Finite Potential Barrier

A finite potential barrier is the standard two-interface scattering problem. It extends the potential step by adding a second boundary, so a wave can reflect many times inside the finite region before emerging to the right or returning to the left.

Use the rectangular barrier

$$V(x)= \begin{cases} 0, & x\lt 0,\\ V_0, & 0\lt x\lt a,\\ 0, & x\gt a, \end{cases} \qquad V_0\gt 0.$$

The same model explains two distinct effects: tunneling when $0\lt E\lt V_0$, and above-barrier reflection when $E\gt V_0$. Both are consequences of wave matching at finite jumps.

Scattering Setup

For a wave incident from the left, the asymptotic regions have the same potential. Define

$$k=\frac{\sqrt{2mE}}{\hbar}.$$

The left and right wavefunctions are

$$\psi_I(x)=e^{ikx}+r e^{-ikx}, \qquad x\lt 0,$$

and

$$\psi_{III}(x)=t e^{ikx}, \qquad x\gt a.$$

The incident amplitude has been set to one. The reflection and transmission probabilities are

$$R=\lvert r\rvert^2, \qquad T=\lvert t\rvert^2,$$

because the left and right wave numbers are equal. If the asymptotic potentials were different, the velocity factor in Reflection and Transmission Coefficients would be required.

Matching Conditions

For a finite jump in the potential, both $\psi$ and $d\psi/dx$ are continuous. At the two interfaces,

$$\psi_I(0)=\psi_{II}(0), \qquad \psi_I'(0)=\psi_{II}'(0),$$

and

$$\psi_{II}(a)=\psi_{III}(a), \qquad \psi_{II}'(a)=\psi_{III}'(a).$$

These four linear equations determine $r$, $t$, and the two interior coefficients. The calculation is algebraically longer than the single-step problem, but it uses the same boundary rules.

Above-Barrier Scattering

For $E\gt V_0$, define the wave number inside the barrier by

$$q=\frac{\sqrt{2m(E-V_0)}}{\hbar}.$$

The interior solution is oscillatory:

$$\psi_{II}(x)=A e^{iqx}+B e^{-iqx}, \qquad 0\lt x\lt a.$$

Solving the four matching equations gives the transmission amplitude, with the phase convention used above,

$$t(E) = \frac{e^{-ika}} {\cos(qa) -i\frac{k^2+q^2}{2kq}\sin(qa)}.$$

Thus

$$T(E) = \left[ 1+ \frac{V_0^2\sin^2(qa)} {4E(E-V_0)} \right]^{-1}.$$

The barrier can reflect even when the particle energy is greater than the barrier height. Classically, a particle with $E\gt V_0$ crosses the region with reduced speed and then continues. Quantum mechanically, the wave number changes at both interfaces, and the reflected amplitudes can interfere.

When

$$qa=n\pi, \qquad n=1,2,3,\ldots,$$

the sine term vanishes and $T=1$. The barrier is then transparent because the interior phase makes the multiple reflected amplitudes cancel on the incident side. This is a single-barrier phase-matching effect; sharper quasi-bound resonance physics is developed in Resonant Transmission.

Tunneling Regime

For

$$0\lt E\lt V_0,$$

define

$$\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.$$

The interior solution is evanescent:

$$\psi_{II}(x)=A e^{\kappa x}+B e^{-\kappa x}.$$

The growing term is not a physical divergence because the barrier has finite width. Both exponentials are needed to satisfy matching at $x=0$ and $x=a$.

The transmission amplitude can be written as

$$t(E) = \frac{e^{-ika}} {\cosh(\kappa a) +i\frac{\kappa^2-k^2}{2k\kappa}\sinh(\kappa a)}.$$

Therefore

$$T(E) = \left[ 1+ \frac{V_0^2\sinh^2(\kappa a)} {4E(V_0-E)} \right]^{-1}.$$

This is the exact rectangular-barrier tunneling formula. Its physical interpretation and opaque-barrier approximation are treated in Rectangular Barrier Tunneling.

Opaque-Barrier Limit

If

$$\kappa a\gg 1,$$

then

$$\sinh^2(\kappa a)\approx \frac14 e^{2\kappa a},$$

and

$$T \approx \frac{16E(V_0-E)}{V_0^2} e^{-2\kappa a}.$$

The exponential factor is the main qualitative lesson:

$$T\propto e^{-2\kappa a}.$$

Increasing the barrier width, the particle mass, or the energy deficit $V_0-E$ suppresses transmission rapidly.

Barrier-Top Limit

The formulas above appear singular at $E=V_0$, because $q$ or $\kappa$ goes to zero. The limit is finite. Using $\sin(qa)\sim qa$ or $\sinh(\kappa a)\sim \kappa a$ gives

$$T(E=V_0) = \left[ 1+\frac{mV_0a^2}{2\hbar^2} \right]^{-1}.$$

The barrier top is therefore not automatically transparent. The finite width still matters.

Conservation and Phase

For the real time-independent barrier, one-channel current conservation gives

$$R+T=1.$$

The transmitted amplitude also contains a phase. The probability $T$ tells how much flux emerges; the phase of $t(E)$ controls interference when barriers are combined, and it matters for wave packets and resonant structures.

The Transfer Matrix Method packages this same calculation into interface and propagation matrices. It becomes especially useful for multiple barriers, wells, and layered potentials.

Physical Interpretation

The finite barrier should be read as a wave-interference problem, not as a particle deciding at one point whether to cross. At each interface, the wave-number mismatch generates reflected and transmitted components. Inside the barrier, those components propagate or decay, then meet the second interface. The observed $r$ and $t$ are the coherent result of all matching conditions at once.

For $E\lt V_0$, the interior is classically forbidden but finite. The evanescent solution can connect the left allowed region to the right allowed region. For $E\gt V_0$, the interior is classically allowed, but the changes in wavelength still produce reflection unless the phases cancel it.

Common Mistakes

• Treating finite-barrier tunneling as the same problem as a semi-infinite step.
• Computing $T$ from an interior coefficient instead of the outgoing current.
• Forgetting above-barrier reflection.
• Assuming $T=1$ whenever $E\gt V_0$.
• Discarding the growing exponential inside a finite forbidden region before applying both boundary conditions.
• Missing the finite limit at $E=V_0$.
• Confusing single-barrier phase transparency with long-lived double-barrier resonances.

Where This Is Used

• Potential Step gives the one-interface calculation.
• Rectangular Barrier Tunneling focuses on the $0\lt E\lt V_0$ tunneling regime and its exponential approximation.
• Reflection and Transmission Coefficients explains why $T$ is a current ratio.
• Transfer Matrix Method systematizes the same matching calculation.
• Resonant Transmission extends the interference idea to double barriers and quasi-bound states.
• Scattering from a Delta Potential is the zero-width, fixed-area limit.
• Finite Square Well uses related matching mathematics for bound states.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.

Exercises

1. For $E\gt V_0$, show that the barrier is perfectly transmitting when $qa=n\pi$.

Solution

The above-barrier transmission probability is

$$T(E) = \left[ 1+ \frac{V_0^2\sin^2(qa)} {4E(E-V_0)} \right]^{-1}.$$

If $qa=n\pi$, then $\sin(qa)=0$. The denominator is $1$, so

$$T=1.$$

The result comes from destructive interference among reflected amplitudes, not from the absence of interfaces.

2. Derive the barrier-top limit of $T$ from the below-barrier formula.

Solution

For $E\lt V_0$,

$$T(E) = \left[ 1+ \frac{V_0^2\sinh^2(\kappa a)} {4E(V_0-E)} \right]^{-1}, \qquad \kappa^2=\frac{2m(V_0-E)}{\hbar^2}.$$

As $E\to V_0^-$,

$$\sinh(\kappa a)\sim \kappa a.$$

Therefore

$$\frac{V_0^2\sinh^2(\kappa a)} {4E(V_0-E)} \sim \frac{V_0^2\kappa^2a^2}{4E(V_0-E)} = \frac{mV_0^2a^2}{2E\hbar^2} \to \frac{mV_0a^2}{2\hbar^2}.$$

Thus

$$T(E=V_0) = \left[ 1+\frac{mV_0a^2}{2\hbar^2} \right]^{-1}.$$

3. Use the exact tunneling formula to recover the leading opaque-barrier dependence.

Solution

When $\kappa a\gg 1$,

$$\sinh^2(\kappa a)\approx \frac14 e^{2\kappa a}.$$

The denominator of $T$ is then dominated by

$$\frac{V_0^2}{4E(V_0-E)} \frac14 e^{2\kappa a} = \frac{V_0^2}{16E(V_0-E)}e^{2\kappa a}.$$

Taking the reciprocal gives

$$T \approx \frac{16E(V_0-E)}{V_0^2}e^{-2\kappa a}.$$