# Hermitian operator <=> observable?

1. Oct 3, 2012

### Arijun

My question is about both sides of the same coin.
First, does a hermitian operator always represent a measurable quantity? Meaning, (or conversely) could you cook up an operator which was hermitian but had no physical significance?
Second, are all observables always represented by a hermitian operator? e.g. (again, conversely) does there exist some operator which is not hermitian but has real eigenvalues and therefore(?) represent an observable?
It's my first day back in school and I pulled an all nighter last night so forgive me if I have had any (possibly serious) brain farts.

2. Oct 3, 2012

### dextercioby

Well, what is <physical significance> ? Does the Hamiltonian classical observable xp_x have a physical significance ?

In the absence of superselection rules, any self-adjoint operator in a complex separable Hilbert space can describe a quantum observable. There are no other types of operators which can describe the observables in the mathematical formalism. In the presence of superselection rules, one is confined to a coherent subspace thus restricting the action of the operators.

3. Oct 11, 2012

### Arijun

Thanks for the reply!

By "have physical significance" I simply meant "correspond to values that could be calculated from measurable quantities". I gather from your answer that all hermitian operators fall under that category.

Quick question: when you say all observables are represented by hermitian operators, does that mean that any non-hermitian operator with real eigenvalues simply has no significance, or that no such operator exists?

4. Oct 12, 2012

### DrDu

There is an active field of research on hamiltonians which are PT-symmetric (i.e. symmetric under joint inversion of space and time) instead of being hermitian. They also have, at least in certain situations, real eigenvectors.

5. Oct 12, 2012

### tom.stoer

There's a generalization, so-called positive-operator valued measures: http://en.wikipedia.org/wiki/POVM

The question is whether you mean "obervable in principle" or "observable in practice". I don't think that there is any procedure to derive a construction of an apparatus related to a hermitean operator.

6. Oct 13, 2012

### Killtech

are you sure about this? i am most interested to read about such a procedure! is there at least a procedure purely within the theory to derive the operator a measurement apparatus corresponds to?

7. Oct 13, 2012

### tom.stoer

as I said, the problem how to describe a measurement im QM is (afaik) not well-understood; the problem is that (afik) a measurement related to a "collaps of the wave function" or alternative interpretations cannot be described by a quantum mechanical operator

8. Oct 13, 2012

### Killtech

this isn't a requirement for such a derivation although it would make things easier. as far as i understand it there are two separate time evolution mechanics that completely exclude each other: the classical like local and deterministic equations of motion (e.g. schrödinger, dirac, ...) and the measurement postulates (containing non local collapse). a consistent mechanism defined purely within terms of the theory to chose between these two time evolution is the main key to get such a correlation between operators and their measurement apparatus. on the other hand if you have an algorithm how to derive the operator for a given detector or the detector for an operator you should be able to derive an exact mathematical description of the mechanism deciding which time evolution is to be used and when. this is why i am eager to understand it because books do not give you much of an explanation. they only give you the impression that the often sloppy formulations are there not only to annoy mathematicians but to hide that QM theory isn't well defined.

and it also makes one wonder, if the definition of an observable - i.e. of what is measurable - isn't somewhat arbitrary if such decisive elements were missing. i mean if i assume to have an arbitrary operator as Hamiltonian i would get c-numbers as energy eigenvalues. but looking on the time evolution you would simply see that the imaginary part of these eigenvalues merely represents a decay time of the corresponding eigenstates while the real part still represent the energy (interpretation is similar for markov processes). thus you still could make perfect sense of them and one could use that interpretation for any other operator thus call all operators observable. but if hermitean is a prerequisite there must be some reasoning that can be broken down to experimentally obtained results, no?

furthermore i am intrigued to understand why non-linear operators are excluded because with such operators one would be able to measure the wave function explicitly.

as far as i understand physics every axiom/definition used has its roots in some experimentally obtained data and is merely an inter/extrapolation of these empirical formulas / relations between physical objects. so i would rather be astonished if any kind of derivations between apparatus and operator were missing because it would mean QM is in most cases unable to provide an answer what is measurable (e.g. wave function).

9. Oct 13, 2012

### strangerep

Killtech,

Have you tried Ballentine's textbook "QM -- A Modern Development" ?

Measurement is (imho) better understood as establishing a correlation between the initial state of a system and the final state of an apparatus, brought about by an interaction between the two. I think this is quite well understood -- the only downside being that one must analyze each experimental situation in careful detail to construct the relevant operators, rather than appealing vaguely to a catch-all collapse mechanism. Ballentine covers all this quite well in his ch9.

The collapse postulate and associated interpretations are just that: interpretations. Ballentine does away with it, emphasizing the statistical features of real measurements, and their relationship to physical symmetries and dynamics. What matters is the dynamical algebra of operators that characterize a (class of) systems. (We don't just make up operators on a whim and expect them to mean something physical.)

10. Oct 13, 2012

### audioloop

first paragraph
right.

http://arxiv.org/pdf/1111.1877.pdf

second paragraph

einstein heinserberg debate
http://web.gc.cuny.edu/sciart/copenhagen/nyc/Holton.doc

"Only the theory decides what one can observe"
Einstein.

i add, how can somebody speak about an observable, if it does not have a definite value ? ha !
i.e. has no CFD

third paragraph

I have written on non-linearity.
remember until now SQM is linear, some people, get indignant when you talk about, linearity is mainstream, imo penrose gravity objectification is an interesting avenue.

https://www.physicsforums.com/showpost.php?p=3973897&postcount=14
http://arxiv.org/pdf/quant-ph/0402180v1.pdf
http://arxiv.org/pdf/0907.1977v1.pdf
http://arxiv.org/pdf/1109.6462v4.pdf
http://arxiv.org/pdf/gr-qc/0610142v2.pdf
http://iopscience.iop.org/1742-6596/343/1/012006/pdf/1742-6596_343_1_012006.pdf

fourth paragraph

(i agree with you, is a sort of operationalist patchwork)

study what is the Quantum State

https://www.physicsforums.com/showpost.php?p=4070884&postcount=12
https://www.physicsforums.com/showpost.php?p=4112229&postcount=512

Last edited: Oct 13, 2012
11. Oct 14, 2012

### tom.stoer

I agree, we don't need an operator which implements the measurement, we only need a procedure which relates an observable with a measurement. But as I said, afaik there is now unique constructive way to derive the apparatus from an observable (or vice versa)

'Observable' is simply a name; it means that there is an hermitean operator with real eigenvalues which can in principle correspond to results of measurements. If you don't like this wording then you may call it 'hermitean operator' in order to make clear that a constructive approach for the measureemnt itself is missing.

I don't understand what you mean by 'non-linear operators'. Using a Hilbert space framework with states |ψ> and operators A acting linearily on the states: A|ψ>. The only way to implement non-linearity in this framework is to use e.g. A². But this is definitly not what you are asking for b/c the operator A² still acts linear on the states (simple example: angular momentum in a problem with spherical symmetry; the operator L²/2mr is a non-linear operator-valued funtion of L and r)

I think you have something different in mind, e.g. A[f] with f = f(ψ); this could be something like A|ψ> = f(<ψ|ψ>)|ψ>. Afaik one has probed several non-linear ansätze but non of them is able to reproduce established results and to solve e.g. the measurement problem.

12. Oct 14, 2012

### Killtech

uh... a awful lot to read for me. this will shut me up for a while i guess, hehe.

this is exactly what one calls an non-linear operator. the reason why i am interested in them is because i have the opposite impression. when i first learned QM and about the H atom i was disappointed because apart of the energy levels calculation it didn't tell me much about what was going on. but when i heard how to acquire the charge density from the wave function (interpretation of dirac equation) and i simply coupled the EM-field classically i got instantly a much clearer view: energy eigenstates have a stationary charge density while mixed states have oscillating density and in case of a simple mix of two eigenstates the oscillation exactly produces an electromagnetic radiation with the frequency of the energy difference of the two states - so it's on the spot of what you see in the experiment. but this is not canonical QM because this is actually pure classical calculation without the need of ANY measurement postulates (and that you only measure decaying mixed states but no pure states). note that this kind of coupling is non-linear thus the total Hamiltonian becomes also a non-linear operator...

also if you look back on pure classical theory you notice that maxwell equations do provide time evolutions for all fields except for one: the charge current (never found that satisfactory because it disallowed em-wave solutions with the amplitude parallel to the movement direction that looks like e+e- wave). but if you look what SE provides it's this missing equation. note that reformulating SE into equations of motion for charge density and current is equivalent except for the global phase goes missing (which has no physical meaning anyway) and one may wonder why the charge current velocity isn't allowed to have rotations (because of missing spin in SE). and also notice that self interaction in classical EM-theory leads to problems when you assume point like particles. but it can be resolved if you let the charge density evolve freely - namely let the point like electron fall apart just like it would in an position eigenstate in QM.

apart from this non-linear dynamics can provide lots of effects natural to quantum mechanics like for example: randomness through chaotic behavior in certain regions, particle-wave duality via soliton solutions and a non-trivial + not stationary vacuum states in inhomogenious PDEs. so i find looking into non-linearity actually the canonical way to go.

Last edited: Oct 14, 2012
13. Oct 14, 2012

### tom.stoer

Hm, I don't think you understood QM and the SE correctly. Everything you are looking for is exactly described by QM (or the relativistic extension, i.e. QFT).

A fully satistactory system of equations for "particles" and their charges and currents is (in the framework of quantum field theory) the Dirac equation coupled to electromagnetic vector potential plus the Maxwell equations coupled to a current which is bilinear in the fermion fields.

Here the fields are translated to field operators; the expressions are partially non-linear in the field operators (the electric 4-vector current density is bilinear in the fermion fields), but these non-linear expressions still act as linear operators on the quantum states.

Last edited: Oct 14, 2012
14. Oct 14, 2012

### Killtech

i know this well. but this is a different approach then the classical coupling in e.g. maxwell-dirac. yeah, the classic approach always lead directly to a QED - but: without the 2nd quantization. that's the difference. naturally if you go the classic way the differences grow when you try to understand how a classical theory would try to describe a two particle system: it would try doing so with only one charge field while in QM needs a separate fields/wave function for each particle (it's a bit different in QFTs but the fock space is even much larger there).

the norm of the states is used to establish the probability interpretation in QM but in a classical approach the natural way would be to use it as a kind of particle number. due to non-linearity of course a state that is just 2 times another state can behave totally different. so in a way you would want dirac-maxwell to quantize in norm too and as far as i know this is unknown if this happens.

in a sense i want to know if QM can be reduced to a (classical) field theory for a classical charge and mass density since these seem to be the underlying real physical objects the theory seems to describe. and it is clear that the equations of motions for these quantities will be everything but linear. QM tries to do it another way by embetting these objects into a wave function and extending the dimensions such that the theory becomes linear (equation of motion acts linearly on the state space). but this comes at a price of an excessively larger state space (in the simplest case of 1 particle schrödinger this already means adding such mathematical artifacts like global phase and it seems to get a lot more worse when it is extended to many particles). thus one can wonder when two states of this space are equivalent (e.g. states differing only by a global phase should be physically equivalent). the formalism of QFT hides these aspects because one usually doesn't want to know how the elements of the fock space look like.

Last edited: Oct 14, 2012
15. Oct 14, 2012

### tom.stoer

Could you post some of the formulas you have in mind to replace standard QM or QFT expressions?

16. Oct 14, 2012

### Killtech

for the start just take Schrödinger equation and

div(E)=<ψ|ψ>

and add it to maxwell equations. obviously the latter equation is non-linear in ψ. but you can do the same with dirac equation which is called dirac-maxwell as i understand it.

but is is also interesting to rewrite the schrödinger equation purely in less abstract terms of the charge density and current. you can do this by using following transformation: ψ = √ρ(x) exp(i u(x)). then similarly like in deriving the continuity equation you multiply <ψ|from the right to both sides of SE. the resulting equations imaginary part is the continuity equation (its also the equation of motion (EoM) of charge density if you multiply it with e) while the real part is the EoM for the current, or more precisely for the phase u which is j = ρ ∇u (just putting the transformation into the definition of the probability current), so it's the potential of the current velocity density and since it has a potential it implies there are no rotations possible (so no spin). the EoM for the charge current can be written in a form similar to navier-stokes and it is also interesting to note that it is already not linear in ρ.

now in the case of two particles i would want to see how this EoM looks like if you again reduce it purely to charge density and current (you leave out more then just the global phase here since the solutions for SE are functions from R^(3+3+1) -> C = R^2 while (x,t) -> (ρ, j=ρ ∇u) is from R^(3+1) -> R^2 (not R^4 because j has not full degree of freedom). my interest is to know how much of the phyisics is still preserved in this reduced EoM or if anything is lost at all. obviously the symmetry arising from indistinguishably of the particles is fully absorbed by this reduction as there is no way to tell which part of the density/current belongs to which 'particle' anymore.

in principle the same transformations into EoM of classic physical fields of charge density and current can also be done to dirac equation but the dimensionallity of the solutions (R^4 -> C^4 = R^8) requires another pair (ρ, j) which is easiest to assume to be the mass density/current. i didn't manage to transform it so far because i would require to extract a positive definite scalar field for the mass density first and make sure it fulfills the continuity equation on each subspaces of particle and ani-particle solutions respectively. and that seems like quite some work just to satisfy my curiosity.

17. Oct 15, 2012

### tom.stoer

But this is standard; with the Dirac equation and jμ = ψ+γ°γμψ it's the well-known expression from QED.

This is standard, too. The fact that it's non-linear is simply due to the choice of functions and does not indicate any new physics.

18. Nov 2, 2012

### A. Neumaier

Remember that words may have multiple meanings.

It is conventional in much of QM to use the words ''observable'' and ''Hermitian operator'' synonymously.

However, the rate of a nuclear reaction and the width of a spectral line are obviously observable (measurable). But they are not given by a Hermitian operator.