WEBVTT
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PROFESSOR: Today we'll
talk about observables
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and Hermitian operators.
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So we've said that an operator,
Q, is Hermitian in the language
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that we've been working
so far, if you find
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that the integral,
dx psi 1 Q psi 2,
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is actually equal to
the integral dx of Q,
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acting this time of
Psi 1 all star psi 2.
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So as you've learned already,
this requires some properties
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about the way
functions far away,
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at infinity, some integration
by parts, some things to manage,
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but this is the
general statement
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for a large class of
functions, this should be true.
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Now we want to, sometimes,
use a briefer notation
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for all of this.
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And I will sometimes
use it, sometimes not,
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and you do whatever you feel.
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If you like to use
this notation, us it.
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So here's the definition.
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If you put up Psi 1,
Psi 2 and a parentheses,
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this denotes a number,
and in fact denotes
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the integral of psi 1
star of x, psi 2 of x dx.
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So whatever you put
in the first input
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ends up complex motivated.
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When you put in
the second input,
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it's like that,
it's all integrated.
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This has a couple of
obvious properties.
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If you put a number times
psi 1 times psi 2 like this,
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the number will appear,
together with psi 1,
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and will complex conjugated.
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So it can go out as
a star psi 1 psi 2.
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And if you put the number
on the second input,
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it comes out as is.
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Because the second
input is not complex
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conjugated in the definition.
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With this definition,
a Hermitian operator,
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Q is Hermitian, has
a nice look to it.
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It becomes kind of
natural and simple.
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It's the statement that if
you have psi 1, Q psi 2,
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you can put the Q
in the first input.
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Q psi 1 psi 2.
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This second term in
the right hand side
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is exactly this integral here.
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And the first tern
in the left hand side
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is the left hand side
of that condition.
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So it's just maybe a
briefer way to write it.
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So when you get tired of writing
integral dx of the first,
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the second, you can use this.
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Now with distance last
time, the expectation
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values of operators.
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So what's the expectation value
of Q in some state psi of x?
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And that is denoted as
these braces here and of psi
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is equal to the integral of psi.
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The expectation value depends
on the state you live in
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and it's psi Q psi.
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Or if you wish, dx in
written notation psi
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Q. I should put the
hats everywhere.
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This is the expectation
value of Q. I'm sorry,
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I missed here a star.
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So so far, so good.
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We've reviewed what a
Hermitian operator is,
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what an expectation value is,
so let's begin with some claims.
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Claim number one.
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The expectation value
of Q, with Q Hermitian.
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So everywhere here,
Q will be Hermitian.
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The expectation
value of Q is real.
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A real number, it belongs
to the real numbers.
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So that's an important thing.
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You want to figure out the
expectation value of Q,
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you have a psi star,
you have a psi.
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Well, it'd better be real
if we're going to think,
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and that's the goal
of this discussion,
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that Hermitian operators
are the things you
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can measure in
quantum mechanics,
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so this better be real.
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So let's see what this is.
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Well, Q psi, that's
the expectation value.
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If I complex conjugate
it, I must complex
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conjugate this whole thing.
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Now if you want to complex
conjugate an integral,
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you can complex
conjugate the integrand.
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Here it is.
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I took this right hand
side here, the integrand.
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I copied it, and now I
complex conjugated it.
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That's what you mean by complex
conjugating an integral.
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But this is equal, integral dx.
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Now I have a product
of two functions here.
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Psi star and Q that
has acted on psi.
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So that's how I think.
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I never think of
conjugating Q. Q is
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a set of operations
that have acted on psi
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and I'm just going
to conjugate it.
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And the nice thing
is that you never
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have to think of what is Q
star, there's no meaning for it.
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So what happens here?
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Priority of two functions,
the complex conjugate
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of the first--
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now if you [INAUDIBLE]
normally something twice,
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you get the function back.
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And here you've got Q psi star.
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But that, these are functions.
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You can move around.
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So this Q hat psi star Q psi.
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And so far so good.
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You know, I've done
everything I could have done.
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They told to come to
complex conjugate this,
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so I complex conjugated it
and I'm still not there.
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But I haven't used that
this operator is Hermitian.
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So because the
operator is Hermitian,
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now you can move the Q from this
first input to the second one.
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So it's equal to integral
dx psi star Q psi.
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And oh, that was the
expectation value of Q on psi,
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so the star of this number is
equal to the number itself,
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and that proves the
claim, Q is real.
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So this is our first claim.
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The second claim that is
equally important, claim two.
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The eigenvalues of the
operator Q are real.
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So what are the
eigenvalues of Q?
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Well you've learned, with
the momentum operator,
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eigenvalues or
eigenfunctions of an operator
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are those special functions
that the operator acts on them
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and gives you a number
called the eigenvalue times
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that function.
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So Q, say, times,
psi 1, if psi 1
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is a particularly
nice choice, then it
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will be equal to some number.
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Let me quote Q1 times psi1.
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And there, I will say
that Q1 is the eigenvalue.
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That's the definition.
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And psi1 is the eigenvector,
or the eigenfunction.
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And the claim is that that
number is going to real.
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So why would that be the case?
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Well, we can prove
it in many ways,
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but we can prove it kind of
easily with claim number one.
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And actually gain
a little insight,
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cold calculate the
expectation value of Q
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on that precise state, psi 1.
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Let's see how much is it.
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You see, psi 1 is
a particular state.
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We've called it an
eigenstate of the operator.
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Now you can ask, suppose
you live in psi 1?
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That's who you are,
that's your state.
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What is the expectation
value of this operator?
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So we'll learn more about
this question later,
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but we can just do it, it's the
integral of dx psi 1 Q psi 1.
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And I keep forgetting
these stars,
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but I remember them
after a little while.
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So at this moment, we can
use the eigenvalue condition,
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this condition
here, that this is
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equal to dx psi 1 star Q1 psi 1.
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And the Q1 can go out,
hence Q 1 integral dx of psi
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1 star psi 1.
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But now, we've proven,
in claim number one,
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that the expectation
value of Q is always real,
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whatever state you take.
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So it must be real if you
take it on the state psi 1.
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And if the expectation
value of psi 1 is real,
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then this quantity, which
is equal to that expectation
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value, must be real.
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This quantity is the
product of two factors.
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A real factor here--
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that integral is not only
real, it's even positive--
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times Q1.
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So if this is real, then
because this part is real,
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the other number must be real.
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Therefore, Q1 is real.
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Now it's an
interesting observation
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that if your eigenstate,
eigenfunction
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is a normalized eigenfunction,
look at the eigenfunction
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equation.
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It doesn't depend on what
precise psi 1 you have,
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because if you put psi 1
or you put twice psi 1,
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this equation still holds.
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So if it hold for psi 1, if psi
1 is called an ideal function,
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3 psi 1, 5 psi 1, minus psi
1 are all eigenfunctions.
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Properly speaking
in mathematics,
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one says that the
eigenfunction is
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the subspace generated by
this thing, by multiplication.
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Because everything is accepted.
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But when we talk
about the particle
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maybe being in the
state of psi 1,
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we would want to normalize it,
to make psi 1 integral squared
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equal to 1.
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In that case, you would
obtain that the expectation
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value of the operator
on that state
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is precisely the eigenvalue.
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When you keep measuring
this operator, this state,
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you keep getting the eigenvalue.
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So I'll think about the
common for a normalized psi
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1 as a true state that you
use for expectation values.
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In fact, whenever we
compute expectation values,
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here is probably a
very important thing.
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Whenever you compute
an expectation value,
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you'd better normalize the
state, because otherwise,
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think of the expectation value.
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If you don't normalize the
state, you the calculation
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and you get some
answer, but your friend
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uses a wave function three
times yours and your friend
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gets now nine times your answer.
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So for this to be a
well-defined calculation,
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the state must be normalized.
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So here, we should really say
that the state is normalized.
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Say one is the ideal
function normalized.
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And this integral would be equal
to Q1 belonging to the reals.
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And Q1 is real.
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So for a normalized psi
1 or how it should be,
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the expectation value
of Q on that eigenstate
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is precisely equal
to the eigenvalue.