WEBVTT
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BARTON ZWIEBACH: Do
normal wave analysis
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to demonstrate that indeed these
things should not quite happen.
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So for that, so ordinary waves
and Galilean transformations.
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So when you have a wave,
as you've probably have
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seen many times before,
the key object in the wave
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is something called
the phaze of the wave.
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Phaze, the phaze.
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And it's controlled by this
quantity kx minus omega t.
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k being the wave number, omega
being the angular frequency
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and we spoke about.
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And the wave may be sine of that
phaze or cosine of that phaze
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or a linear combination of sines
and cosines, or E to this wave,
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any of those things
could be your wave.
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And whenever you have
such a wave, what we say
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is that the phaze of this
wave is a Galilean invariant.
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Invariant.
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What it means is that two
people looking at this wave,
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and they look at the
point on this wave,
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both people will agree on
the value of the phaze,
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because basically, the
reality of the wave
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is based on the phaze, and if
you have, for example, cosine
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of this phaze, the place
where this cosine is 0
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is some of the phaze, and if
the cosine is 0, the wave is 0,
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and everybody should agree that
the wave is 0 at that point.
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So if you have a place where the
wave has a maximum or a place
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where the wave is 0,
this is an ordinary wave,
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everybody would agree that at
that place you have a maximum
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and in that place you have a 0.
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So observers should agree
on the value of this phaze.
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It's going to be an invariant.
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And we can rewrite this phaze
in a perhaps more familiar way
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by factoring the k, and then
you have x minus omega over kt,
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and this is 2 pi over
lambda, x minus--
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this quantity is called
the velocity of the wave,
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and we'll write it this way.
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And I'll write in one last way--
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2 pi x over lambda minus
2 pi V over lambda t.
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And this quantity is omega
and this quantity is k.
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So this is our phaze.
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And we've said that it's
a Galilean invariant, so I
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will say that S should see--
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the observer S prime
should see the same phaze--
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phaze-- as S. So phi prime,
the phaze that S prime sees,
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must be equal to phi when
referring to the same point.
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When referring to the same
point at the same time.
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Let's write this.
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So phi prime should
be equal to phi.
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And phi, we've written there.
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2 pi over lambda x minus Vt.
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And this is so far
so good, but we
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want to write it in
terms of quantities
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that S prime measures.
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So this x should
be replaced by 2 pi
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over lambda x prime plus
Vt minus Vt like this.
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And I could even
do more if I wish.
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I could put t prime
here, because the t and t
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primes are the same.
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So phi prime, by the condition
that these phazes agree,
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it's given by this,
which is by the relation
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between the coordinates and
times of the two frames, just
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this quantity.
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So we can rewrite this as 2
pi over lambda x prime minus 2
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pi over lambda V 1 minus little
v over capital V t prime.
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I think I got the algebra right.
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2 pi over lambda, the sine--
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yes, I grouped those two
terms and rewrote in that way.
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So that is the phaze.
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And therefore, we
look at this phaze
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and see, oh, whenever
we have a wave,
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we can read the wave
number by looking
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at the factor
multiplying x, and we
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can read the
frequency by looking
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at the factor multiplying t.
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So you can do the same
thing in this case
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and read, therefore,
that omega prime,
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this whole quantity
is this, omega prime.
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And this is k prime,
because they can
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respond to the frame as prime.
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So omega prime is
equal to this 2 pi
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V over lambda, which
is omega, times 1
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minus V over V. And
k prime is equal to k
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or, what I wanted to show, that
lambda prime for a normal wave
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is equal to lambda for ordinary
wave moving in the medium.
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So at this moment, one
wonders, so what happened?
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What have we learned?
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Is that this wave function
is not like a sound wave.
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It's not like a water wave.
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We're doing everything
non-relativistic.
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But still, we're
seeing that you're not
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expected to have agreement.
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That is, if somebody
looks at one wave function
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and you look at the
same wave function,
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these two people will not
agree on the value of the wave
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function necessarily.
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So the things that we conclude--
so the conclusions are
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that waves are surprising.
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So size are not
directly measurable--
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measurable-- because if you had
a quantity for which you could
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measure, like a sound
wave or a water wave,
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and you could measure
aspects to it,
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they should agree between
different observables.
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So this is going to
be something that
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is not directly measurable--
not all of psi can be measured.
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Some of psi can be
measured, and you're already
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heard the hints of that.
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Because we said any
number that you multiply,
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you cannot measure, and in
the phase that you multiply,
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you cannot measure.
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So complex numbers
can't be measured,
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you measure real numbers.
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So at the end of the day, these
are not directly measurable,
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per se.
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The second thing is that
they're not Galilean invariant,
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and that sets the stage
to that problem 6.
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You see, the fact
that this phaze that
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controls these waves
is Galilean invariant
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led you to the quality
of the wavelengths,
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but these wavelengths
don't do that.
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The de Broglie wavelengths
don't transform
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as they would do for a
Galilean invariant wave.
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Therefore, this thing is
not Galilean invariant,
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and what does that mean?
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That if you have
two people and you
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ask, what is the value of the
wave function here at 103,
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the two observers might give
you a different complex number
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for the wave function.
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They will just not agree.
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Not all is lost,
because you will
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find how their measurements
can be compared.
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That will be the
task of the problem.
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How-- if you have a wave
function, how does your friend,
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that is moving
with some velocity,
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measure the wave function?
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What does this other
person measure?
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So the end result, if you have
a point here at some time t,
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the wave function
psi of x and t is not
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going to be the same as
the wave function measured
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by the prime
observer at x prime t
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prime, so this point is the
point x and t or x prime and t
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prime.
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These are two different
labels for the same point.
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You're talking about
the wave function
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at the same point
at the same time.
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You still don't agree.
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These two people will not agree.
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If they agreed,
this wave function
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would have a simpler
transformation law
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with a wavelength
that this can serve.
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So by simply discussing
the Galilean properties
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of this wave, we're led to know
that the de Broglie waves are
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not like normal matter
waves that propagate
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in a medium or simple.