M1L5n.txt

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Other questions?
Nicky?
[INAUDIBLE]

Well, a lot of classical counting statistics
leads to shot noise.
If you have independent particles
with a certain probability, then you get shot noise.
But I would sort of press you hard to make
a classical explanation for the last example I gave you,
where we had a strong local oscillator, we reflected it,
and superimposed with squeezed vacuum.
At that point, all the shot noise has disappeared.
So therefore, I would be hard pressed
to connect it with a classical probability of detecting
a photon.
Of course, classically speaking, the empty part
of a beam splitter is empty, and squeezed vacuum
doesn't exist classically.
Yes, Timo?
This might be kind of a silly question,
but so we have a minimal uncertainty
state is when delta x, delta p is zero.
But a lot of times we mostly talk
about the number of phase uncertainty.
So I'm just curious, because a coherent state
has the same circle of-- the area of the circle is
the same, right?
I know this is wrong, but can't you always make the delta n,
delta phi uncertainty very small by having a very large end,
where delta n stays the same but the angle are spread-- always
goes down as you make the circle further away from the origin.
So you're talking about an uncertainty
relation we have not discussed.
Well, it kind of showed up in the last--
In the last homework, then, you looked at it.
So what happens is if you have a coherent state
you would say the face angle phi has this uncertainty.
And then there's an uncertainty relation.

Actually, we talked about this problem yesterday.
I think the uncertainty and the size of the disk is constant.
The size of the disk is uncertain in alpha, but alpha
is proportional not to n but to the square root of n.
We can make the error propagation,
[INAUDIBLE] square root of n.
So if we move the circle further away, the uncertainty of alpha
stays the same, but the uncertainty of m
goes actually up.
But we don't see this [INAUDIBLE].

Yes, you have to be careful.
That's another comment I wanted to make,
but thanks for reminding me of that.
When you have coherent states, it
looks that the uncertainty is always
the same because it's a disk.
But we learned today and in the last classes
that uncertainty of the coherent state when
you measure the photocurrent is square root n.
What happens is the following.
The uncertainty-- and this is what
I was referring to-- the uncertainty, let's say,
of delta a in one [INAUDIBLE] component is constant
for the coherent state, independent of n.
But that would mean what you really
calculate is when you calculate the photocurrent,
it is alpha plus delta a, which is constant,
but then you square it.
In the limit of a strong coherent state,
you get alpha squared, which is the number of photons,
plus 2 alpha times delta a.
So the delta a, which is constant,
gets multiplied with square root n here.

So therefore, when you go to a larger coherent state,
you increase a number uncertainty,
but by the same factor you reduce the phase uncertainty.