# Split operator method

The split operator method (SOM) is a method to avoid manually solving Schrödinger's equation and obtaining an approximate solution for an arbitrary potential of the form $V(\hat{X})$. Later on we shall see what split-operator does mean. For simplicity, our discussion will be regarding the one-dimensional problem. Given the standard form of the time-dependent Schrödinger's equation

$i\hbar\frac{\mbox{d}}{\mbox{d}t} |\psi(t)\rangle = \hat{H} |\psi(t)\rangle,$

its formal solution is as follows:

$|\psi(t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi(0)\rangle.$

Let us consider the following hamiltonian function

$\hat{H} = \dfrac{\hat{P}^2}{2m} + V(\hat{X})$

Since $[\hat{X},\hat{P}]\neq0$, there is no reason to prefer one of the two following options to set the order in which the hamiltonian momentum-dependent and position-dependent part act over the system ket:

1. $e^{f_1(\hat{P}) + f_2(\hat{X})} |\psi(t)\rangle = e^{f_1(\hat{P})} \cdot e^{f_2(\hat{X})} |\psi(t)\rangle,$

2. $e^{f_1(\hat{P}) + f_2(\hat{X})} |\psi(t)\rangle = e^{f_2(\hat{X})} \cdot e^{f_1(\hat{P})} |\psi(t)\rangle.$

In fact, none of the two would be correct.

The SOM consists on somehow setting up an approximation to the exponential. This approximation is

$e^{-\frac{it}{\hbar}\left(\frac{\hat{P}^2}{2m}+V(\hat{X})\right)} \approx e^{-\frac{it}{\hbar}\frac{\hat{P}^2}{4m}} \cdot e^{-\frac{it}{\hbar}V(\hat{X})} \cdot e^{-\frac{it}{\hbar}\frac{\hat{P}^2}{4m}}.$

Thus, we have splitted the hamiltonian operator in a certain way to symmetrize the exponentials. This approximation is valid for a sufficiently small time step.

Now we are going to switch to momentum representation to see how this works for the wave packet. We do not choose position representation because that would require an additional step:

$\bar{\psi}(p,t) = \langle p | \psi(t)\rangle = \langle p | e^{-\frac{it}{\hbar}\frac{\hat{P}^2}{4m}} \cdot e^{-\frac{it}{\hbar}V(\hat{X})} \cdot e^{-\frac{it}{\hbar}\frac{\hat{P}^2}{4m}} | \psi(0)\rangle.$

As we stated before, the time step must be sufficiently small. Apparently, we cannot solve Schrödinger's equation for an arbitrary value of $t$. To solve this problem we are going to make the following substitution:

$\bar{\psi}(p,t) = \bar{\psi}(p,t_0 + \Delta t) = \langle p | \psi(t_0 + \Delta t)\rangle = \langle p | e^{-\frac{i\Delta t}{\hbar}\frac{\hat{P}^2}{4m}} \cdot e^{-\frac{i\Delta t}{\hbar}V(\hat{X})} \cdot e^{-\frac{i\Delta t}{\hbar}\frac{\hat{P}^2}{4m}} | \psi(t_0)\rangle.$

Now we can make $\Delta t$ as small as we need to match convergence requirements.

Given the closure, eigenvalues and dot product relations for the position and momentum basis sets

$\int \mbox{d}x | x \rangle\langle x | = 1,$

$\int \mbox{d}p | p \rangle\langle p | = 1,$

$\hat{X} |x\rangle = x |x\rangle,$

$\hat{P} |p\rangle = p |p\rangle,$

$\langle p |x\rangle = \langle x |p\rangle^* = e^{\frac{-i}{\hbar}p\cdot x},$

and keeping in mind that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, it is true that

$e^{f(\hat{X})} |x\rangle = e^{f(x)} |x\rangle.$

Hence, we can, in a straightforward manner, apply the closure relation twice, once in position and once in momentum, to get to the following result:

$\bar{\psi}(p,t_0 + \Delta t) = \int\mbox{d}p' \int\mbox{d}x \ e^{\frac{i}{\hbar}p'x} \ e^{\frac{-i}{\hbar}px} \ e^{\frac{-i}{\hbar}\frac{p^2+{p'}^2}{4m}\Delta t} \ e^{\frac{-i}{\hbar}V(x)\Delta t} \ \bar{\psi}(p',t_0).$

which is a double Fourier transform, once forward and once backward. To obtain $\psi(x,t)$ we just need to transform again (the additional step I was referring to before). Thus, under the SOM approximation and given that we know the $\bar{\psi}(p,t_0)$ state of the system at a certain $t_0$ time, we can solve Schrödinger's equation by simply using Fourier transforms whatever the shape of the potential and obtain the wave function $\Delta t$ later.

I made the next video with data obtained from a SOM routine I programmed. Both the wave packet and potential barrier are gaussian. In the left side you can see position representation and in the right one you have momentum representation: