Von Neumann Stability Analysis

We know that our wavepacket can be represented by a (discrete) Fourier decomposition:

$$\psi(x_{i}, t) = \sum_{k} E_{k}(t) \exp\left( \frac{\mathrm{i}}{\hbar}p_{k}x_{i} \right)$$

The idea of the von Neumann's stability analysis is to check how an individual mode defined by some value of the momentum \(p_{k}\) propagates in time. For this purpose we inserte a single mode defined by

$$\psi_{i}^{n}=E_{k}^{n}\exp\left( \frac{\mathrm{i}}{\hbar}p_{k}x_{i} \right)$$

into the forward Euler propagation scheme.

In this way we obtain the relation between the single mode amplitudes at neighbouring time-steps:

$$E_{n+1} = \left(1 - \frac{\mathrm{i}}{\hbar}\Delta t \sum_{j}H_{ij} \exp\left( \frac{\mathrm{i}}{\hbar}p_{k}(x_{j}-x_{i}) \right) \right) E_{n}$$

The modulus of the amplitude can be represented by the relation:

$$|E_{n+1}| = g|E_{n}|$$

where

$$ g = |1 - \frac{\mathrm{i}}{\hbar}\Delta t \sum_{j}H_{ij}\exp(\mathrm{i}k(x_{j} - x_{i}))| = \sqrt{1 + \left(\frac{\Delta t}{\hbar}\right)^{2} \left| \sum_{j}H_{ij}\exp(\mathrm{i}k(x_{j}-x_{i})) \right|^{2}}$$

We see that \(g>1\) irrespective of the value of the time step and the momentum. This means that all momentum components will exponentially grow with time, making the forward Euler scheme unconditionally unstable!