V(\hat{X}). This method is called Split-Operator Method (SOM, from now on). Later we'll see what split-operator does mean. For simplicity, our discussion will be regarding the one-dimensional problem.
Given the Schrödinger's equation
i\hbar\frac{\mbox{d}}{\mbox{d}t} |\psi(t)\rangle = \hat{H} |\psi(t)\rangle
It can be formally solved as follows:
|\psi(t)\rangle = e^{-i\frac{\hat{H}}{\hbar}t} |\psi(0)\rangle
Let's consider the following hamiltonian function
\hat{H} = \dfrac{\hat{P}^2}{2m} + V(\hat{X})
Since
[\hat{X},\hat{P}]\neq0, there's 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 them 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 gonna switch to momentum representation to see how this works for the wave packet. We don't 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 can't solve Schrödinger's equation for an arbitrary value of t. To solve this problem we are gonna 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
\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}
Let's remember that, since
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} it's 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 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 potential might be 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:
You are free to distribute and/or modify this work.
