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PROFESSOR: We spoke last
time of the existence

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of symmetric states.

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And for that we were
referring to states

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psi S that belonged to the
M particle Hilbert space.

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And V is the vector space that
applies to the states of the M

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particles.

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And we constructed some states.

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So within the construction,
we postulated,

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there will be some
states that [AUDIO OUT]

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that meant that any permutation
of [AUDIO OUT] state

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will be for the symmetric
state or all alpha.

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So that meant the
state was symmetric.

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Whatever you apply to
it, any permutation,

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the state would be invariant.

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So these are special states,
if they exist, in VN.

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Then there would be
anti-symmetric states.

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And we concluded that an
anti-symmetric state--

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with an A for anti-symmetric--

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also stayed in this
tensor product.

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That state would
react differently

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to the permutation operators.

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It would change up to a
sine epsilon alpha times

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psi A. It would be an eigen
state of all those permutation

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operators.

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But with eigenvalue
epsilon and alpha

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where epsilon alpha
was equal to plus

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1 if P alpha is an
even permutation.

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Or minus 1 if P alpha
is an odd permutation.

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So whether this
permutation is even or odd,

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we also discussed
depends on whether it's

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built with an even or odd
number of transpositions.

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With transpositions being
permutations in which

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one state is flipped
for another state.

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Within the end
states, you pick two,

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and these two are flipped,
that's a transposition.

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All permutations can be built
through transpositions--

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with transpositions--
and therefore you

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can tell from a
permutation whether it's

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an even or an odd one
depending of whether it's

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built with even number of
transpositions or odd number

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of transpositions.

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Now, the number
of transpositions

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you need to build the
permutation is not fixed.

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If we say it's even,
means it's even mod 2.

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So you might have a permutation
is built with 2 transpositions.

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And also somebody
else can write it

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as 4 transpositions
and 6 transpositions.

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It just doesn't matter.

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So a few facts that we
learned about these things

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are that all the P alphas
are unitary operators.

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And we also learned
that all transpositions

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are Hermitian operators.

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Now, transpositions, of
course, are permutations.

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So they're Hermitian, and
they're unitary as well.

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And finally, we learned that
the number of even permutations

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is equal to the number of odd
permutations in any permutation

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group of n objects.

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So these were some of the facts
we learned last time already.

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We have now that symmetric
states form a subspace.

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If you have two
symmetric states,

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you can multiply a
state by a number,

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it will still be symmetric.

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If you add two symmetric
states, will still be symmetric.

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So symmetric states form a
subspace of the full vector

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space V tensor N.

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And anti-symmetric
states also form

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a subspace of the vector
space, V tensor N.

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So let's write these facts.

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So symmetric states
form the subspace,

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and it's called sym N of V
of V tensor N. Anti-symmetric

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states form the subspace
anti N V of V tensor N.

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All right.

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So these are the states.

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But we have not learned how to
build them, how to find them,

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and even more, what
to do with them.

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So main thing is if
this form subspaces,

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there should be a way to
write the projector that

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takes from your big space
down to the subspace.

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So here is the
claim that we have.

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Here are two operators.

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We'll call this S,
a symmetrizer And S

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will be billed as 1 over N
factorial, the sum over alpha

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of all the
permutations, P alpha.

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That's the definition.

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And then we'll have an
anti-symmetrizer, A. This is

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also 1 over N factorial,
sum over alpha.

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Epsilon alpha P alpha.

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Where epsilon alpha,
again, is that sine factor

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for each permutation.

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So here are two
operators, and we're

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going to try to prove
that these operators are

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orthogonal projectors
that take you

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to the subspace's
symmetric states

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and anti-symmetric states.

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So that is our first goal.

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Understanding that these
operators do the job.

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And then we'll see what
we can do with them.

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So there are several
things that we

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have to understand
about these operators.

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If they are projectors, they
should square to themselves.

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So S times S should be
equal to S. Remember, that's

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the main equation
of a projector.

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A projector is P
squared equals P.

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And the second thing,
they should be Hermitian.

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So let's try each of them.

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So the first claim is that
S and A are Hermitian.

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In particular, S dagger equal
S. And A dagger equal A.

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So if you want to prove
something like that,

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it's not completely
obvious at first sight

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that those statements are true.

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Because we saw, for
example, that transpositions

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are Hermitian operators.

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But the general permutation
operator is not Hermitian.

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It's unitary.

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So it's not so obvious.

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You cannot just say each
operator is Hermitian,

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and it just works out.

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It's a little more
complicated than that.

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But it's not extremely
more complicated than that.

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So let's think of the
following statement.

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I claim that if you have the
list of all the permutation

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operators.

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Put the list in front of you.

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All of them.

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And you apply Hermitian
conjugation to that whole list,

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you get another
list of operators.

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And it will be just the
same list scrambled.

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But you will get the same list.

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It will contain all of them.

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And that is kind of obvious
if you think about it a little

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more.

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You have here, for example, the
list of all the permutations.

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And here, you apply Hermitian
conjugation, HC or dagger.

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And you get another list.

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And this list is the same as
this one although reordered.

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In a sense, it permutes
the permutation operators,

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if you wish.

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And the reason is clear.

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If you have two operators
here, and you apply--

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they are different-- and you
apply Hermitian conjugation,

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it should give two
different operators here.

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Because if they were equal,
you could apply again

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Hermitian operator conjugation.

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And you would say,
oh, they're equal.

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But you assume they
were different.

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So two different operators go
to two different operators here.

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Moreover, any operator here--

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you can call it O--

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is really equal to O dagger.

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O dagger.

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Hermitian conjugation
twice gives you back.

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So any operator here is
the Hermitian conjugation.

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Hermitian conjugate of
some operator there.

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So every operator
here will end up

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at the end of an
arrow coming here.

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Maybe those arrows,
we could put them

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like they go in all directions.

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But the fact is
that when you take

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Hermitian conjugation
of all the list,

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you get all the list back.

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Therefore, that takes
care of the symmetrizer.

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When I take the Hermitian
conjugation of P alpha,

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I might get another P beta.

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But at the end of the
day, the whole sum

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will become again the whole sum.

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So that proves the fact
that this map is 1 to 1

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The map Hermitian
conjugation is 1 to 1.

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1 and surjective meaning
that every element

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is reached as well.

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Means that S dagger
is equal to S.

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Because the list doesn't change.

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Here is a slightly
more subtle point.

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This time, the list
is going to change.

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But then maybe when this
changes to another operator,

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maybe the epsilon is
not the right epsilon.

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You would have to
worry about that.

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But that's no worry either.

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That also works out clearly.

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Because of the following thing.

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If a permutation
P alpha is even,

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it's billed with an even
number of transpositions.

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If you take the
Hermitian conjugation

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of an even number
of transpositions,

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you'll get an even
number of transpositions

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in the other order.

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That's what Hermitian
conjugation is.

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Therefore, Hermitian conjugation
doesn't change the fact

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that the permutation
is even or is odd.

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So if P alpha has a sine
factor E of P alpha.

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Let's write it more explicitly.

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And P alpha dagger has as a
sine factor E of P alpha dagger.

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These two are the same.

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If P alpha is even, P
alpha dagger is even.

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If P alpha is odd,
P alpha is odd.

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Therefore, when you take
the Hermitian conjugate

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of this thing, you may map.

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You think the Hermitian
conjugate of this sum.

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You have here a dagger would
be 1 over N factorial, sum

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over alpha, epsilon alpha,
P alpha dagger, which

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is another permutation.

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But this permutation.

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For this permutation,
this sine factor

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is the correct sine factor.

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Because the sine
factor for P alpha

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dagger is the same as the
sine factor of E alpha.

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Since we get the whole sum
by the statement before,

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A is also Hermitian.

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So OK.

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Not completely obvious,
but that's a fact.

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If you think more
abstractly, this

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is something you
could know on a group.

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If you have a group that
has all kinds of elements.

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00:16:13,880 --> 00:16:19,430
Many of you have written papers
in groups in your essays.

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00:16:19,430 --> 00:16:22,160
Then if you take
every group element,

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00:16:22,160 --> 00:16:25,470
and you replace
it by its inverse,

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00:16:25,470 --> 00:16:27,950
you get the same list
of group elements.

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00:16:27,950 --> 00:16:29,750
It's just scrambled.

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00:16:29,750 --> 00:16:35,450
And remember, since this
permutations are unitary,

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taking Hermitian conjugation
is the same thing

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00:16:38,590 --> 00:16:40,710
as taking an inverse.

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00:16:40,710 --> 00:16:42,860
So when you take
a group, and you

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00:16:42,860 --> 00:16:44,810
take the inverse
of every element,

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00:16:44,810 --> 00:16:47,860
you get back the list of
the elements of the group.

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00:16:47,860 --> 00:16:50,470
And that's what's
happening here.