Adjoint of the Quantum Momentum Operator

Question

I'm studying quantum mechanics and I have a question about the momentum operator. We have that the momentum operator is given by

\begin{equation*} p = -i\hbar\nabla \end{equation*}

and so its adjoint is given by

\begin{equation*} p^{\dagger} = i\hbar\nabla^{\dagger}. \end{equation*}

The momentum operator is also self-adjoint, so

\begin{equation*} -i\hbar\nabla = i\hbar\nabla^{\dagger} \quad\Leftrightarrow\quad \nabla^{\dagger} = -\nabla \end{equation*}

in some sense. However, I haven't run across any explanation of in what sense such a relation might hold, or if I'm just completely incorrect. I appreciate any guidance on this!

Edit: My thought process is to examine $\nabla$ in a weak sense in Hilbert space, keeping in mind the boundary conditions imposed on wave functions, but I haven't made any useful conclusions.

Edit: The domain on which I mean the momentum operator to act is the set of quantum states.

Answer 1

Consider a vector space $V$ with an inner product $\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}.$ Given an operator $A:V\rightarrow V$, the adjoint is defined as the unique$^1$ operator satisfying $$\langle A^\dagger\phi,\psi\rangle:=\langle\phi, A\psi\rangle\tag{1}\label{1} \quad\forall\phi,\psi\in V.$$

In OP's case the vector space is the Hilbert space $L^2(\mathbb{R}^3)$ with the inner product $$\langle\phi,\psi\rangle:=\int_{\mathbb{R}^3}\phi^\ast(\vec{r})\psi(\vec{r})d^3\vec{r}\tag{2}\label{2}$$ Let us now rewrite the RHS of \eqref{1} with $A=\nabla$ and see if we can recast it in a form analogous to the RHS. $$\langle\phi, \nabla\psi\rangle=\int_{\mathbb{R}^3}\phi^\ast(\vec{r})\nabla\psi(\vec{r})d^3\vec{r}=\underbrace{\phi^\ast(\vec{r})\psi(\vec{r})d^3\vec{r}\bigg\lvert_{\partial\mathbb{R}^3}}_{=0}-\int_{\mathbb{R}^3}[\nabla\phi^\ast(\vec{r})]\psi(\vec{r})d^3\vec{r}\tag{3}\label{3}:=\langle (-\nabla)\phi\lvert \psi\rangle.$$

Where in the first step we have used intergration by parts together with the requirement that the wavefunctions vanish at infinity. Comparing \eqref{1} and \eqref{3}, we conclude that $$\nabla^\dagger=-\nabla\tag{4}\label{4}$$ in $L^2(\mathbb{R}^3)$.

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$^1$ _{The situation is more delicate for infinite dimensional spaces, in which we should be interested in the present case as $L^2(\mathbb{R}^3)$ is infinite-dimensional and more should be said about the domain of the adjoint operator. In other words, we're not being rigorous here. The definition I gave works fine in the finite dimensional case, though.}