hello in this video we're going to take
a look at the orthonormality
of different energy eigenstates so
we have from the previous video that all
physical states in quantum mechanics are
normalized
but in addition for any two different
energy eigenstates of the infinite
potential well
are also also orthogonal to each other
by which we mean the following
the integral of phi_n* . phi_m
over all of space in this case over
the range of the well
is equal to zero if n doesn't equal m
together these conditions tell us
that the states are
both normalized and orthogonal and we
abbreviate this to 'orthonormal'
so the statement is as follows
so the integral of phi_n* phi_m over
all of space
is equal to the kronecker delta which is
defined to be 1 if n=m
and 0 if n doesn't equal n. This is an
incredibly useful relation
which we'll put to much use in the coming
videos it doesn't just hold for energy
eigenstates of the infinite potential
well it holds
for a much broader class of problems and
we'll see much more of that
when we come to look at matrix mechanics
later on in the course
for now let's look at a worked example
where we show this explicitly
for the eigenstates of the
infinite potential well
who's a good boy
I've got a cameo from Geoffrey in the
background
all right so let's take a look at
the energy eigenstates of the infinite
potential well
another quick reminder as to what the
potential looks like
we have potential going from 0 to L
along x up to infinity here
and 0 within the well and we've seen
that
the normalized
eigenfunctions look like this
and this time we'd like to prove that
two
eigenfunctions are orthogonal to each
other
so what this means is that
the integral from minus infinity to infinity
of phi_n(x)*.phi_m(x) dx
equals delta_{nm}
which by definition equals
1 if n=m 0 if
n doesn't equal m
orthogonal of course implies
something
that there's some similarity to two
vectors being at 90 degrees and we'll
see when we study matrix mechanics that
there's actually a very close analogy
to that idea so let's
show this for the the eigenstates
of the
infinite well let's just substitute this
expression into here
so we get a 2/L out the front
the integral only goes from zero to L
because phi is 0 outside of that range
sin(n pi x/L)sin(m pi x/L)dx
okay we're going to need to use another
trigonometric identity
this time we need to use that
cos(A+B)=cos(A)cos(B)-sin(A)sin(B)
and therefore
cos(A-B)=cos(A)cos(B)+sin(A)sin(B)
okay so if we put A = n pi x/L
B = m pi x/L
we want to
add this one sorry subtract this one
from this one and that'll give us the
two sines
sorry you can't quite see that so we
want to subtract this expression from
this expression that will give us
2sin(A)sin(B) over here
and so overall we'll have so we'll
have a factor of two we bring out the
front so we have 1/L
integral from zero to L
cos(A-B) so it's
cos((n-m) pi x/L)
minus cos((n+m)pi x/L)
this integrates to
these will become sines
sin((n-m)pi x /L)/((n-m)pi)
and multiplied by L; minus
sin((n+m)pi x/L)/((n+m)pi)
multiply by L
between zero and L okay well the Ls
cancel so that's good
news when we stick in
sin(0)=0 so the zero
limit is always zero when we substitute
the L in
this one is always going to be zero
because n+m
for integer n and m is always an
integer as the sum of two integers is an
integer
and so this is ... substitute in
here these two cancel
so it's an integer times pi. Sine of that
is always zero so this one
disappears this one also almost
disappears
the zero limit disappears when we put L
in so we have sin((n-m)pi)
well n-m is also an integer for
integer n and m
the only problem is when n equals m this
is zero. sin(0)=0 -- good --
but n-m is also 0 on the bottom
and 0 divided by 0 is undefined
so let's substitute it in the only one
we need to worry about
sin((n-m)pi)/((n-m)pi)
so to work out what that is we use
l'Hopital's rule differentiate the top
and bottom
and we can differentiate them with
respect to, say, n-m
so this thing must equal
as we multiply this we get
cos((n-m)pi)
differentiate with respect to n-m
and we just get a pi on the bottom
cancel those this is evaluated at n
equals
m and so this thing equals one
and so we found that if we'd put
anything else anything other than n
equals m into here it would have
disappeared because if n
and m are different integers this is
non-zero this is zero
the whole thing is zero if n equals m it
evaluates to one and that's precisely
what we wanted to show
we've shown the orthogonality of
these eigenstates and in fact we've
shown that they're orthonormal because
we've already normalized the eigenstates
so the normalized eigenstates
of the states within the infinite
well
are orthonormal okay thanks for your
time