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Why doesn't a quantum particle in an attractive 1D potential accumulate at the center?

+ 6 like - 0 dislike
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I have two questions regarding (possibly counter intuitive results) Schrodinger equation and its application to two (strictly hypothetical) scenarios.

  1. Consider the 1D potential $V(x) = - \frac{\alpha}{|x|}$, which is an attractive one. As it is attractive, after a sufficiently large amount of time I'd expect the probability density of finding the particle at $x=0$ (center) to be higher than that anywhere else. I hope it is reasonable to expect such a thing.

    But the solution according to Schrodinger equation is (as given in this answer) is $$u_n(x,t) \sim \lvert x\rvert e^{-\lvert x\rvert/na} ~L_{n -1}^1\biggl(\frac{2\lvert x\rvert }{na}\biggr) e^{-E_nt/\hbar}$$ Consider the amplitude at $x=0$ $$\lvert u_n(0,t)\rvert = 0$$ which I think is contradictory to our reasonable expectations based on intuition. I'd appreciate some comments on why it is counter intuitive.

  2. Consider the particle in box problem and according to Schrodinger's equation, in almost all energy states the probability of finding the particle close to the boundaries is zero. Now instead of intuition (which is not predicting anything) I'd give a practical example where it is quite contradictory. (Pardon me if this example is not suitable, I am learning and I'd love to know why it isn't intuitively not suitable). Consider a metal conductor, I guess its an example for particle in a box as the potential inside is zero and we know that the charge is usually accumulated at the edges of the conductor.

EDIT

An interesting thing to add. The probability density 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.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
asked Jul 4, 2013 in Theoretical Physics by Rajesh D (55 points) [ no revision ]
For 1: Even, if the probability is exactly zero at $x=0$, With the negative exponential, the probability decreases quickly for $|x|> na$. So you can say that the particle is approximatively in the region $|x|< na$.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Trimok
@Trimok : I know that exponential decreases much quickly than $O(x)$, but that is not the point of the question. (check the original title of the question before edit by David), the current title is what is somewhat misleading.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
@DavidZaslavsky : IMHO, the original heading of the question was more appropriate than the current one. Especially see the comment by Trimok. Also see the second part of the question.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
You'll note that the 1D finite square well has solution with antinodes at the center, so the question is reasonable. I suspect it comes down to the strange feature of this potential: $V'$ is discontinuous at $x=0$ which seems to make it hard to write a good solution with nonzero value of the wave function there.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user dmckee
@Rajesh the problem is that the original title said almost nothing about the actual content of the question. If you think the new title is inappropriate, feel free to change it, just make sure it actually says what you want to ask.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user David Z
A quick comment on your edit. You're mixing up wavefunction and probability. The wavefunction does not go to zero, but the probability will do because of the jacobian factor. That is to probability that it lies between $r$ and $r + dr$ is $4 \pi r^2 a^2 e^{-2kr}dr$, which clearly IS zero at the origin.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user VolatileStorm
@VolatileStorm : Agreed. Thanks for pointing out. I meant the probability density going to zero, but ended up saying probability is zero, but probability of finding at any given single point on real line is always zero. I have edited it now.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D

5 Answers

+ 1 like - 0 dislike

Question 1:

Your intuition that in the probability density of finding the particle should be the highest at $x=0$ (when it is in the ground state, to be precise) must be correct, but the solution you quoted is faulty.

I read the reference from the answer you linked. There, what the authors really have done is to solve the Schrodinger's equation for $V(x)=- \frac{1}{x}$ and $V(x)=+\frac{1}{x}$ separately (with no restriction on the range of $x$ for either case) to obtain the bound-state solutions $\psi_{+}(x)$ and $\psi_{-}(x)$. Then, they claim that the full solution is such that $\psi(x)=\psi_{+}(x)$ for $x>0$ and $\psi(x)=\psi_{-}(x)$ for $x<0$.

However, what they have solved is a totally different problem than they had originally intended to. Most importantly, the potentials $V(x)=\mp\frac{1}{x}$ have infinite barriers at $x=0^{\mp}$. Therefore, we can expect that a particle bound to the potential $V(x)=- \frac{1}{x}$ ($+\frac{1}{x}$) on the $x>0$ ($x<0$) side can never penetrate into the other side. You can check this fact mathematically by Fourier transforming the momentum-space wave function from Eq. 13 in the reference. You'll see that $\psi_{+}(x)$ completely vanishes for $x\le 0$ and $\psi_{-}(x)$ for $x\ge 0$. This is certainly not what would happen for $V(x)=-\frac{1}{|x|}$.

In the same paper, other people's results are also mentioned. Most notably, some of them apparently obtained eigenfunctions associated with a continuum of negative-energy spectrum, although the authors of this paper consider them unphysical. IMHO, this sick negative-energy continuum probably does exist, and what it tells us is not that we should discard the corresponding solutions while keeping more well-behaved ones, but that Schrodinger's equation for $V(x)=-\frac{1}{|x|}$ is a fundamentally sick problem in the sense that the electron can really fall indefinitely down to the infinitely deep potential well.


Question 2:

Coulomb interaction between electrons in a metal is screened and becomes short-ranged. This is at least a part of the reason why non-interacting electrons in an infinite potential well can be a reasonable model of a metal for certain purposes.

On the other hand, charges accumulating at the boundary of a metal are "excess" charges. Coulomb repulsion between them is not screened. The non-interacting model by no means contains this kind of physics, but may still be used to describe the interior of the metal in which the Coulomb interaction is screened.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user higgsss
answered Jul 6, 2013 by higgsss (50 points) [ no revision ]
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Thanks for the answer. The reference in deed may be faulty which apparently is. But what puzzles me is that, instead of trying to find the correct solution for the 1-d problem, you go on to claim that it is a fundementally sick problem. All I am interested is in a correct solution, and I am not pre-hand care whether it is sick problem or not!. No problem is sick by itself.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
What do you mean by "No problem is sick by itself"? After reading this I am fairly convinced that an infinitely deep potential well will not yield physically valid solutions, and therefore trying to interpret the solutions will probably be problematic. Are you saying that there is a way to phrase this problem in a "non-sick" way?

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user VolatileStorm
I didn't mean that this problem is worthless to study because it is "sick". (If you don't like this word, you can call it "singular".) My point was that the seemingly unphysical negative-energy continuum obtained in certain papers (cited in the paper I linked) is probably due to the singular nature of this problem, and we shouldn't simply discard them.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user higgsss
Probably the right way to think about this problem is to regulate the singular 1-D Coulomb potential as $V(x) = -1/(|x|+x_{0})$ or $V(x) = -1/|x| \, (|x|>x_{0}), \, -1/x_{0} \,(|x|<x_{0})$ for some x_{0}>0, solve this problem, and consider study the limit $x_{0} \rightarrow 0$, as did in at least one of the papers trying to rationalize the negative-energy continuum.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user higgsss
@VolatileStorm : The main motivation is that it is Coloumb's potential! My interests are from theoretical nature and not practical, so I am sure I want interested in this solution. i look at it as a potential problem to open up the intricacies of QM.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
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Some thoughts - you state that you want to explore the intricacies of QM, but I'd suggest that if you're willing to throw away one of the necessary parts of QM - that the system is bounded from below - you're no longer studying QM anymore, just solutions to a D.E. so it's quite likely that your intuition will break. Secondly, if you're wanting to ask the question of "non accumlation", maybe it would be more instructive to use the QHO potential. The odd eigenfunctions show the property of being 0 at the origin as well.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user VolatileStorm
One can see the "sickness" of this problem by considering the expectation value of the potential w.r.t. some wave function, i.e., $\langle V \rangle = -\int dx \frac{\alpha}{|x|} |\psi(x)|^{2}$. Any $\psi(x)$ which is differentiable everywhere and such that $\psi(0) \neq 0$ yields $\langle V\rangle \rightarrow - \infty$ and some finite $\langle\frac{p^{2}}{2m}\rangle$. This shows that the energy of the 1-D "Hydrogen atom" is unbounded from below. On the other hand, the 2-D and higher-dimensional Coulomb potentials are free of this problem.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user higgsss
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Just to show that the first situation is not that uncommon: the gravitational force is attractive, yet the probability of finding a planet inside the sun is tiny. (of course, who would be brave enough to go check, right?)

The semi-intuitive explanation is this: the particle has a certain amount of energy and the force field is conservative. So no energy is lost. However, if the particle were at the center of the system, at rest, it would have the lowest energy possible. No potential energy (if we set the zero for potential energy at the center, which is common) and no kinetic energy since the particle isn't moving. So basically it's energy conservation that prevents a particle in an attractive potential from settling down at the center.

For your second case: the potential inside a metal isn't completely zero. The conduction electrons are still influenced to some degree by the ions (and each other). Otherwise, every metal would be a superconductor at any temperature. And in the absence of an exterior potential, I don't think the electrons accumulate at the edges of the metal. They move through the lattice of ions as a nearly-free gas, their motion modulated (slightly) by the (motions of the) ions.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Wouter
answered Jul 4, 2013 by Wouter (30 points) [ no revision ]
Your are considering Planets and sun, then they are in 3-D. 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.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
Yes I know, I only mentioned it to make it seem less counterintuitive. I also thought about mentioning the nonvanishing probability in the hydrogen atom, because that would seem the more counterintuitive to me. The rest of my discussion doesn't assume a sun-planet system though, only the first paragraph is about that.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Wouter
In the second paragraph, you had mentioned,"basically it's energy conservation that prevents a particle in an attractive potential from settling down at the center", but I have never said that the particle should settle down at the center, i wanted some non zero probability and more than that at other places, thats not any close to saying the particle should settle down at the center. right?

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Rajesh D
Well, it's not a rigorous discussion, I agree. The main thing I was trying to get across is that it isn't that surprising to have a higher probability in other places than the center. Energy conservation suggests this. Perhaps it's not what you were looking for.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Wouter
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Once again I would like to emphasize that when we go to the quantum regime, the concept "particle" does not carry one to one from the classical regime.

In the classical regime a particle is defined by its center of mass and the position and velocity of the center of mass, but those are not its only attributes. It is also defined by the volume it occupies and the mass density per unit volume.

In the quantum regime two things happen, an "elementary particle" can be particle like, defined by the first sentence in the above paragraph, with and (x,y,z,t) and a (p_x,p_y,p_z,E) four vector ( in units where c=1), identical to the definition of a classical regime particle. BUT it has no volume or mass density, it is a point particle. The second difference is that an elementary particle may display a wave behavior, characterized by experimentally observed interference patterns. This is not the behavior of macroscopic particles.

The quantum mechanical equations developed to fit this observed behavior, the particle like and the wave like nature, and the solutions of the equations give us the probabilities of finding a "particle" at a specific space time point, which probability has wave like properties. These are static solutions for time independent potentials. They do not develop over time, fortunately, otherwise we would not be here since atoms depend on this stability. ( rather the solutions of the equations describe the stability of atoms).

So yes, there is a probability for an electron of an S state to find itself in the middle of the nucleus, and that is how electron capture happens and nuclei are transformed to lower Z. It is a tiny probability, considering that the size of the nucleus to the size of the atom is orders of magnitude smaller, but it happens, and is well described by the appropriate solutions of the QM system.

Now conductors are complicated. The conduction electrons are shared by all the nuclei because there are common energy layers. It cannot be modeled by a square well potential.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user anna v
answered Jul 6, 2013 by anna v (1,875 points) [ no revision ]
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Intuitively (I.e. classically), for a particle in an attractive 1D potential one would expect the probability density to be maximal at positions where the particle comes to a stop, i.e. at positions where it bounces back. The opposite is also true: where the potential is deepest, the particle will spend less time and one would intuitively expect the probability density to be minimized.

For particles with precisely zero angular momentum in a rotationally symmetric 2D (or higher dimensional) potential (the alternative case considered by OP) the behavior is different. This is due to the fact that the zero angular momentum constraint causes the origin (the point where the potential is deepest) to act as a "focal point": the particle moves quickest at this point but is guaranteed to pass through this point whereas most positions away from the origin will be missed by the particle.

Considering either the strictly 1D classical case or any higher dimensional classical case in which the angular momentum can impossibly be defined with infinite precision, I would defend the position opposite to that stated in the question: observing a maximum probability density for finding a particle at the deepest point of the potential would be counterintuitive.

Now turning to the quantum case: it is not correct to assume that the ground state probability distribution for a point particle in a potential will always attain a non-zero value at the location of the classical ground state. A straightforward counter example is provided by any 1D potential V(x) that is unbounded for negative x, and monotonically increasing for positive x.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Johannes
answered Jul 6, 2013 by Johannes (280 points) [ no revision ]
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In one dimension, motion in a Coulomb-like attracting potential $U=-\frac{\alpha}{|x|}$ is quite ill-defined. I'll now try to explain this for classical particles, then will say some words about quantum mode.

Let's consider two cases:

  • 3D motion of a particle nearly free falling into the attracting center, but with nonzero angular momentum $L$.

In this case we know from Kepler laws that the solution is an elliptic trajectory. Its area will go to $0$ as $L\to0$, leaving major axis length finite. Thus, in the limit of $L\to0$ we have a free falling particle with trajectory being a straight line, and, surprisingly, the particle will effectively repulse from the attracting center after reaching it. But this motion is nothing than 1D motion in Coulomb-like potential, and the particle doesn't ever go past attracting center, having asymmetric trajectory.

  • 1D motion in a more general potential $U=\frac{-\alpha}{\sqrt{x^2+\varepsilon^2}},$ which will become usual Coulomb-like potential when $\varepsilon=0$.

Let $\varepsilon\ne0$ and starting speed of particle $v_0=0$. At $x\approx0$ the potential behaves like a harmonic oscillator, so the particle will go through it. Eventually the particle will slow down at the point opposite to starting point, and continue moving in the opposite direction. Now make $\varepsilon$ smaller. The particle will still move in a trajectory, symmetric with respect to origin. As you take the limit $\varepsilon\to0$, you'll get your particle moving in a symmetric trajectory.

You can see that both cases show one-dimensional motion in Coulomb-like potential, but they arrive at contradicting results. So, one should draw the conclusion that the solution of the problem depends on from which side you take your limit: either nearly free-falling particle with $L\to0$, or nearly Coulombic potential with $\varepsilon\to0$.

When you take quantum equation, i.e. Schrödinger equation, you will in fact arrive at similar results with both approaches because you no longer have definite trajectories, and your approach in the first case is putting all the 3D system into a quantum wire with attracting Coulombic center in the middle and shrinking its walls.

The result in both cases in quantum mode will be that your ground state energy is unbounded from below as you approach 1D Coulombic regime, and the wavefunction will converge to something like a Dirac delta function, though the limit might not be strictly it. Of course, this result is quite nonphysical: you need infinite energy to excite the particle.

As for the solution given in the answer you link to, I'd say it's at least incomplete: it doesn't include boundary conditions on joining of $u^+$ and $u^-$ functions, and even if it does give you the solution, it doesn't include ground state.

This post imported from StackExchange Physics at 2014-08-22 04:59 (UCT), posted by SE-user Ruslan
answered Nov 19, 2013 by Ruslan (85 points) [ no revision ]

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