Multistate Quantum Dynamics

After studying quantum dynamics in one state, we now enter the realm of multistate dynamics. This time, a wave function is represented by a vector $$ \vec{\psi}(x, t) = \begin{pmatrix} \psi_{0} \\ \psi_{1} \\ \vdots \\ \psi_{n-1} \end{pmatrix} $$ for a total of \(n\) states. The operators for kinetic and potential energy must be adjusted, too. Since the kinetic energy operator for a single state, which we denote as\(\hat{t}\) to avoid confusion, does not vary with different electronic states, the total kinetic energy operator \(\hat{T}\) can be obtained by simply forming a direct product between \(\hat{t}\) and an unit operator: $$\hat{T} = \hat{t} \otimes \mathbb{1}_ {n}$$

The potential energy operator does depent on the electronic state. We shall therefore write it as $$\hat{V} = \begin{pmatrix} \hat{v}_ {1, 1} & \hat{v}_ {1, 2} & \cdots & \hat{v}_ {1, n-1} \\ \hat{v}_ {2, 1} & \hat{v}_ {2, 2} & \cdots & \hat{v}_ {2, n-1} \\ \vdots & \vdots & \ddots & \vdots \\ \hat{v}_ {n-1, 1} & \hat{v}_ {n-1, 2} & \cdots & \hat{v}_ {n-1, n-1} \end{pmatrix} $$ where \(\hat{v}_ {s, s}\) is the potential energy operator for state \(s\) and \(\hat{v}_ {s, t}\) the doupling between states \(s\) and \(t\).

On a discretized grid, the wave function is represented by a 2-dimensional array \(\psi{i, s}\), where index \(i\) denotes space and index \(s\) denotes state. Since the kinetic energy operator is the same for all electronic states, we can just use the old representation and apply it for all states separately. The potential energy operator can be represented as a 3-dimensional array \(V_{i, s, t}\), where the potential energy or couling \(v_{s, t}(x)\) for all states and state-paris are discretize on the space grid.

While the computation of the kinetic propagator $$\exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right)$$ stays roughly the same, the potential energy propagator makes some problems.

For a single state, we calculated \(\exp\left(-\frac{\mathrm{i}}{\hbar}\frac{V_{i}}{2}\Delta t\right)\) for every point in the space grid. Now, we have \(\exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\mathbf{V_{i}}}{2}\Delta t\right)\) with the matrix \(\mathbf{V_{i}}\) containing potential energies and couplings at grid point \(i\). We have already seen that a matrix exponential is defined by the Taylor expansion in chapter 7, but a direct computation with this would be very time consuming, since higher matrix powers must be calculated. Assuming matrix \(\mathbf{A}\) is diagonalizable, i.e. \(\mathbf{A} = \mathbf{S} \mathbf{D} \mathbf{S}^{-1}\), the expression \(\exp(\mathbf{A})\) can be expanded as $$ \begin{align} \exp(\mathbf{A}) &= \sum_{n=0}^{\infty} \frac{1}{n!} \mathbf{A}^n \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} (\mathbf{S} \mathbf{D} \mathbf{S}^{-1})^n \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \underbrace{ (\mathbf{S} \mathbf{D} \mathbf{S}^{-1})\ (\mathbf{S} \mathbf{D} \mathbf{S}^{-1}) \ \cdots \ (\mathbf{S} \mathbf{D} \mathbf{S}^{-1}) }_ {n \text{times}} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \mathbf{S} \mathbf{D} \underbrace{\mathbf{S}^{-1}\ \mathbf{S}}_ {\mathbb{1}} \mathbf{D} \underbrace{\mathbf{S}^{-1}\ \cdot}_ {\mathbb{1}} \cdot \underbrace{\cdot \ \mathbf{S}}_ {\mathbb{1}} \mathbf{D} \mathbf{S}^{-1} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \mathbf{S} \mathbf{D}^n \mathbf{S}^{-1} \\ &= \mathbf{S} \left( \sum_{n=0}^{\infty} \frac{1}{n!} \mathbf{D}^n \right) \mathbf{S}^{-1} \\ &= \mathbf{S}\ \exp(\mathbf{D})\ \mathbf{S}^{-1} \end{align} $$

Since the matrix exponential of a diagonal matrix is just the exponential of its diagonal elements, it is very easy to compute. Therefore, for diagonalizable matrices, we can at first diagonlize it to obtain the diagonal matrix \(\mathbf{D}\) and the diagonalizing matrix \(\mathbf{S}\), from which the matrix exponential can be computed easily. Since in our case, the potential energy matrices \(\mathbf{V_{i}}\) are all Hermitian, they are diagonalizable with unitary diagonalizing matrices \(\mathbf{U_{i}}\), whose inverse are easily computed by taking its adjoint. Hence, we can use $$ \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\mathbf{V_{i}}}{2}\Delta t\right) = \mathbf{U}\ \exp\left( -\frac{\mathrm{i} \Delta t}{2 \hbar} \mathbf{D} \right) \mathbf{U}^{\dagger} $$ to calculate the potential energy propagator.