# 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 $\nabla$.

• That's a pretty good idea! The thing is that the gradient is anti-self-adjoint, that is, $\nabla^\dagger=-\nabla$, so your proposed operator is just minus the laplacian. – Javier Sep 27 '16 at 2:28
• This could be of interest to you...see Bochner Laplacian en.wikipedia.org/wiki/… – ClassicStyle Sep 27 '16 at 4:04
• @TylerHG and on that Wikipedia article you even hear that "the connection Laplacian and Bochner Laplacian differ only by a sign", which is in agreement with what Javier was saying – QuantumFool Sep 27 '16 at 4:28
• Would Mathematics be a better home for this question? – Qmechanic Sep 27 '16 at 6:59
• On smooth rapidly decreasing functions and sticking to the one-dimensional case, we have $\langle (-\nabla) \psi| \phi \rangle = \langle \psi| \nabla \phi \rangle$ just using integration by parts. This identity says that $\nabla^\dagger = - \nabla$ (dealing with the afore-mentioned class of functions). Therefore $-\Delta = -\nabla \nabla = \nabla^\dagger \nabla$. – Valter Moretti Sep 27 '16 at 7:27

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 $\Delta f = \mathrm{d}^\dagger\mathrm{d} f$.

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).

• Yep! Alternately, the fact that $\hat{p}=-i\hbar\nabla$ is self adjoint implies $-i\hbar\nabla=i\hbar\nabla^\dagger$, or $\nabla=-\nabla^\dagger$. (Really the implication goes the other way, I think, but it shows consistency at least) – user12029 Sep 27 '16 at 4:40
• @NeuroFuzzy I actually think that could suffice for a proof, if you use properties of physical reality as mathematical fact. We know measured values of momentum are real, therefore $\hat{p}$ is Hermitian, and the proof follows. Nice insight! – QuantumFool Sep 27 '16 at 5:18

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.