Free Time Dependent Schrodinger Equation with Inhomogeneous Dirichlet boundary

Question

There exists a FFT-based method to solve the poisson equation in inhomogeneous Dirichlet boundary condition using the sine-transform. For example, Which fourier series is needed to solve a 2D poisson problem with mixed boundary conditions using Fast Fourier Transform?

I expect similar approach could be applied to Schrodinger equation, since they both have a Laplacian operator. After looking around, I failed to find a similar method to solve the Free Schrodinger equation with Dirichlet boundary condition. So either such method is not possible, since wave function is complex valued, or I must miss some of the papers.

Is there a FFT/DFT-based numerical method that can solve the Free Schrodinger equation with inhomogeneous Dirichlet Boundary condition?

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Update based on the answer, and also one of the answer from Testing the time dependent Schrodinger Equation with an analytical solution? we could solve the following with inhomogeneous BC $$u_t = iu_{xx}\\ u(x,0) = u_0\\ u(0,t)=a(t)\\ u(L,t)=b(t)$$
by performing a shifting data $v=u-\eta$ to form a homogeneous problem where $\eta=(1-x/L)a+bx/L$. Note that $a$ and $b$ are complex values, and are often changing with time.

$$v_t = iv_{xx}\\ v(x,0) = v_0\\ v(0,t)=0\\ v(L,t)=0$$

The problem now can be solved using the FFT/DFT sine-transform. However, in my 1D Gaussian wave test, I found that the solution has numerical oscillations (green). Does anyone know how could we reduce the numerical oscillations when using the sine-transform based method? Note: y-axis is $|u|^2$ .

Answer 1

The TDSE is given by $$i\partial_t|\phi(t)\rangle = \hat H |\phi(t)\rangle\,.$$

Expanding the wavefunction $|\phi \rangle $ into a set of eigenfunctions of the Hamiltonian, $$ \hat H |\psi_i\rangle = E_i |\psi_i\rangle\,, $$ one gets \begin{align} |\phi(t)\rangle = \sum_n c_n(t) |\psi_n\rangle\,.\tag{1}\label{1} \end{align}

Insertion into the TDSE yields a decoupled set of one-dimensional equations $$ i {\dot c}_n(t) = c_n(t_0)\; e^{-iE_n (t-t_0)} \tag{2}\label{2} $$ which only depend on the system's eigenenergies and the initial coefficients at time $t_0$.

Now, the case of Dirichlet boundary conditions in 1D (wavefunction at $\pm L/2$) is well known as the "particle-in-a-box" model. The eigennergies and -functions (in position representation) can be calculated analytically as

$$ E_n = \frac{\hbar^2 k_n^2}{2m}\,, \quad k_n=\frac{n \pi}{ L}\,\\ \psi_n(x,t) = \begin{cases} A \sin\left(k_n \left(x-x_c+\tfrac{L}{2}\right)\right) e^{-i\omega_n t}\quad & x_c-\tfrac{L}{2} < x < x_c+\tfrac{L}{2}\\ 0 & \text{otherwise} \end{cases}$$

With this one observes that one procedure for solving the TDSE is the following: Given an initial condition $\psi_0(x)$ satisfying the Dirichlet conditions, obtain the expansion coefficients in Eq. (\ref{1}). Once you have found them, the solution to the TDSE can be obtained exactly in the chosen basis according to Eq. (\ref{2}).

Now, the first step, finding the coefficients in a series of $\sin(nx)$, is exactly what the sine-transform is going to give you. And as usual, a fast sine-transform can be obtained via the FFT. So yes, the TDSE with Dirichlet conditions can be solved via the FFT.

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The above is usually considered with homogeneous boundary conditions. If you want inhomogeneous BCs, e.g. $\psi(-L/2) = 0$, $\psi(L/2) = a$, you can make the following ansatz to your wavefunction: $$ \phi(x,t) = g(x) + \eta(x,t) $$ where $g(x)$ is a straight line satisfying the Dirichlet BCs, and $\eta(x,t)$ satisfies homogeneous BCs.

In the example above, you would have to expand the linear function $g(x)$ in the Fourier basis series, which corresponds to the fourier series of a sawtooth wave. This will then further introduce cosine terms, and thus require a full Fourier transform instead of a sine transform, but it'll work as well.

EDIT2: and thinking again, in the Fourie setting you can also directly set up the term $g(x)$ using cosines ... the concept stays the same, but the function representation is likely easier.