The time-independent Schrödinger equation:
$$\ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$
is second order, so we should expect the solution to have two "degrees of freedom" which can be fixed by specifying boundary conditions. However in at least some cases, imposing the wavefunction normalization requirement determines those conditions.
For example, the infinite square well potential has a general solution $A\sin(kx) + B\cos(kx)$. The constant k is determined by the width of the well, so we get to pick two values (A and B) as expected. However continuity requires that B = 0, and the normalization requirement ends up fixing the value of A.
The quantum harmonic oscillator ends up similarly: the general solution has a term that goes like $Ae^{-x^2}$ and another that goes like $Be^{x^2}$, but the normalization requirement forces B to 0 and ultimately determines A.
My questions:
-
What physical interpretation can we assign to the choice of boundary conditions for the time-independent Schrödinger equation? Is there anything like the "initial position and velocity" interpretation for Newton's second law?
-
Under what circumstances does imposing the wavefunction normalization requirement determine the boundary conditions?
Best Answer
In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation.
Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. There's no time evolution happening in these states, so an initial condition doesn't make sense!
However, if you solve the time-independent Schrodinger equation using a potential $V(x)$ that does not go to infinity as $x$ goes to infinity, then you get solutions $\psi(x)$ that aren't bound states: they'll have some complicated behavior inside the potential well, then look like $e^{ikx}$ at infinity. Imposing the boundary condition "look like $e^{ipx}$ at infinity" physically means you want to consider scattering of particles with incoming momentum $k = p$.