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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.