Hamilton Operator for particle on a circle -- Matrix representation....

[JonnyMaddox]

Hey JO.

The Hamiltonian is:
H= \frac{p_{x}^{2}+p_{y}^{2}}{2m}

In quantum Mechanics:
\hat{H}=-\frac{\hbar^{2}}{2m}(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial x^{2}})

In polar coordinates:
\hat{H}=-\frac{\hbar^{2}}{2m}( \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \phi^{2}})

Now I want to write this operator in matrix form. What is an appropriate basis? I thought a good would be {sin(x),sin(2x),sin(3x),...,sin(nx)} Now how do I do that in a two dimensional space? What is the basis for that? Something like {sin(x),sin(y),...sin(nx),sin(ny)} ?

 

[ShayanJ]

Your Hamiltonians are wrong. The first two Hamiltonians are for a free particle in two dimensions, not for a particle on a circle. Anyway, its not reasonable to do this in Cartesian coordinates.
Going to polar coordinates, and noticing that there is only one degree of freedom(the azimuthal angle), the Hamiltonian is $\hat H=-\frac{\hbar^2}{2mR^2}\frac{\partial^2}{\partial \phi^2}$.

Now consider the set of basis vectors $\{|\psi_n\rangle\}$. The matrix elements of an operator $\hat S$ w.r.t. this set of basis vectors is defined as $S_{mn}=\langle \psi_m|\hat S|\psi_n\rangle$. The basis vectors don't have to be related to the operator but a convenient basis for this job is a basis in which the matrix $S_{mn}$ is diagonal. So the set of eigenvectors of the operator $\hat S$($\hat S|\psi_m\rangle=\lambda_m|\psi_m\rangle$) is a convenient basis.

When the operator is a differential operator and the vectors are functions, the matrix elements can be calculated using an integral $S_{mn}=\int_{\zeta_1}^{\zeta_2} \phi_m(\zeta) \hat S \phi_n(\zeta) d\zeta$. For the above Hamiltonian, $\zeta$ is the azimuthal angle $\phi$ and the range of integration is 0 to 2π and the eigenfunctions are exponential functions with imaginary exponents which can also be written as the sum of sines and cosines if you want. Note that there are an infinite number of eigenfunctions and so the matrix $H_{mn}$ will be infinite dimensional.