanics have very different
connotations from identical
particles in classical
mechanics.
 
So in classical mechanics,
suppose you have identical
twins, okay?
 
They're born and they're
separated, they're moving
around.
 
Let's say they look identical
in every way.
We can still follow them.
 
We know this is Joe and this is
Moe.
Follow the two characters no
matter how identical they are,
because we can keep track of
them continuously.
So let's do the following
experiment involving these
twins.
 
There are 4 doors in this room
and 1 twin comes out like this,
the other twin comes out like
that.
Then they do one of 2 things.
 
Either they cross over like
this, or this guy goes back to
that door, that guy goes back to
this door.
Now suppose you saw them
running like this in the
beginning,
and you left the room when this
was happening,
and you saw them enter these 2
doors,
you cannot tell which of the 2
happened,
because you've just got 2
identical twins and these 2
identical exit doors.
But somebody knows what happens.
 
Somebody in that room who was
watching them can clearly tell
you, if this happened or that
happened, because you can follow
the twins at all times.
 
So even though they're
identical, they're
distinguishable.
 
They cannot swap roles without
your knowing.
But imagine now that these are
not classical particles but
quantum particles,
like electrons.
For an electron,
you don't have a definite
location or a trajectory.
 
You only have probabilities.
 
So you know an electron was
emitted here and emitted here,
and later on was absorbed here
and absorbed here,
and you cannot tell who really
came here.
Was it this guy or was it that
guy?
There's no way to tell.
 
So when you have identical
particles whose trajectories you
cannot follow,
when you catch a particle here
and a particle there,
you cannot say Joe was here and
Moe was there.
 
It's not allowed,
because you're not following
their names.
 
You can only say,
"I found a particle here,
I found a particle there."
 
Therefore the theory cannot
give different probabilities for
finding particle 1 here and
particle 2 there,
and particle 2 there and
particle 1 here,
because the outcomes are
indistinguishable,
so the probabilities must be
equal.
So for 2 identical
particles, p (x_1
x_2) must =
p( x_2
x_1).
 
That's not true for these
functions.
Now I'll write a function for
which it's actually true.
Can you make a guess on what it
may be, what kind of function
will respect the fact that if
you swap the coordinates,
probabilities don't change?
 
We know there's particle in
state a and a particle in state
b and they're identical.
 
This says particle 1 is in and
2 is in b, but you're not
allowed to say who is who,
so the correct answer is,
you can also do this.
 
Look at this function now.
 
Here it's a superposition,
quantum mechanical
superposition of --
1 doing something,
2 doing something else,
the opposite.
You add them together.
 
Now I invite you to check that
if you exchange
x_1 and
x_2,
see what happens.
 
This becomes
Y_a(
x_2),
that guy sitting here.
This becomes
Y_b(
x_1),
that guy sitting here.
This becomes
Y_b(
x_2),
that sitting here.
So these 2 terms exchange roles
when you exchange the particles.
Y(x) to
x_1 is in fact
Y(x)--
Y(x_1,
x_2) is the same as
Y(x_2,
x_1).
 
So that is an allowed state in
quantum mechanics,
where you add them and you
symmetrize the product.
You guys following that?
 
Now I'm going to try another
combination.
I'm going to put a - sign here.
 
If you put a - sign,
perhaps you can see if I
exchange the 2 guys and this
goes into that,
that goes into this,
I don't come back to where I
am.
 
I come back to where the - of
this guy.
If this looked like x -
y, that looks like y -
x.
 
So you might say,
"Hey, things are not the
same when I exchange the
particles."
But remember,
the physical quantities are not
given by Y but the square
of Y.
The probabilities are not
Y;
the probabilities are
Y^(2).
So even with the - sign,
the probability for
x_1,
x_2 here is also the
same as the probability for
x_2 ,x_1
here,
because when you find the
absolute value,
the - sign goes away.
So in quantum mechanics,
there are 2 options for
identical particles.
 
Either you can take the product
function and add to it the
product with the reversed
coordinates, or subtract from
it.
 
And the particles in the world,
all of them decide to go with
one camp or the other.
 
Particles called bosons always
choose that sign,
and particles called fermions
always choose the - sign.
And every particle is either
boson or fermion.
P mesons are bosons.
 
Photons are bosons.
 
Electrons are fermions.
 
Quarks are fermions.
 
Gravitons are bosons.
 
So everybody is one or the
other.
If you put 2 bosons in a box,
their wave function,
when you exchange them,
will remain the same.
If you put 2 fermions in a box,
the wave function will change
sign.
 
But now here is the beautiful
result.
Take this case for 2 fermions
and ask yourself--this is a
fermionic wave function for 2
particles.
Ask yourself the following
question - can they both be in
the same quantum state?
 
In other words,
can the state a be the same as
the state b?
 
Remember, here a was this and b
was that.
I'm asking, can they both be in
the same state?
So I invite you to put a = b
here.
Let's see what happens.
 
That's a.
 
That's also a.
 
What do you get?
 
Can you see something when both
are a?
You get 0.
 
So you cannot write a wave
function in which the 2 fermions
are in the same state,
and that's the Pauli principle.
Pauli principle says,
for some of the particles,
they cannot be in the same
state.
Bosons, on the other hand,
can be in the same state and
like to be in the same state.
 
I don't have time to talk about
that, but this is the Pauli
principle.
 
And you can show if you've got
3 fermions and 4 fermions and so
on,
you will find out the quantum
mechanical wave functions never
allow any 2 of them to be in the
same state.
 
And that is the origin of the
entire periodic table of atoms,
because what you do when you've
got many electrons in an atom
is,
you find these energy levels
that you did,
the 1/n^(2) that you
got,
but the nuclear charge of
course may not be 1.
 
It's some non 0 number,
depending on how many charges
are in the nucleus.
 
You take these levels,
I don't know how many levels
there are here,
then you start putting
electrons into them.
 
The first electron will go here.
 
It turns out they have
something called spin,
so I will just say 1 way goes
with down and 1 way goes with up
and down.
 
The third electron,
you have to put here.
It has to go to higher energy
state.
If they were bosons,
you can put them all in the
lowest energy state.
 
That world will look completely
different from the world we live
in.
 
In the world we live in,
the levels keep filling up as
you put more and more and more
electrons.
They've got to go to higher and
higher energy levels.
You go to higher and higher
energy levels,
what happens once in a while
is, you fill all of this right
now, then that atom becomes very
passive.
It is not interested in either
giving up electrons or taking
electrons.
 
Whereas if you had one more
electron here,
it's very happy to lose this to
some other atom,
which may have just one
electron in its lowest state.
It's a waste of it to be here;
it will go sit there.
Sometimes they like to give an
electron to another atom,
or if they do,
this will become positively
charged and that will become
negatively charged and they have
an attraction and they stay
together as a molecule.
So you can understand that as
you go on piling more and more
electrons, a time will come when
this level is completely full.
That atom, whatever its place
in the periodic table is,
will be also very passive.
 
So you will find things which
are electrically very active
with loose electrons in the
upper,
called valence states,
and the inner shells are
filled,
they are not.
They are inert.
 
So the behavior of active,
inert, active,
inert is periodic,
and that's the periodic table
that is observed.
 
But you get that from quantum
mechanics in great detail by
solving for the energy levels,
then using the Pauli principle
to put only 1 electron there
every quantum state,
and you can see a lot of this
behavior can be anticipated.
So anyway, these are things you
will learn if you learn
chemistry, if you learn physics.
 
You will see where everything
comes from.
All right, so this is the end
of the quantum mechanics part.
I'm just going to tell you that
I want to stop here and I'll see
you for the discussion section.
 
I'm really going to miss my
Mondays and Wednesdays because
for me, that's the best time of
the week.
So really good to be with you
guys.
Thank you.
 
>
 
 
 
