Split Operator

The idea of the split-split operator method is to split the exact propagator into a part that is dependent only on the kinetic energy operator and a part that is dependent only on the potential energy operator:

$$ \exp\left( -\frac{\mathrm{i}}{\hbar}\left( \hat{T}+\hat{V} \right)\Delta t \right) \approx \exp\left( -\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t \right) \exp\left( -\frac{\mathrm{i}}{\hbar}\hat{V}\Delta t \right)$$

Unfortunately, such splitting does not really work for operators since they in general do not commute. It can be shown that such splitting in the quantum propagation would be equivalent to a propagation of a system with a different effective Hamiltonian! However, if the difference between the effective Hamiltonian and the system Hamiltonian are in some sense small, this still may lead to an accurate propagation scheme. In the following we will seen how the simple split operator idea may be improved.

In order to analyze the error that is introduced by a simple operator splitting scheme, we utilize the Baker-Campbell-Hausdorff (BCH) formula, which you have derived in a lectures on the theory of quantum dynamics.

We know that for operators in general the following relation:

$$\exp(\hat{A})\exp(\hat{B}) = \exp\left(\hat{A}+\hat{B}\right)$$

does not hold.

Instead, the BCH-formula can be proven to hold in the following form:

$$ \exp\left( \hat{A} \right) \exp\left( \hat{B} \right) = \exp\left( \hat{A} + \hat{B} + \frac{1}{2} \left[ \hat{A}, \hat{B} \right] + \frac{1}{12} \left( \left[ \hat{A}, \left[ \hat{A}, \hat{B} \right] \right] + \left[ \hat{B}, \left[\hat{B}, \hat{A} \right] \right] \right) + \cdots \right) $$

If we now insert:

$$\hat{A} = -\frac{\mathrm{i}\Delta t}{\hbar}\hat{T}$$ and $$\hat{B} = -\frac{\mathrm{i}\Delta t}{\hbar}\hat{V}$$

we can show that the product of the exponential operators for the kinetic and potential part of the Hamiltonian can be transformed into:

$$ \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{V}\Delta t\right) = \exp\left( -\frac{\mathrm{i}}{\hbar}\left(\hat{T}+\hat{V}\right)\Delta t -\frac{\Delta t^{2}}{2\hbar^{2}}[\hat{T},\hat{V}]+\cdots \right) $$

This expression can be now rewritten in the following form:

$$ \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{V}\Delta t\right) = \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{H}_ {eff}\Delta t\right)$$

where we have defined an effective Hamiltonian that is dependent on the time-step:

$$\hat{H}_ {eff}=\hat{H}+\frac{\Delta t}{2\hbar}\left[\hat{T},\hat{V}\right]$$

We see that the propagation scheme based on the operator splitting corresponds to an exact propagation with an effective Hamiltonian that differs from the original system Hamiltonian by a leading term proportional to the time-step \(\Delta t\) and the commutator between the kinetic energy and the potential energy. Since the kinetic and the potential operators in general do not commute, we can expect that the propagation using the split-operator scheme will differ from the exact propagation.

It turns out that the propagation error can be significantly reduced by using a symmetric operator splitting defined by:

$$\hat{H}=\frac{\hat{V}}{2}+\hat{T}+\frac{\hat{V}}{2}$$

We can now apply the generalization of the BCH formula that can be straightforwardly derived from the original BCH relation:

$$ \exp\left(\hat{A}\right)\exp\left(\hat{B}\right)\exp\left(\hat{C}\right) = \exp\left(\hat{A}+\hat{B}+\hat{C} +\frac{1}{2} \left(\left[\hat{A},\hat{B}\right]+\left[\hat{A},\hat{C}\right] +\left[\hat{B},\hat{C}\right]\right) +\frac{1}{12} \left(\left[\left[\hat{A},\hat{B}\right],\hat{A}+\hat{B}+\hat{C}\right]\right)+\cdots\right) $$

Again, we define operators as:

$$ \begin{align} \hat{A} &= -\frac{\mathrm{i}\Delta t}{2\hbar}\hat{V} \\ \hat{B} &= -\frac{\mathrm{i}\Delta t}{\hbar}\hat{T} \\ \hat{C} &= -\frac{\mathrm{i}\Delta t}{2\hbar}\hat{V} \end{align} $$

and insert into the generalized BCH relation. The term containing the first order operators now vanishes since:

$$ \left[\hat{A},\hat{B}\right]+\left[\hat{A},\hat{C}\right]+\left[\hat{B},\hat{C}\right] \rightarrow -\left(\frac{\Delta t}{\hbar}\right)^{2} \cdot \left(\frac{1}{2}\left[\hat{V},\hat{T}\right] +\frac{1}{2}\left[\hat{V},\hat{V}\right] +\frac{1}{2}\left[\hat{T},\hat{V}\right]\right) = 0 $$

Thus, the symmetric splitting reduces the propagation error since now the error in the effective Hamiltonian will be proportional to \(\Delta t^2\):

$$ \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V}}{2}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V}}{2}\Delta t\right) = \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{H}\Delta t -\frac{1}{2}\Delta t^{3}[\hat{H},[\hat{T},\hat{V}]]+\cdots\right) $$

In the following we implement the split-operator method using such symmetric splitting:

$$ |\psi(t+\Delta t)\rangle = \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V}}{2}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V}}{2}\Delta t\right) |\psi(t)\rangle $$

The numerical implementation of the split operator propagation scheme will again rely on the application of the FFT algorithm. The action of the exponential operators will be calculated either by direct multiplication in the coordinate space (potential part) or by performing the FFT and multiplication in the momentum space followed by a inverse FFT (kinetic part).