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PROFESSOR: How about
the expectation value
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of the Hamiltonian in
a stationary state?
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You would imagine,
somehow it has
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to do with energy ion
states and energy.
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So let's see what happens.
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The expectation value
of the Hamiltonian
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on this stationary state.
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That would be integral
dx stationary state
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Hamiltonian stationary state.
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And we're going to see
this statement that we made
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a few minutes ago become clear.
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Well what do we get
here? dx psi star
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of x e to the i Et over
h bar H e to the minus i
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Et over h bar psi of x.
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And H hat couldn't care less
about the time dependence,
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that exponential is
irrelevant to H hat.
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That exponential of time can be
moved across and cancelled with
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this one.
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And therefore you
get that this is
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equal to dx psi star of x H hat
psi of x, which is a nice thing
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to notice.
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The expectation value of H
on the full stationary state
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is equal to the
expectation value of H
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on the spatial part of
the stationary state.
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That's neat.
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I think it should be noted.
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So it's equal to the
H of little psi of x.
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But this one, we can
evaluate, because if we
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are in a stationary state, H hat
psi of x is E times psi of x.
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So we get an E
integral the x psi
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star of psi, which
we already show
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that integral is equal to
one, so we get the energy.
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So two interesting things.
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The expectation value
of this quantity
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of H in the stationary
state is the same
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as it's quotation value
of H in the spatial part,
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and it's manually
equal to the energy.
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By the way, you know, these
states are energy eigenstates,
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these psi of x's, so you
would expect zero uncertainty
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because they are
energy eigenstates.
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So the zero uncertainty of the
energy operator in an energy
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eigenstate.
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There's zero uncertainty even
in the whole stationary state.
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If you have an H squared here,
it would give you an E squared,
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and the expectation
value of H is equal to E,
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so the expectation value of H
squared minus the expectation
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value of H squared
would be zero.
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Each one would be
equal to E squared.
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Nothing would happen, no
uncertainties whatsoever.
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So let me say once
more, in general, being
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so important here is the
comment that the expectation
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value of any time independent
operator, so comments 1,
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the expectation value of any
time-independent operator
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Q in a stationary state
is time-independent.
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So how does that go?
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It's the same thing.
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Q hat on the psi of
x and t is general,
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now it's integral dx capital Psi
of x and t Q hat psi of x and t
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equals integral dx-- you have to
start breaking the things now.
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Little psi star of x E
to the i et over H bar.
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And I'll put the
whole thing here.
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Q hat Psi of x E to
the minus i et over H.
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So it's the same thing.
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Q doesn't care about time So
this factor just moves across
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and cancels this factor.
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The time dependence
completely disappears.
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And in this case,
we just get-- this
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is equal to integral dx psi star
Q psi, which is the expectation
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value of Q on little psi
of x, which is clearly
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time-independent, because
the state has no time anymore
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and the operator has no time.
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So everybody loves their
time and we're in good shape.
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The second problem is
kind of a peculiarity,
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but it's important to
emphasize superposition.
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It's always true,
but the superposition
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of two stationary states is
or is not a stationary state?
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STUDENT: No.
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PROFESSOR: No, good.
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It's not a stationary
state in general
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because it's not factorizing.
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You have two stationary states
with different energies,
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each one has its
own exponential,
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and therefore, the whole
state is not factorized
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between space and time.
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One time-dependence has one
space-dependence plus another
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time-dependence and
another space-dependence,
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you cannot factor it.
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So it's not just a plain fact.
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So the superposition of
two stationary states
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of different energy
is not stationary.
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And it's more than just saying,
OK, it's not stationary.
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What it means is that if
you take the expectation
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value of a
time-independent operator,
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it may have time-dependence,
because you are not anymore
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guaranteed by the
stationary state
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that the expectation value
has no time-dependence.
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That's how, eventually, these
things have time-dependence,
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because these things
are not [INAUDIBLE]
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on stationary states.
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On stationary
states, these things
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would have no time-dependence.
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And that's important,
because it would
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be very boring,
quantum mechanics,
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if expectation
values of operators
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were always time-independent.
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So what's happening?
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Whatever you measure never
changes, nothing moves,
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nothing changes.
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And the way it's
solved is because you
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do have those stationary
states that will give you
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lots of solutions.
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And then we combine them.
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And as we combine them,
we can get time-dependence
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and we can get the most
[INAUDIBLE] equation.