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.

Helpful background. Potential Step to match open and evanescent scattering channels.

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. For matrix composition, measure position from a local reference point $x_*$ in each region: $\psi(x)=A(x_*)e^{ik_j(x-x_*)}+B(x_*)e^{-ik_j(x-x_*)}$. Global coefficients are constant within a region; the local coefficients change when the reference point moves.

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

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

The mass is constant throughout, and the interface has no delta term. At an exact threshold $k_R=0$, the traveling-wave basis becomes singular. Use the regular endpoint data $(\psi,\psi^\prime)$ and the zero-wavenumber solutions $1,x$, or take a controlled limit.

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:

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

Here $k_L,k_R$ must be positive real open-channel wavenumbers. The transmitted coefficient is referenced to the right endpoint: if that point is at $x=d$ and the left reference is zero, the global convention $\psi=t_{\mathrm{global}}e^{ik_Rx}$ has $t_{\mathrm{global}}=t e^{-ik_Rd}$. Keep this phase when comparing different derivations or packet delays.

Since $\det I_{R\leftarrow L}=k_L/k_R$ and $\det P=1$, a full stack has $\det M=k_L/k_R$; equal exterior wavenumbers give determinant one.

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. Passing it does not certify the relative accuracy of a tiny $T$: replacing $T=10^{-17}$ by zero is invisible to a conservation test with an absolute tolerance of $10^{-12}$. Compare $T$, or $\log T$ for very opaque barriers, with a stable formulation and with increased precision.

Use the known analytic determinant $\det M=k_L/k_R$ when extracting $t$; computing it by subtracting products of large entries introduces avoidable cancellation. This helps for a single barrier but does not make arbitrary long transfer-matrix products stable. Scattering-matrix composition instead joins the incoming and outgoing amplitudes of smaller pieces. Yin and Tao (2008) discuss the growing evanescent factors and stabilized alternatives. Exercise 3 makes the loss of transmission explicit; Conditioning and Stability explains why a small residual alone is insufficient.

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. Consider one rectangular barrier with equal exterior potentials, $E=V_0/2$, and $s=\kappa a>0$. Then the exterior wavenumber is $k=\kappa$. Derive its transfer matrix in the local endpoint convention, find $R$ and $T$, and explain how an opaque barrier can pass $R+T=1$ numerically while its computed transmission is completely wrong. Give a formula for $\log T$ that avoids growing exponentials.

Solution

Multiplying the two interface matrices and the intervening propagation matrix gives

$$M= \begin{pmatrix} \cosh s&-i\sinh s\\ i\sinh s&\cosh s \end{pmatrix}, \qquad \det M=1.$$

Thus

$$r=-i\tanh s, \qquad t=\operatorname{sech}s, \qquad R=\tanh^2s, \qquad T=\operatorname{sech}^2s.$$

The global transmitted coefficient is $t_{\mathrm{global}}=e^{-ika}\operatorname{sech}s$; the endpoint phase does not affect $T$. Both evanescent components remain necessary inside the finite barrier.

The matrix is Hermitian and positive, with eigenvalues $e^s,e^{-s}$. Its Euclidean condition number is therefore $e^{2s}$. At $s=20$, it is about $2.35\times10^{17}$, while the exact transmission is

$$T=\operatorname{sech}^2 20\approx1.69934\times10^{-17}.$$

In double precision, $\cosh s$ and $\sinh s$ can round to the same value. Evaluating $\det M=\cosh^2s-\sinh^2s$ then gives zero, so the determinant formula returns $t=0$. Yet $R$ rounds to one, and the computed $R+T$ is also one. The transmission has a $100\%$ relative error despite this perfect conservation residual.

For this barrier, using the exact determinant gives $t=1/M_{22}$. An expression that also avoids overflow in $\cosh s$ is

$$\log T=2\log2-2s-2\log(1+e^{-2s}).$$

Evaluate the last term with a numerical $\operatorname{log1p}$ function when available. The logarithm remains useful even when $T$ itself is too small to represent. The exponential ill conditioning of the transfer representation does not prevent this directly evaluated transmission from being accurate.

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.
• H. Yin and R. Tao, “Improved transfer matrix method without numerical instability,” EPL 84, 57006 (2008), doi:10.1209/0295-5075/84/57006, arXiv:0709.3681.