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!

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    $\begingroup$ by the way: as far as the convention for the antilinearity in the second argument is concerned: does this also come from your physicist lecturer? because physicists usually take it to be antilinear in the first^^ $\endgroup$ – user135041 May 10 '14 at 0:17
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    $\begingroup$ Yes it does! But he decided not to tell us, confused me for ages when I saw his proof for Reisz representation theorem! $\endgroup$ – Ellya May 10 '14 at 5:33

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).

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    $\begingroup$ "No mathematician uses this word?" Seriously? "Hermitian" is standard terminology for a matrix that is equal to its own conjugate-transpose. $\endgroup$ – mweiss May 9 '14 at 22:32
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    $\begingroup$ @Herbert I am sincerely wondering whether you are a troll. If not, I would like to assure you that the term "Hermitian" is in common usage within the mathematical community. $\endgroup$ – Ben Grossmann May 9 '14 at 22:34
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    $\begingroup$ For somebody adamant about everyone else's language usage, you should be more careful about your own. You've opened your answer with "What do you mean by Hermitian? No mathematician uses this word". I do understand that the term "Hermitian" is generally reserved for matrices and not operators. $\endgroup$ – Ben Grossmann May 9 '14 at 22:38
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    $\begingroup$ @ellya: I will write a more careful answer in few minutes. I totally agree with you, I will give you the standard definitions of the terms. $\endgroup$ – user135041 May 9 '14 at 22:49
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    $\begingroup$ @Omnomnomnom: its okay, I didn't realize that what I said about "Hermitian operators" was so polemic, I just have never seen this word in a functional analysis textbook. I have given a reference at the very end. There are also other excellent functional analysis books, such as the one by Kato. By the way don't trust wikipedia that much. $\endgroup$ – user135041 May 9 '14 at 23:41

The way these terms are generally used, there is some sort of chain:

Suppose that $A$ is an operator on Hilbert space $H$, which has inner product $(\cdot,\cdot )$. We say that $A$ is self-adjoint if (and only if) for all $x,y \in H$, we have $$ (Ax,y) = (x,Ay) $$ $A$ is a matrix that, when considered as an operator on $\mathbb{C}^n$ with the standard inner product, is self-adjoint, if and only if it is Hermitian. The definition of Hermitian is that a matrix $A$ is Hermitian if and only if $A = A^*$. That is, $A$ is equal to its conjugate-transpose. Generally, the terms "Hermitian" and "symmetric" are not applied to operators.

If $A$ is a matrix over the same space with real entries, then $A$ is symmetric if and only if it is Hermitian. The definition of symmetric is that $A = A^T$; that is, that $A$ is equal to its own transpose. Note that for real matrices, $A^T = A^*$.

There is a technicality here: sometimes (though rarely in the context of functional analysis), people discuss complex matrices that are symmetric rather than Hermitian. Note that $A^* = \overline{A^T}$, so that if $A$ is a complex matrix, it will not generally be the case that $A^* = A^T$, so that it is possible to have $A = A^T \neq A^*$.

  • $\begingroup$ I really wanted to consider things in terms of operators rather than matrices (though the analogies are nice to get your head around concepts), I'm really looking for a solid definition of each type of operator in the original post, and if possible a chain of implications :) $\endgroup$ – Ellya May 9 '14 at 22:39
  • $\begingroup$ @ellya I have yet to see the terms "Hermitian" and "symmetric" applied to operators rather than matrices outside of your question. Are you sure that's how they are being used? $\endgroup$ – Ben Grossmann May 9 '14 at 22:41
  • $\begingroup$ en.wikipedia.org/wiki/Self-adjoint_operator#Symmetric_operators $\endgroup$ – Ellya May 9 '14 at 22:44
  • $\begingroup$ @ellya it seems that in this context, a "symmetric operator" is an operator $A$ that satisfies $(Ax,y) = (x,Ay)$ over its domain. If $A$ is defined over the entire space, then $A$ is a self-adjoint operator. So, it seems that self-adjoint $\implies$ symmetric here. $\endgroup$ – Ben Grossmann May 9 '14 at 22:47
  • $\begingroup$ @ellya from this definition, it seems that Hermitian $\iff$ self-adjoint. $\endgroup$ – Ben Grossmann May 9 '14 at 22:50

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.

  • $\begingroup$ @Herbert I have posted my own answer, see what you think.. $\endgroup$ – Ellya May 9 '14 at 23:17
  • $\begingroup$ as for the hermitian case I cannot say, in my understanding physicists use it as meaning the same as symmetric. (basically physicists don't care about domains at all.) but what you say is correct, apart from symmetric: you need to assume in addition that $D(T)\subset D(T^\ast)$ $\endgroup$ – user135041 May 9 '14 at 23:29
  • $\begingroup$ @Herbert yes thank you, I will edit :) it does seem that the whole confusion stems from a physicists lack of rigor! I shall stick with this chain of defenitions for now as my lecturer is a physicist, but I thank you for your clarity on the matter :) $\endgroup$ – Ellya May 9 '14 at 23:33
  • $\begingroup$ in physics perhaps it is better not to know the difference. apparently heisenberg once said, when asked about symmetric and self-adjoint operators, "what's the difference?" $\endgroup$ – user135041 May 9 '14 at 23:46

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