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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 1D. 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 1D potential.

Roger V.
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Rajesh D
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  • 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. – Qmechanic Jul 03 '13 at 12:40
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    @Qmechanic: the potential is the same, but the radial part of the Laplacian is not the same as $\partial_r^2$, right? – Vibert Jul 03 '13 at 12:51
  • @Vibert: Right, therefore the wave function has to be redefined accordingly. – Qmechanic Jul 03 '13 at 13:23

2 Answers2

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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 $20b$

Trimok
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  • @Trimok : I have two questions. What is the 'r', and how did you introduce 'n', and why we need to introduce 'n'? – Rajesh D Jul 04 '13 at 03:31
  • @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. – Trimok Jul 04 '13 at 05:35
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    @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 $1D$) – Trimok Jul 04 '13 at 05:47
  • @Trimok : I'd find it hard to see the amplitude at $x=0$ is zero even though it is an attractive force field. – Rajesh D Jul 04 '13 at 06:47
  • 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 $x=0$. – Trimok Jul 04 '13 at 06:55
  • What surprises me is that we don't seem to have solutions with odd parity. For all $n$, we have $u_n(-x) = u_n(x)$, whereas I'd expect some solutions with $u_n(-x) = -u_n(x)$ because the Hamiltonian commutes with the parity operator. Is there a simple reason why we don't see odd solutions? – Lagerbaer Jul 04 '13 at 06:57
  • @Lagerbaer : Here $Pu_n(x) = u_n(-x)=u_n(x)$. So it seems to me that there is no new solution. – Trimok Jul 04 '13 at 07:05
  • Trimok : I suspect any quantization should happen in 1-D. I guess you have adopted the quantization from 3-D rather blindly. Looking at the differential equation with hardly any boundary conditions and perfect symmetry, I doubt there is any quantization necessary. – Rajesh D Jul 04 '13 at 09:19
  • @RajeshD : I have made an edit to the answer, because there are subtelties: see this reference page $192$ formulae $20a$ and $20b$. In fact, there are 2 set of functions $u^+_n$ and $u^-_n$, which are only defined, respectively, on $x>0$ and $x<0$ – Trimok Jul 04 '13 at 09:20
  • @Trimok : "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", good one, but I hardly want to put my money on it. I think its rather dubious at best. – Rajesh D Jul 04 '13 at 09:21
  • @RajeshD : " I think its rather dubious at best." You are probably right ...But the fact that the wave function is exactly zero at $x=0$ is one thing, and the fact that the wave function is concentrated in some interval (of size $na$) around $x=0$, is another thing. – Trimok Jul 04 '13 at 09:28
  • @Trimok : $u_n(x)$ is continuous at $x = 0$, which means give me any probability $P$, however small, I can find an $\epsilon$ such that the probability of finding the particle in the interval $(-\epsilon,\epsilon)$ is smaller than $P$. – Rajesh D Jul 04 '13 at 09:41
  • @Trimok : I am reading your reference. Meanwhile an interesting thing. The probability vanishing at center does not seem to arise in 3-D hydrogen atom. Check the wave function of 1s orbital of Hydrogen atom, it of the form $\psi_{1s}(r) = a e^{-kr}$, where $k$ and $a$ are some constants and $r$ is the radial distance from nucleus. – Rajesh D Jul 04 '13 at 09:49
  • @RajeshD : $\psi_{1s}$ correspond to $l=0$. For the general case $l>0$, you have a $r^l$ coefficient, so the wavefunction is 0 at $x=0$ – Trimok Jul 04 '13 at 14:44
  • Btw, every one-dimensional attractive potential has a bound state. – Lagerbaer Jul 04 '13 at 15:08
  • @Lagerbaer ; (I have made a edit to my previous answer which was not rigourous enough). Yes, but this should be true for other dimensions, no ?. It seems to me that there is an equivalence between attractive potential and bound states ? – Trimok Jul 04 '13 at 15:16
  • Thank you for posting the paper. It's really interesting and not at all intuitive that the singularity of the Coulomb potential is, essentially, a potential barrier. Regarding bound states, in 2D and 3D you can have attractive potentials that are too weak to trap an electron. – Lagerbaer Jul 04 '13 at 15:36
  • @Lagerbaer : Could you give an example of such a 2D or 3D potential ? Thanks. – Trimok Jul 04 '13 at 15:41
  • Here is one example, at the end, for the spherically symmetric potential well: http://en.wikipedia.org/wiki/Finite_potential_well – Lagerbaer Jul 04 '13 at 15:45
  • If you want to continue a discussion, please move it to [chat] (you can create a room) – Manishearth Jul 04 '13 at 15:58
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One-dimensional Schrödonger equation with Coulomb potential is known to be problematic, as the ground state energy diverges. It can however be solved by introducing an appropriate cutoff. Here is the abstract from the classical Loudon's paper One-dimensional hydrogen atom, which solves this problem:

The quantum-mechanical system which consists of a particle in one dimension subjected to a Coulomb attraction (the one-dimensional hydrogen atom) is shown to have a ground state of infinite binding energy, all the excited bound states of the system having a twofold degeneracy. The breakdown of the theorem that a one-dimensional system cannot have degeneracy is examined. The treatment illustrates a number of properties common to the quantum mechanics of one-dimensional systems.

Although initially published in American Journal of Physics, as being of only academical interest, the paper has become a rather cited one in the field of carbon nanotubes, which are effectively one-dimensional systems, and where the exciton energies were predicted to be anomalously large (of course, in the nanotubes there is a natural cutoff parameter — the nanotube diameter). The problem is less manifest in other one-dimensional structures, such, e.g., as semiconductor quantum wires, since there is screening of the Coulomb potential by the electrons present in the material outside of the wire.

Ruslan
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Roger V.
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