Interpretation of boundary conditions in time-independent Schrödinger equation

Question

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:

1. 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?

2. Under what circumstances does imposing the wavefunction normalization requirement determine the boundary conditions?

Answer 1

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 $k = p$.

Answer 2

1. The choice of boundary conditions fixes the domain and thus a self-adjoint extension of your Schroedinger operator $$-\frac{\hbar^2}{2m} \Delta + V\:.$$ In turn, this choice determines the spectrum of that operator, i.e., the eigenvalues $E$. (There is no analogy with initial conditions here.)

2. No circumstances! Boundary conditions are uncorrelated with normalization as they just define a subspace of the Hilbert space, the domain of the operator above, and if $\psi$ belongs to that subspace also $c\psi$ does for every $c\in \mathbb C$.

Answer 3

As far as I know, and I'm only an undergraduate student, the boundary conditions in Schrodinger equations are there to hold some special subspace of the Hilbert space of the system or the Hilbert space as a whole.

Bound states, for example, form a subspace on the Hilbert space. The boundary condition to that is that $\psi\sim e^{-r}$ at infinity, for every time.

Scattering states, are wave packets that at $t=\pm\infty$ behave as a free wave packet (get envolve on time by $\frac{-\hbar ^2}{2m}\nabla ^2$ instead of $H$).

Turns out that energy eigenstates, solutions of the time-independent Schrodinger equations, are always bound states. This is so because plane waves cannot be normalized - but you can always construct wave packets of plane waves - . This wave packets are the scattering states. You can consider sometimes a state like $|k\rangle$ for a plane wave and for practical purposes this work well all the time. But, some times, like the proof of the Lippmann-Schwinger theorem, things get little trick and is good to remember this stuff of wave packets.

For time-independent Schrodinger equation, the boundary condition could be simple the requirement that the wave functions can be normalized. This is a special feature of the time independent equation. The physical explanation is simple: the eigenstates that are bounded are the only states that is independent of time (get only a phase) and can be normalized. Other normalizable state that is not bounded is necessarily time dependent, because it need to be a wave packet for be normalizable, plane waves are not.

The quantization of energy in bound states comes from the boundary conditions of the Schrodinger equation. The boundary condition imposes that some numbers need to be quantized like angular momentum and energy. The energy, is due the fact that the wave function is bounded in space, and angular mometum is due symmetries of the space.

Actually, this is the nature of the quantization on physical quantities. The fact that a particle, or something, is localized at some finite region sufficiently narrow, the interference on the amplitudes of probability is important and there would be only some stationary states for this interferences.