How to derive the free-particle propagator in 2D?

Question

I am trying to derive the following expression of the free-particle propagator in 2D, given by

$$ \rho_0(\mathbf{r},\mathbf{r'}, \beta) = \langle \mathbf{r'} \rvert e^{-\beta \hat{H}} \lvert \mathbf{r} \rangle = \frac{1}{4\pi \beta}e^{-\frac{(\mathbf{r}-\mathbf{r'})^2}{4\beta}}$$ where $\hat{H} = -\nabla^2_\mathbf{r}$ is the free-particle Hamiltonian in 2D, and where we assumed $\frac{\hbar^2}{2m} = 1$.

In a first attempt, I tried assuming rotational invariance of the wavefunction $\psi(r, \theta) = \frac{1}{\sqrt{2\pi}}R(r)$ (thus only considering s-wave solutions), which gives solutions of the Schroedinger equation in the form of Bessel functions of zero-th order. Unfortunately, I cannot reconcile these results with the equation above.

Answer 1

You have also taken $\beta= it/\hbar$. The two Cartesian parts of the problem do not communicate, so they factor. Why, on earth, would you be going to polar coordinates?

Your answer is a trivial 2D Gaussian Fourier transform, so the mere product of two completely disconnected 1D propagators! $$ \rho_0(\mathbf{r},\mathbf{r'}, \beta) = \langle \mathbf{r'} \rvert e^{-\beta \hat{H}} \lvert \mathbf{r} \rangle \\ = \langle \mathbf{r'} \rvert e^{-\beta \hat{H}} \int\!\! d\mathbf{k}~ | \mathbf{k}\rangle \langle \mathbf{k}| \mathbf{r} \rangle = \int\!\! d\mathbf{k}~\frac{ e^{i\mathbf{k}\cdot (\mathbf{r'}-\mathbf{r})}}{4\pi^2} e^{-\beta \mathbf{k}^2}\\ = \frac{1}{4\pi \beta}e^{-(\mathbf{r}-\mathbf{r'})^2 /4\beta} \qquad \therefore $$