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PROFESSOR: And let me
I assume, for example,

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that I'll put the
state alpha beta in.

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Alpha and beta.

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What do I get out?

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So you have this
state, alpha beta.

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What do you get out?

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Well, state comes in and is
acted by beam splitter 1.

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So you must put the
beam splitter, 1 matrix.

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And then it comes the mirrors.

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And lets assume
mirrors do nothing.

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In fact, mirrors--
the two mirrors

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would multiply by minus 1,
which will have no effect.

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So lets ignore mirrors.

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And then you get
to beam splitter 2

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and you must multiply by the
matrix of beam squared 2.

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And that's the output.

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And that output is a
two-component vector.

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That gives you the amplitude
up and the amplitude down.

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So I should put BS2 here,
BS1 over here, alpha beta.

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The numbers move away, 1
over square root of 2 and 1

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over square root of 2.

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Commute in matrix
multiplication.

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Then you multiply
these two matrices.

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You get 0, 2, minus
2, and 0, alpha beta.

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And you put the 2 in so
you get beta minus alpha.

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So here is the rule.

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If you have alpha
and beta, you get,

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here, beta and minus alpha
here, or a beta minus

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alpha photon at the end.

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Good.

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So let's do our first
kind of experiment.

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Our first experiment is to
have the beam splitters here.

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D0-- detector D0 and
detector D1 over there.

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And let's send in a
photon over here only--

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1 or input 0, 1.

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Well this photon,
0, 1, splits here.

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You act with BS1--

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the matrix BS1.

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You get two things.

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You act with the matrix
BS2, and it gives you this.

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But we have the rule already.

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If you have an alpha beta,
out comes a beta minus alpha.

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So it should have as 1, here,
and minus 0, here, which is 0.

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So you get a 1, 0.

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So what is really happening?

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What's really happening is
that your photon that came in

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divided in two,
recombined, and, actually,

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there was a very interesting
interference here.

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From the top beam
came some amplitude

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and gave some reflected
and some transmitted.

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From the bottom beam,
there was some transmission

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and some reflection.

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The transmission from the top
and reflection from the bottom

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interfered, to give 0.

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And this, too, the reflection
from the top and transmission

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from the bottom, were
coherent and added up to 1.

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And every single
photon ends up in D0.

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If you would put the beam--

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well, Mach and
Zehnder were working

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in the late 1800s, 1890s.

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And they would shine light.

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They had no ability
to manipulate photons.

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But they could put
those beam splitters

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and they could get this
interference effect,

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where everything goes to D0.

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So far, so good.

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Now let me do a slightly
different experiment.

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I will now put the same
thing, a BS1 and a beam

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going in, mirror,
mirror, BS2 here.

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But now, I will put a block
of concrete here on the way.

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I'll put it like this.

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So that if there
is any photon that

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wants to come in this
direction, it will be absorbed.

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Photon could still go
like this, but nothing

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would go through here.

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And here, of course,
there might be D0 and D1.

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And here are the
mirrors, M and M.

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Now the bottom
mirror is of no use

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anymore because there is
a big block of concrete

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that will stop any photon
from getting there.

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And we are asked,
again, what happens?

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What do the detectors see?

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And this time, we
still have a 01.

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Now I would be tempted
to use this formula,

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but this formula was right
under the wrong assumption--

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that there was no block here.

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So I cannot use that formula.

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And certainly, things are
going to be different.

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So I have to calculate things.

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And we're doing a quantum
mechanical calculation.

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Well, up to here,
before it reaches here,

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I can you do my
usual calculation.

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Certainly, we have BS1
acting on the state, 01,

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and this is 1 over square root
of 2, I think, minus 1, 1, 1,

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1.

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Yup, that B is 1, acting on 01.

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And that gives me 1
over square root of 2,

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1 over square root of 2.

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So, yes, here I have one over
square root of 2 amplitude.

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And here I also have 1 over
square root of 2 amplitude.

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OK.

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Now that's the end
of this amplitude.

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It doesn't follow.

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But on the other
hand, in this branch,

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the mirror doesn't change the
amplitude, doesn't absorb.

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So you still have 1 over
square root of 2 here.

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And now you're reaching BS2.

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Now what is the input for BS2?

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The input is a one over
square root 2 in the top beam,

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and nothing in the lower beam
because nothing is reaching BS2

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from below.

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This is blocked.

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So yes, there was some
times when something reached

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from below, but nothing here.

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So to figure out the
amplitudes, here, I

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must do BS2 acting on 1 over
the square root of 2, 0.

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Because 1 over square
root of 2 is coming in,

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but nothing is
coming in from below.

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And, therefore, I get 1
over the square root of 2,

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1, 1, 1, minus 1, 1 over
square root of 2, 0.

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This time, I get 1/2 and 1/2.

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OK, we must trust the math.

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1/2 here and 1/2 there, so
1/2 a column vector, 1/2, 1/2.

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OK, let me maybe tabulate
this result, which

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is somewhat strange, really.

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So what is strange about
it is the following.

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In the first case,
where the interferometer

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was totally clear, nothing
in the middle, everything

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went to D2.

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And nothing went into D1.

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But now, you do something that
should block some photons.

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You block some photons
in the lower path,

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and yet, now you seem to be
able to get something into D1.

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There is an amplitude
to get into the D1.

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So by blocking a source,
you're getting more somewhere.

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It's somewhat counterintuitive.

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You will see by the end of
the lecture in 10 minutes,

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that it's not just
somewhat counterintuitive,

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it's tremendously
counterintuitive.

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Let's summarize
the result here--

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the outcome in the blocked lower
branch case and the probability

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for those events.

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So photon at the block--

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the photon can end
in three places.

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It can end on the block.

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It can end on the D0.

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Or it can end on D1.

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So photon at the block--

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well, the amplitude to be here
is one over square root of 2.

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The probability should be 1/2.

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Photon at D0, probability
amplitude, 1/2, probability,

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1/4--

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photon at D1, probability, 1/4.

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You could put another
table here-- outcome

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all open, probability.

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And in this case, there's
just photon at D0.

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That's 1.

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And photon at D1 was 0.