How do you show that the Laplacian is the square of the (Euclidean) Dirac operator?

If I understand correctly, the Euclidean Dirac operator is given by

$$D=\sum_{i=1}^n e_i \frac{\partial}{\partial x_i},$$

where $e_i$ are bases for $Cl_{0,n}(\mathbb{R})$, i.e., the $n$-dimensional Clifford algebra with negative-definite signature over the reals (so $e_i^2=-1$), and $x_i$ are the corresponding coordinates. Several sources state that $D^2 = -\Delta_n$ where $\Delta_n$ is the standard Euclidean Laplace operator

$$\Delta_n = \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$$

When I write out $D^2 f$ explicitly for some function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, scalar terms from the Laplacian certainly appear, e.g.,

$$e_1 \frac{\partial}{\partial x_1}\left( e_1 \frac{\partial}{\partial x_1} f \right) = e_1 \left( e_1 \frac{\partial^2}{\partial x_1^2}f + \left(\frac{\partial}{\partial x_1}e_1\right)\frac{\partial}{\partial x_1}f \right)=e_1^2 \frac{\partial^2}{\partial x_1^2}f = -\frac{\partial^2}{\partial x_1^2}f.$$

But I also end up with bivector cross terms that shouldn't be there:

$$e_1 \frac{\partial}{\partial x_1}\left( e_2 \frac{\partial}{\partial x_2} f \right) = e_1 \left( e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f + \left(\frac{\partial}{\partial x_1}e_2\right)\frac{\partial}{\partial x_2}f \right)=e_1 e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f = e_{12}\frac{\partial^2}{\partial x_1 \partial x_2}f.$$

Should I only be considering the scalar part of $D^2$, or am I simply doing something wrong here?

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Are you perhaps just forgetting that $e_{12} + e_{21} = 0$ and second derivatives commute (for $f$ smooth enough)? – Marek Nov 13 '10 at 21:33
Yes Marek, that is exactly what I'm forgetting. Thanks for pointing out this obvious blunder! – funarharpsichord Nov 13 '10 at 21:40