Transfer Matrix Method

The transfer matrix method is a compact way to solve one-dimensional scattering problems made from constant-potential regions. Instead of writing a large system of matching equations from scratch, one stores right-moving and left-moving amplitudes in a vector and multiplies matrices for interfaces and propagation through layers.

This page gives a practical wave-mechanics convention. Full scattering theory introduces related $S$-matrix and Green-function methods, but the transfer matrix is often the quickest way to handle finite stacks of steps, wells, and barriers.

Region Amplitudes

In a region where the potential is constant,

$$V(x)=V_j,$$

the stationary Schrödinger equation has plane-wave solutions when $E\gt V_j$:

$$\psi_j(x) = A_je^{ik_jx} + B_je^{-ik_jx}, \qquad k_j=\frac{\sqrt{2m(E-V_j)}}{\hbar}.$$

The amplitude vector is

$$\mathbf v_j = \begin{pmatrix} A_j\\ B_j \end{pmatrix}.$$

Here $A_j$ multiplies the right-moving component and $B_j$ multiplies the left-moving component.

If $E\lt V_j$, then $k_j=i\kappa_j$ with

$$\kappa_j=\frac{\sqrt{2m(V_j-E)}}{\hbar}.$$

The same algebra works, but the two exponentials become growing and decaying evanescent terms. This is useful analytically and potentially dangerous numerically.

Propagation Through a Constant Region

Across a constant-potential segment of length $a$, the amplitudes acquire phases:

$$\mathbf v_j(x+a) = P_j(a)\mathbf v_j(x),$$

where

$$P_j(a) = \begin{pmatrix} e^{ik_ja} & 0\\ 0 & e^{-ik_ja} \end{pmatrix}.$$

For an evanescent region with $k_j=i\kappa_j$,

$$P_j(a) = \begin{pmatrix} e^{-\kappa_ja} & 0\\ 0 & e^{\kappa_ja} \end{pmatrix}.$$

The second entry grows with $a$. That growth does not mean the physical wavefunction diverges; it means this amplitude basis is tracking a component that must cancel appropriately when matched to neighboring regions.

Interface Matrix

At a finite potential jump, both $\psi$ and $d\psi/dx$ are continuous. Put the interface at $x=0$, with wavenumber $k_L$ on the left and $k_R$ on the right. Then

$$\psi_L=A_L+B_L, \qquad \psi_R=A_R+B_R,$$

and

$$k_L(A_L-B_L)=k_R(A_R-B_R).$$

Solving for the right-side amplitudes in terms of the left-side amplitudes gives

$$\mathbf v_R = I_{R\leftarrow L}\mathbf v_L,$$

with

$$I_{R\leftarrow L} = \frac12 \begin{pmatrix} 1+\frac{k_L}{k_R} & 1-\frac{k_L}{k_R}\\ 1-\frac{k_L}{k_R} & 1+\frac{k_L}{k_R} \end{pmatrix}.$$

This matrix is just the potential-step matching calculation written once and reused.

Composition

For a stack of layers, multiply the matrices in the order encountered from left to right. For example, suppose the wave crosses an interface from region $0$ to region $1$, propagates across region $1$ for length $a$, and then crosses into region $2$. With the convention above,

$$\mathbf v_2 = I_{2\leftarrow1}P_1(a)I_{1\leftarrow0}\mathbf v_0.$$

The total transfer matrix is

$$M=I_{2\leftarrow1}P_1(a)I_{1\leftarrow0}, \qquad \mathbf v_2=M\mathbf v_0.$$

Longer stacks are handled by inserting more propagation and interface matrices.

Extracting r and t

For left incidence, set the incoming amplitude on the far left to $1$:

$$\mathbf v_L = \begin{pmatrix} 1\\ r \end{pmatrix}.$$

If there is no incoming wave from the far right, then on the far right

$$\mathbf v_R = \begin{pmatrix} t\\ 0 \end{pmatrix}.$$

Let

$$\mathbf v_R = M\mathbf v_L, \qquad M= \begin{pmatrix} M_{11} & M_{12}\\ M_{21} & M_{22} \end{pmatrix}.$$

Then

$$0=M_{21}+M_{22}r,$$

so

$$r=-\frac{M_{21}}{M_{22}}.$$

The transmitted amplitude is

$$t=M_{11}+M_{12}r.$$

Equivalently,

$$t=\frac{\det M}{M_{22}}.$$

If the left and right asymptotic wavenumbers differ, the transmission probability includes the velocity factor:

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

The reflection probability is

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

for the standard one-channel left-incident setup.

Single Step Check

For one interface, $M=I_{R\leftarrow L}$. The no-incoming-from-right condition gives

$$r=\frac{k_L-k_R}{k_L+k_R},$$

and

$$t=\frac{2k_L}{k_L+k_R}.$$

These are exactly the amplitudes found in the Potential Step calculation. The transfer matrix therefore reproduces the elementary result before being trusted on more complicated stacks.

Multiple Barriers

For a double barrier, the total matrix has the schematic form

$$M= I_{\mathrm{out}\leftarrow b} P_b(a) I_{b\leftarrow w} P_w(L) I_{w\leftarrow b} P_b(a) I_{b\leftarrow \mathrm{in}},$$

where $b$ denotes barrier regions and $w$ denotes the intermediate well region. The transmission probability can show sharp peaks when the phase accumulated in the well makes multiple reflected amplitudes interfere constructively. This is the transfer-matrix route to Resonant Transmission.

Numerical Stability

Transfer matrices can be poorly conditioned for thick barriers or long stacks. Evanescent propagation contains factors such as

$$e^{\kappa a},$$

which may become enormous even when the physical transmission is tiny. Multiplying many such matrices can produce overflow, underflow, or severe cancellation.

For robust numerical work, one often uses:

• scattering matrices instead of transfer matrices;
• logarithmic derivatives;
• stable recursive algorithms;
• rescaling after each layer;
• arbitrary-precision arithmetic only after the formulation is checked.

A stable result should be checked against current conservation when the potential is real and the scattering is one-channel:

$$R+T=1.$$

Failure of this check can signal physical absorption, but in a nominally conservative calculation it usually signals a numerical or convention error.

Common Mistakes

• Multiplying matrices in the wrong order.
• Mixing amplitude conventions from different texts.
• Forgetting the velocity factor in $T$ when asymptotic wavenumbers differ.
• Treating evanescent growth in the transfer basis as a physical divergence.
• Trusting raw transfer-matrix multiplication through very opaque barriers.
• Comparing $R+T$ before checking whether the model includes absorption or extra channels.

Exercises

1. Derive the interface matrix $I_{R\leftarrow L}$ from continuity of $\psi$ and $d\psi/dx$ at $x=0$.

Solution

Continuity gives

$$A_R+B_R=A_L+B_L,$$

and derivative continuity gives

$$k_R(A_R-B_R)=k_L(A_L-B_L).$$

Thus

$$A_R+B_R=A_L+B_L, \qquad A_R-B_R=\frac{k_L}{k_R}(A_L-B_L).$$

Adding and subtracting these equations yields

$$A_R = \frac12 \left[ \left(1+\frac{k_L}{k_R}\right)A_L + \left(1-\frac{k_L}{k_R}\right)B_L \right],$$

and

$$B_R = \frac12 \left[ \left(1-\frac{k_L}{k_R}\right)A_L + \left(1+\frac{k_L}{k_R}\right)B_L \right].$$

This is exactly the stated matrix.

2. Use the single-step transfer matrix to recover $r=(k_L-k_R)/(k_L+k_R)$.

Solution

For left incidence,

$$\mathbf v_L= \begin{pmatrix} 1\\ r \end{pmatrix}, \qquad \mathbf v_R= \begin{pmatrix} t\\ 0 \end{pmatrix}.$$

The lower component of $\mathbf v_R=I_{R\leftarrow L}\mathbf v_L$ gives

$$0= \frac12 \left[ \left(1-\frac{k_L}{k_R}\right) + \left(1+\frac{k_L}{k_R}\right)r \right].$$

Solving,

$$r = \frac{k_L-k_R}{k_L+k_R}.$$

3. In a barrier region with $E\lt V_0$, why can a transfer matrix contain $e^{+\kappa a}$ even though the physical wavefunction should not blow up in the final answer?

Solution

Inside a finite barrier, the general solution contains both $e^{-\kappa x}$ and $e^{+\kappa x}$ terms. The growing term is allowed inside a finite interval because the interval has finite length and because matching at both interfaces determines coefficients that combine to give the physical solution.

Numerically, however, $e^{+\kappa a}$ can become very large. The final physical amplitudes may require cancellation between large intermediate numbers, which is unstable in floating-point arithmetic. This is why transfer matrices should be used carefully for opaque barriers.

Where This Is Used

• Potential Step is the single-interface check.
• Finite Potential Barrier is the simplest two-interface barrier calculation.
• Rectangular Barrier Tunneling is the simplest two-interface barrier problem.
• Reflection and Transmission Coefficients gives the current-based interpretation of $R$ and $T$.
• Resonant Transmission applies matrix composition to double barriers and quasi-bound states.
• Anderson Localization turns random transfer-matrix products into Lyapunov exponents and one-dimensional localization lengths.
• Conditioning and Stability explains why raw matrix products can become unreliable.
• One-Dimensional Scattering Revisited derives the associated $S$-matrix and identifies bound and resonance poles through the transfer denominator.

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.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• P. Yeh, Optical Waves in Layered Media, Wiley, 1988.