Can the Laplacian be thought of as $ \nabla^{\dagger} \nabla $? I'm trying to read Peskin and Schroder's book on Quantum Field Theory. I was trying to justify transitioning from one line to the next by using the self-adjoint property of the Laplacian, $ \Delta = \nabla \cdot \nabla $, which I had proved by using an un-insightful integration by parts trick. This got me thinking:


*

*The Laplacian is self-adjoint: $ \Delta^{\dagger} = \Delta $.

*The Laplacian is an operator dotted with itself.

*Is there a sense in which we can write $ \Delta = \nabla^{\dagger} \nabla $?


By this, I mean that each element of the vector-operator $ \nabla^{\dagger} $ is the dagger of the element of the vector-operator $ \nabla $. This would make the self-adjointess manifest: $ \left ( \nabla^{\dagger} \nabla \right )^{\dagger} = \left ( \nabla \right )^{\dagger} \left ( \nabla^{\dagger} \right )^{\dagger} = \nabla^{\dagger} \nabla $.
Is there a way to formalize this idea?  
My first thought: Perhaps if we thought of $ \left | \psi \right \rangle $ as a vector of infinitely many components, and each element of the vector-operator $ \nabla $ as a matrix acting on the input $ \left | \psi \right \rangle $, then each element of $ \nabla^{\dagger} $ could be thought of as the adjoint of the matrix in the corresponding element of $ \nabla $. 
 A: This is somewhat well-known as the generalization of the Laplacian as the Laplace-deRham operator. Given the exterior differential $\mathrm{d}$ (your $\nabla$, but acting to produce forms, not vectors) and its adjoint, the codifferential $\delta$ or $\mathrm{d}^\dagger$, the general Laplacian acting on a form is
$$ \Delta = \mathrm{d}^\dagger \mathrm{d} + \mathrm{d}\mathrm{d}^\dagger$$
and when acting on an ordinary function the latter term vanishes, meaning $\Delta f = \mathrm{d}^\dagger\mathrm{d} f$.
A: Okay, I think my idea works with the negative sign proposed by Javier.
Let's work in one dimension, so the gradient operator is really just a one-element ket containing the derivative operator. Let's write:
\begin{align}
\left | \psi \right \rangle & = \begin{pmatrix} ... \\ \psi \left ( -2 \epsilon \right ) \\ \psi \left ( - \epsilon \right ) \\ \psi \left ( 0 \right ) \\ \psi \left ( \epsilon \right ) \\ \psi \left ( 2 \epsilon \right ) \\ ... 
\end{pmatrix}
\end{align}
Where $ \epsilon $ is very small. We can now write:
\begin{align}
\partial_x & = \frac{1}{2 \epsilon} \begin{pmatrix} ... & ... & ... & ... & ... & ... & ... \\
... & 0 & 1 & 0 & 0 & 0 & ... \\ ... & -1 & 0 & 1 & 0 & 0 & ... \\ ... & 0 & -1 & 0 & 1 & 0 & ... \\ ... & 0 & 0 & -1 & 0 & 1 & ... \\ ... & 0 & 0 & 0 & -1 & 0 & ... \\ ... & ... & ... & ... & ... & ... & ... \end{pmatrix} \\
\partial_x^{\dagger} & = \frac{1}{2 \epsilon} \begin{pmatrix} ... & ... & ... & ... & ... & ... & ... \\
... & 0 & -1 & 0 & 0 & 0 & ... \\ ... & 1 & 0 & -1 & 0 & 0 & ... \\ ... & 0 & 1 & 0 & -1 & 0 & ... \\ ... & 0 & 0 & 1 & 0 & -1 & ... \\ ... & 0 & 0 & 0 & 1 & 0 & ... \\ ... & ... & ... & ... & ... & ... & ... \end{pmatrix} \\
\Delta_{\text{proposed}} & = - \partial_x^{\dagger} \partial_x \\
& = \frac{1}{4 \epsilon^2} \begin{pmatrix} ... & ... & ... & ... & ... & ... & ... \\ ... & -2 & 0 & 1 & 0 & 0 & ... \\ ... & 0 & -2 & 0 & 1 & 0 & ... \\ ... & 1 & 0 & -2 & 0 & 1 & ... \\ ... & 0 & 1 & 0 & -2 & 0 & ... \\ ... & 0 & 0 & 1 & 0 & -2 & ... \\ ... & ... & ... & ... & ... & ... & ... \end{pmatrix}
\end{align}
Which is indeed the second derivative limit formula (this exact matrix product doesn't work out near the boundary, hence the $ ... $'s).
In more than one dimension, we can treat it as a sum of one-dimensional derivatives, and for each such one-dimensional derivative we can pretend we are working in the above case. Therefore, I think we do have $ \Delta = - \nabla^{\dagger} \nabla $ in a fairly formal sense.
P.S. I guess it also checks out that $ - \Delta $ is a positive definite operator, while eigenvalues of most operators in quantum mechanics of the form $ A^{\dagger} A $ can be thought of as positive real quantities (the number operator, for instance).
A: A third idea (the first and second being integration by parts and infinite matrices) is inspired by NeuroFuzzy's comment to my first answer (with matrices). Again in one dimension, we have $ \hat{p} = -i \hbar \nabla $, where $ \hat{p} $ is Hermitian because momentum must be real (in other words, $ \hat{p} $ is an observable). This tells us that:
\begin{align}
\left ( \hat{p} \right )^{\dagger} = i \hbar \nabla^{\dagger} & = \hat{p} = -i \hbar \nabla \\
\nabla^{\dagger} & = - \nabla
\end{align}
From which it follows that (again in one dimension) $ \Delta = \nabla^2 = - \nabla^{\dagger} \nabla $, again with the negative sign suggested by Javier. The extension to higher dimensions is then natural.
