# Can this kind of TDSE be solved by series?

I have a particular kind of time dependent Schrodinger equation:

\begin{align} i \hbar \frac{\partial}{\partial t} \Psi (t) &= \left(\hat{H_0}+ \frac{t}{\tau} \hat{H_1} \right) \Psi (t) \\ \Psi (0) &= \Psi_0 \end{align}

$$\tau$$ is some suitable time scale. The initial state is normalized $$\langle \Psi_0 | \Psi_0 \rangle=1$$. With appropriate coordinate transformation we can write $$\hat{H_1} = i\frac{\partial}{\partial x} \, ,$$ So the full Hamiltonian appears to be Hermitian.

Let's assume the wavefunction can be expressed as Taylor series in time (I don't see why not).

$$\Psi(t)=\sum_{n=0}^\infty C_n \left( \frac{t}{\tau} \right)^n \\ C_0 = \Psi_0$$

Then by direct substitution we obtain

$$i \frac{\hbar}{\tau} \sum_{n=0}^\infty C_{n+1} (n+1) \left( \frac{t}{\tau} \right)^n =\hat{H_0} \sum_{n=0}^\infty C_n \left( \frac{t}{\tau} \right)^n + \hat{H_1} \sum_{n=1}^\infty C_{n-1} \left( \frac{t}{\tau} \right)^n$$

Collecting the terms, we obtain the recurrence relation:

$$i \frac{\hbar}{\tau} (n+1) C_{n+1} = \hat{H_0} C_n + \hat{H_1} C_{n-1}, \qquad n \geq 1 \\ i \frac{\hbar}{\tau} C_1= \hat{H_0} C_0$$

This seems like a full solution, assuming we can apply the operators $$\hat{H}_{0,1}$$ as many times as we want.

However, I have two questions:

• Is there any problem with my derivation? In general $$[\hat{H}_0, \hat{H}_1] \neq 0$$, could this lead to any issues?

• What about normalization? Would the series preserve the normalization of the initial state? If not, how do I fix that?

I know this is not the usual method for solving the TDSE, but I wanted to give it a try, since for my particular problem is seems the most suitable, as opposed, for example, to numerical scheme.

Edit:

About the explicit form of $$\hat{H}_0$$: I'm mostly interested in the case of exciton equation with confinement in one dimension, so it would look something like:

$$\hat{H}_0=- \frac{\partial^2}{\partial X^2}- \frac{\partial^2}{\partial x^2}- \frac{\partial^2}{\partial \rho^2}- \frac{1}{\rho}\frac{\partial}{\partial \rho}+U(X,x)-\frac{1}{\sqrt{x^2+\rho^2}}$$

Where I omitted all the coefficients for brevity. $$X$$ is the center of mass coordinate in one direction, $$x$$ is the distance between electron and hole in the same direction, $$\rho$$ is the distance in plane.

Update 2:

For what it's worth, I tried some numerical experiments in different representations, and it doesn't seem like this series converges at all. Maybe for some potentials it could be possible, but not for my case.

• Expressing the solution as a series is only useful if you know 1) if the series actually converges at all and 2) what the radius of convergence of the series is. – probably_someone May 8 '19 at 17:52
• @probably_someone, does it work if I'm pretty sure the series converges for any $t$? I'm fine with checking any particular cases numerically. I wanted to know if this solution is allowed at all (formally), provided the series converges – Yuriy S May 8 '19 at 17:53
• That depends on what "pretty sure" means. How did you come to this conclusion? And checking numerically is dangerous - how do you know that there isn't some really high-order term with a huge coefficient that makes your solution very inaccurate unless it's included? – probably_someone May 8 '19 at 17:54
• @probably_someone, if you really want, I can add more details about what I'm doing, but let's just assume the series converges, I'm interested in whether or not I can formally solve the equation this way – Yuriy S May 8 '19 at 17:56
• Is $\hat{H}_{0}^{}\propto -\left[\frac{\partial}{\partial x}\right]_{}^{2}$? If so,it can be easily solved in Fourier domain. – Sunyam Oct 1 '19 at 14:36

Simply cancelling the "i" before $$\hat H_1$$. Why did you put it there?
• The operator contains a first derivative only, so I'm pretty sure it's Hermitean because of $i$. That is, $$\hat{H_1} = A \frac{\partial}{\partial x}$$ where $A$ is a constant and $x$ one of the coordinates. – Yuriy S May 8 '19 at 19:18
• I remind you that $i\hat{H_1}$ looks like the $x$-component of a particle linear momentum $$p_{x}=\boldsymbol{-}i\hbar\frac{\partial}{\partial x}$$ so it's hermitian. In other words the hermitian conjugate of $\frac{\partial}{\partial x}$ is $\boldsymbol{-}\frac{\partial}{\partial x}$. – Frobenius Oct 1 '19 at 15:45