The difference between hermitian, symmetric and self adjoint operators.

I am struggling with the concept of hermitian operators, symmetric operators and self adjoint operators. All of the relevant material seems quite self contradictory, and the only notes I have do not quite seem to do the job.

Overall I would like to know if there is a specific chain of implications, I.e

self adjoint $\Rightarrow$ symmetric $\Rightarrow$ Hermitian.

Is there such a chain?

Also when I consider an inner product $\langle u,v\rangle$, I would consider the antilinearity in the second argument, i.e $\langle u,av+bw\rangle= \overline{a}\langle u,v\rangle +\overline{b}\langle u,w\rangle$

Thank you all!

Solution 1:

Let $\mathscr{H}$ be a Hilbert space. The domain of an operator $A$ on $\mathscr{H}$ is denoted $D(A)$; by an extension of $A$ is meant an operator $B$ with $D(A)\subset D(B)$ and $B|D(A)=A$ (where $B|D(A)$ denotes the restriction of $B$ to $D(A)$). If $B$ is an extension of $A$ it is very standard to write $A\subset B$.

Now take $A$ to be densely defined, meaning that the linear subspace $D(A)$ of $\mathscr{H}$ is dense in $\mathscr{H}$. This condition allows defining the adjoint operator $A^\ast$ of $A$; its definition is such that $D(A^\ast)$ is the set of all $y\in\mathscr{H}$ such that the map $D(A)\ni x\mapsto (Ax,y)\in\mathbf{C}$ is continuous (by a theorem of Hahn-Banach this map extends then to $\mathscr{H}$, and does so uniquely since we assumed $A$ to be densely defined). By the Riesz representation theorem for every $y\in D(A^\ast)$ there is a unique element of $\mathscr{H}$, which is denoted $A^\ast y$, such that $(Ax,y)=(x,A^\ast y)$ for all $x\in D(A)$. Thus $A^\ast$ is so defined as to guarantee $(Ax,y)=(x,A^\ast y)$ for all $x\in D(A)$ and $y\in D(A^\ast)$.

$A$ is symmetric (or formally self-adjoint, apparently physicists call them also Hermitian, but no mathematician would do this) if $A\subset A^\ast$; self-adjoint if $A=A^\ast$. Thus every self-adjoint operator is symmetric, but the converse need not hold. However, if $A$ is continuous and $D(A)=\mathscr{H}$ then $A$ symmetric implies $A$ self-adjoint. (In the finite-dimensional case every linear map is continuous.)

Something that is striking in quantum mechanics (theorem of Hellinger-Toeplitz): if $A$ is symmetric and $D(A)=\mathscr{H}$ then $A$ is continuous. Hence if $A$ is not continuous and symmetric, it cannot be defined on the whole of $\mathscr{H}$. (This shows that you cannot discuss quantum-mechanics without worrying about domains, since one can show that if operators $A$ and $B$ satisfy the canonical commutation relation $AB-BA=iI$ then at least one of $A$ and $B$ cannot be continuous). Also a symmetric operator $A$ is self-adjoint iff its spectrum is a subset of the real line (this is important in quantum mechanics as the spectrum is there given a physical interpretation).

There is another notion which is frequently useful, particularly in mathematical physics. $A$ is essentially self-adjoint if $A$ is symmetric and its closure is self-adjoint (such an operator admits a unique self-adjoint extension, namely its closure). This appears e.g. in the standard representation of the canonical commutation relations: $\mathscr{H}=L^2(\mathbf{R})$, $A=-id/dx$ and $B$ be multiplication by $x$. $A$ and $B$ are essentially self-adjoint on the Schwartz space $\mathscr{S}(\mathbf{R})$.

Remark: Even if $A$ is densely defined, $A^\ast$ need not be densely defined (in fact it is densely defined iff $A$ is closable, e.g. if $A$ is symmetric).

If you want a reference, the standard reference is Reed/Simon: methods of modern mathematical physics, volume I (this is perfectly rigorous by the standards of mathematics).

Solution 2:

From the observations I have made from what people have helpfully answered with and from the material I have been given, I believe these definitions hold unity and a chain of implications. Please feel free to comment or adjust any of it:

An operator $T:D(T)\to H$ is hermitian if:

$\langle Tu,v\rangle=\langle u,Tv\rangle$, $\forall u,v\in D(T)$

An operator is symmetric if the above holds and $D(T)\subset D(T^*)$ is Densly defined in $H$.

An operator is self adjoint if both of the above hold, and $D(T)=D(T^*)$.

So clearly there is a hierarchy moving back up the list.