The transfer matrix method

Last section ended by laying the problem on the table: every extra slab of potential means one more region, two more undetermined coefficients, two more matching equations. The single square barrier of section 2.11 took 4 equations and plenty of sweat; a resonant-tunneling diode is a three-layer barrier-well-barrier stack, 8 equations; a semiconductor superlattice starts at a hundred layers. The brute-force route of solving simultaneous equations has hit its end.

Look at the problem from a different angle. The slab-by-slab matching is in fact highly repetitive: the same thing happens in every slab — two waves propagate some distance, then get converted at an interface, via the continuity conditions, into the next slab’s two waves. For a repeated linear operation, mathematics has a ready-made packaging: matrices. One $2\times2$ matrix per slab, and the whole potential is the matrices multiplied together in order. A hundred layers? A hundred matrix multiplications — the blink of an eye for a computer, and barely a page by hand.

This is the transfer matrix method.

Two building blocks

In a piecewise-constant potential, the solution in slab $j$ (potential $V_j$, width $d_j$, left endpoint $x_j$) is always a superposition of two waves:

$$\psi_j(x)=A_j\,\ee^{\ii k_j(x-x_j)}+B_j\,\ee^{-\ii k_j(x-x_j)},\qquad k_j=\frac{\sqrt{2m(E-V_j)}}{\hbar}\tag{4.3.1}$$

$A_j$ is the amplitude of the right-moving wave, $B_j$ of the left-moving one (when $E<V_j$, $k_j$ is purely imaginary and the two terms automatically become growing and decaying exponentials — the inside-the-barrier solutions of section 2.11, no separate rule needed). The entire content of a slab is one pair of numbers $(A_j,B_j)$. The method’s core question is single: given this slab’s $(A_j,B_j)$, how do we get the next slab’s?

Derivation: the propagation matrix and the interface matrixbasic~8 minexpand

Getting from $(A_j,B_j)$ to $(A_{j+1},B_{j+1})$ takes two gates.

Gate one: cross the slab. The right-moving wave travels from the left end to the right end (distance $d_j$) and accumulates phase $\varphi_j=k_jd_j$; the left-moving wave accumulates $-\varphi_j$. At the right endpoint, the two waves’ “local amplitudes” are $A_j\ee^{\ii\varphi_j}$ and $B_j\ee^{-\ii\varphi_j}$. As a matrix:

$$P_j=\begin{pmatrix}\ee^{\ii\varphi_j}&0\\0&\ee^{-\ii\varphi_j}\end{pmatrix}\tag{4.3.2}$$

Diagonal — propagation never mixes left- and right-movers; it only rotates each one’s phase (or, inside a barrier, grows and shrinks them).

Gate two: cross the interface. At an interface, $\psi$ and $\psi'$ are continuous ($V$ is finite — the old rule from section 2.2). Using the local amplitudes from the previous step (write $\tilde A=A_j\ee^{\ii\varphi_j}$, $\tilde B=B_j\ee^{-\ii\varphi_j}$), the two conditions read

$$\tilde A+\tilde B=A_{j+1}+B_{j+1}\tag{4.3.3}$$$$k_j(\tilde A-\tilde B)=k_{j+1}(A_{j+1}-B_{j+1})\tag{4.3.4}$$

Adding and subtracting, solve for the new amplitudes (write $r=k_j/k_{j+1}$):

$$A_{j+1}=\frac{1+r}{2}\tilde A+\frac{1-r}{2}\tilde B,\qquad B_{j+1}=\frac{1-r}{2}\tilde A+\frac{1+r}{2}\tilde B\tag{4.3.5}$$

That is, the interface matrix

$$S_{j\to j+1}=\frac12\begin{pmatrix}1+r&1-r\\1-r&1+r\end{pmatrix},\qquad r=\frac{k_j}{k_{j+1}}\tag{4.3.6}$$

Check: when $k_{j+1}=k_j$ (no step) we get $r=1$ and $S$ collapses to the identity — no interface, no conversion, correct. When $r\ne1$ the off-diagonal entries are non-zero: the interface stirs left- and right-movers together, which is exactly where reflection comes from, matching section 2.11’s picture that “reflection needs an impedance mismatch”.

Assembly. Pass through the gates in order, leftmost slab to rightmost:

$$\begin{pmatrix}A_N\\B_N\end{pmatrix}= \underbrace{S_{N-1\to N}P_{N-1}\cdots S_{1\to2}P_1S_{0\to1}}_{\displaystyle M} \begin{pmatrix}A_0\\B_0\end{pmatrix}\tag{4.3.7}$$

Mind the order of multiplication: the matrix the wave passes through first sits rightmost, like walking through a string of doors. The physics of the entire potential is condensed into one $2\times2$ complex matrix $M$.

Reading the scattering quantities off M

Let the wave come in from the left: $A_0=1$ (incident), $B_0=r_{\text{amp}}$ (reflection, to be found), and in the rightmost slab $B_N=0$ (no wave source on the right — the problem’s one and only physical input, same as section 2.11). Expand the second row of $M$:

$$0=M_{21}+M_{22}\,r_{\text{amp}} \;\Longrightarrow\; r_{\text{amp}}=-\frac{M_{21}}{M_{22}}\tag{4.3.8}$$

Substituting into the first row gives the transmission amplitude $t=A_N=\det M/M_{22}$. When the two outer regions share the same potential, one can verify block by block that $\det P_j=1$ and $\det S=r$; in the product all the step ratios cancel telescopically, $\det M=1$, and so

$$T=\frac{1}{|M_{22}|^2},\qquad R=\frac{|M_{21}|^2}{|M_{22}|^2},\qquad T+R=1\tag{4.3.9}$$

The last step uses a universal structural fact about $M$ (time-reversal symmetry gives $|M_{22}|^2-|M_{21}|^2=\det M=1$). Current conservation, once again, comes free.

The first trophy: double-barrier resonance

Let’s practise on a real device — the core structure of a resonant-tunneling diode (RTD): two AlGaAs barriers sandwiched in GaAs. Typical parameters: barrier height $V_0=0.3$ eV, each barrier 2 nm wide, a 6 nm well in between, and electron effective mass $m^*=0.067\,m_e$ (that is $m^*c^2\approx34.2$ keV — electrons in the crystal come out “lighter”, a gift of the band structure).

Five slabs, four interfaces, $M$ a product of 8 matrices. Scanning energy by energy for $T(E)$:

$E$ (eV)	0.050	0.068	0.100	0.200	0.275
$T$	0.035	1.000	0.071	0.167	1.000

At $E=0.068$ eV the transmission reaches exactly 1 — even though a single 2 nm barrier at this energy manages only $T_1\approx0.20$, and naive intuition says two barriers “in series” should give $T_1^2\approx0.04$. The resonance amplifies that by a factor of 24, all the way to complete transparency. Move slightly off resonance and $T$ collapses back to a few percent.

Incidentally, the table above can be reproduced in about twenty lines of code (the full numerical toolkit unfolds over the next two sections):

import numpy as np

hbarc, mstar = 197.3, 0.067 * 511000.0     # eV·nm; GaAs effective mass

def transmission(E, widths, heights):
    """widths/heights: widths (nm) and potentials (eV) of the inner slabs; the two ends are V=0 electrodes."""
    Vs = [0.0] + list(heights) + [0.0]
    ks = [np.sqrt(2 * mstar * (E - V + 0j)) / hbarc for V in Vs]
    ds = [0.0] + list(widths)               # slab 0 needs no propagation
    M = np.eye(2, dtype=complex)
    for j in range(len(ds)):
        phi = ks[j] * ds[j]
        P = np.diag([np.exp(1j * phi), np.exp(-1j * phi)])
        r = ks[j] / ks[j + 1]
        S = 0.5 * np.array([[1 + r, 1 - r], [1 - r, 1 + r]])
        M = S @ P @ M                       # first traversed, first multiplied (on the right)
    t = (M[0, 0] * M[1, 1] - M[0, 1] * M[1, 0]) / M[1, 1]
    return abs(t) ** 2

# Double barrier: 2 nm barrier / 6 nm well / 2 nm barrier
print(transmission(0.0682, [2, 6, 2], [0.3, 0.0, 0.3]))   # → 1.000

The complex square root hands the $E<V$ slabs an imaginary wavenumber automatically, so barriers and wells need no separate treatment — the manual “rewrite everything with $\kappa$” chore of section 2.11 is swallowed by one 0j.

From two layers to infinitely many: the birth of energy bands

The transfer matrix’s most beautiful application is the periodic structure. Take “barrier + well” as one period with matrix $M_1$; a superlattice repeating it $N$ times has total matrix $M_1^N$ — no equations to re-match.

Whether a wave can still get through as $N\to\infty$ comes down to whether $M_1^N$ stays bounded. Linear algebra gives a crisp criterion: with $\det M_1=1$, the two eigenvalues of $M_1$ are reciprocals $\lambda,1/\lambda$, and $\lambda+1/\lambda=\operatorname{Tr}M_1$. Hence —

• $|\operatorname{Tr}M_1|\le2$: the eigenvalues are unit-modulus phase factors $\ee^{\pm\ii qL}$, and the wave propagates without decay — these energies form the allowed bands;
• $|\operatorname{Tr}M_1|>2$: one eigenvalue exceeds 1 and the other falls below it, and the wave decays exponentially — the forbidden gaps.

The trace of a $2\times2$ matrix undulates with energy, sweeping inside and outside the range between $\pm2$, and the energy axis gets sliced into alternating allowed bands and gaps — this is the Kronig–Penney model, the simplest edition of solid-state band theory. The discrete levels of an isolated well broaden into bands under multilayer coupling; the difference between semiconductors, insulators, and conductors traces back to whether the Fermi energy lands in an allowed band or a gap. The RTD’s single resonance peak can be viewed as “an allowed band with only one period” — the two perspectives merge here.

∎Key formulas

• Propagation matrix

  $$P_j=\begin{pmatrix}\ee^{\ii k_jd_j}&0\\0&\ee^{-\ii k_jd_j}\end{pmatrix}$$

  Diagonal: propagation never mixes left- and right-movers; for E<V it turns into growing/decaying exponentials automatically

• Interface matrix

  $$S=\frac12\begin{pmatrix}1+r&1-r\\1-r&1+r\end{pmatrix},\quad r=\frac{k_j}{k_{j+1}}$$

  Non-zero off-diagonal entries mean reflection; equal k collapses it to the identity

• Scattering quantities

  $$M=S\cdots PS,\quad T=\frac{1}{|M_{22}|^2},\quad R=\frac{|M_{21}|^2}{|M_{22}|^2}$$

  With equal potentials at both ends, det M = 1 and T+R=1 holds automatically

• Band criterion

  $$|\operatorname{Tr}M_1|\le2\ \Rightarrow\ \text{allowed band},\qquad |\operatorname{Tr}M_1|>2\ \Rightarrow\ \text{gap}$$

  M₁ is the matrix of a single period; the Kronig–Penney model

?Self-check4 questions

1. 1.

   The transfer matrix method turns a multilayer potential problem into a product of matrices. The fundamental reason it can do so is:

   • Each slab's solution is fully determined by two coefficients, and both propagation and interface matching are linear transformations of that pair
   • The Schrödinger equation is a second-order differential equation
   • Every potential can be approximated as piecewise constant
   • Matrix multiplication is faster than solving simultaneous equations

   Submit
2. 2.

   A double barrier has T = 1 at resonance, while a single one of its barriers has T₁ ≈ 0.2 at that energy. The correct explanation of this "1 exceeds 0.2²" is:

   • Waves making multiple round trips in the well interfere constructively in transmission and destructively in reflection, just as in a Fabry–Pérot cavity
   • One of the two barriers has stopped working
   • The particle gained extra energy from the well
   • The T₁² calculation forgot a factor of 2

   Submit
3. 3.

   In a periodic potential the single-period matrix M₁ satisfies det M₁ = 1. An energy in a forbidden gap means: (select all that apply)

   Select all that apply

   • |Tr M₁| exceeds 2
   • One eigenvalue of M₁ has modulus above 1 and the other below 1
   • The wave decays exponentially through the structure and cannot propagate in the infinite-period limit
   • The Schrödinger equation has no solution at that energy

   Submit
4. 4.

   Treating the RTD's 6 nm well crudely as an infinite well, estimate the ground-state level (the zeroth-order estimate of the resonance energy), in eV. Electron effective mass m*c² = 0.067×511000 ≈ 34240 eV, ħc = 197.3 eV·nm.

   eV1% relative tolerance

   Submit

What comes next

The transfer matrix sweeps up the whole class of piecewise-constant potentials. But its signature skill is exactly what exposes its limit: real potentials are mostly smooth — interaction potentials in molecules, confinement potentials in quantum dots, potentials tilted by an applied field — none of them looks like a staircase. Force a staircase approximation and, as the slab count climbs, even assembling matrices by hand loses its elegance.

Time to hand the whole thing to the computer. The next section chops the $x$ axis itself into a dense grid of points, and the differential equation transforms into a tridiagonal matrix eigenvalue problem — chapter 3’s slogan “operators are matrices” becomes completely literal for the first time: the Hamiltonian is a matrix you can print out and look at, the energy levels are its eigenvalues, and it all fits in ten lines of code.