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PROFESSOR: SHO algebraically.
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And we go back to the
Hamiltonian, p squared over 2m
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plus 1/2 m omega
squared x hat squared.
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And what we do is observe
that this some sort
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of sum of squares plus
p squared over m--
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p squared over m
squared omega squared.
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So the sum of two
things squared.
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Now, the idea that
we have now is
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to try to vectorize
the Hamiltonian.
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And what we call vectorizing is
when you write your Hamiltonian
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as the product of two vectors,
V times W. Well actually,
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that's not quite
the vectorization.
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You want kind of the same
vector, and not even that.
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You sort of want this to be the
Hermitian conjugate of that.
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And if there is a
number here, that's OK.
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Adding numbers to a Hamiltonian
doesn't change the problem
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at all.
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The energies are all
shifted, and it's
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just how you're defining
the zero of your potential,
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is doing nothing but that.
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So vectorizing the Hamiltonian
is writing it in this way,
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as V dagger V. And you
would say, why V dagger V?
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Why not VV dagger or VV
or V dagger V dagger?
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Well, you want the
Hamiltonian to be Hermitian.
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And this thing is Hermitian.
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You may recall that AB dagger.
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The Hermitian conjugate of AB
dagger is B dagger A dagger.
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So the Hermitian
conjugate of this product
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is V dagger times the
dagger of V dagger.
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A dagger of a dagger is the same
operator, when you dagger it
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twice, you get the same.
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So this is Hermitian.
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V dagger times V is
a Hermitian operator,
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and that's a very good thing.
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And there will be
great simplifications.
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If you ever succeed in writing
a Hamiltonian this way,
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you've gone 90% of the way
to solving the whole problem.
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It has become infinitely easier,
as you will see in a second,
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if you could just write
this vectorization.
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So if you had x minus--
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x squared minus this, you
would say, oh, clearly that's--
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A squared minus B squared
is A minus B time A plus B,
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but there's no such thing here.
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It's almost like A
squared plus B squared.
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And how do you sort
of factorize it?
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Well, actually, since
we have complex numbers,
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this could be A minus
IB times A plus IB.
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That is correctly A
squared plus B squared,
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and complex numbers are supposed
to be friends in quantum
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mechanics, so having Is, there's
probably no complication there.
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So let's try that.
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I'll write it.
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So here we have x
squared plus p squared
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over m squared omega squared.
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And I will try to write it as
x minus i p hat over m omega
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times x plus I p
hat over m omega.
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Let's put the question
mark before we
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are so sure that this works.
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Well, some things work.
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The only danger here is
that these are operators
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and they don't commute.
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And when we do this, in one
case, in the cross-terms,
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the A is to the left of B,
but the other problem the B
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is to the left of A. So we
may run into some trouble.
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This may not be exactly true.
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So what is this?
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This x with x, fine.
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x squared.
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This term, p with ps, correct.
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Plus p squared over m
squared omega squared.
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But then we get
plus i over m omega,
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x with p minus p with x,
so that x, p commutator.
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So vectorization of operators
in quantum mechanics
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can miss a few concepts
because things don't commute.
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So the cross-terms
give you that,
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and this x, p is I h bar, so
this whole term will give us
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the following statement.
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What we've learned is
that what we wanted,
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x squared plus p
squared over m squared
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omega squared is equal to--
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so I'm equating this
line to the top line--
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is equal to x hat minus i
p hat over m omega times
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x hat plus i p hat over m omega.
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And then, from this whole
term, i with i is minus,
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so it's h bar over m omega.
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So I'll put it in--
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it's a minus h bar
over m, [INAUDIBLE].
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So here is plus h bar
over m omega times
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a unit vector, if you wish.
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OK.
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So this is very good.
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In fact, we can call this
V dagger and this V. Better
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call this V first
and then ask, what
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is the dagger of this operator?
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Now, you may remember that,
how did we define daggers?
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If you have phi with psi
and the inner product--
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with an integral
of five star psi--
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if you have an A
psi here, that's
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equal to A dagger phi psi.
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So an operator is acting
on the second wave
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function, moves as A dagger
into the first wave function.
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And you know that x moves
without any problem.
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x is Hermitian.
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We've discussed that p
is Hermitian as well,
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moves to the other side.
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So the Hermitian
conjugate of this operator
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is x, the p remains means p,
but the i becomes minus i.
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So this is correct.
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If this second
operator is called V,
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the first operator should
be called V dagger.
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That is a correct statement.
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One is the dagger
of the other one.
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So the Hamiltonian
is 1/2 m omega
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squared times this
sum of squares,
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which is now equal to V dagger
V plus h bar over m omega.
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So h hat is now
1/2 m omega squared
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V dagger V plus a sum, which
is plus 1/2 h bar omega.
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So we did it.
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We vectorized the
Hamiltonian V dagger V,
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and this is quite useful.
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So the Vs, however, have units.
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And you probably are aware that
we like things without units,
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so that we can see
the units better.
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This curve is perfectly nice.
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It's a number added
to the Hamiltonian.
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It's h omega, it
has units of energy,
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but this is still
a little messy.
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So let's try to clean up those
Vs, and the way I'll do it
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is by computing their
commutator, to begin with.
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So let's compute the
commutator of V and V dagger
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and see how much
is that commutator.
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It's a simple commutator,
because it involves vectors
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of x and V. So
it's the commutator
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of x plus ip over
m omega, that's V,
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with x minus ip over m omega.
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So the first x talks
only to the second piece,
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so it's minus i
over m omega x, p.
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And for the second
case, you have plus
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i over am omega p with x.
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This is i h bar, and
this is minus i h bar.
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Each term will contribute
the same, i times minus i
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is plus, so h bar over
and, omega times the 2.
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That is V dagger V.
VV dagger, I'm sorry.
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2 h bar over m omega.
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So time to change
names a little bit.
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Let's do the following.
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Let's put square root of
m omega over 2 h bar V.
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Have a square root of m
omega over 2 h bar V dagger,
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commute to give you 1.
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That's a nice commutator.
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It's one number-- or an
operator is the same thing.
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So I brought the square
root into each one.
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And we'll call the first term--
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because of reasons
we'll see very soon--
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the destruction operator, A
square root of m omega over 2 h
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bar V. It's called the
destruction operator.
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And the dagger is
going to be A dagger.
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Some people put hats on them.
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I sometimes do too,
unless I'm too tired.
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2h bar V dagger.
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And those A and A daggers
are now unit-free--
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and you can check That--
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Because they have
the same units.
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And A with A dagger is
the nicest commutator, 1.
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Is A a Hermitian operator?
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Is it?
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No.
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A is not Hermitian.
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A dagger is different from
A. A is basically this thing,
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A dagger is this thing.
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So not Hermitian.
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So we're going to work
with these operators.
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They're non-Hermitian.
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I need to write the
following equations.
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It's very-- takes a
little bit of writing,
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but they should be
recorded, they will always
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make it to the formula sheet.
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And it's the basic relation
between A, A dagger, and x
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and p.
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A is this, A
dagger, as you know,
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is x minus ip hat over m omega.
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Since I'm copying, I'd
better copy them right.
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x, on other hand,
is the square root
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of h bar over 2m
omega A plus A dagger,
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and p is equal to i square
root of m omega h bar over 2
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A dagger minus A.
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So these four equations,
A and A dagger
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in terms of x and p and
vice versa, are important.
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They will show up all the time.
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Here are the things to notice.
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A and A dagger is
visibly clear that
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on is the Hermitian
conjugate of the other.
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Here, x is Hermitian.
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And indeed, A plus A
dagger is Hermitian.
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When you do the Hermitian
conjugate of A plus A dagger,
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the first A becomes an A dagger.
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The second A, with another
Hermitian conjugation,
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becomes A. So this is Hermitian.
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But p is Hermitian, and here we
have A dagger minus A. This is
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not Hermitian, it changes sign.
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Well, the i is there for that
reason, and makes it Hermition.
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So there they are, they're
Hermitian, they're good.