What is the Interpretation of Observable Operators in Quantum Theory?

[Killtech]

TL;DR Summary

How to interpret the large variety of observables QT allows us to construct?

So generally in most literature observables are represented via self-adjont (or equivalently real-valued) linear operators in QT. But that definition leaves it open for a wide variety of operators that can be view as observables. I was always a bit uncertain how to understand this freedom, so i want to scrutinize a bit its implication.

For that let's assume we have some observable $O$ and we pick some orthonormal basis of the Hilbert space $|\psi_i\rangle$, then we can construct a new operator $O_{\psi} :=\sum_i \langle \psi_i |O| \psi_i \rangle | \psi_i \rangle \langle \psi_i|$ via a sum of projection operators multiplied by real valued constants. But on the other hand these constants $\langle \psi_i |O| \psi_i \rangle$ are the expectation values of $O$ for a state $|\psi_i \rangle$ and so in general we can check that $\langle \phi |O| \phi \rangle = \langle \phi |O_{\psi}| \phi \rangle$ for any state $|\phi\rangle$. Hence, if we are just interested in getting the expectation value of $O$ we have actually a large family of operators at our disposal to do that. While this family has in common to always yield the same expectations, the uncertainty $\Delta O_{\psi}$ differs as per construction the choice of basis makes it certain/sharp for states of that basis. This also means that while $O_{\psi}$ will match in expectation, it will have a different spectrum i.e. represent the same variable via different quantum numbers. However this makes each a real valued operator and therefore self-adjoint, hence they are technically valid observables.

Here is the thing, that I don't know anything in QT that would forbid calling $O_{\psi}$ an observable, but then what is its interpretation? If we were just associate observables by their expectation to a variable they represent, then this leaves us with some ambiguity for that variable. On the other hand there is the correspondence principle that specifically picks out one operator from its family for each variable. But while it does make an unique correspondence, does it actually disqualify other possibilities? The correspondence principle was historically intended to bridge the gap between the new and old theory and i haven't heard it making any assumptions about the uniqueness of this mapping - nor anyone considering that it might not be.

When it comes to conservation laws written in terms of expectation values, we can exchange any observable by one from its same-expectation-family of operators and it changes nothing. So for this purpose all observables of the same expectation can be viewed as equivalent.

Going back to the spectrum of $O_{\psi}$, isn't quantization in QT actually a part of the measurement process specifically? If a quantum system is undisturbed, then it supports superposition of states like Schrödinger's cat and the quantization towards a "dead" or "alive" cat quantum number only happens when the system is measured. So wouldn't it make sense to consider that observable operators don't just represent what variable they measure but also how they acquire it/how its measurement interacts with the system and therefore governs how it quantizes? However that would also mean that we might figure out a way to measure the cats state via a different method/observable and potentially have a scenario where we find that it has a cat quantum number of 0.5*dead + 0.5*alive with 100% certainty after repeated measurements (yet consequently such a measurement won't be able to measure the cat in a alive state with certainty).

EDIT: typo in formula

 

[PeterDonis]

For that let's assume we have some observable $O$ and we pick some orthonormal basis of the Hilbert space $|\psi_i\rangle$, then we can construct a new operator $O_{\psi} :=\sum_i \langle \psi_i |O| \psi_i \rangle \langle \psi_i|$

This isn't an operator, it's a bra vector (since it's a weighted sum of bra vectors).

via a sum of projection operators multiplied by real valued constants.

That is not a correct description of what you have written.

 

[Killtech]

This isn't an operator, it's a bra vector (since it's a weighted sum of bra vectors).
That is not a correct description of what you have written.

Oups, you are right, thanks. It should be $O_{\psi} :=\sum_i \langle \psi_i |O| \psi_i \rangle | \psi_i \rangle \langle \psi_i|$
Don't know why but latex preview didn't work when i originally posted this (now it seems to work) and so i didn't see it. Fixed my original post.

 

[Nullstein]

$\langle \phi |O| \phi \rangle = \langle \phi |O_{\psi}| \phi \rangle$

That's not true for $| \phi \rangle = | \psi_1\rangle + | \psi_2\rangle$.

 

[PeterDonis]

If we were just associate observables by their expectation to a variable they represent

We don't. An observable is certainly not defined just by its expectation value. The full spectrum of the observable needs to be taken into account.

 

[anuttarasammyak]

Oups, you are right, thanks. It should be

You wrote the case when basis {$\psi_i$} are eigenvectors of O. In general
$$O_\psi:=\sum_{i,j}|\psi_j><\psi_j|O|\psi_i><\psi_i|=O$$

 

[anuttarasammyak]

Summary:: How to interpret the large variety of observables QT allows us to construct?

the uncertainty ΔOψ differs as per construction the choice of basis makes it certain/sharp for states of that basis.

Standard deviation of O for state $|\phi>$,
$$\sqrt{<\phi|O^2|\phi>-<\phi|O|\phi>^2}$$
does not depend on choice of basis {$\psi_i$}.

 

[Killtech]

That's not true for |ϕ⟩=|ψ1⟩+|ψ2⟩.

Hmm, yeah now i see. The expectations would only agree for every state if the basis is the basis for which the operator is diagonal, in which case this yield just the diagonalized representation of it... yeah, $\langle \psi_i |O_{\psi}| \psi_j \rangle$ doesn't null out.

We don't. An observable is certainly not defined just by its expectation value. The full spectrum of the observable needs to be taken into account.

Hmm. Which observable corresponds to which variable is kind of given by axiom which makes it harder to scrutinize it. There is a loose derivation from following Schrödinger's motivation of his equation. For the energy though, the spectrum played historically obviously a central part.

Anyhow, the problem is this doesn't work well in reverse: starting with an observable operator and asking what it represents.

Standard deviation of O for state |ϕ>,
<ϕ|O2|ϕ>−<ϕ|O|ϕ>2
does not depend on choice of basis {ψi}.

That's true, but $O_{\psi} \neq O$, hence the standard deviation differs for them - but to my mistake, so can the expectation.I guess however the projection operators onto the states of an orthonormal basis do themselves form a basis of all observables that can be simultaneously measured, since they all commute amongst each other and so will any linear combination of them. I guess I should first try to understand what variables these simple observables represent before looking at anything else. From there it shouldn't be too hard to get an interpretation the entire rest.

Such projection ops are merely constrained by the expectation of their sum which must add up to 1, but are otherwise independent. An infinite number of elements minus a single constraint means they represent an infinite number of degrees of freedom. This is quite a change, since classically a particle is described by merely 6 degrees of freedom whereas here we have an infinity of variables.

 

[PeterDonis]

Which observable corresponds to which variable is kind of given by axiom

What do you mean by "variable"? As far as I can tell it's just a synonym for "observable", which makes your statement a trivial tautology.

starting with an observable operator and asking what it represents

Yes, in general there is no way of telling what the physical meaning of some arbitrary Hermitian operator is. That's why QM in general does not look at arbitrary Hermitian operators and try to assign them physical meanings; it works in the other direction, obtaining particular Hermitian operators from considerations of what we would expect to have physical meaning. So even though QM in principle does say that any Hermitian operator should define an observable, in practice no use is made of this (and of course no proof of any such statement is given in any formulation of QM, it's just sort of assumed in the background).

 

[anuttarasammyak]

That's true, but Oψ≠O, hence the standard deviation differs for them - but to my mistake, so can the expectation.

If you admit the formula in #6, because
$$I=\sum_i |\psi_i><\psi_i|$$
$$O_\psi:=\sum_{i,j}|\psi_j><\psi_j|O|\psi_i><\psi_i|=IOI=O$$
We are not able to introduce $O_\psi$ as a different one from O.

 

[Killtech]

What do you mean by "variable"? As far as I can tell it's just a synonym for "observable", which makes your statement a trivial tautology.

I attempted to differentiate between the formal "observable" definition as in QM (as a linear operator) and "variable" as a general concept shared between all physical theories.

Yes, in general there is no way of telling what the physical meaning of some arbitrary Hermitian operator is. That's why QM in general does not look at arbitrary Hermitian operators and try to assign them physical meanings; it works in the other direction, obtaining particular Hermitian operators from considerations of what we would expect to have physical meaning. So even though QM in principle does say that any Hermitian operator should define an observable, in practice no use is made of this (and of course no proof of any such statement is given in any formulation of QM, it's just sort of assumed in the background).

Which i would find is a very odd situation.

If there are observables that can be measured simultaneously (as in commute) with the usually set of observables, yet mathematically are additional degrees of freedom of the system, then it feels odd just to ignore them. After all it means there is further information about the system the other observables are entirely blind towards - in the sense that values of such observables cannot be derivable from the usual observables.

It either means it's actual physical information about the system that we could potentially measure yet just chose to ignore - in which case it would be quite significant. Or this information turns out to be not measurable in reality. In which case these observables represent mathematical artifacts of the theory that are not realized by nature and if possible should be eliminated from a minimal framework - i.e. like hidden variables of some extended interpretation which the theory doesn't actually need, or hidden assumptions which require further artificial assumptions to eliminate them (through artificial symmetries).

It does not seem particularly difficult to classify all observable operators that are technically possible since due to their linear nature we can reduce the problem down to a set of basis elements and focus on those. But even after doing so it appears there is still many more degrees of freedom then what one might naively expect.

If you admit the formula in #6, because
$$I=\sum_i |\psi_i><\psi_i|$$
$$O_\psi:=\sum_{i,j}|\psi_j><\psi_j|O|\psi_i><\psi_i|=IOI=O$$
We are not able to introduce $O_\psi$ as a different one from O.

Ah, okay. Yes, your sum has two indices, mine had just one leaving out the off diagonal terms. I chose this form for a different reason and did not start with the general representation of an operator within a basis. I mistook the impact which leaving out the off-diag-terms has on the expectation value - though in my head i made the same calculation you wrote down here but failed to notice that it relies on the double indices over which the sum goes - hence my mistake.

leaving the off diagonal elements represents the most general form of all operators that commute amongst each other. I very specifically wanted to get a picture of what those are. But the idea to approach the problem from that starting point turned out not fruitful.

 

[PeterDonis]

I attempted to differentiate between the formal "observable" definition as in QM (as a linear operator) and "variable" as a general concept shared between all physical theories.

The term "variable" as a general concept shared between all physical theories means something like "an element in a mathmatical expression", which I doubt is what you mean by it. So I still think you should clarify what you mean by it.

If there are observables that can be measured simultaneously (as in commute) with the usually set of observables, yet mathematically are additional degrees of freedom of the system, then it feels odd just to ignore them.

Observables that operate on different degrees of freedom (for example, spin-z on particle A and spin-z on particle B) always commute. That's not the issue.

The issue is that there could be an arbitrary number of possible Hermitian operators that all operate on the same degrees of freedom, and most of them won't have any discoverable physical meaning. That's why QM doesn't start from the Hermitian operators in general and try to interpret them.

It does not seem particularly difficult to classify all observable operators

Have you looked to see if anyone has tried?

due to their linear nature we can reduce the problem down to a set of basis elements and focus on those. But even after doing so it appears there is still many more degrees of freedom then what one might naively expect.

I think you are using the term "degrees of freedom" incorrectly in this context. See my comments above (after my second quote from you).

 

[Killtech]

The term "variable" as a general concept shared between all physical theories means something like "an element in a mathmatical expression", which I doubt is what you mean by it. So I still think you should clarify what you mean by it.

Since i was talking about the correspondence, i used "observable" strictly in the technical sense of QM and "variable" in whatever it is held against in the correspondence limit. I think that's one process that enables us to determine the interpretation of an observable in QM.

The issue is that there could be an arbitrary number of possible Hermitian operators that all operate on the same degrees of freedom, and most of them won't have any discoverable physical meaning. That's why QM doesn't start from the Hermitian operators in general and try to interpret them.

whether observables describe the same degree of freedom is a fair question. Yet it is something we can check:

If one finds a situation where for two scenarios all observables always yield the same expectations, yet one can further process the state before measurement/rearrange measurement in such a way that yields a clear distinction for the two cases, then the system has more degrees of freedom then the observable expectations are able to describe and whatever was measured that enabled the distinction, represents an observable that contains an independent degree of freedom.

With that in mind if we have a state that isn't an element of a selected basis, then it is impossible to derive the expectation value of each projection operator $|\psi_i\rangle \langle \psi_i|$ observable from whatever finite number of observable expectations one might have measured (for observables compatible with that basis), yet QM suggests that such information is obtainable due to all observables in question commuting.

Have you looked to see if anyone has tried?

My starting point is usually to ask around people that have a good chance to know such things, like in these forums. I don't expect everyone tackle an issue the same way as I would, meaning that what i am looking for might be found under quite a different approach/name. From the answers i get here, i can often take some good directions what to search for.

 

[A. Neumaier]

I guess however the projection operators onto the states of an orthonormal basis do themselves form a basis of all observables that can be simultaneously measured, since they all commute amongst each other and so will any linear combination of them. I guess I should first try to understand what variables these simple observables represent before looking at anything else. From there it shouldn't be too hard to get an interpretation the entire rest.

Such projection ops are merely constrained by the expectation of their sum which must add up to 1, but are otherwise independent. An infinite number of elements minus a single constraint means they represent an infinite number of degrees of freedom. This is quite a change, since classically a particle is described by merely 6 degrees of freedom whereas here we have an infinity of variables.

These projections just measure whether or not the measurement result has a given value. This is not so different from the classical variable that assigns to a particular measurement result and zero to the others.

It does not seem particularly difficult to classify all observable operators that are technically possible

For a system whose Hilbert space is small enough, all Hermitian operators correspond to recipes for observing them to a certain accuracy. See for example the discussion and references in Section 3.1 of my paper Quantum Physics via Quantum Tomography.

As the Hilbert space gets larger, this gets more and more difficult (or the accuracy gets lower and lower) since the number of possibilities for operators grows much more rapidly than the number of feasible experimental arrangements.

 

[PeterDonis]

My starting point is usually to ask around people that have a good chance to know such things, like in these forums.

But you aren't asking, you're making claims. You aren't asking around here to see if anyone has any relevant information about the classification of observables. You're making the claim that such work "does not seem particularly difficult" so it should already be done.

 

[Killtech]

But you aren't asking, you're making claims. You aren't asking around here to see if anyone has any relevant information about the classification of observables. You're making the claim that such work "does not seem particularly difficult" so it should already be done.

Well, if i have a question i always make thoughts of my own about it and often combine both things in my posts. Isn't that even a rule of these forums? Anyhow, the post from @A. Neumaier just shows a good answer and a way to do it, though taking a somewhat different angle.

These projections just measure whether or not the measurement result has a given value. This is not so different from the classical variable that assigns to a particular measurement result and zero to the others.

For a system whose Hilbert space is small enough, all Hermitian operators correspond to recipes for observing them to a certain accuracy. See for example the discussion and references in Section 3.1 of my paper Quantum Physics via Quantum Tomography.

As the Hilbert space gets larger, this gets more and more difficult (or the accuracy gets lower and lower) since the number of possibilities for operators grows much more rapidly than the number of feasible experimental arrangements.

That seems like a fair approach for the special case of polarized states - even giving a concrete recipe of how to measure the the observable in question. However the interpretation part becomes a little tedious since it needs to be derived always from setup used to measure it.

But as that example discusses mainly photons, i just figured that the classic correspondence approach works quite good in this situation, doesn't it? Since for a photon the correspondence limit would be taken against a classic field theory instead of a point particle.

Sure, the individual projectors don't really measure much when looking at a single measurement as only one of them must yield a value while all else must be zero. Yet on the other hand, we can just look at what information the projectors pick up in the correspondence limit. If we take the momentum basis, then each projector agrees with the Fourier transform of the classical field at the specific $k$ 4-vector. Or rather, due to the Hermitian nature of observables they only pick up the absolute square of the transform and thereby lose the phase information. So the total of all projectors pick up all these values which in the correspondence limits would suggest that the absolute square of a Fourier transform of a field is itself an observable.

So while for a single measurement the total of momentum projectors will just yield a Kronecker delta type result, the expectation of all the projectors combined becomes interesting as it should correspond to the absolute square Fourier transform of a field. As all projectors do commute, the expectation should give an accurate account in the sense of that measuring any part of the transform won't disturb measurement of the rest. In my understanding, won't a single-photon state that is a specific superposition of momentum states will have almost the same physical behavior as the classical field with the Fourier transform matching that superposition? I mean of course the major difference will be measurement where we need to aggregate the individual observations into expectation values whereas the classical field theory allows to observe the transform in a single try.

 

[A. Neumaier]

That seems like a fair approach for the special case of polarized states - even giving a concrete recipe of how to measure the the observable in question. However the interpretation part becomes a little tedious since it needs to be derived always from setup used to measure it.

Of course the interpretation of a measurement always (whether measuring classical or quantum observables) depends on the setup used to measure.

The point is that one can start from the operator and always design a setup that measures it. See, e.g.,

• U. Leonhardt, Quantum physics of simple optical instruments, Rep. Progress Phys. 66 (2003), 1207–1248.
• U. Leonhardt and A. Neumaier, Explicit effective Hamiltonians for linear quantum-optical networks, J. Optics B: Quantum Semiclass. Opt. 6 (2004), L1–L4.

But as that example discusses mainly photons, i just figured that the classic correspondence approach works quite good in this situation, doesn't it? Since for a photon the correspondence limit would be taken against a classic field theory instead of a point particle.

It also works for operators acting on photon pair (or triple, etc.) states, which do not have a classical field interpretation.