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PROFESSOR: This
definition in which

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the uncertainty of the
permission operator

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Q in the state psi.

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It's always important to
have a state associated

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with measuring the uncertainty.

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Because the uncertainty will be
different in different states.

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So the state should
always be there.

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Sometimes we write
it, sometimes we

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get a little tired of writing
it and we don't write it.

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But it's always implicit.

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So here it is.

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From the analogous discussion
of random variables,

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we were led to this
definition, in which we

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would have the expectation
value of the square

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of the operator minus the
square of the expectation value.

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This was always-- well, this
is always a positive quantity.

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Because, as claim 1 goes, it can
be rewritten as the expectation

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value of the square of the
difference between the operator

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and its expectation value.

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This may seem a little strange.

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You're subtracting from
an operator a number,

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but we know that numbers can be
thought as operators as well.

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Operator of minus a
number acting on a state

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is well defined.

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The operator acts on the state,
the number multiplies a state.

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So this is well defined.

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And claim 1 is proven
by direct computation.

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You certainly indeed prove.

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You can expand what is
inside the expectation value,

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so it's Q hat squared.

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And then the double product
of this Q hat and this number.

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Now, the number
and Q hat commute,

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so it is really
the double product.

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If you have A plus B times A
plus B, you have AB plus BA,

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but if they commute
it's 2AB, so this

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is minus 2 Q hat Q. Like that.

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And then, the last term
is the number squared,

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so it's plus Q squared.

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And sometimes I don't
put the hats as well.

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And all this is the
expectation value

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of the sum of all these things.

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The expectation value
of a sum of things

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is the expectation
value of the first

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plus the expectation
value of the second,

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plus the expectation
value of the next.

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So we can go ahead and do
this, and this is therefore

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expectation value of Q
squared minus the expectation

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value of this whole thing.

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But now the expectation value
of a number times an operator,

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the number can go out.

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And this is a number,
and this is a number.

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So it's minus 2 expectation
value of Q, number went out.

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And then you're left with
expectation value of another Q.

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And the expectation
value of a number

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is just the number,
because then you're

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left within the world of psi
star psi, which is equal to 1.

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So here is plus Q hat squared.

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And these two terms, the
second and the third,

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are the same really.

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They are both equal to
expectation value of Q squared.

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They cancel a little bit,
and they give you this.

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So indeed, this is equal
to expectation value

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of Q squared minus expectation
value of Q squared.

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So claim 1 is true.

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And claim 1 shows in particular
that this number, delta Q

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squared, in the expectation
value of a square of something,

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is positive.

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We'll see more
clearly in a second

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when we have claim number 2.

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And claim number 2
is easily proven.

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That's another expression
for uncertainty.

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For claim number 2, we will
start with the expectation

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value of Q minus Q squared, like
this, which is the integral dx

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psi star of x and t, Q minus
expectation value of Q, Q

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minus expectation
value of Q, on psi.

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The expectation value
of this thing squared

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is psi star, the
operator, and this.

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And now, think of this as an
operator acting on all of that.

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This is a Hermitian operator.

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Because Q hat is Hermitian, and
expectation value of Q is real.

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So actually this real
number multiplying something

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can be moved from
the wave function

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to the starred wave
function without any cost.

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So even though you might
not think of a real number

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as a Hermitian operator, it is.

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And therefore this whole
thing is Hermitian.

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So it can be written as dx.

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And now you have this
whole operator, Q minus Q

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hat, acting on psi of x and t.

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And conjugate.

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Remember, the operator,
the Hermitian operator,

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moves to act on psi, and
the whole thing [INAUDIBLE].

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And then we have here
the other term left over.

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But now, you see that you
have whatever that state is

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and the state
complex conjugated.

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So that is equal
to this integral.

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This is the integral dx of the
norm squared of Q hat minus Q

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hat psi of x and t squared,
which means that thing, that's

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its complex conjugate.

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So this completes
our verification

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that these claims are
true, and allow us

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to do the last step on
this analysis, which

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is to show that if you
have an eigenstate of Q,

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if a state psi is an eigenstate
of Q, there is no uncertainty.

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This goes along with our
measurement postulate that

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says an eigenstate
of Q, you measure Q

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and you get the eigenvalue of
Q and there's no uncertainty.

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In particular, we'll
do it here I think.

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If psi is an eigenstate
of Q, so you'll

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have Q psi equal lambda psi,
where lambda is the eigenvalue.

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Now, this is a nice thing.

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It's stating that the state
psi is an eigenstate of Q

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and this is the eigenvalue,
but there is a little bit more

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than can be said.

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And it is.

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It should not surprise
you that the eigenvalue

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happens to be the expectation
value of Q on the state psi.

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Why?

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Because you can take this
equation and integrate dx times

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psi star.

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If you bring that in into
both sides of the equation

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then you have Q psi equals
integral dx psi star psi,

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and the lambda goes up.

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Since my assumption whenever
you do expectation values,

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your states are normalized,
this is just lambda.

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And by definition, this is
the expectation value of Q.

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So lambda happens to be equal
to the expectation value of Q,

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so sometimes we can say
that this equation really

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implies that Q hat psi is equal
to expectation value of Q psi

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times psi.

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It looks a little
strange in this form.

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Very few people write
it in this form,

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but it's important to recognize
that the eigenvalue is

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nothing else but the
expectation value

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of the operator of that state.

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But if you recognize
that, you realize

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that the state satisfies
precisely Q hat minus Q on psi

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is equal to 0.

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Therefore, if Q hat minus
Q on psi is equal to 0,

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delta Q is equal to 0.

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By claim 2.

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Q hat minus Q expectation
value kills the state,

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and therefore this is 0.

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OK then.

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The other way is also true.

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If delta Q is equal to 0, by
claim 2, this integral is 0.

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And since it's
the sum of squares

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that are always positive, this
state must be 0 by claim 2.

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And you get that Q minus
Q hat psi is equal to 0.

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And this means that psi
is an eigenstate of Q.

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So the other way
around it also works.

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So the final
conclusion is delta Q

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is equal to 0 is
completely equivalent of--

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I'll put in the psi.

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Psi is an eigenstate of Q.

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So this is the main conclusion.

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Also, we learned some
computational tricks.

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Remember you have to
compute an expectation

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value of a number,
uncertainty, you

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have these various
formulas you can use.

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You could use the
first definition.

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Sometimes it may
be the simplest.

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In particular, if the
expectation value of Q

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is simple, it's the easiest way.

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So for example, you can have
a Gaussian wave function,

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and people ask you, what is
delta of x of the Gaussian wave

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function?

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Well, on this Gaussian
wave function,

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you could say that
delta x squared

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is the expectation value of x
squared minus the expectation

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value of x squared.

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What is the
expectation value of x?

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Well, it would seem reasonable
that the expectation value of x

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is 0.

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It's a Gaussian
centered at the origin.

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And it's true.

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For a Gaussian it would be 0,
the expectation value of x.

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So this term is 0.

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You can also see 0
because of the integral.

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You're integrating x
against psi squared.

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Psi squared is even,
x is odd with respect

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to x going to minus x.

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So that integral
is going to be 0.

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So in this case, the
uncertainty is just

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the calculation of the
expectation value of x squared,

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and that's easily done.

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It's a Gaussian integral.

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The other good thing
about this is that

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even though we have not
proven the uncertainty

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principle in all generality.

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We've only [? multivated ?] it.

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It's precise with
this definition.

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So when you have
the delta x, delta p

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is greater than or
equal to h bar over 2,

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these things are computed
with those definitions.

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And then it's precise.

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It's a mathematically
rigorous result.

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It's not just hand waving.

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The hand waving is good.

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But the precise result
is more powerful.