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.

  • $\begingroup$ If you want to speak about self-adjointness, you need to specify domains, to start with... For example, on the Schwartz space $\mathscr S \subsetneq L^2(\mathbb R^3)$ the three momentum operators $P_k$ with $k=1,2,3$ (or $x,y,z$) are symmetric, but not self-adjoint; however, they are essentially self-adjoint. $\endgroup$ Mar 18 at 10:10
  • $\begingroup$ Regarding your edit: No, this is not true. The momentum operator cannot be defined on the whole Hilbert space... A self-adjoint everywhere defined operator is necessarily bounded (Hellinger-Toeplitz); but one can prove that the momentum operator is unbounded. $\endgroup$ Mar 18 at 10:13
  • $\begingroup$ @TobiasFünke You're correct of course, and more often than not, I guess the domain is $L^{2}(\mathbb{R}^{3})$, but the Schwartz space on $\mathbb{R}^{n}$ is a subspace of $L^{p}(\mathbb{R}^{n})$ for $1\leq p \leq \infty$. $\endgroup$
    – kandb
    Mar 18 at 10:15
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    $\begingroup$ Let me try to rephrase everything a bit: We work with the Hilbert space $L^2(\mathbb R^3)$. A linear operator $A$ is, in general, only defined on a (dense) subspace $D(A) \subset L^2(\mathbb R^3)$. An operator is self-adjoint if $A=A^*$ and $D(A)=D(A^*)$, where $A^*$ denotes the adjoint of $A$. So every function in the domain of $A$ is in $L^2$ automatically, but the converse is not true in general. Yet, for bounded operators $A$, we can make use of the BLT theorem and extend $A$ uniquely to $L^2$. Not all operators are bounded. $\endgroup$ Mar 18 at 10:29
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    $\begingroup$ @TobiasFünke You do get my question, and you answered it thoroughly and precisely. If you had written an answer, I would've chosen it. tschüß! (I also appreciate the Arrested Development reference!) $\endgroup$
    – kandb
    Mar 18 at 10:45

1 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 $L^2(\mathbb{R}^3)$.

$^1$ 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.

  • $\begingroup$ Yes, of course! Thank you for your clear explanation! My thought process was correct, I just didn't reach the right answer somehow. The key is that wave functions vanish on the boundary. Somehow, I wasn't assuming that the test function in a weak sense vanished on the boundary, only the wave function. $\endgroup$
    – kandb
    Mar 18 at 10:02
  • $\begingroup$ The argument that general wave functions vanish at infinity is not correct. There are easy counter examples of (equivalence classes of) $L^2$ function which do not vanish at infinity. $\endgroup$ Mar 18 at 10:06
  • $\begingroup$ @TobiasFünke Interesting. I suppose that for situations of physical interest it's about normalizable wavefunctions. Are we really considering a subspace of $L^2$ for physical applications? $\endgroup$ Mar 18 at 10:33
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    $\begingroup$ Yes, for example in 1D, by construction of the domain of the momentum operator, all function in this said domain vanish at infinity, see e.g. this thread. Regarding the claim from my previous comment, see e.g. this. $\endgroup$ Mar 18 at 10:35

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