Repeling particles on a ring, minimum angular momentum.

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  • #1
Spinnor
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Main Question or Discussion Point

Say we have two particles of mass m which repel each other, V = V(seperation). Let these particles be constrained to move on a circle of radius r. The particles want to stay at opposite sides of the circle because they repel each other. We want to treat this as a quantum problem so the particles may tunnel past each other depending on how the particles repel each other at short distances. (Seems like this system could "vibrate", rotate, tunnel, and resonate? By varying the potential, mass, and circle diameter we might get interesting quantum states?)

What is the l = 1 angular momentum state of these two particles when they rotate together? Will it be L = [2]^.5hbar = 2mvr where v is the velocity of the particles?

v = [2]^.5hbar/2mr

If V is turned off and the particles no longer interact the l = 1 angular momentum state for each particle now L = [2]^.5hbar = mvr?

v = [2]^.5hbar/mr

Two interacting particles of mass m and velocity v on a circle is similar to a single particle of mass 2m and the same velocity v constrained to a circle as far as angular-momentum is concerned?

Can you say anything about the angular-momentum of a system of N mutually repelling particles of mass m constrained to move on a circle of radius r?

Thanks for any help!
 
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Answers and Replies

  • #2
tom.stoer
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You have to state this problem exactly.

First of all the dynamical variable for particles on a ring is just the angle, so momentum and angular momentum are somehow identical

[tex]p_\phi \sim L_\phi = -i\partial_\phi[/tex]

with a wave function

[tex]\psi(\phi)[/tex]

Then the potential cannot be just the separation V ~ \phi b/c this would violate the symmetry of the problem. So you have to use some periodic function

[tex]V(\phi) = \sum_n v_n\,e^{in\phi}[/tex]

(with appropriate vn to make the imaginary part of V vanish).

I think this is the setting for which you have to calculate the eigenfunctions of

[tex]H = -\partial_\phi^2 + V(\phi)[/tex]


EDIT: for two particles with interaction and w/o external potential you have to use

[tex]H = -\partial_1^2 - \partial_2^2 + V(\phi_1 - \phi_2)[/tex]

which you can rewrite in order to find c.o.m. motion and relative motion.


Last but not least there are unitarily inequivalent representations with "twisted boundary" conditions. For a discussion refer to this thread https://www.physicsforums.com/showthread.php?t=488860
 
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  • #3
tom.stoer
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The next step would be to separate into c.o.m. and relative motion

[tex]\phi_\pm = \frac{1}{\sqrt{2}}(\phi_1 \pm \phi_2)[/tex]

[tex]\partial_\pm = \frac{1}{\sqrt{2}}(\partial_1 \pm \partial_2)[/tex]

[tex]\psi(\phi_1,\phi_2) \;\to\; \psi(\phi_+,\phi_-)[/tex]

[tex]H(\phi_1,\phi_2) \;\to\; H(\phi_+,\phi_-) = -\partial_+^2 - \partial_-^2 + V(\phi_-)[/tex]

An obvious ansatz is

[tex]\psi(\phi_+,\phi_-) = e^{iM_+\phi_+}u(\phi_-)[/tex]

with

[tex][H(\phi_+,\phi_-) - E]\psi(\phi_+,\phi_-) = 0 \;\to\; [M_+^2 - \partial_-^2 + V(\phi_-) - E]u(\phi_-) = 0[/tex]

Here M+ with M = 0, ±1, ±2, ... represents the c.o.m momentum (or the total angular momentum, if you like), and the wave function u contains the relative motion which is the interesting part. What I have done so far is valid for arbitary V (satisfying periodic boundary conditions). So your next step is to study reasonable (repelling) potentials V.

Let me summarize: the problem can be reduced to an effective one-particle problem with the Schrödinger equation

[tex][- \partial_-^2 + V(\phi_-) - (E-M_+^2)]u(\phi_-) = 0[/tex]
 
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  • #4
Spinnor
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Thank so much! You set this up quite nicely, I will try and come up with some solutions.
 
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  • #5
tom.stoer
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I think you should start with the simplest potential like

[tex]\cos(\phi_1 - \phi_2) = \cos(\sqrt{2}\phi_-)[/tex]
 
  • #6
Spinnor
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I think you should start with the simplest potential like

[tex]\cos(\phi_1 - \phi_2) = \cos(\sqrt{2}\phi_-)[/tex]
I was thinking that but of all the simple one dimensional Q.M. problems does a cos(phi) potential uccur? In order of complexity, delta function, square well or square wall, linear, harder and harder? The equation you wrote down looks like it will separate into the plus and minus phi coordinates and the equation with the cos potential looks so simple!

Thanks again. Time to install some trim.
 
  • #7
DrDu
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Consider a molecule like CO2 in two dimensions. It can rotate, that's the [itex]\phi_+[/itex] motion. The molecule also can perform a bending motion, that is the [itex] \phi_-[/itex] motion with a [itex]\cos (\phi_2-\phi_1)[/itex] potential, approximately.
 
  • #8
tom.stoer
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... delta function, square well or square wall, linear, ...
These potentials violate the boundary conditions.

The Cosine potential is simple b/c the Hamiltonian will contain a term

[tex]\cos(\sqrt{2}\phi_-) = \frac{1}{2}\left(\exp(\sqrt{2}i\phi_-) + \exp(-\sqrt{2}i\phi_-)\right)[/tex]

which will shift the Fouriermodes

[tex]\exp(\sqrt{2}in\phi_-) [/tex]

by ±1; that means kinetic term is diagonal with n², the Cosine term couples n with n±1.
 
  • #9
Spinnor
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These potentials violate the boundary conditions.

...
Maybe I'm not clear, but why wouldn't a "square hill" function work? If the particles get within some angular distance of each other they feel a large repulsive force over some very small distance? You could have,

V = some value if abs(phi_1 - phi_2) is less then or equal to θ,
V = 0 otherwise, with θ less then ∏.

Thank you for your help!
 
  • #10
tom.stoer
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  • #11
Spinnor
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Looks like this simple second order direrential equation,

[tex][- \partial_-^2 + cos(\phi_-) - (a)]u(\phi_-) = 0[/tex]

can have complicated solutions. With help from Tom and Wolfram Alpha we get,

http://www.wolframalpha.com/input/?i=d^2f(theta)/dtheta^2+-+cos[theta]f(theta)+++a*f(theta)+=+0

http://www.wolframalpha.com/input/?i=Mathieuc[4a,2,theta/2]

http://www.wolframalpha.com/input/?i=Mathieus[4a,2,theta/2]

This differential equation has a history, see,

http://en.wikipedia.org/wiki/Mathieu_function

My brain is too fuzzy to interpret the graphs above. If the mass and the circle is large enough the problem is that of a simple H.O. for the lowest energy levels? For larger energies we should expect band gaps as in solid state physics and for even larger energies we should expect a continuum of energy levels?
 
  • #12
tom.stoer
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You have to restrict the Mathieu functions to solutions respecting periodicity in [0,2π] which singles out the characteristic values for the Mathieu cosine and sine functions.

And of course you can do more complicated things like scattering of two wave pakets constructed from Fourier series (for which the time evolution is non-trivial) or from series in terms of Mathieu functions (for which the time evolution is trivial b/c they are energy eigenfunctions).
 
  • #13
DrDu
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The solutions of the cosine potential are approximately harmonic oscillations at low energies while at higher energies they can accurately be calculated quasi-classically.
When giving statistical mechanics classes calculation of the heat capacity this way used to be an exercise. The point is that the quasi-classical distribution function becomes a Bessel function, so all works out quite nicely.
 
  • #14
tom.stoer
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The periodic solutions for a repelling cosine potential are the so-called elliptic cosine and sine functions ce(n,q,x) and se(n,q,x) with x in [0,π], q related to the coefficient V and n=0,1,2,...
 
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