Deriving Stationary State Wave Functions for a Free Particle on a Ring

[MichalXC]

Hi everyone.

I want to derive for fun the stationary state wave functions for FREE a particle of mass m on a ring of radius R. The question seems trivial, but I am getting hung up on something silly.

What I think I know:

Since \psi can be written as a function of the radial angle \phi,

\hat{H} \psi (\phi)= -\frac{\hbar^2}{2m} \nabla^2 \psi (\phi)=E \psi (\phi)

My problem:

I am unsure of how to modify the Laplacian (del-squared) in the above equation. I know that in two-dimensional Cartesian coordinates the Laplacian is written as:

\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}

How can I change the Laplacian to polar coordinates so that it properly acts on \psi(\phi)? (Without changing coordinates, I have:

\nabla^2 \psi(\phi) = \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) \psi(\phi)

which does not make sense mathematically.)

Any help is appreciated! Also, please let me know if I am thinking in the wrong direction.

Finally, this is one of my first posts, so I apologize if I posted in the wrong section.

 

[MichalXC]

Judging by the lack of understandable material on the Internet concerning this question, I am beginning to think that I am either (1) in over my head, or (2) misunderstanding the problem.

I am still, however, thinking.

Since the specified potential V(\phi) is a function of only one variable, can we not simply assume separable solutions for the TDSE, and obtain the TISE by separation of variables? I would get

- \frac{\hbar^2}{2m} \frac{d^2 \psi(\phi)}{d \phi^2} + V(\phi) = E \psi(\phi)

which is simply the TISE for a particle in one-dimension (with the variable x replaced with \phi). No?

 

[Matterwave]

Yes, but you have no V, and you have periodic boundary conditions:

\psi(\phi)=\psi(\phi+2\pi)

Technically, there should be a 1/R^2 in front of your derivative term, but it's just a constant, so it doesn't matter too much (it makes the units work out).

 

[MichalXC]

Very cool. Thanks for the reply.

The 1/R^2 comes from changing the Laplacian to spherical polar coordinates... and crossing out the terms that don't affect \psi(\phi)... right? I saw a very scary, brute-force derivation of this somewhere online.

I also understand that V(\phi) = 0 and that \psi(\phi) = \psi(\phi + 2 \pi).

Now my question is:

How will this wave function be different than the wave function for a free particle in (not on) a circular ring (see below)? It seems like I would just get the same answer...

Wikipedia: http://en.wikipedia.org/wiki/Particle_in_a_ring

 

[Matterwave]

The particle on a ring obeys periodic boundary conditions, and therefore is easily normalizable. A free particle does not have boundary conditions, and the plane wave solutions of the Schrodinger equation are not normalizable, extending to infinity in both directions. One must use wave-packets to normalize a regular free particle.

Additionally, because of the periodic boundary condition, the particle on the ring is confined to certain quantized energy levels; while a free particle can take on any continuum range of energies.

 

[tom.stoer]

It's similar to a free particle on a sphere. You drop the r-derivatives i.e. fix r = R = const. and you get the spherical harmonics w/o the radial part as solution to your modified Schrödinger equation.

 

[MichalXC]

Thanks for your replies, guys.

Tom, I haven't yet studied quantum mechanics in three-dimensions, but a free particle on a sphere certainly seems like a natural analog: I imagine letting a particle move freely on the outside of a hollow sphere centered at the origin of the 3D Cartesian coordinate system, then restricting the motion of the particle to one plane (the x,y-plane, for example). Is that close?

I will post my attempted solution soon.

 

[tom.stoer]

It's close.

The only problem is that you must not "confine" the particle. This would mean that you add something like a one-dim. square-well potential which becomes narrower - resulting in growing (ground state) energy as you already know.

 

[MichalXC]

It's close.

The only problem is that you must not "confine" the particle. This would mean that you add something like a one-dim. square-well potential which becomes narrower - resulting in growing (ground state) energy as you already know.

Right, "squeezing" the particle between two infinitely close walls of infinite potential would give the particle an extremely high energy, which obviously does not correspond to this situation. (I am looking at the equation for the infinite square well. As the size of the well approaches zero, the allowed energies approach infinity.)

So how should we define the potential at those points not on the circle? Would we say V = 0 for all space? Or is the question simply nonsensical, since the "universe" is simply limited to the path of the circle?

 

[tom.stoer]

One must not start with an 3-dim. problem and reduce it (confine the particle) to 2 dimensions; instead one must start with a classical particle moving on a curved manifold (e.g. a 2-sphere) and construct the kinetic energy of the quantum Hamiltonian which is the generalized Laplace-Beltrami operator on the curved manifold.

I guess a constraint quantization approach a la Dirac which implements the dimensional reduction is possible as well, but I have to work it out.

 

[MichalXC]

So the TISE on the path of the ring is:

-\frac{\hbar^2}{2m} \frac{1}{R^2} \frac{ \partial^2 }{ \partial \phi^2 } \psi = E \psi

which has a solution of the form

\psi(\phi) = A e^{ik \phi} + B e^{-ik \phi}

where

k^2 = \frac{2 m E R^2}{\hbar^2}

Is this correct so far?

 

[Matterwave]

Looks good to me. Now just implement your boundary conditions, and normalization condition. For this, it may be simpler to express the wave function as a sum of sines and cosines instead of exponentials.

 

[MichalXC]

Cool. Here are my calculations.

Imposing the periodic boundary condition \psi(\phi) = \psi(\phi + 2 \pi) let's us find B in terms of A:

B = A e ^{2 i k (\phi + \pi)}

The wave function then becomes:

\psi(\phi) = A e^{i k \phi} \left( 1 + e^{2 i k \pi} \right)

Applying the periodic boundary condition once more gives us:

e^{2 \pi i k} =1

which implies that

k = 0, \pm 1, \pm 2 ... = n, where n \in \mathbb{Z}

We can now find the allowed energies:

k^2 = \frac{2 m E R^2}{\hbar^2} = n^2 \Longrightarrow E_n = \frac{n^2 \hbar^2}{2 m R^2}

Finally, normalizing \psi(\phi) = A e^{i k \phi} \left( 1 + e^{2 i k \pi} \right) gives us:

\psi(\phi) = \sqrt{\frac{1}{2 \pi}} e^{i n \phi}

Thanks for all your help, guys!

 

[Matterwave]

Looks good to me. Just remember that now n can take negative and positive values as well as 0.

 

[PraTux]

I've a problem accepting that, applying the periodic boundary condition \psi(\phi) = \psi(\phi + 2 \pi) twice, as MichalXC does would yield the values of A, B and k in this equation. I think s/he is treating \phi as a variable (angle) while imposing the boundary condition, this results in the value of B = A e ^{2 i k (\phi + \pi)} , a function dependent on \phi which we clearly defined as a constant while taking a trial solution for TISE. If we treat \phi as a constant angle, and hence not substitute it with \phi + 2\pi while imposing the boundary conditions, we end up getting a not-so-useful equation, 1=1. The idea that one can find values for 3 variables from a single linear equation doesn't seem right to me. The fact that MichalXC got the right answer must be just a fluke.

I think one correct approach to this problem would be as follows: to get rid of one of the two constants, A and B, we can use the fact that each base of a second order homogeneous differential equation, satisfies the differential equation individually. So, we can use \psi_{one singlek}(\phi) = A e^{ik \phi} , as a solution and then impose the boundary condition to find k as a function of A. Next, we can normalize the wave function over interval 0 to 2Pi, to eliminate A and find k. The rest is described my MichalXC.

Well, I could totally be wrong with my claims (I'm taking my first quantum class right now). So, please correct me.

 

[MichalXC]

This results in the value of B = A e ^{2 i k (\phi + \pi)}, a function dependent on \phi, which we clearly defined as a constant while taking a trial solution for TISE.

After thinking about it for a while, I agree with PraTux. Can anyone clarify the confusion?

 

[tom.stoer]

Fixing A and B doesn't have anything to do with the boundary conditions. The solution to the equation is an arbitrary exponential; restricting it to the circle = using periodic boundary conditions results in discrete k-values. For given |k| one can make a basis transformation which results in sine and cosine as a special case, but in principle any |k>' = A|k> + B|-k> is allowed.

 

[MichalXC]

Applying the boundary condition \psi(\phi) = \psi(\phi + 2 \pi) implies that B changes as a function of \phi:

B = A^{2 i k (\phi +\pi)}

But the general solution \psi(\phi) = A e^{i k \phi} + B e^{-i k \phi} requires that A and B be constant, right?

Could you explain it again?

 

[tom.stoer]

You start with two solutions

\psi_{\pm k}(\phi)=e^{\pm ik\phi}

The periodicity condition reads

\psi_{\pm k}(\phi+2\pi) = e^{\pm 2\pi i k}\,e^{\pm ik\phi} \stackrel{!}{=}\,e^{\pm ik\phi}

That means

e^{\pm 2\pi i k}=1

Now consider solutions (of course with fixed A, B)

\psi_{AB}(\phi)=A \psi_{+k}(\phi) + B \psi_{-k}(\phi)

Of course we have

\psi_{AB}(\phi+2\pi) = A \psi_{+k}(\phi+2\pi) + B \psi_{-k}(\phi+2\pi) = A \psi_{+k}(\phi) + B \psi_{-k}(\phi) = \psi_{AB}(\phi)

So using periodicity nothing follows for A, B; they are still arbitrary!

One must consider additional requirements to fix them
a) normalization fixes one constant
b) requiring that the functions are eigenfunctions of the momentum operator fixes either {A arbitrary, B=0} or {B arbitrary, A=0}
c) requiring parity does the same

 

[naffin]

I'm having some difficulties trying to understand this problem. In particular, I don't understand why we have to use the boundary condition \psi(\phi) = \psi(\phi + 2 \pi) .

In what Hilbert space we are ?

 

[tom.stoer]

The particle is living on a circle S¹. That means that instead of having an x-coordinate in ]-∞, +∞[ the real line is compactified to [0,L[ plus periodic boundary conditions. That means that two positions x and x+L are identified.

We want to have a single-valued wave function for every x, which means we cannot accept having ψ(x) = a and at the same time ψ(x) = b with b≠a. Now we could describe this x using some value for angle, and of course this value for the angle +2π. These two different angles refer to the same x on on S¹. Not having such a boundary condition would allow for wave functions having different values for different angles mod 2π and therefore different values at the same x. This is of course nonsense.

So the boundary condition must hold for every wave function ψ. Applied to plane wave states one immediately finds a discretization of the allowed k-values.

 

[lugita15]

I wonder, what are the implications of the gauge symmetry in this problem, i.e. |θ>=|θ+2π>? I know the practical consequence is that it gives rise to periodic boundary conditions, but at a more formal level what can we say about it?

 

[tom.stoer]

It's not really a gauge symmetry b/c is rigid.

You can gauge it by introducing a 'vector potential' or a B-field with a flux through the ring; this will create a kind of theta-angle or 'winding number'. I have to look for some papers ...

 

[atyy]

Unitarily inequivalent quantizations (like LGQ)?

http://arxiv.org/abs/quant-ph/0510234
http://arxiv.org/abs/quant-ph/0611116

 

[naffin]

The particle is living on a circle S¹. That means that instead of having an x-coordinate in ]-∞, +∞[ the real line is compactified to [0,L[ plus periodic boundary conditions. That means that two positions x and x+L are identified.

The problem is that in fact we are not only requiring \psi(\phi) = \psi (\phi + 2 \pi), but also the continuity \lim_{\phi \rightarrow 2 \pi ^{-}} \psi(\phi) = \psi (2 \pi).
In other words: we are not just searching for functions defined on [0, 2 \pi [ and then exteding them on ]- \infty , + \infty[ periodically. We are requiring continuous peridiocity.

Without this continuity we can't figure out a discretization of k: any k would be possible.

 

[tom.stoer]

The problem is that in fact we are not only requiring \psi(\phi) = \psi (\phi + 2 \pi), but also the continuity \lim_{\phi \rightarrow 2 \pi ^{-}} \psi(\phi) = \psi (2 \pi).
In other words: we are not just searching for functions defined on [0, 2 \pi [ and then exteding them on ]- \infty , + \infty[ periodically. We are requiring continuous peridiocity.

Without this continuity we can't figure out a discretization of k: any k would be possible.

Yes

(and no, b/c we all know that this will not work in general: first we require continuity, then we write the wave function as a Fourier series - and we know that in general L² wave functions and Fourier series need not be continuous; that means that we cannot safely guarantuee that we deal only with C^k, especially under time evolution; this depends crucially on the Hamilton operator)