# Particle in a ring

1. Oct 13, 2015

### maverick_76

So I am working on the problem of the particle bound to a ring of radius R. I am trying to solve it two ways, as a standing wave and as a running wave. I'm stuck right now solving for the standing wave. So far I have:

ψ(x)=Asin(kx) + Bcos(kx)

I know that it is periodic from 0 to 2π so if I integrate the modulus of ψ(x) squared I can set it equal to 1 and solve for A & B.

So my question is this, would the B value be zero? Since it is periodic around the ring, would the only way to make cos(kx) zero be to make coefficient B zero?

2. Oct 13, 2015

### The_Duck

No, there's no reason to set B to zero.

Assuming $x$ here is the angle around the circle, your constraints are:

* $\psi(0) = \psi(2 \pi)$
* $\frac{d}{dx}\psi(0) = \frac{d}{dx}\psi(2 \pi)$
* $\psi(x)$ should solve the Schrodinger equation

If you impose these constraints you will find:
* k can only take on certain discrete values
* $A^2 + B^2$ must equal a certain value, but $A$ and $B$ are not individually constrained

Why aren't $A$ and $B$ constrained invididually? This corresponds to the fact that if you have a solution $\psi(x)$ and then rotate it around the circle by any angle, you get another solution. So you have a bunch of solutions, all of which are linear combinations of $\sin(kx)$ and $\cos(kx)$.

3. Oct 13, 2015

### maverick_76

Okay so solving the integral of modulus psi squared, I find

A^2 + B^2 = (1/pi)

How exactly do I find A & B values with this, If I can't assume one is zero? I know:

psi(0) = B = psi(2pi)
&
psi'(0) = Ak = psi'(2pi)

but I am rusty here, can I use this info to solve for A & B?

4. Oct 14, 2015

### The_Duck

Any values for $A$ and $B$ that satisfy $A^2 + B^2 = 0$ give a solution. For each $k$ there is an infinite family of solutions. You can think of them all as linear combinations of the "basic" solutions $\sin(kx)$ and $\cos(kx)$.

5. Oct 14, 2015

### maverick_76

How is that possible? If k=n, n=1,2,3.... then it has to equal 1/sqrt(pi) right?

A^2 + B^2 = 1/sqrt(pi) is what I mean, how can it equal 0???

6. Oct 15, 2015

### vanhees71

It's much easier to find the energy eigenstates in terms of exponential rather than cos and sin functions. The reason that you have this "ambiguity" is that the energy eigenvalues are degenerate.

The Hamiltonian in position representation is
$$\hat{H}=-\frac{\hbar^2}{2m R^2} \partial_{\varphi}^2.$$
$$\hat{H} u_E(\varphi)=E u_E(\varphi) \; \Rightarrow \; \partial_{\varphi}^2 u_E(\varphi)=-{2m R^2 E}{\hbar^2}u_E(\varphi)=-\alpha^2 u_E(\varphi).$$
The appropriate square integrable solutions on $L^2([0,2 \pi])$ are
$$u_E(\varphi)=A \exp(\mathrm{i} \alpha \varphi).$$
Since the function must be periodic with period $2 \pi$ you have
$$\alpha=\alpha_k:=2 \pi k, \quad k \in \mathbb{Z}.$$
The corresponding energy eigenvalues are
$$E_k=\frac{\hbar^2 \alpha^2}{2m R^2}=\frac{\hbar^2}{2mR^2} (2 \pi k)^2.$$
As you see, except for the ground state for $k=0$ all energy eigenstates are degenerate, because for any $k$ also $-k$ gives the same energy.

From the theory of Fourier series, we know the above set of eigenstates
$$u_k(\varphi)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} \alpha_k \varphi).$$
are complete, and thus this is the complete solution of the energy-eigenvalue problem. It's also easy to verify that the Hamiltonian is really self-adjoint.

7. Oct 15, 2015

### maverick_76

Thank you, I was talking with my professor this morning and he said the same thing that tackling the problem with exponentials is much easier than using trig functions. The problem also is much clearer to understand this way too.