5
$\begingroup$

Consider an unbounded self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$. Let $\mathcal{J} \subset \mathcal{H}$ a closed subspace reducing $A$, i.e. such that $P A \subset A P$ where $P$ denotes orthogonal projection onto $J$. Equivalently, $P \mathcal{D}(A) \subset \mathcal{D}(A)$ and $P A \psi = A P \psi$ for all $\psi \in \mathcal{D}(A)$. Then the restriction $A|\mathcal{J}$ is a densely defined operator on $\mathcal{J}$ with domain $\mathcal{D}(A) \cap \mathcal{J}$.

My question is this: is $A|\mathcal{J}$ again self-adjoint?

The reason I am interested in this question is the following: Take $\mathcal{H} = L^2(\mathbb{R}^{3N})$, and $\mathcal{J} = \Lambda L^2(\mathbb{R}^{3N})$, where the $\Lambda$ denotes the totally antisymmetric subspace, i.e. those functions $\psi(\vec{x_1},...,\vec{x_N})$ with the property that for any permutation $\sigma \in S^N$,

\begin{equation} \psi(\vec{x_{\sigma(1)}},...,\vec{x_{\sigma(N)}}) = sign(\sigma) \psi(\vec{x_1},...,\vec{x_N}) \end{equation}

Then $\mathcal{H}$ is the phase space of an atom consisting of $N$ electrons, with the nucleus fixed at the origin, and $\mathcal{J}$ is the phase space for the same system, but respecting the Pauli principle. My operator on $\mathcal{H}$ is the self-adjoint operator given by

\begin{equation} H^N = - \sum_{j=1}^{N} \Delta_j + \sum_{j = 1}^{N} V_{en}(x_j) + \sum_{i < j} V_{ee}(x_i - x_j) \end{equation}

where the $V_{ee}$ terms denote electron-electron repulsion, and $V_{en}$ electron-nucleus attraction.

In fact, in this case I know of a proof: since the Fourier transform maps antisymmetric functions to antisymmetric functions, one can first show that $H_0 = - \Delta$ is self-adjoint when restricted to $\mathcal{J}$. Then use the fact that the remaining terms are $H_0$-bounded with $H_0$-bound $0$, which remains true for the restriction.

I am aware of the related question at Selfadjoint operators. However, the case I am interested in is very different, in the sense that the restricted operator $A|\mathcal{J}$ is considered an operator on $\mathcal{J}$ rather than on the full Hilbert space $\mathcal{H}$. Indeed, considering $A|\mathcal{J}$ to be an operator on $\mathcal{H}$, it is in general (and certainly in my case) not even densely defined.

$\endgroup$
12
  • $\begingroup$ A symmetric operator $A$ is selfadjoint iff $(A\pm iI)$ are surjective. No restriction of $A$ can have that property. A restriction could be essentially selfadjoint, but only if its closure is $A$. $\endgroup$ Commented Jan 31, 2017 at 18:43
  • $\begingroup$ @TrialAndError Yes, but is this also true if we restrict the operator to the subspace $\mathcal{J}$? I think the statement is true at least in the above example with $-\Delta$. $\endgroup$
    – Max
    Commented Jan 31, 2017 at 22:40
  • $\begingroup$ If $A : \mathcal{D}(A)\subseteq H \rightarrow H$ is selfadjoint, then $A\pm iI$ are injective and surjective. If you restrict $A$ to a subspace $J \subset\mathcal{D}(A)$, then $(A\pm i I)J \ne H$, which means that the restriction of $A$ to $J$ cannot be selfadjoint, though it might be essentially self-adjoint, with a selfadjoint closure. $\endgroup$ Commented Jan 31, 2017 at 22:51
  • $\begingroup$ I see what you mean. However the condition on $\mathcal{J}$ is precisely such that $A$ can be restricted to an operator $A| \mathcal{J}: \mathcal{D}(A) \cap \mathcal{J} \rightarrow \mathcal{J}$. So it might still be true that $(A \pm i I) \mathcal{J} = \mathcal{J}$. Clearly, in general $A | \mathcal{J}$ as an operator on $\mathcal{H}$ wont make sense because it wont even be densely defined. That's why I was referring to the similar question I linked to. $\endgroup$
    – Max
    Commented Feb 1, 2017 at 16:35
  • $\begingroup$ $A\pm iI$ are surjective and injective for a selfadjoint $A$. So $D_1 \ne D_2$ implies $(A+iI)D_1 \ne (A+iI)D_2$. $\endgroup$ Commented Feb 1, 2017 at 16:42

1 Answer 1

4
$\begingroup$

The answer is yes.

Since $A|\mathcal{J}$ certainly remains symmetric, it suffices to show that $(A \pm i I)(\mathcal{D}(A) \cap \mathcal{J}) = \mathcal{J}$. But we know that $A \pm i I : \mathcal{D}(A) \rightarrow \mathcal{H}$ are surjective since $A$ is self-adjoint. Hence for $\psi \in \mathcal{J} \subset \mathcal{H}$ there exist elements $\phi_{\pm} \in \mathcal{D}(A)$ such that $(A \pm i I)\phi_{\pm} = \psi$. Now replacing $\phi_{\pm}$ by $P \phi_{\pm}$ where $P$ is the orthongonal projection to $\mathcal{J}$, we see that $P \phi_{\pm} \in \mathcal{D}(A) \cap \mathcal{J}$ and $(A \pm i I)P \phi_{\pm} = P(A \pm i I) \phi_{\pm} = P \psi = \psi$.

Hence $A | \mathcal{J}: \mathcal{D}(A) \cap \mathcal{J} \rightarrow \mathcal{J}$ is self-adjoint.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .