Boundary Conditions

The time-independent Schrödinger equation,

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi,$

is a second-order ordinary differential equation. Its general solution in any region where $V$ is constant contains two free constants, which are fixed by imposing boundary conditions — rules that $\psi$ must satisfy wherever the potential changes.

Why continuity is required

The Schrödinger equation fixes $d^2\psi/dx^2 = \frac{2m}{\hbar^2}(V-E)\psi$ . Wherever $V$ is finite, the right-hand side is bounded, so $d^2\psi/dx^2$ is bounded. Integrating a bounded function across an infinitesimal interval gives a change that vanishes as the interval shrinks, so $d\psi/dx$ is continuous; and because $d\psi/dx$ is finite, integrating once more makes $\psi$ itself continuous. These requirements follow from the equation's structure:

$\psi(x)$ must be continuous everywhere.
$d\psi/dx$ must be continuous wherever $V$ is finite (see the caveat below).

Together, conditions 1 and 2 are the standard boundary conditions applied at every point where two regions of different potential meet.

Matching across a boundary

Consider a particle at a boundary $x = a$ between two regions. Label the solution in the left region $\psi_L$ and in the right region $\psi_R$ . Continuity of $\psi$ requires

$\psi_L(a) = \psi_R(a).$

Continuity of the derivative requires

$\psi_L'(a) = \psi_R'(a).$

These two equations — sometimes called the junction conditions — are the practical tool for constructing solutions that span multiple regions. They reduce the total number of free constants from (2 per region) to a manageable set, and combined with normalization they fix the energy eigenvalues.

The infinite-potential exception

What happens when the potential is infinite, as at the hard walls of an infinite square well? Rewriting the Schrödinger equation as

$\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}(V - E)\psi,$

reveals that if $V \to \infty$ inside a region, the only way to keep $d^2\psi/dx^2$ finite is to require $\psi = 0$ throughout that region. At the wall itself, continuity of $\psi$ then demands

$\psi = 0 \quad \text{at the wall.}$

The derivative $d\psi/dx$ , however, need not be continuous at an infinite-potential wall, because the second derivative is singular there anyway. This is the one exception to condition 2 above.

Physical meaning of the conditions

The two conditions encode real physics. Continuity of $\psi$ is required because $|\psi|^2$ is a probability density: a discontinuous probability density would assign different probabilities to the same point from the left and the right, which is meaningless. Continuity of $\psi'$ follows from the finite-kinetic-energy condition: the kinetic energy operator involves $d^2\psi/dx^2$ , and a jump in $\psi'$ would make $d^2\psi/dx^2$ a Dirac delta, which would require an infinitely sharp spike in the potential to balance it.

Worked example: finite step potential

As a concrete illustration, consider a particle of energy $E$ travelling from left to right, with $V = 0$ for $x < 0$ and $V = V_0$ (finite, with $E > V_0 > 0$ ) for $x \geq 0$ .

Left region ( $x < 0$ ): The Schrödinger equation reads

$\frac{d^2\psi}{dx^2} = -k_1^2\psi, \quad k_1 = \frac{\sqrt{2mE}}{\hbar}.$

The general solution is $\psi_L = Ae^{ik_1 x} + Be^{-ik_1 x}$ , combining an incident wave and a reflected wave.

Right region ( $x \geq 0$ ): With potential $V_0$ ,

$\frac{d^2\psi}{dx^2} = -k_2^2\psi, \quad k_2 = \frac{\sqrt{2m(E-V_0)}}{\hbar}.$

The transmitted wave moving to the right is $\psi_R = Ce^{ik_2 x}$ .

Applying junction conditions at $x = 0$ :

Continuity of $\psi$ : $A + B = C.$

Continuity of $\psi'$ : $ik_1(A - B) = ik_2 C.$

These two equations relate $B$ and $C$ to $A$ . Solving,

$\frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2}, \qquad \frac{C}{A} = \frac{2k_1}{k_1 + k_2}.$

The reflection coefficient $R = |B/A|^2$ and transmission coefficient $T = (k_2/k_1)|C/A|^2$ satisfy $R + T = 1$ , confirming probability is conserved. Notice that even when $E > V_0$ — the classically unrestricted case — there is a non-zero reflected amplitude whenever $k_1 \neq k_2$ . This quantum reflection has no classical analogue and is a direct consequence of the wave nature of $\psi$ .

Summary

| Condition | When it applies | Mathematical form | |-----------|----------------|-------------------| | Continuity of $\psi$ | Always | $\psi_L(a) = \psi_R(a)$ | | Continuity of $\psi'$ | $V$ finite at $a$ | $\psi_L'(a) = \psi_R'(a)$ | | $\psi = 0$ at wall | $V = \infty$ inside | $\psi(a) = 0$ |

Applying these conditions at every interface is how quantum mechanics turns a differential equation into discrete energy levels, transmission coefficients, and the other quantitative predictions that can be tested against experiment.