May be this question might have already been asked but i couldn't find it, so let me know if its already there.

Consider a potential, $V(x) = -\frac{1}{|x|}$ and if we apply this to a one dimensional Schrodinger's equation, I'd like to know the solution for the wave function in 1-D. Does it have a simple analytical solution? Does it have any oscillatory behavior like $$\psi(x,t) = P(x) e^{ikx}e^{i\omega t}$$ I mean will there be a factor like $e^{ikx}$ ? From the internet search, looking at one-dimensional hydrogen atom, first of all i am not sure whether there is any analytical solution, but I guess it was suggested that an exponential decay, something like $$P(x) = e^{-\alpha x}$$ is present. But I am not sure about presence of oscillations like $e^{ikx}$. Hence I'd appreciate some suggestions and clarification.

PS : I am not interested in Hydrogen atom, but in this specific 1-D potential.

  • $\begingroup$ One possible method: A 1D Schr. problem with wave fct. $\phi$ can be mapped to an equivalent 3D radial Schr. problem with radial wave fct. $R(r)=r\phi(r)$, whose solution can be found in any QM textbook. $\endgroup$ – Qmechanic Jul 3 '13 at 12:40
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    $\begingroup$ @Qmechanic: the potential is the same, but the radial part of the Laplacian is not the same as $\partial_r^2$, right? $\endgroup$ – Vibert Jul 3 '13 at 12:51
  • $\begingroup$ @Vibert: Right, therefore the wave function has to be redefined accordingly. $\endgroup$ – Qmechanic Jul 3 '13 at 13:23

With a potential $V(x) = - \frac{\alpha}{|x|}$, with the notation $a = \large \frac{\hbar^2}{m \alpha}$, solutions are :

$$u^+_n(x,t) \sim x e^{ - \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}) e^{ -\frac{1}{\hbar} \large E_nt}~~for~~ x>0$$

$$u^+_n(x,t) = 0~for~~ x\le0$$

and :

$$u^-_n(x,t) \sim x e^{ + \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}) e^{ -\frac{1}{\hbar} \large E_nt}~~for~~ x<0$$

$$u^-_n(x,t) = 0~for~~ x\ge0$$

whose energy is : $$E_n = - \frac{1}{n^2} (\frac{m \alpha^2}{2 \hbar^2})$$

$L_n^\gamma$ is the Generalized Laguerre Polynomial

[EDIT] There are 2 different set of basis functions, see this reference page $192$ formulae $20a$ and $20b$

  • $\begingroup$ @Trimok : I have two questions. What is the 'r', and how did you introduce 'n', and why we need to introduce 'n'? $\endgroup$ – Rajesh D Jul 4 '13 at 3:31
  • $\begingroup$ @RajeshD : Sorry, there was a typo, it is an $|x|$ (not $r$). I have edited the answer. The indice $n$ corresponds to the different possibilities for the solution. The solution $u_n$ is an eigenvector for the energy operator (the hamiltonian) with the eigenvalue $E_n$. The set of $u_n$ is a basis for any general solution, that is, any general solution $u(x,t)$ can be written $u(x,t) = \sum \lambda^n u_n(x,t)$, where the $\lambda^n$ are complex coefficients. $\endgroup$ – Trimok Jul 4 '13 at 5:35
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    $\begingroup$ @RajeshD : For the practical introduction of $n$, the first idea (as indicated by Qmechanic) is to begin with a "3D" problem, that is hydrogen atom We know, that, in this case, energy is quantized, and the energy levels are indiced by an integer $n$. The $1D$ solution is roughly $r$ times the radial part of the $3D$ solution, with taking $l=0$ (because there is not angular momentum in $1D$) $\endgroup$ – Trimok Jul 4 '13 at 5:47
  • $\begingroup$ @Trimok : I'd find it hard to see the amplitude at $x=0$ is zero even though it is an attractive force field. $\endgroup$ – Rajesh D Jul 4 '13 at 6:47
  • $\begingroup$ This means that the probability to find the particle at $x=0$ is zero. In fact, we may try to think classically. If the potential is very big at $x=0$, classically that means that the speed is very big at $x=0$, so a very little time is spend by the particle around $x=0$, so the probability of finding the particle is very tiny around $x=0$. $\endgroup$ – Trimok Jul 4 '13 at 6:55

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