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TDSE Solution: Implicit Methods
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Program 1

Implicit Methods

Output 1

Implicit Method for TDSE

The implicit method uses backward differences for time derivatives, expressed as:

$$\frac{\psi^{n+1}_k - \psi^n_k}{\Delta t} = \frac{\partial \psi}{\partial t}\bigg|_{t=t_{n+1}}$$

Equation Formulation

Substituting into the TDSE gives:
$$\small i \frac{\psi^{n+1}_k - \psi^n_k}{\Delta t} = -\frac{1}{2} \frac{\psi^{n+1}_{k+1} - 2\psi^{n+1}_k + \psi^{n+1}_{k-1}}{(\Delta x)^2} + V \psi^{n+1}_k$$ we take $$m = \hbar = 1$$ and let $$\mu = \frac {i \Delta t}{2 \Delta x^2}$$;
Rearranging leads to:
$$\small \psi^n_k = \left({1 + 2\mu + i\Delta t V_k^{n+1}} \right) \psi^{n+1}_{k} - \mu \psi^{n+1}_{k+1} + - \mu \psi^{n+1}_{k-1} $$

Matrix Representation

This can be expressed in matrix form as:
$$\mathbf{\psi}^n = B \cdot \mathbf{\psi}^{n+1}$$
where:
$$\tiny B = \begin{bmatrix} 1 + 2\mu + i\Delta t V^{n+1}_1 & -\mu & 0 & \ldots & 0 \\ -\mu & 1 + 2\mu + i\Delta t V^{n+1}_2 & -\mu & \ldots & 0 \\ 0 & -\mu & 1 + 2\mu + i\Delta t V^{n+1}_3 & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & -\mu \\ 0 & 0 & 0 & -\mu & 1 + 2\mu + i\Delta t V^{n+1}_{N-2} \end{bmatrix} $$
Inverting gives:
$$\mathbf{\psi}^{n+1} = B^{-1} \mathbf{\psi}^n$$

Pros and Cons of Implicit Method

  • Stability: More stable for small time steps; eigenvalues of \(M\) approach 1.
  • Norm Conservation: Ensures stability in the wavefunction norm across steps.
  • Efficiency: More accurate for larger time steps, but requires solving a system of equations, making it computationally intensive.