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I am trying to learn about the split-step Fourier method for solving partial differential equations, in terms of the Schrodinger equation:

$$ \frac{\partial\Psi(x,t)}{\partial t} = -\frac{i}{\hbar}\hat{H}\;\Psi(x,t) \tag{1} $$ where $$ \hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x). \tag{2} $$

The solution of (1) for advancing the wavefunction by a time step $\Delta t$ should be given by $$ \begin{align} \Psi(x,t+\Delta t) &= \textrm{exp}\Big(-\frac{i}{\hbar}\hat{H}\;\Delta t\Big)\Psi(x,t)\\ &\approx \textrm{exp}\Big(-\frac{i}{2\hbar}\hat{V}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{\hbar}\hat{T}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{2\hbar}\hat{V}\;\Delta t\Big)\;\Psi(x,t)\\ &=\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\;\Delta t\Big)\;\textrm{exp}\Big(\frac{i\hbar}{2m}\frac{\partial^2}{\partial x^2}\;\Delta t\Big)\;\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\;\Delta t\Big)\;\Psi(x,t)\tag{3} \end{align} $$ where I have used the symmetrised split-step approximation.

Now from reading, it seems like Eq. (3) can be written in terms of Fourier transforms: $$ \Psi(x,t+\Delta t) = \textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\Delta t\Big)\;\mathcal{F}^{-1}\Bigg\{\textrm{exp}\Big(-\frac{i\hbar k^2}{2m}\Delta t\Big)\;\mathcal{F}\Big\{\textrm{exp}\Big(-\frac{i}{2\hbar}V(x)\Delta t\Big)\;\Psi(x,t)\Big\}\Bigg\} \tag{4} $$ but how can I show this explicitly?

Note, I am aware that a second derivative can be taken spectrally like so $$ \mathcal{F}\Bigg\{\frac{\textrm{d}^2y}{\textrm{d}x^2}\Bigg\} = -k^2 \mathcal{F}\big\{y\big\},\tag{5} $$ but I am not sure how to handle the second derivative inside the exponential in Eq. (3). It seems I will need to evaluate something of the form $$ \mathcal{F}^{-1}\Bigg\{ \mathcal{F}\Big\{ \textrm{exp}\bigg(\frac{\partial^2}{\partial x^2}\bigg) \cdot A \Big\} \Bigg\} = \mathcal{F}^{-1}\Bigg\{ \textrm{exp}\big(-k^2\big) \cdot\mathcal{F}\big\{ A \big\} \Bigg\} $$

Thank you.

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Because you want to use the Strang splitting method, you first need to solve the potential term for $\tau \in [t, t + \Delta t/2]$ with timestep $\Delta t/2$ and initial condition $\psi_{0}$

$$\dot{\psi} = -\frac{i}{{\hbar}} V \psi \implies \hat{\psi}^{*} = \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\} \tag{1}$$

where the hat denotes the Fourier transform. Note that at this iterate we could have left the solution as

$$\psi = \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}$$

However, because at the next iterate (solving the diffusion part) you want to solve the problem using the FT, and the solution to the potential part acts as an initial condition for this next iterate, the initial condition needs to be FTed as well.

You then solve the diffusion problem for $\tau \in [t, t + \Delta t]$ with initial condition $\hat{\psi}^{*}$ and timestep $\Delta t$

\begin{align} \dot{\psi} = \frac{i \hbar}{2m} \psi_{xx} \implies \hat{\dot{\psi}} &= -\frac{i \hbar k^2}{2m} \hat{\psi} \\ \implies \hat{\psi} &= \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right)\hat{\psi}^{*} \\ &= \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{\hbar}} V \Delta t\right) \psi_{0}\right\} \\ \implies \psi^{**} &= \mathcal{F}^{-1}\left\{ \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\}\right\} \tag{2} \end{align}

where we have inverted the FT now so that the solution in is physical space because the next iteration is the last in the procedure. Finally, you want to solve the potential term for $\tau \in [t + \Delta t/2, t + \Delta t]$ with timestep $\Delta t/2$ and initial condition $\psi^{**}$

\begin{align} \dot{\psi} = -\frac{i}{{\hbar}} V \psi \implies \psi &= \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi^{**} \\ &= \exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \mathcal{F}^{-1}\left\{ \exp \left( -\frac{i \hbar k^2}{2m} \Delta t\right) \mathcal{F}\left\{\exp\left( -\frac{i}{{2\hbar}} V \Delta t\right) \psi_{0}\right\}\right\} \end{align}

which is the result you were after.

If you are looking for more information, I have also written a solution that uses splitting methods to solve the KdV problem here.

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  • $\begingroup$ Thank you for your answer. Is it not possible to obtain my Eq.(4) directly from my Eq.(3), using my Eq.(5) ? $\endgroup$
    – teeeeee
    Nov 11 '20 at 18:37
  • $\begingroup$ Not that I can see, no. $\endgroup$
    – mattos
    Nov 16 '20 at 11:51

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