Reflection and Transmission Coefficients

Reflection and transmission coefficients are probability-current ratios, not merely squared wavefunction amplitudes. In one-dimensional conservative scattering, they measure what fraction of the incident flux is reflected back and what fraction is transmitted through the scattering region.

For a single incoming beam,

$$R=\frac{\lvert j_{\mathrm{ref}}\rvert}{j_{\mathrm{inc}}}, \qquad T=\frac{j_{\mathrm{trans}}}{j_{\mathrm{inc}}}.$$

The absolute value in $R$ appears because the reflected current flows opposite to the incident current.

Plane-Wave Current

For a one-dimensional wavefunction, the probability current is

$$j =\frac{\hbar}{2mi} \left( \psi^*\frac{d\psi}{dx} -\psi\frac{d\psi^*}{dx} \right).$$

For a right-moving plane wave,

$$\psi(x)=A e^{ikx}, \qquad k\gt 0,$$

the current is

$$j=\frac{\hbar k}{m}\lvert A\rvert^2.$$

For a left-moving plane wave $A e^{-ikx}$, the current is negative:

$$j=-\frac{\hbar k}{m}\lvert A\rvert^2.$$

This is why amplitudes alone are not always probabilities. A wave with a larger amplitude but a smaller velocity can carry the same flux as a smaller-amplitude faster wave.

General One-Dimensional Setup

Suppose the potential approaches constants on the left and right:

$$V(x)\to V_L \quad\text{as }x\to-\infty, \qquad V(x)\to V_R \quad\text{as }x\to+\infty.$$

For energy above both asymptotic potentials, define

$$k_L=\frac{\sqrt{2m(E-V_L)}}{\hbar}, \qquad k_R=\frac{\sqrt{2m(E-V_R)}}{\hbar}.$$

With a wave incident from the left,

$$\psi(x)\sim A_{\mathrm{inc}}e^{ik_Lx} +A_{\mathrm{ref}}e^{-ik_Lx} \quad (x\to-\infty),$$

and

$$\psi(x)\sim A_{\mathrm{trans}}e^{ik_Rx} \quad (x\to+\infty).$$

Define scattering amplitudes by

$$r=\frac{A_{\mathrm{ref}}}{A_{\mathrm{inc}}}, \qquad t=\frac{A_{\mathrm{trans}}}{A_{\mathrm{inc}}}.$$

Then

$$R=\lvert r\rvert^2,$$

while

$$T=\frac{k_R}{k_L}\lvert t\rvert^2.$$

The factor $k_R/k_L$ is the velocity factor. If the left and right asymptotic potentials are equal, then $k_R=k_L$ and $T=\lvert t\rvert^2$.

Flux Conservation

For a real, time-independent potential with one open channel on each side, probability current is conserved:

$$j_{\mathrm{inc}} =\lvert j_{\mathrm{ref}}\rvert +j_{\mathrm{trans}}.$$

Dividing by $j_{\mathrm{inc}}$ gives

$$R+T=1.$$

This statement is not a definition; it is a consequence of unitary time evolution and real conservative dynamics. It can fail if the model includes absorption, gain, complex optical potentials, explicit time dependence, or additional outgoing channels not included in the one-dimensional bookkeeping.

Closed Channels And Evanescent Tails

If $E\lt V_R$, then the right side is classically forbidden and the right-region solution is evanescent rather than propagating. It can have nonzero amplitude near the interface, but it carries no current to $x\to+\infty$.

For a semi-infinite step with $E\lt V_R$,

$$T=0, \qquad R=1.$$

This does not contradict finite-barrier tunneling. A finite barrier has a second interface, allowing the evanescent solution inside the barrier to match onto a propagating transmitted wave on the far side.

Flux-Normalized Amplitudes

Scattering theory often uses flux-normalized waves,

$$\frac{1}{\sqrt{v}}e^{ikx}, \qquad v=\frac{\hbar k}{m}.$$

With this convention, each unit-amplitude incoming or outgoing channel carries unit flux. The velocity factor is built into the basis, so transmission probabilities can be written as squared magnitudes of flux-normalized matrix elements.

This convention is useful in scattering matrices, but it should not obscure the physical definition: probabilities are current ratios.

Worked Example: Above A Potential Step

For a step with $E\gt V_0$, the left and right wave numbers are

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

Matching at the step gives

$$r=\frac{k-q}{k+q}, \qquad t=\frac{2k}{k+q}.$$

The reflection coefficient is

$$R=\lvert r\rvert^2 =\left(\frac{k-q}{k+q}\right)^2.$$

The transmission coefficient is

$$T=\frac{q}{k}\lvert t\rvert^2 =\frac{4kq}{(k+q)^2}.$$

Then

$$R+T=1.$$

If one incorrectly used $T=\lvert t\rvert^2$, the result would generally not conserve flux.

Common Mistakes

• Computing $T$ as $\lvert t\rvert^2$ when the incident and transmitted wave numbers differ.
• Calling an evanescent tail a transmitted current.
• Forgetting that reflected current is negative, then mishandling the sign of $R$.
• Assuming $R+T=1$ in a problem with absorption, a complex potential, or extra channels.
• Comparing amplitudes before specifying the normalization convention.
• Treating stationary plane-wave coefficients as detector probabilities without converting to currents.

Where This Is Used

• Conductance Quantization promotes flux-normalized transmission probabilities to a multichannel terminal-conductance ledger.
• Potential Step is the simplest example where the velocity factor matters.
• Finite Potential Barrier is the standard two-interface barrier calculation.
• Rectangular Barrier Tunneling has equal wave numbers on both sides in the standard setup, so $T=\lvert t\rvert^2$ there.
• Free Particle gives the plane-wave current and group velocity.
• Normalization Conventions explains box, delta, and flux normalization.
• Wave Packets connects stationary scattering amplitudes to localized incoming beams.
• Wave Packets and Scattering explains how these current coefficients become late-time packet probabilities.
• One-Dimensional Scattering Revisited embeds $r$ and $t$ in a flux-normalized two-channel $S$-matrix.

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.

Exercises

1. A scattering solution has $k_L=2k_R$, $r=1/3$, and $t=4/3$. Compute $R$ and $T$.

Solution

The reflection coefficient is

$$R=\lvert r\rvert^2=\frac{1}{9}.$$

The transmission coefficient includes the velocity factor:

$$T=\frac{k_R}{k_L}\lvert t\rvert^2 =\frac{1}{2}\frac{16}{9} =\frac{8}{9}.$$

Thus $R+T=1$.

2. For a plane wave $A e^{-ikx}$ with $k\gt 0$, show that the current is negative.

Solution

Using

$$j =\frac{\hbar}{2mi} \left( \psi^*\psi'-\psi(\psi^*)' \right),$$

with $\psi=Ae^{-ikx}$ gives $\psi'=-ik\psi$ and $(\psi^*)'=ik\psi^*$. Therefore

$$j =\frac{\hbar}{2mi} \left( -ik\lvert A\rvert^2 -ik\lvert A\rvert^2 \right) =-\frac{\hbar k}{m}\lvert A\rvert^2.$$

3. Why is $R+T$ not necessarily equal to $1$ for a complex absorbing potential?

Solution

A complex absorbing potential is not a conservative Hamiltonian for the one-channel probability current. It removes norm from the explicit scattering channel, modeling loss into untracked degrees of freedom. The incident current can exceed the sum of reflected and transmitted currents, so $R+T\lt 1$ in the effective one-dimensional description.