Odd parity solutions:
This means we can only have the sin(kx) term, and we have
(I need that minus sign in the last wave function, with the same coefficient as the first, otherwise, to assure it's odd) We again require that these three different looking functions will have to hook up, so that the wave function and its first derivative is continuous and finite everywhere.
From continuity of the wave function at -L/2:
From continuity of the derivative at -L/2:
Once again, the relations at +L/2 don't provide anything new, because we already used symmetry. This time, taking the ratio of the two equations at -L/2 gives
Once
again, the upper curve depends on the value of a. I have used the same value
(just below 3 pi) as in the previous plot. As a grows, the upper curve moves
higher, and we get more solutions. This is true for both the even and odd
plots. Also, as a grows, V0 (which is proportional to a^2) also grows. So, the
deeper the well, the more solutions. If a goes to infinity, we get an infinite
number of solutions. Notice that when a is big, the solutions for the even
cases are all very close to (but slightly less than) half odd integer multiples
of Pi (Pi/2, 3Pi/2, 5Pi/2, etc). The odd solutions in this case are close to,
but just less than, integer multiples of Pi (Pi, 2 Pi, 3 Pi, etc) In the limit
of a going to infinity, then, the solutions would be exactly on these values.
In that case:
These are just exactly our infinite well energies,
!
Also, the solutions are even, odd, even, odd, ... etc, just like our infinite well solutions.
For finite V0, x is always a little less, so our finite well energies are all a little lower than the corresponding infinite well energies. We found this qualitatively when we sketched the solutions earlier.
For any given value of "a", stare at the plots. You'll see that:
No matter how small "a" is, there is always at least one even solution. (Neat! Even weak wells can bind a particle)
The solutions always alternate even-odd, no matter what "a" is
There are a finite number of solutions. That number can end with an even or odd solution, depending on what "a" is exactly.
Deeper wells mean larger a mean more solutions.
Very deep wells mean the solutions go smoothly to the infinite box case. (go
back, and note that for a deep well,
,
so our exponentials cut off very fast, and the wave function at the
boundary is
.
Since E goes to the E for an infinite box, k goes to a zero of cos or sin, so
the wave function indeed goes to 0 at +/- L/2.
Notice also, once we have found x, we know E, and thus k and
.
That means our constants A and C in the wave function are related, because continuity of the wave function at either boundary gives
So we are left with only one overall unknown constant, say "C". But we
still haven't normalized! If we use
,
we will fix C, and we've determined the entire wave function, completely.
The bottom line is, you give me L, m, and V0, and I have found that
certain energies
are possible, where x is no longer simply an integer, but is discrete. We
obtain it by the graphical technique discussed above. These energies have
corresponding wave functions. As E moves up, the wave functions are even, odd,
even, odd, etc. And, there is now a maximum energy E(final) which is less than
V0, above which no more states exist. For E>V0, of course, we have free
particles. (Well, sort of. They can certainly run anywhere from -infinity to
+infinity) We'll discuss this case later.
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