Gradient and Laplacian in $S^1$ I'm trying to solve the particle in a ring problem without embedding the circle in $\Bbb R^3$, by instead taking the entire space to be $S^1$.  Unfortunately, I haven't taken differential geometry yet and am unsure of how the gradient (or really the laplacian) is found in this space.  I assume it should still come out to being $\nabla^2 = \frac 1{r^2} \frac {\partial^2}{\partial \theta^2}$, but I don't how to derive that.  Is this something simple enough that you guys could explain it in a post, or do I just need to study differential geometry before attempting to solve a problem like this?

Edit:
What I've got so far is $$\frac {\partial^2 \psi(x)}{\partial x^2}=\frac{-2mE}{\hbar^2}\psi(x)$$ where $\psi(x)=\psi(x+2\pi Rn),\ n \in \Bbb Z$ and $\int_{-\infty}^\infty |\psi(x)|^2 dx = 1$.
$$\implies \psi(x)=Ae^{ikx}+Be^{-ikx}=Ae^{ik(x+2\pi Rn)}+Be^{-ik(x+2\pi Rn)}$$
$$\psi'(x)=ikAe^{ikx}-ikBe^{-ikx}=ikAe^{ik(x+2\pi Rn)}-ikBe^{-ik(x+2\pi Rn)}$$
$$\psi''(x)=-k^2Ae^{ikx}-k^2Be^{-ikx}=-k^2Ae^{ik(x+2\pi Rn)}-k^2Be^{-ik(x+2\pi Rn)}$$
$$\implies -k^2(Ae^{ikx}+Be^{-ikx})=\frac{-2mE}{\hbar^2}(Ae^{ikx}+Be^{-ikx})$$
$$\implies k=\sqrt{\frac {2mE}{\hbar^2}}$$
but I don't know how to find $A$ and $B$.

Edit 2:
@CameronWilliams, is the periodic boundary condition, $\psi(x)=\psi(x+2\pi Rn),\ n \in \Bbb Z$, not enough to specify $A$ and $B$ then?  Because I haven't really found a way to use that condition, yet.
 A: This answer addresses the issue of the Laplacian on $S^1$ and not the issue of whether you are solving the differential equation correctly.
Circles are completely classified by their radius $r$, or if you prefer, by their circumference $C$, where of course $C = 2 \pi r$. What I mean by this is that whether or not your circle is originally given to you as being embedded in $\mathbb{R}^2$ in the usual way, you can always view it as being embedded in $\mathbb{R}^2$, in the sense that there is an isometry (a length-preserving bijective smooth map) between the original circle and the usual one in $\mathbb{R}^2$. For your purposes, I would guess that the distinction between the original circle and the one in $\mathbb{R}^2$ is not too important.
This means that your circle can always be parametrized by the coordinate $\theta$, where $0 \leq \theta < 2\pi$. If you are familiar with tangent vectors in an abstract setting, the tangent vector $\frac{\partial}{\partial \theta}$ has constant length $r$ (which, remember, is a fixed constant). You can also parametrize the circle with respect to arc length, say, using the coordinate $t$, where $0 \leq t < 2\pi r$, and the tangent vector $\frac{\partial}{\partial t}$ has constant length $1$.
The Laplacian in $\mathbb{R}^n$ is $\Delta u = \sum_{k=1}^n \frac{\partial^2 u}{\partial x_k^2}$. This formula relies on the fact that the coordinates $x_1, \dots, x_n$ are the usual Euclidean coordinates, or, in other words, that the tangent vectors $\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}$ are orthonormal. On a more general Riemannian manifold, the key thing to remember is that the same formula for $\Delta$ applies, but only as long as one is working in a coordinate system that is orthonormal (to second order, meaning the first derivatives of the metric tensor vanish).  In a general coordinate system, the Laplacian is given by
$$ \Delta u = \sum_{j,k} \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x_j} \left( \sqrt{|g|}   g^{jk} \frac{\partial u}{\partial x_k}  \right), $$
where $|g|$ is the determinant of the metric tensor $(g_{jk})$ and the $g^{jk}$'s are the components of the inverse of the metric tensor. If the coordinate system is orthonormal to second order at a point, then this formula (at that point) reduces to the familiar formula from $\mathbb{R}^n$. (Note it is not generally possible to choose such a coordinate system globally, or even in a small open set, but we can always choose one so that those properties hold at a single point at the center of the coordinate system.)
You can read more about the Laplacian on Riemannian manifolds on Wikipedia, for example. That article also contains links to articles about the gradient and divergence, including discussion of those operators on manifolds. I suppose this may not be too useful if you don't know any differential geometry, but you still might want to take a look.
Now, back to the circle: Recall the two coordinates $\theta$ and $t$ that we could use to parametrize the circle. Since the circle is one-dimensional, there is no sum on $j$ and $k$ in the formula for $\Delta$. If we use $t$ (an orthonormal coordinate system), $g_{tt} = g^{tt} = |g| = 1$, so we obtain simply $\Delta u = \frac{\partial^2 u}{\partial t^2}$. If we use $\theta$ (not orthonormal but still quite simple), $g_{\theta \theta} = |g|= r^2$ and $g^{\theta \theta} = \frac{1}{r^2}$ (remember, $r$ is a constant), so we obtain $\Delta u = \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}$, as you claimed.
To summarize, you have to divide by $r^2$ in the $\theta$ formula for $\Delta$ to account for the fact that $\frac{\partial}{\partial \theta}$ has length $r$, and that you are taking two derivatives with respect to $\theta$. This correction is not necessary when you use the coordinate $t$, since $\frac{\partial}{\partial t}$ has length $1$.
