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I have a particular kind of time dependent Schrodinger equation:

$$i \hbar \frac{\partial}{\partial t} \Psi (t) =(\hat{H_0}+i \frac{t}{\tau} \hat{H_1}) \Psi (t) \\ \Psi (0) = \Psi_0 $$

Both $\hat{H}_{0,1}$ are real and don't depend on time. $\tau$ is some suitable time scale. The initial state is normalized $\langle \Psi_0 | \Psi_0 \rangle=1$.

In my case with appropriate coordinate transformation $\hat{H_1} = A \frac{\partial}{\partial x}$ where $A$ is a constant. 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).

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

Then by direct substitution we obtain:

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

Collecting the terms, we obtain the recurrence relation:

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

This seems like a full solution, assuming we can apply the operators $\hat{H}_{0,1}$ as many times as we want (that shouldn't be a problem if we use infinitely differentiable functions everywhere).

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.

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  • $\begingroup$ 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. $\endgroup$ – probably_someone May 8 at 17:52
  • $\begingroup$ @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 $\endgroup$ – Yuriy S May 8 at 17:53
  • $\begingroup$ 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? $\endgroup$ – probably_someone May 8 at 17:54
  • $\begingroup$ @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 $\endgroup$ – Yuriy S May 8 at 17:56
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Would the series preserve the normalization of the initial state?

Surely it won't. Your Hamiltonian isn't Hermitean.

If not, how do I fix that?

Simply cancelling the "i" before $\hat H_1$. Why did you put it there?

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  • $\begingroup$ I put $i$ there because it is present in my Hamiltonian. That is, in position representation $H_1$ is a real differential operator and there's $i$ in front of it. Due to vector potential (which is where the time dependence comes from) $\endgroup$ – Yuriy S May 8 at 19:08
  • $\begingroup$ 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. $\endgroup$ – Yuriy S May 8 at 19:18

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