BARTON ZWIEBACH: De Broglie,
as we discussed last time, we
spoke about waves.
Matter waves.
Because people
thought, anyway light
is waves so the surprising
thing that would
be that matters are waves.
So a free particle with momentum
p can be associated to a wave--
to a plane wave, in fact--
plane wave-- with
wavelength lambda
equals Planck's constant over p.
So this wave is what eventually
becomes a famous wave function.
So de Broglie was
writing the example
or trying to write the example
of what eventually would become
wave functions, and the
equations for this wave
would become the
Schrodinger equation.
So really, this is a pillar
of quantum mechanics.
You're getting there when
you talk about this wave.
So Schrodinger's equation is a
wave equation for these matter
waves and this plane
wave eventually
will become the wave
function and there is
a Schrodinger equation for it.
So it's a wave of what?
he was asking-- de Broglie had
little idea what that wave was.
When you have waves, like
electromagnetic waves,
you have polarization, you
have directional properties,
the electric field
points in some direction,
the wave is polarized.
Is there a same property
for the wave function?
The answer is yes.
We'll have to wait a
little in 8.04 to see it,
but it has to do with spin.
When the particles
have spin, there
are directional
properties of the wave
and typically, you use
several wave functions
that correspond to directional
components of this wave.
So photons are spin
1 particle electrons
or spin 1/2 particles,
so there will be
directional properties to it.
But to begin with,
let's consider cases
where this directional
properties don't matter
so much, and for the
case of electrons,
if the electrons
have small velocities
or they are inside
small magnetic fields
where some of these properties
of the spin is important,
we can ignore that and work
with a wave function that
will be a complex number.
So it will be a wave function--
we'll denote it by the letter
psi, capitol psi--
that depends on
position and time,
and that's the wave function.
And to begin with, simplicity
will be one of them,
and it's a complex number.
And it's just one wave function.
And the obvious questions
about this wave function are,
is it measurable and
what it's meaning is?
So is it measurable?
And what is its meaning?
But to understand some
of that-- in fact,
to get to realize that these
waves are no ordinary waves,
we're going to think
a little about what
it means to have a wave
whose wavelength is inversely
proportional to the
momentum of a particle.
That's certainly a
strange statement
and probably these are
strange waves as you will see.
And by understanding that
these are strange waves,
we are ready to admit later on
that the interpretation could
be somewhat surprising as well.
And the nature of this number
is, again, a little strange
as you will see.
So all of that will come by just
looking a little more in detail
at this formula of
de Broglie and asking
a very simple question--
you have this particle
moving with some momentum
and I say, OK, it has
this much wavelength.
How about the person?
If one of you is moving
relative to me, like you usually
do with Einstein,
these observers
that are boosted, but let's just
do non-relativistic physics,
what is called the Galilean
transformation, in which there
will be another observer
moving with constant velocity
with respect to you and
you and that other observer
compare the results on the
momentum and the wavelength
and see if you find a reasonable
agreement or things make sense.
So we're going to try to
think of p is h over lambda.
And 2 pi's are very
useful sometimes.
So you put an h over
2 pi here and a 2 pi
over lambda and you rewrite
this in terms of quantities
that are a little more common--
one is h-bar and the other
is called the wave number k.
So these are these two
constants and this one
is called the wave number.
The 2 pi's are all
over the place.
If you have a wave
with some frequency nu,
there's also a frequency
omega, which is 2 pi nu.
So we're going to
look at this wave,
and it has some momentum
and some wave number,
therefore it has
some wavelength,
and let's see-- if
we compare things
between two different
frames, what do we find?
So we'll put the
frame S and a frame S
prime moving with some velocity
plus v in the x direction.
So the setup is
relatively common.
We'll have one frame
here that's the S frame,
and it's the x-axis
of the S frame.
And the S prime frame coincided
with the S frame at time
equals 0-- now it's moving,
so it's now over here,
it's S prime.
It has moved a distance of vt--
it's moving with
velocity v and there's t.
And S prime has--
and x is x prime.
On this, we're going
to write a few things.
We're going to say we
have a particle of mass m.
It has velocity v
underbar, otherwise
I'm going to get all
my velocities confused.
So this velocity v is the
velocity of the frame,
v underbar is the
velocity of the particle,
and v underbar prime,
because the velocity depends
on the frame of reference.
Similarly, it will
have a momentum--
and all the things we're
doing are nonrelativistic,
so momentum p or p prime.
Here is the particle.
And that's the position x
prime with the particle.
And that's a position
x of the particle.
So that's our system.
This particle is moving with
some velocity over here,
and we're going to compare
these observations.
So it's simple to
write equations
to relate the coordinates.
So x prime, for
example, is the value
of the corner at x of the
particle minus the separation.
So x minus vt.
And I should say it
here, we're assuming
that t prime is
equal to t, which
is good nonrelativistically.
It's fairly accurate.
But that's the exact
Galilean answer--
when you talk about
Galilean transformations
and Galilean physics,
it's very useful.
Even in condensed
matter physics,
people write these days lots of
papers about Galilean physics,
so when you have particles
moving with low velocities,
it's accurate enough, so
might as well consider it.
And these are the two ways
you transform coordinates,
coordinates and time.
So from this, we can take a
time derivative talking about
the particle-- so we have
dx prime and dt prime or t,
it's your choice--
I guess I should
put dt prime here,
dx/dt minus v, which means that
the velocity v prime underbar
is equal to v underbar
minus little v.
And that's what you expect.
The difference of
velocities is given
by the subtraction
of the velocity
that the frame is moving.
So if this particular has some
high velocity with respect
to the lab frame with
respect to this frame,
it will have a smaller velocity.
So this sine seems right.
And therefore,
multiplying by m, you
get that p prime is
equal to p minus mv.
So if you have
that, we would have
that lambda prime, the de
Broglie wavelength measured
by either running person,
is equal to h over
p prime is equal to
h over p minus mv,
and it's quite different, quite
substantially different from h
over p, which is equal to the
de Broglie wavelength seen
in the lab.
So these two de
Broglie wavelengths
will differ very substantially.
If this would be a
familiar type of wave--
like a sound wave
that propagates
in the medium, any kind of wave
that propagates in a medium,
like a water wave or
any wave of that type--
this would simply not happen.
In the case of those waves,
you get a Doppler shift--
omega is changed-- but
the wavelength really
doesn't change.
The wavelength is
almost like something
you look at when
you take a picture
and whether you take
a picture of the wave
as you run or you take
a picture of the wave
as you are sitting still, you'll
measure the same wavelength.
Let me convince you of that.
It's an opportunity to
just do a little more
formal transformations,
because these
are going to be Galilean
transformations,
simple transformations.
So our first observation is
that the de Broglie wavelength
don't agree, which
pretty much, I think,
intuitively is saying that
if you could just sort of see
those waves and measure
the distance between peaks,
they should agree, but they
don't, so there's something
very strange happening here.
