Why is $ \psi = A \cos(kx) $ not an acceptable wave function for a particle in a box with rigid walls at $x=0$ and $x=L$ where
$$ k = \frac {(2mE)^{1/2}} {\hbar} \, ?$$
I had plugged the wave function into the time-independent Schrodinger equation for a particle in a box. By solving the Schrodinger equation from both sides I saw that the left hand side equaled to the right hand side, hence the function is a solution to the Schrodinger Equation.
I don't understand why this is not an acceptable wave function for a particle in a box with finite rigid walls.
Best Answer
Indeed the cosine function is valid for instance for other boundary conditions $$\psi'(0)=\psi'(L)=0$$ Your goal when you look for a set of solutions to the Schroedinger equation is to be able to decompose any general wavefunction as a sum over this set, and to do it consistently your boundary conditions must be the same for all solutions. So it is unlikely that you are going to find a 1-D box with boundary conditions so retorted than both the sine and the cosine functions are solutions.
Mathematically it can be seen that there is a four-parameter family of conditions that you can impose in the values of the wavefunction and its derivative evaluated at points $0$ and $L$. If you add the physical condition of locality, so that probability is not magically transferred from one extreme to other -your box transformed then in a circle-, you have only two parameters to play with:
$$\psi(0) = -\alpha \ \psi'(0)$$ $$\psi(L) = \beta \ \psi'(L)$$
where the derivatives are the ones taken inside the interval. The conditions can be seen as "bouncing" the probability wave with some extra phase, and you can think the whole set of four parameters as defining reflection and transmission (from one extreme to other) coefficients for the waves, but I will not complicate the answer writing them. Instead let me point out what it is happening technically: the hamiltonian for an one dimensional box is hermitian (symmetric) but not self-adjoint because its dual is defined in a different set of integrable functions, and one must consider their self-adjoint extensions, restrictions on the valid wavefunctions that grant the equality of the definition domains, the set of functions where the operators (momentum, position, energy hamiltonian...) and their adjoints act. A lot of papers can be found in the internet googling for this concept of "self-adjoint extensions", and I do hesitate about what to recommend, but for instance in page 21 of http://arxiv.org/abs/quant-ph/0103153 you can find your cosine :-)
Of course one must still explain with the usual conditions are choosen to be $\alpha = \beta=0$ and not infinite, or any finite value, or zero in one extreme and infinite in the other. The paper I have cited argues that it is because they proceed to build the system by raising the walls from a finite well.