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Okay, so we already saw in a previous
video that now that we talk about

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observers and their real outcomes to each
measurement, we could also talk about the

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expected value of a measurement. So of
course, a measurement outcome is

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probabilistic. And so, we could talk about
the statistics of what, what we expect to

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see. So in this video we are going to look
at this question a little more precisely

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and closely. So let's say that we have a
an observed M so M is a summation matrix

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so M = M diagram. It's k by k summation
matrix and lets say that we have a quantum

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state of our system which we now. Right in
the eigenbasis of M. So you write is as

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summation alpha phi and, so now if you
measure this observable you get a random

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outcome. And let's, let's the random
variable X denote the outcome of this

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measurement. Okay. So what we want to do
is, we want to understand what's the

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distribution of X? So, well, of course,
this is sort of easy to do because we know

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that, what, what, what are going, what are
the outcomes going to be? Well, they are

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supported on the eigenvalues of, of M. So
if the eigenvalues are lamnda-naught,

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lambda one through lambda K-1, then we
know that. Lamda I will occur so, so the

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probability that, that, that x equal to
lambda sub j is just what? Well, the j

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outcome will occur with probability alpha
sub j magnitude squared. Because, because

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our state psi, which is the state of our
quantum system, we have decomposed in

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terms of, of the eigenbasis of M. And the
amplitude of phi sub j is exactly alpha

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sub j. So we got a probability
distribution of this kind on j different

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outcomes, on our k different outcomes,
lamdba naught lamdba k minus one. Every

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other outcome has probability zero. So now
we can, we can try to, well off-course,

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ideally we would want to know the entire
probability distribution that, but usually

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we collect some statistics arriving this
distributions and that tells us, gives us

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a rough idea about what the distribution
looks like. And so the two most, useful

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you know, the first order statistics are,
the mean of the distributions, the

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expected value of x, which be might denote
by mu. And the variance of x, which, or

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sigma squared. The sigma is the standard
deviation. The variance of x is, is the

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expected square of the deviation for the
mean. So, it's expected value of x minus

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the mean, the whole square. Okay so. I'm
sure you all know about, about these

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quantities. But You know, let's, let's
just try to make sure that you understand

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them you know, you have a feel for what
these two quantities are. So we are given

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some, some probability distribution and we
might you know, you might imagine that,

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that you actually make a wooden cut out of
this, of this distribution. So you make a

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wooden shape which looks like this. Okay,
so then what's the meaning of the

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distribution? Well it's, it's the point at
which you can balance this distribution.

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Right, if you, if you think of this as a
see-saw, then this is exactly the middle

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point. The, the point across which, if you
hold it at this point, the point on the

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right and the point on the left exactly
balances. And what about this standard

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deviation or the square root of the
variance. How do you interpret that? Well

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there's also a very nice physical
interpretation of that which is, that if

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you were to, rotate this, this shape about
this, about this main position. So you

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rotate it about a vertical, about this
vertical axis. And then you, you try to

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fill, how much moment of inertia is? How
difficult is it to rotate this shape?

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Right? Well then. This, the standard
deviation is, is exactly sort of the, the

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you know if, if you were to replace this
complicated shape by this simple shape,

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where you put half the mass, half the
total weight of this, of this shape on the

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left here, half on the right here at
distance exactly sigma over two on each

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side. Then, sorry, distance exactly sigma
+ sigma and - sigma, then this would be a

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shape which has exactly the same
resistance to rotation. Right, and the

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reason is that the moment of inertia comes
from, from the expected square of the

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distance from, from, the mean, from the,
from the center of gravity, and in this

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case, you know, we, we, we are achieving
the same square distance on average, from

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the center. Okay, so, so now let's try to,
approach this from a more qunatitative

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point of view. So we have an observable M,
state sigma. And. We want to understand

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what the mean value of this, you know of
the, of the outcome of this measurement

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is. And I claim that it's given by
[inaudible] and [inaudible]. Okay so, so

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let's, let's try to prove this. So let's
say that M had eigennvectors phi sub j

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with eigenvalue lambda sub j. And let's
say, as before, that psi, we wrote as

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summation of phi sub j. So we, we wrote,
we wrote, quantum state out in the

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eigenbasis of M. And these were,
amplitudes so now, what's the expected

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value of X? So, so min which is the
expected value of X, is what? Well, we

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have outcome j, outcome lambda sub j with
what probability? Well, the, the outcome

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is lambda sub j if, if we happen to
project on to the j-th eigenvector, which

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happens with probability alpha sub j
magnitude squared. And so the expected

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value is the sum of all j of this. Okay,
so now what we want to show is that this

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is exactly the same as that quantity. So
let's write out that quantity. Okay, so

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what's get psi? Well, if you were writing
in the, you know, let's write out you

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know, it depends upon which basis we write
it out in. So let's use, let's write out

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get psi in the basis of eigenvectors of M.
So, what does get psi look like? Well, it

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looks, it looks like, exactly, exactly
this. It looks like alpha naught, alpha

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one, alpha K - one. What is M look like in
the same basis? If you're writing out M in

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the eigenbasis, well then it looks like
lambda naught, lambda one, lambda K - one.

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And zero, zero so it's a diagonal matrix,
in its, in its own eigenbasis. And what's

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bra phi? Well, bra phi is the raw vector
corresponding to this. It's alpha naught

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alpha one complex conjugate, alpha K - one
<i>. Okay, so you< /i> multiply this out.</i>

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What do you get? Well, you follow your r
ules of multiplication and you get, you

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get exactly, you get summation lambda sub
j okay sorry. Let's, let's, let's write it

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out in, in proper order. It's alpha sub j
lambda sub j alpha sub j. These are all

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complex numbers. They can, sorry. These
are, well, these are complex numbers.

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That's real. They commute with each other.
So you can just write it as summation

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lambda j sorry alpha j alpha j lambda sub
j. Now whats this? Whats alpha j x alpha

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j? It's just the square of the magnitude
of alpha j. Okay so, so this is more

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equal. Okay, what about the variance
effects? So, sigma squared well, I'm sure

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you know the variance of X is, is you can,
you can write it has the expected value of

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X squared - the expected value of X
holding squared, which is. Okay and this

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is just expected value of x squared - mew
squared. Okay. So, so what are they

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claiming about this? We're claiming it's,
it's, it's this bilinear form. It's, it's

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bra psi M squared, ket psi - bra psi M ket
psi whole squared. Well you can see where

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this second piece comes from, this is just
mew squared. So what about this first

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piece? Well, okay. So what's the expected
value of X^2? So what's X^2 going to be

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equal to? Well if we have a j-th outcome
then, then, then the square of the outcome

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is just lambda j^2. And this what
probability does it happen? Well its still

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the same probability. It's probability
that it project onto phi sub j which is

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alpha j magnitude squared. Okay. Sum of
all j. So that what the expected value of

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X^2 is. Now whats, whats this bilinear
form? Well this binary form is again it's

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bra phi. Sorry, alpha naught alpha one
alpha k - one times, what's M^2? And

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written out in its eigenbasis? Well you
can square this matrix, what will it look

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like? Well it would just look like, its
diagonal matrix, it look like lambda one,

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lambda naught^2, lambda one^2, lambda K -
one^2. So when you, you square a matrix,

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its eigenvectors remain unchanged, and its
eigenvalues get squared. And so, you know

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of course now you can see that this is
clearly root of that.