# Why do we not require higher derivatives to match at boundary when solving the Schrödinger equation in a given potential?

When solving the time independent Schrödinger equation for a given potential in 1D, the main part of the solving involves matching boundary conditions. Usually, we require the value and the first derivative to match at boundary. This intuitively make sense as we would like the wave function to have matching value and slope. However, why do we not enforce the curvature, and indeed, higher derivative to match?

• Actually the first derivative is not required to be continuous. This is only true for the special and unrealistic case, if the effective mass in both regions is the same. The correct boundary condition would be 1/m1 dPsi1/dx = 1/m2 dPsi2/dx – engineer Apr 23 '15 at 6:13

## 3 Answers

Because in ODE you only need $n$ conditions according the $n$-order of ODE, but these conditions could be on the same point, value of function and derivative, or two points, value of function first and last point of interval.

If I am correctly interpreting your question, you have a 1D time independent S. equation with $V$ which is discontinuous in some points. You solve that equation for a fixed eigenvalue $E$ separately in each continuity interval obtaining functions $\psi_E$ which are $C^2$ in every open interval and depends on two arbitrary constants. Finally you mach the found functions at the boundaries of the various intervals. You are asking why only continuity and first-derivative continuity is required on the singular point and not continuity of the second derivative.

I consider proper eigenvalues and associated eigenvectors in the following.

The reason is technical. You have an operator of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x) \tag{1}$$ and you are looking for eigenfunctions $\Psi_E$, so that $$H\psi_E= E\psi_E\:.\tag{2}$$ As you probably know, the operator $H$ must be self-adjoint, otherwise it does not have a Hilbert basis of eigenvectors. On the other hand, the domain of the operator, $D(H)$, is not the whole space, so $\psi_E$ must belong to that domain. The point is that operators with the form (1) are not self-adjoint when defined on spaces of differentiable functions. They are however essentially self-adjoint, i.e. $H^\dagger$ is self-adjoint and $H^\dagger$ is the true energy observable. So, the physically correct Schroedinger equation is not (2), but is $$H^\dagger \psi_E = E\psi_E\tag{3}$$ where $\psi_E$ belongs to the larger domain of $H^\dagger$ which, in turn, is not a differential operator. Using the defintion of adjoint operator, (3) can equivalently be written $$\langle H \phi| \psi_E \rangle = E\langle \phi| \psi_E \rangle \quad \forall \phi \in D(H)\:.$$ Explicitly $$\int_{\mathbb R} \frac{d^2 \phi}{dx^2}\psi_E(x) dx =\int_{\mathbb R} \frac{2m}{\hbar^2}(V(x)-E)\phi(x) \psi_E(x)\: dx\quad \forall \phi \in D(H)\tag{4}$$ As usually $D(H)\supset C_0^\infty(\mathbb R)$, (4) says that $\psi_E$ admits second weak derivative. At this point an analysis based on Weyl elliptic regularity theorems proves that, if $\psi_E \in L^2$ satisfying (4) exists it must be $C^2$ in the intervals where $V$ is continuous and it must be $C^1$ on the remaining points.

Apart from (physically motivated) asymptotic prescribed behaviour at infinity $|x|\to\infty$, we don't actually impose/demand/require any conditions on the wave function $\psi$ beyond the TISE (understood in weak sense). Continuity and (possibly higher) smooth conditions are instead derived from a standard bootstrap argument, see e.g. my Phys.SE answer here for details.