States & Observables: Are They Really Different?

[Demystifier]

Usually states and observables are treated as fundamentally different entities in quantum theory. But are they really different? A state can always be represented by a density "matrix", which is really a hermitian (or self-adjoint) operator. Since observables are also hermitian (or self-adjoint) operators, what exactly is the difference? Can we think of density "matrix" as an observable, and if not, why not? I emphasize that this is not meant to be a philosophical question, but a practical and/or a formal one.

 

[anuttarasammyak]

I do not have clear idea to answer your question. I just observe :

an observable $\in$ {Hermitian}
a density matrix $\in$ {Hermitian}
are not enough to conclude that they have no difference.

If density matrix is observable, what physical quantity does it correspond ?
If observable is density matrix, what state does it correspond ?

 

[malawi_glenn]

what physical quantity does it correspond

What is a physical quantity?

 

[anuttarasammyak]

What is a physical quantity?

In QM textbooks I learned they are coordinates, momentum, energy, etc which correspond to Hermite operators. Does something similar correspond to density matrix which is Hermitian ?

tr(\rho A)=<A>
Both matrices $\rho$ and A are necessary to procuce number <A>.
They are Hermitians but seem to have different characters or meanings in physics.

 

[gentzen]

Usually states and observables are treated as fundamentally different entities in quantum theory. But are they really different?

Yes, they are different. There are basic observables (+basic "observable like" constituents of possible Hamiltonians) that implicitly define the locality structure of "a given" system under consideration. For example, in the typical circuit model of a quantum computer, you have the measurement operators in the computational base, plus basic one- and two-qubit operations (i.e. the basic "observable like" constituents of possible Hamiltonians). You can build more complicated observables from those basic observables, and even approximate an arbitrary observable arbitrarily close in that way, but the basic observables exists nevertheless, and define a "complexity measure" for each specific way to build a more complex observable from them.
The state(s) on the other hand have much less structure that would allow to measure locality or complexity in a similar way. Of course, typically there is a special state, namely the ground state, which one would like to regard as especially simple, but its relation to the locality structure of "a given" system is often not guaranteed, sometimes (say for adiabatic quantum computation) it is even explicitly the opposite of a local state (because it is the solution to some complicated problem).

In a less computer science oriented context, Neumaier's thermal interpretation gives examples of basic physical observables, namely the generators of some representation of some "relevant" symmetry groups.

A state can always be represented by a density "matrix", which is really a hermitian (or self-adjoint) operator. Since observables are also hermitian (or self-adjoint) operators, what exactly is the difference? Can we think of density "matrix" as an observable, and if not, why not? I emphasize that this is not meant to be a philosophical question, but a practical and/or a formal one.

The density matrix of the current state should not be thought of as an observable, because it is not built from basic building blocks in a similar way as actual physical observables.

 

[vanhees71]

It's in the different meaning of the mathematical objects. Of course states are described by the statistical operator, i.e., a self-adjoint positive semidefinite operator with trace 1 and also observables are described by self-adjoint operators, but their physical meaning is different.

The state describes a preparation procedure and it's given as an initial condition. The observables are quantities that can be measured on the system by making the system interact with a measurement device. Given the state at the initial time and the Hamiltonian of the system you get in the usual way the probabilities for getting one of the possible values when measuring an observable.

Also in the formalism the statistical operator (describing the state) and the operators that are associated with observables are treated differently. First you can choose an arbitrary picture of time evolution, and the results are independent of this choice. That's the invariance of the QT formalism under (maybe time-dependent) unitary transformations. For me the most "natural" choice of picture is the Heisenberg picture, because it mathematically follows the above interpretation of "states" and "observables": In the Heisenberg picture the state is represented by the time-independent statistical operator $\hat{\rho}(t)=\hat{\rho}_0$ and the operators representing observables have a time dependence induced by the full hamiltonian, $\dot{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]$.

 

[Fra]

Maybe in a generalized sense?

Usually states and observables are treated as fundamentally different entities in quantum theory. But are they really different? A state can always be represented by a density "matrix", which is really a hermitian (or self-adjoint) operator. Since observables are also hermitian (or self-adjoint) operators, what exactly is the difference? Can we think of density "matrix" as an observable, and if not, why not? I emphasize that this is not meant to be a philosophical question, but a practical and/or a formal one.

If we think that observable is the simple inference process - "PVM measurement" of inferring a certain values, given the quantum state (or the density matrix for that matter).

But how is the quantum state inferred in the first place? Above, it's given, but the observer has to learn about it, by which process?

One can perhaps think of the state (or density matrix) as resulting from the little bit more complex inference process - quantum tomography, given data from a tomographically complete set of measurements.

One similariy is that both are "inferences" or "observations" in the informal sense, but of different types. I think this association is interesting as from the perspective of inference, learning about a value via a simple measurement, is from the perspective of inference, no different than learning about the state/density matrix from a preparation?

/Fredrik

 

[Demystifier]

In QM textbooks I learned they are coordinates, momentum, energy, etc which correspond to Hermite operators. Does something similar correspond to density matrix which is Hermitian ?

tr(\rho A)=&lt;A&gt;
Both matrices $\rho$ and A are necessary to procuce number <A>.
They are Hermitians but seem to have different characters or meanings in physics.

Entropy
$$-{\rm Tr}(\rho\, {\rm ln}\, \rho)$$

 

[anuttarasammyak]

Thanks.
tr(\rho \ln \rho)=&lt;\ln \rho&gt;=-S
Following your #1 can we say that there exist an observable $\hat{S}$ which corresponds to density matrix with
\rho = e^{-\hat{S}}
where
&lt;\hat{S}&gt;=S?

 

[vanhees71]

No! $S$ is a measure of the missing information relative to the situation of complete knowledge. It's a property of the state and as such not an observable, which are independent of the state.

 

[Demystifier]

Thanks.
tr(\rho \ln \rho)=&lt;\ln \rho&gt;=-S
Following your #1 can we say that there exist an observable $\hat{S}$ which corresponds to density matrix with
\rho = e^{-\hat{S}}
where
&lt;\hat{S}&gt;=S?

Yes and no. In some formal sense yes, but we also know that the standard answer (e.g. by @vanhees71) is no. I would like to understand it more deeply, which is exactly what this thread is supposed to be about.

 

[Demystifier]

No! $S$ is a measure of the missing information relative to the situation of complete knowledge. It's a property of the state and as such not an observable, which are independent of the state.

Consider the following protocol. You prepare the system either in an eigenstate of $\sigma_x$ (spin operator in the $x$-direction), or in an eigenstate of $\sigma_y$, and tell me what you did. If you tell me that you prepared it in a $\sigma_x$ eigenstate , I measure $\sigma_x$. If you tell me that you prepared it in a $\sigma_y$ eigenstate , I measure $\sigma_y$. By this protocol, the measured observable depends on the state.

 

[vanhees71]

To prepare, e.g., a silver atom in an eigenstate of $\vec{n} \cdot \hat{\vec{\sigma}}$ I use a Stern-Gerlach magnet and block the atoms at the places of the other eigenstate.

If I prepare a $\sigma_y$ state, you are completely free to measure the spin in any direction you like. There's no dependency on the state about what you can measure on the system. The meaning of the preparation in this state is that $\sigma_y$ has a determined value $1/2$ or $-1/2$, which you get when measuring $\sigma_y$ with 100% probability. Whether or not you decide to always measure $\sigma_y$, because I send you an atom prepared in an eigenstate of $\hat{\sigma}_y$ is irrelevant to the fact that you can measure, whatever you like, independent of the state the atom is prepared in.

 

[hilbert2]

I believe you could, for example, describe a quantum state $\left|\psi\right.\rangle$ by giving an operator $\hat{O}$ that converts the ground state $\left|0\right.\rangle$ to $\left|\psi\right.\rangle$: $\hat{O}\left|0\right.\rangle = \left|\psi\right.\rangle$, and then use the Heisenberg picture for time development.

With the harmonic oscillator and non-interacting quantum fields, this kind of operator could be easily written with creation and annihilation operators, but that's not the only way because the condition merely says how $\left|0\right.\rangle$ has to change when acted on by $\hat{O}$. If it's some other type of system, you can symbolically play with "raising and lowering operators" for each quantum number, as those operators exist even when it's not possible to write a simple analytical formula for them. But I'm not sure if this kind of description would be any more useful for anything compared to the vector definition of state.

 

[Demystifier]

There's no dependency on the state about what you can measure on the system.

True, but in my protocol there is a dependency on the state about what I do measure (not what I can measure).

 

[vanhees71]

Of course, you are free to choose anything you want to measure, but this choice doesn't change the generally valid physical laws.

 

[Demystifier]

Of course, you are free to choose anything you want to measure, but this choice doesn't change the generally valid physical laws.

So you are saying that the laws of physics don't depend on our choices. What about the opposite, do our choices depend on the laws of physics?

 

[vanhees71]

Of course, they don't depend on our choices. That's what makes them objective.

Our choices depend on the laws of physics in the sense that of course we cannot violate any natural law, provided it's really generally valid. E.g., if you'd objectively and reproducibly discover a particle collision, which violates the conservation of electric charge, then this fundamental law were proven wrong, and with it Maxwell's equations and QED. Then one would have discovered something fundamentally new, and theorists have to think hard about, how to describe the new finding (and at the same time explain by the new theory, why so far standard (Q)ED was so successful). So far such a thing has not been observed, and it's pretty sure that you cannot violate the fundamental law of electric-charge conservation.

 

[Morbert]

A state can only be determined by tomography. Even if a system is prepared in a pure state, that preparation cannot be determined by a measurement on that system. So entropy being a property of a state independent of measurement implies entropy can only be determined by tomography.
-

Following your #1 can we say that there exist an observable $\hat{S}$ which corresponds to density matrix with
\rho = e^{-\hat{S}}

What would the spectral decomposition of this matrix be? I.e. Could you write it as \hat{S}= \sum S|S\rangle\langle S|?
[edit] - Expanded.

 

[Morbert]

As an aside: Consistent histories provides us with a schema for framing a choice of measurement as contingent on a preparation. We expand our state space to include the lab and use dynamics that reproduce the protocol outlined by @Demystifier . But I suspect this is against the spirit of the question asked by OP.

 

[Morbert]

Usually states and observables are treated as fundamentally different entities in quantum theory. But are they really different? A state can always be represented by a density "matrix", which is really a hermitian (or self-adjoint) operator. Since observables are also hermitian (or self-adjoint) operators, what exactly is the difference? Can we think of density "matrix" as an observable, and if not, why not? I emphasize that this is not meant to be a philosophical question, but a practical and/or a formal one.

The eigenvalues of the matrix will have a different interpretation depending on whether the matrix is doing work as an observable or a state right? If the matrix is an observable, then the eigenvalues are possible measurement results of that observable. If the matrix is a state, then the eigenvalues are the probabilities of the outcomes of a measurement in the eigenbasis of the state.

Similarly, the eigenbasis has a different interpretation: The eigenbasis of an observable is the measurement basis of the measurement carried out. The eigenbasis of a state is the basis with minimal Shannon entropy.

 

[A. Neumaier]

No! $S$ is a measure of the missing information relative to the situation of complete knowledge. It's a property of the state and as such not an observable, which are independent of the state.

In statistical thermodynamics, $S=\langle\hat S\rangle$ (the formula in post #9) is clearly measurable, hence observable.

But thermodynamics is derived from quantum statistical mechanics. If entropy is not an observable in quantum mechanics, how then does entropy become an observable in thermodynamics???

 

[A. Neumaier]

Usually states and observables are treated as fundamentally different entities in quantum theory. But are they really different?

Yes, at least with the traditional meaning of states = density operators and observables = Hermitian operators. They are dual objects, and transform differently in time evolution (related by the time reversal transformation).

 

[gentzen]

In statistical thermodynamics, $S=\langle\hat S\rangle$ (the formula in post #9) is clearly measurable, hence observable.

But thermodynamics is derived from quantum statistical mechanics. If entropy is not an observable in quantum mechnaics, how then does entropy become an observable in thermodynamics???

I would argue that the following quotes from "Foundations of quantum physics II. The thermal interpretation" (https://arxiv.org/abs/1902.10779) provide the key to an answer:

Extrapolating from the macroscopic case, the thermal interpretation considers the functions of the state (or of the parameters characterizing a state from a particular family of states) as the beables, the conceptual equivalent of objective properties of what really exists.

By definition, all functions of q-expectations are beables, and all beables arise in this way.

On the other hand, in the thermal interpretation of quantum physics, the theoretical observables (the beables in the sense of Bell) are the expectations and functions of them.

For some suitable basis (of size N), an approximation to the density matrix (wrt to that basis) can be determined by measuring the expectation value of N*N-1 suitable observables. The (approximation) to the entropy (wrt the subspace defined by the basis) is then a non-linear function of those N*N-1 expectation values.

So we have a recipe for measuring the entropy (or at least the entropy wrt some specific subspace) that cannot be translated back into a single observable represented by some Hermitian operator.

 

[Demystifier]

In statistical thermodynamics, $S=\langle\hat S\rangle$ (the formula in post #9) is clearly measurable, hence observable.

But thermodynamics is derived from quantum statistical mechanics. If entropy is not an observable in quantum mechnaics, how then does entropy become an observable in thermodynamics???

That's an interesting question. I'm not even sure that thermodynamic entropy is "measurable". We have a thermometer that measures temperature, we have a calorimeter that measures heat, but there is no such thing as "entropymeter". In thermodynamics, entropy is a quantity that we can reproduce by combining measurements with certain calculations, but it's not a quantity that we can measure directly. And it's quite similar to the state in quantum mechanics, which can also be reproduced by combining measurements with certain calculations (tomography), but it's not measured directly.

 

[A. Neumaier]

entropy is a quantity that we can reproduce by combining measurements with certain calculations, but it's not a quantity that we can measure directly.

This is quite similar to how most observables except for the simplest ones are measured. One measures raw data and computes from these the observables of interest.

Well-known examples are the Greek measurements of the diameter of the Earth, and modern measurements of the masses of the Sun, the Moon, the planets, and many stars.

High precision measurements of positions under an electron microscope, and of particle momenta in collider experiments also belong to this class of measurements - the latter are inferred from track data.

In fact, many measurements (e.g., those of speed) are taken by taking the quotient (or a limit of the quotient) of two more accessible raw measurements. Precisely the same happens for entropy. According to Wikipedia,

In 1865, German physicist Rudolf Clausius, one of the leading founders of the field of thermodynamics, defined it as the quotient of an infinitesimal amount of heat to the instantaneous temperature.

 

[A. Neumaier]

I would argue that the following quotes from "Foundations of quantum physics II. The thermal interpretation" (https://arxiv.org/abs/1902.10779) provide the key to an answer:

Yes, of course. My thermal interpretation is where I got the insights from.

It solves most elegantly almost all interpretation problems in the microscopic domain!

 

[A. Neumaier]

We have a thermometer that measures temperature, we have a calorimeter that measures heat,

Then take the quotient to get the entropy.

Note also that temperature is not measured directly; what is measured is the height of a quicksilver column or something equivalent, which is then related to temperature by applying the laws of thermodynamics.

Furthermore, I never saw a discussion of a temperature operator in statistical mechanics, though it is clearly an observable.

 

[vanhees71]

In statistical thermodynamics, $S=\langle\hat S\rangle$ (the formula in post #9) is clearly measurable, hence observable.

But thermodynamics is derived from quantum statistical mechanics. If entropy is not an observable in quantum mechanics, how then does entropy become an observable in thermodynamics???

How do you "observe" entropy in thermodynamics?

 

[A. Neumaier]

How do you "observe" entropy in thermodynamics?

See my posts #26 and #28.