Is there a known analytical solution for the following Schrödinger equation

$$i \partial_t \psi=-\frac{1}{2}\partial^2_{x} \psi + \psi x.$$

  • 3
    $\begingroup$ Airy functions. $\endgroup$
    – march
    Feb 18, 2022 at 3:49

1 Answer 1


The potential that you are given is a constant force. So that $$H=\frac{p^2}{2m}+Fx$$ There exists an analytic solution to this problem. It's easy to solve this problem in the momentum basis so that TISE is given by

$$\left[\frac{p^2}{2m}+F\left(i\hbar \frac{d}{dp}\right)\right]\tilde{\phi}(p)=E\tilde{\phi}(p).$$ It's easy to solve this differential equation. Next you can find position space wavefunction using $$\phi(x)=\int dp\ e^{-2\pi i px/\hbar}\ \tilde{\phi}(p).$$ It turns out that in position basis we have Airy's function as the solution. $$\phi_E(\xi)\sim \text{Ai}(-\xi),$$ where $$\text{Ai}(z)=\frac{1}{\sqrt{\pi}}\int_0^\infty du\cos\left(\frac{u^3}{3}+uz\right),$$ where $$\xi\equiv \frac{2m}{\hbar^2}(E+Fx).$$

Suppose you are given the initial state $\psi(x,0)$, then you have to expand this state as $$\psi(x,0)=\sum_n c_n\phi_n(x)$$ where $$c_n = \int \phi_n(x)\psi(x,0) dx$$

Then $$\psi(x,t)=e^{-iHt/\hbar}\psi(x,0)=\sum_n c_n e^{-iE_nt/\hbar}\phi_n(x)$$ Done!

  • $\begingroup$ You should clarify what happens with the time dependence. $\endgroup$
    – Mauricio
    Feb 18, 2022 at 15:57
  • $\begingroup$ $\phi(p)=C \cdot exp[-i/F(Ep-\frac{p^3}{2m})+i(\frac{p^2}{2m}+Fx)t]=C \cdot exp[-i/F(\frac{-2p^3}{3m}+gxp)+i(\frac{p^2}{2m}+Fx)t]$ $\phi(x,t)=\int dp^3 \phi(p,t)e^{ipx}$ is my solution. But how do I implement the initial condition, i.e. $\phi(x,0)$. $\endgroup$
    – NicAG
    Feb 18, 2022 at 18:07
  • $\begingroup$ Above $g$ is $F$. $\endgroup$
    – NicAG
    Feb 18, 2022 at 18:17
  • $\begingroup$ @NicAG See the edit $\endgroup$ Feb 19, 2022 at 8:40
  • $\begingroup$ But the eigenvalues of the state are continuous here. So the sum is replaced by a momentum integral. But does $c(p)=\int \phi_p(x)\psi(x,0)dx$ still hold? $\endgroup$
    – NicAG
    Feb 21, 2022 at 14:29

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