Matching Conditions at Potential Steps
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You should be comfortable solving the time-independent Schrödinger equation in regions of constant potential and understand the physical meaning of the wavefunction's amplitude and phase.
Why this matters
Matching conditions allow us to stitch together local solutions into a global wavefunction that respects the physics of quantum transitions.
When a particle encounters a sudden change in potential energy, such as a step or a barrier, the mathematical form of the wavefunction changes. To determine the transmission and reflection probabilities, we must connect the solutions on either side of the discontinuity. These connections are not arbitrary; they are dictated by the requirement that the probability current remains conserved and that the wavefunction remains physically meaningful (single-valued and finite) across the entire domain.
The Physical Intuition
Imagine a wave travelling along a rope that is joined to a heavier rope. The wave does not vanish at the junction; part of it reflects and part is transmitted. The rope cannot break at the junction, and the knot is massless, so the rope must also have the same slope on both sides; these two conditions fix how much is reflected. In quantum mechanics the wavefunction \(\psi(x)\) obeys the same two conditions at a finite step. If \(\psi\) jumped, \(\psi\prime\) would contain a delta function and \(\psi\prime\prime\) its derivative, so the kinetic-energy term of the Schrödinger equation could not be balanced by any finite potential. If \(\psi\prime\) jumped, \(\psi\prime\prime\) would contain a delta function, which only a delta-function (or infinite) potential can balance. For any finite potential step, both the wavefunction and its slope are continuous.
Mathematical Derivation of Continuity
Consider a potential \(V(x)\) that has a finite jump at \(x=a\). The time-independent Schrödinger equation is:
Rearranging for the second derivative:
Since \(V(x)\) is finite and \(\psi(x)\) is assumed to be finite, the right-hand side is finite everywhere. If the second derivative is finite, the first derivative \(\psi'(x)\) must be continuous. If \(\psi'(x)\) is continuous, then \(\psi(x)\) must also be continuous. Therefore, at \(x=a\):
These are the standard matching conditions. They hold for any finite potential discontinuity. Note that if \(V(x)\) contained a delta function, the second derivative would have a delta function, allowing a discontinuity in \(\psi'\), but this is a special case not covered by the standard finite-step conditions.
Applying Conditions to a Potential Step
Consider a particle of energy \(E\) incident from the left on a potential step \(V_0\) at \(x=0\). Region I (\(x < 0\)): \(V=0\). The wavefunction is \(\psi_I = A e^{ik_1 x} + B e^{-ik_1 x}\), where \(k_1 = \sqrt{2mE}/\hbar\). Region II (\(x > 0\)): \(V=V_0\). Case 1: \(E > V_0\). The particle is transmitted. \(k_2 = \sqrt{2m(E-V_0)}/\hbar\). The wavefunction is \(\psi_{II} = C e^{ik_2 x}\) (no reflected wave in region II). Case 2: \(E < V_0\). The particle is evanescent. \(\kappa = \sqrt{2m(V_0-E)}/\hbar\). The wavefunction is \(\psi_{II} = C e^{-\kappa x}\) (decaying exponential).
In both cases, we apply the matching conditions at \(x=0\):
For \(E > V_0\), the derivative condition is \(ik_1(A - B) = ik_2 C\). For \(E < V_0\), the derivative condition is \(ik_1(A - B) = -\kappa C\). Solving these two linear equations for \(B\) and \(C\) in terms of \(A\) yields the reflection and transmission amplitudes. The probability current is conserved, ensuring that the sum of reflection and transmission probabilities equals unity.
Probability Current and Conservation
The physical justification for the continuity of \(\psi'\) lies in the conservation of probability. The probability current density in one dimension is given by:
The continuity equation states that \(\frac{\partial |\psi|^2}{\partial t} + \frac{\partial J}{\partial x} = 0\). For a time-independent state, this implies that \(J\) is constant in space. Continuity of \(\psi\) and \(\psi'\) guarantees that \(J\) has the same value on both sides of the step, so no probability piles up at the boundary. This is why the transmission and reflection coefficients, defined from the currents, always add up to one.
The derivation
Continuity Conditions at Potential Steps
Across any point x = a where the potential V(x) has a finite jump discontinuity, a solution of the one-dimensional time-independent Schrödinger equation and its first derivative are both continuous: ψ(a+) = ψ(a−) and ψ′(a+) = ψ′(a−). These are not extra postulates but consequences of the equation itself, valid whenever V stays bounded.
A common error is to assume that the wavefunction must be zero at a potential step. This is only true for an infinite potential barrier. For a finite step, the wavefunction penetrates into the classically forbidden region (if \(E < V_0\)) or continues with a different wavelength (if \(E > V_0\)). The value of \(\psi\) at the boundary is generally non-zero and is determined by the matching conditions. C
Worked example
Problem. An electron with energy \(E = 5.0\ \mathrm{eV}\) is incident from the left on a potential step of height \(V_0 = 3.0\ \mathrm{eV}\). Calculate the reflection coefficient \(R\).
- Identify the wave numbers. Since \(E > V_0\), the particle is transmitted. \(k_1 = \frac{\sqrt{2mE}}{\hbar}\) and \(k_2 = \frac{\sqrt{2m(E-V_0)}}{\hbar}\). Note that \(k_2/k_1 = \sqrt{\frac{E-V_0}{E}} = \sqrt{\frac{5.0-3.0}{5.0}}\), with \(E-V_0 = 2.0\ \mathrm{eV}\). Ratio \(r_k = \frac{k_2}{k_1} = \sqrt{\frac{2.0}{5.0}} = \sqrt{0.4}\).
- Use the matching conditions at \(x=0\): \(A + B = C\) \(ik_1(A - B) = ik_2 C \implies A - B = \frac{k_2}{k_1} C\).
- Solve for \(B/A\). From the first eq, \(C = A + B\). Substitute into the second: \(A - B = \frac{k_2}{k_1} (A + B)\). \(A (1 - \frac{k_2}{k_1}) = B (1 + \frac{k_2}{k_1})\). \(\frac{B}{A} = \frac{1 - k_2/k_1}{1 + k_2/k_1}\).
- Calculate the reflection coefficient \(R = |\frac{B}{A}|^2\). Let \(u = k_2/k_1 = \sqrt{0.4} \approx 0.6325\). \(\frac{B}{A} = \frac{1 - 0.6325}{1 + 0.6325} = \frac{0.3675}{1.6325} \approx 0.2251\). \(R = (0.2251)^2 \approx 0.0507\).
Units check. The reflection coefficient is a dimensionless ratio of probabilities, so it is unitless. The input energies were in eV, but the ratio \(k_2/k_1\) depends only on the ratio of energies, so the units cancel correctly.
Three quick tests
- What happens to the wavefunction \(\psi(x)\) at a finite potential step?
Answer
It remains continuous, and its first derivative \(\psi'(x)\) also remains continuous. - If \(E < V_0\), what is the form of the wavefunction in the region \(x > 0\)?
Answer
It is an exponentially decaying function, \(\psi(x) = C e^{-\kappa x}\), where \(\kappa\) is real. - Why must the probability current \(J\) be continuous across the boundary?
Answer
Because probability is conserved; there are no sources or sinks of probability at a finite potential step.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.