2
$\begingroup$

I am a physicist and I am trying to get a grasp on the following terms from functional analysis:

  1. As I understand, an operator is Hermitian if it is symmetric and bounded (domains of A and A* don't need to be equal in this case.)

  2. An operator is selfadjoint if it is symmetric and the domains of A and A* are equal, D(A) = D(A*), so A = A*.

My question is, is the boundedness here an artefact of the finiteness of the Hilbert space? I.e., in the finite Hilbert spaces we can 'safely' work with the Hermitian operators, but in the infinite Hilbert spaces we lose the notion of boundedness, so we need to start to work with self-adjoint operators?

Or, if I am completely wrong here, what is the difference between the self-adjoint and Hermitian operator, in the context of finite vs. infinite Hilbert space?

$\endgroup$
2
$\begingroup$

I am not a fan of the term 'hermitian', because it is so often used haphazardly (sometimes it means symmetric, sometimes it means self-adjoint, sometimes it presupposes that the vector space is finite-dimensional). The term 'self-adjoint' seems to be used much more consistently, to mean a densely defined operator which satisfies $A=A^*$ (equivalently $A$ is symmetric and $\mathcal D(A)=\mathcal D(A^*)$). Basically, what you call a hermitian operator is a bounded self-adjoint operator. The point is that a self-adjoint operator is always closed, and the domain of a densely defined closed operator which is bounded must the entire space, so you get the domain assumption for free in the bounded case.

Now, why do we care that the Hamiltonian is self-adjoint and not just symmetric? One explanation goes as follows: When the Hamiltonian is self-adjoint, the spectral calculus allows us to define the strongly continuous one-parameter group of unitaries $\mathcal U(t)=e^{-itH}$. Omitting some details, this group allows us, for any wave function $\psi_0$, to construct a solution to the Cauchy problem for the Schrödinger equation, namely $\psi(t)=e^{-itH}\psi_0$. On the other hand, given a strongly continuous one-parameter group of unitaries $\mathcal U(t)$, there is a self-adjoint operator $H$ such that $\mathcal U(t)=e^{-itH}$ (this is Stone's theorem) and $\psi(t)=\mathcal U(t)\psi_0$ is a solution to the corresponding Cauchy problem for the Schrödinger equation with this $H$. Thus, if you want to have a reasonable time-evolution (in the sense above), and you want $\psi(t)$ to solve the Schrödinger equation, the Hamiltonian must be self-adjoint.

Edit: Perhaps more on topic, in the finite dimensional setting, all linear operators are bounded, so the (subtle) domain problems disappear completely (i.e. symmetric is equivalent to self-adjoint)!

$\endgroup$
2
$\begingroup$

In the finite case every operator is bounded and since the domain is the whole Hilbert space, every hermitian operator is self-adjoint. For the infinite case, the good definition of a hermitian operator is a densely defined operator $T$ such that $T\subset T^*$, i.e., it is symmetric and the adjoint $T^*$ is an extension, we have $\mathcal{D}(T)\in\mathcal{D}(T^*)$. The use of self-adjoint and not just hermitian operators in physics is a subtle one, but it has reasons. One is the fact that we want operators with real spectrum, and that's true in general only for self-adjoint, not for hermitian operators. So what we have in practice is a hermitian operator $T$ defined in some domain in the Hilbert space, and we want to find an extension of that domain such that $T$ becomes self-adjoint. This problem may have infinite answers, just one or none. This depends on the von Neumman deficiency index theory. For the case where it's has only one self-adjoint extention, we call the operator essentially self-adjoint and the extension is simply the closure of $T$. For the case of many different extentions, we must look to the physical properties of the system to chose the one that suits best.

$\endgroup$
2
$\begingroup$

The commutation relation $[A,B]=iI$ (i.e., $AB-BA=iI$) is not satisfied by two linear operators on any finite-dimensional linear space. So Quantum Mechanics requires infinite-dimensional spaces. Even in infinite-dimensional spaces, the commutation relation cannot be satisfied by two bounded operators $A$ and $B$; at least one of them must be discontinuous. The basic problem is that a differentiation-like property such as $\frac{d}{dx}x^{n}=nx^{n-1}$ causes the norm to increase without bound. Quantum Mechanics requires infinite-dimensional inner-product spaces. It is matter of convenience that the inner-product spaces are assumed to be complete (Hilbert) spaces.

John von Neumann was probably the first to recognize that the operators of Quantum Mechanics would have closed graphs in $\mathcal{H}\times\mathcal{H}$ because they would be symmetric $$ (Ax,y) = (x,Ay),\;\;\; x,y\in\mathcal{D}(A). $$ Symmetry is required in order to give real values for observation; in fact, symmetry in this setting is equivalent to the requirement that $(Ax,x)$ is real. That imposes an interesting restriction on $\mathcal{D}(A)$ because of the Closed Graph Theorem, which states that every linear operator with a closed graph that is defined everywhere on a Hilbert (or Banach) space is automatically bounded. Hence, some of the operators of Quantum are not going to be defined on the entire Hilbert space, because some of them cannot be bounded. But it is generally assumed that the domain $\mathcal{D}(A)$ of an observable $A$ is dense, which is good as you can do for the operators which must be unbounded.

John von Neumann studied adjoint for a densely-defined closed linear operator $A$ on a Hilbert space $\mathcal{H}$ by looking at the orthogonal complement of the graph $\mathcal{G}(A)$, which was a natural fit because of how one could interpret $(Ax,y)=(x,A^{\star}y)$ as an inner-product on the graph space in $\mathcal{H}\times\mathcal{H}$: $$ \left( \langle x,Ax\rangle, \langle A^{\star}y,-y \rangle \right)_{\mathcal{H}\times\mathcal{H}} = 0. $$ The above is an inner-product on $\mathcal{H}\times\mathcal{H}$ of the ordered pairs (not inner-product) $\langle x,Ax\rangle$ and $\langle A^{\star}y,-y\rangle$. So he used orthogonal complement in the Hilbert space $\mathcal{H}\times\mathcal{H}$ to prove the existence of an adjoint $A^{\star}$ for a densely-defined $A$ with a closed graph $\mathcal{G}(A)\subseteq\mathcal{H}\times\mathcal{H}$. A closed densely-defined linear operator $A$ has a closed densely-defined adjoint $A^{\star}$, but the domains may not be the same at all. The additional requirement that $A$ be symmetric forces the adjoint $A^{\star}$ to an extension of $A$, which means that $\mathcal{D}(A)\subseteq\mathcal{D}(A^{\star})$ and $A^{\star}=A$ on $\mathcal{D}(A)$; this is written as $A\preceq A^{\star}$. Hermitian for you is "symmetric" or, equivalently, $A\preceq A^{\star}$. Selfadjoint is $A=A^{\star}$.

Because the graph of $A$ and the graph of $A^{\star}$ are related through orthogonal complement in $\mathcal{H}\times\mathcal{H}$, enlarging the domain of a symmetric $A$ shrinks the domain of $A^{\star}$ and can lead to a situation where one obtains $A\preceq A_{0}=A_{0}^{\star}\preceq A^{\star}$. von Neumann characterized the symmetric operators with selfadjoint extensions, and he described all possible selfadjoint extensions of a symmetric operator $A$, which are restrictions of the adjoint $A^{\star}$. This has to do with abstract "boundary conditions". Quantum Mechanics formulations require selfadjoint observables; symmetric is not enough.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.