#### kof9595995

We know in general a Hermitian operator is not guaranteed to have eigenvalues, but self-adjoint operator is(if I remember correctly). Then why we still claim all observables are hermitian instead of claiming them to be self-adjoint?

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#### DrDu

I think real observables corresponding at least remotely to what can be measured have to be bounded so that there is no difference as to whether we talk of them being hermitian and self-adjoint.

#### kof9595995

Err, are you saying that all bounded self-adjoint operators are Hermitian? But last time I checked it's said any self-adjoint operator is Hermitian, but an Hermitian operator is not necessarily self-adjoint, because some subtle issues on the domain of the operator.

Edit: or are you saying all bounded Hermitian operators are self-adjoint? Anyway I wasn't thinking about the boundness, I was thinking about the domain issue.

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#### kof9595995

Also I'd like to know what is physically unacceptable about unbounded operator?

#### cgk

In practice these issues are typically ignored, and calculations are done as if all the states in question could be expanded within a finite basis to any required accuracy (typically true). In that case the question of self-adjointness and hermicity becomes irrelevant. That is, in physics self-adjointness and hermiticity are usually used as interchangable terms.

In principle this interchangable use is wrong of course. There are plenty of observables which which have canonical definitions in terms of non-self-adjoint differential operators (think of momentum, kinetic energy, etc), which obviously have no eigenvalues or eigenvectors in the classical sense (but they do in a distribution sense, although this leads to other mathematical problems.) However, much of the physicists algebra can be transformed into more functional analytic concepts by transforming the relations involving the hermitian operators into other relations involving their resolvent. The latter can often be made compact (and by this self-adjoint) even if the original operator is not (e.g., relations involving differential operators could be transformed into equivalent relations involving integral operators, where the functional analysis is much more friendly). However, from the physical perspective, such transformations just make everything (look) more complicated without granting additional insight into the physical aspects of the problem. So there is little reason to do it.

The best approach really usually is "treat everything as if it could be expanded within a finite basis to any desired accuracy" unless you have very good reason to suspect that this would not work.

#### A. Neumaier

I think real observables corresponding at least remotely to what can be measured have to be bounded so that there is no difference as to whether we talk of them being hermitian and self-adjoint.
Indeed for bounded operators being Hermitian and being self-adjoint is equivalent.

But most observables appearing in the theory of QM or QFT are unbounded - position, momentum, angular momentum, energy, boost generators, (smeared) values of bosonic quantum fields, etc..

With your ''simplification'', all these are deprived of the status of being observable, and only messy ''observables'' without a nice theoretical description are left.

The importance of the distinction of being Hermitian and being self-adjoint lies in the fact that it is needed for those observables that figure in the theoretical equations (i.e., all those mentioned above).

On the importance to distinguish between hermiticity and self-adjointness see
http://arxiv.org/pdf/quant-ph/9907069
for a gentle introduction and counterexamples. An in depth discussion is given in Vol. 1 of the math physics treatise by Reed and Simon, or Vol.3 of the math physics treatise by Thirring. See also the entry ''How much functional analysis is needed to understand quantum mechanics?'' in Chapter A1 of my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html#selfadjoint

#### dextercioby

Homework Helper
We know in general a Hermitian operator is not guaranteed to have eigenvalues, but self-adjoint operator is(if I remember correctly). Then why we still claim all observables are hermitian instead of claiming them to be self-adjoint?
We don't. Only introductory books don't make the difference. Mathematically minded texts use terms as: unbounded, closed, symmetric, essentially self-adjoint and self-adjoint. For example Thirring's text, Teschl's text, Prugovecki's text, Reed & Simon 4 volume text and Galindo & Pascual QM text.

#### Landau

To me, Hermitian and self-adjoint are synonyms. Could you give your definitions? Perhaps one of them means symmetric in the sense that <Ax,y>=<x,Ay> for all x,y in the domain, which need not be the whole Hilbert space? And the other want the domain to be all of the Hilbert space? Or want boundedness?

#### dextercioby

Homework Helper
The definitions are discussed in at least 100 books and certainly appear on wikipedia. Hermitean is a term used by mathematicians only when speaking about finite dimensional spaces endowed with a scalar product. When the dimension of the underlying vector space is infinite, then issues such as boundedness and continuity of endomorphisms occur. Mathematicians will call a linear operator in a Hilbert space 'symmetric' (NOT hermitean) if it's included in its adjoint, while calling it 'self-adjoint' if it's equal to it. The existence of the adjoint is granted by Riesz's theorem, while the uniqueness of it is achieved if the operator is densly-defined (a theorem whose proof I remember seeing in Akhiezer and Glazman's old text on Hilbert spaces, but might be found in Dunford and Schwartz as well, I guess).

So one has bounded operators and unbounded ones. Bounded ones can be extended through continuity on all the vectors in a Hilbert space. So they are defined on all Hilbert space, thus symmetric bounded operators are automatically self-adjoint.

Passing to unbounded operators, we encounter the theorem of Hellinger & Toeplitz which claims that: Let A be a symmetric operator defined everywhere on a Hilbert space. Then A is bounded. Using logics, one sees that an unbounded symmetric operator cannot be defined on a whole Hilbert space, but only on a dense everywhere subset of it.

Both bounded and unbounded cases have their own spectral theorems, discussed in books, and admit generalizations to a rigged Hilbert space where the spectral theorem of Gelfand and Kostyuchenko (or Gelfand and Maurin as others name it) applies.

The preprint article Arnold posted a link to gives examples when quantum mechanics must use the finesse of functional analysis as not to make erroneous claims or reach wrong conclusions.

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#### Landau

The definitions are discussed in at least 100 books and certainly appear on wikipedia.
I am sorry if my question annoyed you. In fact I am familiar with the mathematical terminology; I was under the impression some people in this thread use different terminology in the context of physics. By the way:
Hermitean is a term used by mathematicians only when speaking about finite dimensional spaces endowed with a scalar product.
Like I said, for me hermitian and self-adjoint are synonyms (although I personally prefer self-adjoint as it says what it is). This claim about what "mathematicians use" is not true. I don't want to call myself a mathematician, but you probably won't argue about Paul Halmos.

#### kof9595995

So the dilemma is, self adjoint operators have nice mathematical properties, but in physics we have some observables which can only be Hermitian but not self-adjoint, so it's still lack of a classification that covers all physical observables meanwhile maintains nice mathematical properties, is it?

#### dextercioby

Homework Helper
I am sorry if my question annoyed you. In fact I am familiar with the mathematical terminology; I was under the impression some people in this thread use different terminology in the context of physics. By the way:
Like I said, for me hermitian and self-adjoint are synonyms (although I personally prefer self-adjoint as it says what it is). This claim about what "mathematicians use" is not true. I don't want to call myself a mathematician, but you probably won't argue about Paul Halmos.
It didn't annoy me. And my remark with 100 books is justified, as you picked one of the very many which used a different terminology than the majority. It's a book published in 1951, which is exactly 60 years ago. However, the same decade saw the book of Riesz and Nagy which used the <symmetric> vs <self-adjoint> distinction. So it just might be that Halmos is only an exception...

#### A. Neumaier

So the dilemma is, self adjoint operators have nice mathematical properties, but in physics we have some observables which can only be Hermitian but not self-adjoint, so it's still lack of a classification that covers all physical observables meanwhile maintains nice mathematical properties, is it?
No. Hermitian operators that are not self-adjoint don't qualify as observables in the conventional sense. For example, their spectrum is not real.

#### dextercioby

Homework Helper
So the dilemma is, self adjoint operators have nice mathematical properties, but in physics we have some observables which can only be Hermitian but not self-adjoint
No, not really. Actually, one achieves more. Observable would normally mean 'self-adjoint', however a set of different operators such as position and momentum cannot be self-adjoint on the same domain of L^2(R). They can be self-adjoint on D(x) and D(p), but D(x) is different than D(p). They don't contain exactly the same vectors. The fundamental commutation relation [x,p]=1 holds only on a dense subset of L^2(R) which is neither D(p), nor D(x). It turns out that we actually have 2 different sets of operators: x,p which are self-adjoint on D(x) and D(p), respectively, and x',p' which are not. The first set doesn't obey the commutation relation, the second does. What's the relation between the 2 sets ? Well, x,p are the closures in the norm topology of x',p'. As x,p are self-adjoint, x',p' are esentially self-adjoint, which is the nicest feature one can expect from 2 symmetric operators obeying commutation relations, as the operators are being forced to share a common domain which is imposed to be dense-everywhere in L^2(R).

So one can 'relax' the self-adjointness property to mere essential self-adjointness, in order to be able to make computations with the operators and at the end apply Stone's theorem or the spectral theorem to the closures of the operators one had been using.

QUOTE=kof9595995 said:
, so it's still lack of a classification that covers all physical observables meanwhile maintains nice mathematical properties, is it?
It isn't. E.s.a. operators are in a way equally nice to the s.a. ones, because they almost have the same domain (which is anyway dense everywhere in the Hilbert space) and on the common domain they have the same range.

So I would formulate the postulate of QM as:

<Observables are described by linear self-adjoint operators in the Hilbert space of states.>

One is then forced to work with e.s.a. operators, because they must obey the fundamental commutation relations (and operators such as the Hamiltonian or angular momentum are derived from x,p which are only essentially self-ajoint). Obeying the postulate means that observables are actually the unique self-adjoint extensions of the operators used.

#### Physics Monkey

Homework Helper
Hi kof9595995,

I would say it is important as a matter of mathematical clarity to understand the formal difference between hermitian/symmetric and self-adjoint.

However, any claim that this difference is physically meaningful is almost certainly wrong. We must always be able to account for physical phenomena using a finite number of degrees of freedom, and in such a setting there is no difference. Indeed, how important can an infinite dimensional description be when a sufficiently large but finite quantum system cannot even explore a non-trivial fraction of its finite dimensional Hilbert space in the age of the universe?

Of course, this is not to say that infinite dimensional notions are unimportant. Only that their importance is a matter of personal taste and mathematical convenience. Personally, I find the finite dimensional description much more physical, useful, and elegant, but I know people who think the other way.

Hi kof9595995,

I would say it is important as a matter of mathematical clarity to understand the formal difference between hermitian/symmetric and self-adjoint.
Well, I would disagree. It is not only matter of a mathematical rigor.

In this discussion there was no mention of the problem of finding an actual self adjoint operator. That is, one usualy starts with some symmetric (some authors call it hermitian, even it is not everywhere defined) operator and tries to find its self adjoint extension. There might by many of those, or none. These are then typically given by some particular boundary conditions and might describe different physical situation.

See for example the case of energy of 1d particle confined to a half line (i.e. with an infinite barrier somewhere). There is a 1-parameter family of self-adjoint operators given by the same "differential expression", but with different boundary condition imposed in its domain. This parameter describes the quality of the barrier. In fact it gives the elasticity of the barrier (or reflection coefficient).

For those who are interested i would recommend: http://ajp.aapt.org/resource/1/ajpias/v69/i3/p322_s1?isAuthorized=no [Broken]

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#### dextercioby

Homework Helper
The article you have reccomended fortunately has a free preview on arxiv.org.

#### DrDu

I think it is also merely a matter of nomenclature: John von Neumann, whose "Mathematische Grundlagen der Quantenmechanic" was very influential on the development of early quantum mechanics, does not speak of "self-adjoint" operators but of "hypermaximal hermitean"operators while he uses self-adjoint sometimes instead of what today is called only hermitean.

See:
John von Neumann: selected letters
von John Von Neumann,Miklós Rédei, p. 14

The distinction between hermitean, maximal hermitean and hypermaximal hermitean was often dropped in introductory qm texts.

#### dextercioby

Homework Helper
Of course the terminology has suffered changes along the decades, still the meaning of each term should be crystal clear, irrespective whether we read it in a physics book, or in a mathematics book.

The distinction between hermitean, maximal hermitean and hypermaximal hermitean was often dropped in introductory qm texts.
Yes, and I think that in introductory qm courses it is quite OK to skip this.

I come from a mathematical physics community and the usual terminology over there is:

A is hermitian = bounded and A = A* (where A* in this case is the adjoint of A, which is defined as a bounded operator using the Riesz lemma)

A is symmetric = A is densely defined and A is a restriction of its adjoint A* (which is now defined in a different way than for bounded operators, if A is densely defined and linear everything is ok). This condition is equivalent to the the fact that for any f,g in the domain of A one has <f, A g> = <A f, g>.

A is self-adjoint = A is densely defined and A = A* (i.e. both A and A* has the same domain and Af = A*f for any f from this domain).

Notes:

1) spectrum of a self adjoint operator is a subset of the real line. The spectral theorem (essential for solving the Schrodinger equation and other stuff) holds *only* for self adjoint operators, not symmetric ones.

2) the spectrum of a (closed) symmetric operator is either the entire complex (upper/lower) half plane or a subset of the real line (and in this case it is self adjoint).

3) symmetric + defined everywhere => hermitian (Hellinger-Toeplitz theorem)

4) If one has symmetric operator, than the question whether it has any self-adjoint extension is solved by von Neuman's theory of deficiency indices (or in more modern terms by boundary triples: http://arxiv.org/pdf/math-ph/0611088).

5) If i remember correctly, von Neuman's terminology is as follows:

"hermitian" = symmetric
"maximal hermitian" = symmetric with no self-adjoint extensions (i.e. operators useless for physics, they do not correspond to any observable, classical example is the momentum on the half line (which, to be honest, is quite a puzzle for me :-))).

#### Physics Monkey

Homework Helper
Well, I would disagree. It is not only matter of a mathematical rigor.

In this discussion there was no mention of the problem of finding an actual self adjoint operator. That is, one usualy starts with some symmetric (some authors call it hermitian, even it is not everywhere defined) operator and tries to find its self adjoint extension. There might by many of those, or none. These are then typically given by some particular boundary conditions and might describe different physical situation.

See for example the case of energy of 1d particle confined to a half line (i.e. with an infinite barrier somewhere). There is a 1-parameter family of self-adjoint operators given by the same "differential expression", but with different boundary condition imposed in its domain. This parameter describes the quality of the barrier. In fact it gives the elasticity of the barrier (or reflection coefficient).

For those who are interested i would recommend: http://ajp.aapt.org/resource/1/ajpias/v69/i3/p322_s1?isAuthorized=no [Broken]
I think you're missing my main point. I agree that if you want to use the continuum language, then you may resort to this kind of formal technology at various points.

However, this can all be avoided by regulating the problem: consider a discrete lattice instead of the real line and make the number of lattice points finite to deal with IR issues. There is no further mathematical subtlety. There is only the physics of precisely how you terminate the half line. It is clear that there is now a lot of choice "at the lattice scale" and that the "barrier" can have structure including non-trivial reflection, resonances, etc.

I won't say that no one should use the infinite dimensional language, but I do maintain that it is a matter of convenience and personal preference.

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