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PROFESSOR: Now we
prove the other thing
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that we used in order to
solve the square well.
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So this is property number 3.
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It's so important that I
think I should do it here.
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If a potential is even--
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here comes again the
careful statement--
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the energy eigenstates can
be chosen to be even or odd
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under x goes to minus x.
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So that's analog of the first
sentence in property number 2.
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But then comes the
second sentence,
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that you can imagine what it is.
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For 1D potentials the bound
states are either even or odd.
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So look again at this freedom.
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You have a general
problem-- you're not
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talking bound states.
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You have a wave
function that solves
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a problem of a potential of the
symmetric around a mid-point.
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Then you find an arbitrary
solution, no need
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to work with that solution.
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You can work with a
solution that is even,
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and a solution that is odd.
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You can always choose
to be even or odd.
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But if you have
one-dimensional potential,
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there is no such solution
that is neither even nor odd.
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You cannot find it.
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It will be, automatically,
even or odd--
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which is kind of remarkable.
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It's sufficiently subtle that
in the general exam at MIT
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for graduate quantum mechanics,
the professor that invented
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the problem forgot this property
and the problem had to be
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cancelled.
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It's a very interesting.
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So let's just try to prove it.
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So complete proof in this case.
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So what is the equation?
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Proof.
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What is the equation
we have to solve?
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Sine double prime of x,
plus 2M over h squared.
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E minus v of x.
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Equals 0.
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Now, the proof,
actually, is very simple.
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I just do it and
I elaborate on it,
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because it's possible to get
a little confused about it.
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I think it's kind
of interesting.
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So here's equation one.
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And sine double
prime of x notation
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means the second derivative
of psi evaluated at x.
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You see, what you want to do
is to show that psi of minus x
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solves the same equation.
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Right?
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It's kind of clear.
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Well, you sort of put psi of
minus x here, minus x here.
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Well, you would do minus x here.
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But if the potential is even it
will solve the same equation.
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Now, the only
complication here is
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that there are a few X's
in the derivatives here.
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But whether there's a
complication or not there's
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two derivatives, so the
[? signs ?] should not matter.
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But I want to make
this a little clearer,
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and in order to do that I
will just define phi of x
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to be equal to psi at minus x.
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So if you have that,
the derivative of phi--
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with respect to x--
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you must differentiate this
with respect to the argument.
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You evaluate at the
argument, and then
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differentiate the argument
with respect to x.
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So that leaves you a minus 1.
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On the second derivative of
phi with respect to x squared--
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at x-- will be yet
another derivative.
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So you now get a
second derivative
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evaluated at the thing.
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And then differentiate
the thing inside again.
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So minus 1 plus another minus 1.
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So this is just psi
double prime of x--
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of minus x.
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Now, evaluate equation
one at minus x.
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Well, it would be the second
derivative of psi evaluated
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at minus x, plus 2m h squared,
E minus V of minus x--
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but that's the same as v of x--
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psi at minus x equals 0.
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And then you'll realize
that this thing is just
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the second phi of x.
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[? h squared ?] of x,
plus 2m over h squared,
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e minus v of x,
phi of x equals 0.
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So actually you've
proven that phi
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defined this way solved
the same Schrodinger
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equation with the same energy.
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So if one is true--
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this thing-- I guess we
could call it [? three ?]
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or [? two. ?]
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So you've proven that both
psi of x and phi of x--
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which is equal to
psi of minus x are
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solutions of the Schrodinger
equation with the same energy.
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And, therefore, if you
have two solutions--
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and now I emphasize this psi
of minus x, and the psi of x.
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If you have two
solutions then you
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can form the symmetric part of
the wave function, which is 1/2
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psi of x, plus psi of minus x.
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And the anti-symmetric
part of the wave function,
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which would be 1/2 of psi
of x minus psi of minus x.
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And notice that by
definition psi s of minus
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x is indeed psi s of x.
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It's symmetric.
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If you change x for minus
x on the left hand side,
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this goes into this,
this goes into that.
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So it's unchanged.
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Here it's changed by a psi.
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So psi a of minus x is equal
to psi minus psi a of x.
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And these two are solutions
with the same energy--
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psi s and psi a.
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You see, if you have--
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remember that key
fact-- if you have
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two solutions of the Schrodinger
equation with the same energy,
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any linear combination
of them is a solution
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with the same energy.
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So we form two
linear combinations
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and they have the same energy.
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And, therefore, the
theorem has been proven.
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The first part of the theorem--
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the wave functions
that you work with--
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can be chosen to be even or odd.
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And that's pretty nice.
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But now we go to the second
part of the statement.
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So for one-dimensional
bound states--
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1-D bound states.
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Again, there cannot
be two solutions.
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So it cannot be that there
are two degenerate solutions.
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So after all, psi of
x and psi of minus x,
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we have two solutions.
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This and psi of minus x.
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But if you're in a
one-dimensional bound state you
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cannot have two solutions.
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So they must be
proportional to each other.
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Now, if you started
with a solution--
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I want to say this.
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You start with a
solution from there,
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from the beginning
you can assume now--
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because of property two--
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that the solution is real.
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[? You can work those ?]
with real solutions.
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So in here, I can
assume that psi is real.
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Just simpler.
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So these two solutions-- that
would be two real solutions--
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would be degenerating energy--
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there's no degeneracy
for bound states.
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Therefore, these
two must be the same
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up to a constant, that again--
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because psi is real c--
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is real.
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There cannot be two solutions.
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Let x goes to minus
x in this equation.
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So you would get psi of x
equals c of psi of minus x.
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But psi of minus x uses
this equation again.
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You get c times c psi of x.
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But from comparing
these two sides,
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you get that c squared
must be equal to 1.
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But c is real.
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Therefore, there's
only two solutions.
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Two options.
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C is equal to plus
1-- in which case--
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psi is even-- automatically.
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Or c is equal to minus
1 and psi is odd.
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You have no option.
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You may think that the general
solution of a bound state--
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of a symmetric potential--
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could be arbitrary.
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But no.
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The solutions come out
automatically symmetrical
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or anti-symmetrical.
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And that's why-- when
we decided to search
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for all the solutions of
the finite square-- well,
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we could divide
it into two cases.
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Let's find the
symmetric solutions
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and the anti-symmetric
solutions.
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There is no other solution
of the Schrodinger equation.
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But what if you add a symmetric
to an anti-symmetric solution?
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Don't you get the
general solution?
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Well you cannot add them,
because for bound states they
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are different energies.
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And adding two solutions
with different energies
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is pointless.
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It's not a energy
eigenstate anymore.
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So a very powerful theorem.
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We'll be using it a
lot, and I thought
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you really ought to see it.