maverick280857 said:
Lets go through this step by step. The Time Independent Schrodinger Equation inside the well reduces to
-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} = E\psi
or
\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0
Define
k^2 = \frac{2mE}{\hbar^2}
The "general" solution can be written as
\psi(x) = A e^{ikx} + B e^{-ikx}
Note that I have not made any statement about the sign of E or equivalently, about the real or complex nature of k. (If you will, k = \sqrt{2mE/\hbar^2} and the two solutions correspond to +k and -k)
Now, let's impose the boundary condition at x = 0 and x = a. At both these points, the wavefunction must vanish. This yields
A + B = 0
A e^{ika} + Be^{-ika} = 0
With B = -A, the second equation reduces to
A(e^{ika}-e^{-ika}) = 0
For a nontrivial solution, A\neq 0. This means that
e^{ika}-e^{-ika} = 0
or equivalently
sin(ka) = 0
Now, if the energy E were to be negative, k would be purely imaginary, of the form
k = i\kappa
so
sinh(\kappa a) = 0
the only solution of which is \kappa = 0 or equivalently, E = 0. This shows that there is no nontrivial (nonzero energy) solution of negative energy. So, we return to our old friend sin(ka) = 0 for real values of k and positive energies. This equation gives you the allowed energies that the well can support.
Its always better to reason ab-initio before you can get a feel for the solutions. That will come after you analyze the solutions to potentials which can be solved exactly. As you can see, this systematic approach of defining k^2 instead of k directly and considering the two possibilities separately is better than losing track of the k's and K's.
Physically, a simple exponentially decaying (or increasing) solution cannot possibly vanish at both walls, which is what your wavefunction must do in this problem. So, at the outset, the possibility of real exponentials is ruled out in this problem. (In contrast with a "finite" square well -- where V doesn't shoot off to infinity at the walls -- where you will encounter both kinds of solutions.)
Thank you for this lengthy response, I understand all these but through your solution right before we jump to
sinh(\kappa a) = 0
We can still go on in this manner
sin(ka) = 0
k = i\frac{\sqrt{2m\left|E\right|}}{\hbar} = \frac{n \pi}{a}
Then
\left|E\right| = -\frac{(n\pi\hbar)^2}{2ma^2}
Which taking into account the absolute value leads to a contradiction again, in terms of the Energy.
What I am trying to say is that one can show E<0 solutions are not applicable
only by reaching a contradiction.
That is to say "not satisfying the boundary conditions"
only depends on your way of solving the equation, if you define k sloppily you will end up with the real exponential solutions where you cannot satisfy the boundary conditions, it has nothing to do with E>0 or E<0.
As in:
I am not putting a restriction on E right now.
Define: k = \frac{\sqrt{-2mE}}{\hbar}
\frac{d^2\psi}{dx^2} =k^2\psi
Here I end up with the real exponential solutions, and I realize that only for E<0 the real exponentials can be regarded as complex exponentials and go on from there. Thus I can say that E<0 solutions are not applicable because they do not satisfy the boundary conditions. Or I can still go on assuming k has imaginary factors and reach a contradiction for E<0.
You see my question is not looking for an answer, what I am struggling through is that in order to show E<0 solutions are not applicable by showing that they do not satisfy the boundary conditions I
must be sloppy about defining k and be aware of it. Whereas showing E<0 solutions are not applicable by contradiction is physically and mathematically more satisfying because it does not depend on my definition of k.
What I am asking is if you are seeing anything that is wrong with this argument.
Thanks in advance