Finite rectangular barrier matching

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

The tunnelling decay constant is the inverse length controlling the exponential decay of an evanescent wave in a constant potential barrier with $E<V_0$. The wavefunction amplitude decay length is $1/\kappa$; the squared-amplitude decay length is $1/(2\kappa )$.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

For a thick rectangular barrier with equal exterior potentials, constant mass and $0<E<U$, the exact matched transmission probability has the displayed limit when $\kappa a\gg 1$. The exponential is fixed by the evanescent action; the interface prefactor remains energy-dependent. A leading-exponential estimate can therefore differ by an order-one multiplier from the matched result.

For fixed $0<E<V_0$ and $\kappa a\gg 1$, the exact rectangular barrier transmission probability givesThis is the leading term of an asymptotic expansion: the wavefunction amplitude scale is $e^{-\kappa a}$, and the transmission probability scale is $e^{-2\kappa a}$, with an interface prefactor. Increasing width by $\Delta a$ multiplies this approximation by $e^{-2\kappa \Delta a}$. At fixed incident energy, increasing barrier height increases the tunnelling decay constant as $\sqrt{V_0-E}$; heavier nonrelativistic particles also have stronger suppression. For the numerical example in rectangular barrier transmission probability, the approximation gives $0.01397$, compared with the exact $0.01388$.

The regime condition matters. Near $E=V_0$ at fixed width, $\kappa a$ tends to zero and the opaque barrier approximation fails; use the exact threshold value. For $\kappa a\ll 1$, expansion of the hyperbolic sine givesThis tends to one as $a\to 0$ at fixed $E,V_0$, but the condition $\kappa a\ll 1$ alone does not imply nearly perfect transmission if the remaining factor is large. At $E>V_0$, use the oscillatory formula and its scattering resonances, rather than an exponential decay law.