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Do operators in QM represent in some fashion the action of the measurement apparatus on a state being measured? Usually operators in QM are introduced as abstract transformations whose eigenvectors/eigenvalues are axiomatically the possible results of measurement, with an explanation along the lines of "because it works". However it seems like a coincidence that the operators that determine possible measurement results are, well, operators that transform states on which they act, as though the dynamical act of measurement itself were being modeled by a coarsely-grained apparatus-state interaction during the process of measurement, and the possible results of measurement are those fixed-point states for which the operator isn't "scrambling things up" during measurement (i.e the eigenfunctions). For example the momentum operator is associated with infinitesimal spatial translations, which makes sense because an apparatus that measures momentum has to in some fashion probe how a state translates in space without changing it. Has a view like this been fleshed out? It seems like it could shed some light on the measurement problem; it would make sense for the dynamical evolution of states being acted on by operators to eventually settle down (collapse) to the fixed points of the operator.

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I do not think that operators represent the action of the measurement apparatus. Operators are introduced because they are the only "sensible" way to represent quantum observables. Let me explain in more detail the last assertion.

It is customary to model mathematically physical observables, i.e. measurable quantities, as maps that associate to a configuration of the system a given value (the outcome of the hypothetical measurement). It is then possible to manipulate mathematically these observables to obtain new ones, especially by summing, rescaling and multiplying them.

However, the aforementioned maps are commutative objects, since it makes no difference in which order we consider them. Therefore, these maps are a sensible way only to model classical observables. In fact, the experimental evidence is that quantum observables do not commute: the outcome of measurements may be different depending on the order the observables are considered.

To solve this problem one could try to define things more abstractly, trying to capture mathematically the basic properties that the collection of quantum observables shall satisfy. If one considers as such properties the one mentioned above, i.e. the possibility of summing, rescaling and multiplying the observables to again obtain an observable, and introduces the easiest possible element of noncommutativity (as the experimental evidence suggests), namely the noncommutativity of the product of observables, one obtains a so-called noncommutative algebra of observables.

Another useful notion one could put on observables is the concept of the largest value (in "magnitude", i.e. in modulus) that the observable could yield as a measurement outcome. Such largest value has mathematically the properties of a norm. An algebra with a norm is a Banach algebra. Finally, one could also assume some other (physically very reasonable) properties on the interplay between taking the norm and the multiplication of observables, as well as the concept of complex observables (admittedly mostly for mathematical convenience, nonetheless this is something very natural also at the classical level, think e.g. of the convenience of complex quantities in classical electromagnetism). With these additional properties, the collection of observables is modeled mathematically by a so-called noncommutative C*-algebra.

Now, noncommutative C*-algebras have a very interesting mathematical property:

A C*-algebra is always equivalent (in mathematical terms, isomorphic) to an algebra of linear operators acting on some complex Hilbert space.

Therefore, representing quantum observables as operators acting on a Hilbert space is the most natural thing to do, in view of the noncommutativity present in quantum physics. Let me also remark that the classical observables are also modeled naturally by a C*-algebra, albeit a commutative one. And commutative C*-algebras are always equivalent to algebras of complex-valued functions. Therefore at the classical level it is natural to model observables by functions.

The mathematical description of a physical state also comes very naturally once the observables have been modeled as described above, and the corresponding probability theories (noncommutative at the quantum level, commutative at the classical one) concerning, e.g., expected values and averages in measurements, are also readily established.

The measurement problem (or more precisely, the measurement process problem) is in my opinion somewhat tangential to the observables being operators (that, as discussed, is naturally related to noncommutativity). It is in my opinion more of a physical interpretation issue (albeit a very interesting and deep one, for which there is not yet a completely satisfactory solution).

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    $\begingroup$ An interesting approach, but I don't think it contradicts the idea that the operators also represent the measurement process in a sense. We use operators on a vector space to represent the fact that measurements do not commute, OK. But at the same time the choice of a specific operator to represent a specific measurement process is motivated by how it relates to the process. $\endgroup$ – PhysicsTeacher Jan 31 at 12:29
  • $\begingroup$ @PhysicsTeacher Ontologically, I prefer to think about observables as the intrinsic quantities, fundamentally defining a physical system, and the measurement process as a procedure done to study such observables, i.e. to characterize the features of the given system. This is, of course, just an interpretation. In accordance with this interpretation, there is an interesting mathematical result that says that for every quantum observable there always exist at least one measurement process (defined as the interaction with an apparatus yielding the outcome on a scale). $\endgroup$ – yuggib Jan 31 at 13:06
  • $\begingroup$ I agree, however, that this does not exclude the possibility that the observable and measurement process are actually more fundamentally related, in a way we do not yet fully understand, and that this could help to solve some of the ambiguities in the problem of taking measurements in quantum mechanics. $\endgroup$ – yuggib Jan 31 at 13:07
  • $\begingroup$ I really like your write-up, and it makes sense to me with regard to why a noncommutative algebra of observables is useful to relate in some way to measurement outcomes, but it doesn't address what I see as the elephant in the room, which is: why are measurement outcomes specifically and only the spectrum of those specific operators? There is a leap in going from the vague statement that it's nice to relate noncommutative algebra of observables to measurements, and the more specific claim that e.g. the possible p states are eigenvectors of the generator of space translations. $\endgroup$ – user1247 Feb 1 at 17:12
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I have two answers: answer 1, for the OP who is clearly at the level of someone already knowledgable about QM and probing the nature of measurement quite carefully, and answer 2, for anyone learning QM.

Answer 1. The observable being measured produces just a multiple when one acts with that operator on a possible measured output state (that's the eigenvalue equation after all), and such an action always translates to a displacement or something like a displacement (i.e. a sequence of phase factors) when one looks at states in a conjugate basis. This has been looked into but I don't know a good reference except to say there is a huge literature on puzzling over measurement and how to describe it best.

Answer 2. The rest of my post is an educational post for anyone learning QM. I have written it because I know that the hunch that "operators do measurements" is a common pitfall in learning the subject. I am going to advise readers to avoid that pitfall. I hope it helps someone!

The relation of operators to measurement is one which I think everyone who ever learned quantum mechanics has asked themselves. It is also one which most textbooks don't address very clearly in my opinion.

Operators are called "operators" mainly because of their role in the mathematics rather than the physical operations that happen in a lab (or anywhere). When looking into physical things behaving, the operators are more like properties of the physical things, only they are not quite that either; this is why in QM they are called "observables". The word means "something you can observe" and here that means "sort of like a property of the system---only not quite; wait a bit and I will tell you more".

Mathematically an operator is defined as "that which can act on a ket (or a wavefunction) so as to give another ket (or possibly the same ket)". The point is that a combination such as $$ \hat{Q} | \psi \rangle $$ (where $\hat{Q}$ is some arbitrary operator) produces, or is equal to, not an operator nor a complex number nor a giraffe nor a snake but a ket: $$ \hat{Q} | \psi \rangle = | s \rangle . $$ My point here is merely to say why operators are called operators. It is a suitable mathematical term. And, by the way, the ket (or state-vector) given here does not have to be normalized, and usually will not be normalized even when $| \psi \rangle$ is normalized.

Now you may be thinking, "hang on, the Hamiltonian operator gives as output not a state but a rate of change of a state": $$ \hat{H} | \psi \rangle = i \hbar \frac{d}{dt} | \psi \rangle $$ This is correct, but it is a special case: it is the equation of motion; it tells you that the ket you would get from $\hat{H} | \psi \rangle$ is the same as the one you could get from $i \hbar | \dot{\psi} \rangle$.

Now to come back to the question. It is v. useful to notice that operators, especially Hermitian ones, have eigenstates and eigenvalues, so the first thing to do with an operator is often to find out what the eigenstates and eigenvalues are. Once you have found out those (and notice this can be a purely mathematical task, which need not involve physical experiments) then you have a sort of "menu", like a menu in a restaurant. It says "here are the available states the system can adopt if it is going to have a precisely defined value of the physical property associated with $\hat{Q}$". This mathematical task involves the equation $$ \hat{Q} | u_q \rangle = q | u_q \rangle $$ where $q$ is an eigenvalue (a real number assuming $\hat{Q}$ is Hermitian) and $| u_q \rangle$ is an eigenvector. Now this expression looks a lot like we took a system and acted on it or operated on with $\hat{Q}$. Is that right? (after all, we have learned that eigenvalues are possible outcomes of measurements, so isn't that what is going on here?) The answer is no. That is a false impression. In the above equation nothing at all happened to the physical system. Rather, the chef in the restaurant was preparing the menu. It was a purely mathematical investigation, like solving for the normal modes of a set of classical oscillators, or finding the spherical harmonic functions which are solutions of Laplace's equation. Finding those says nothing at all about what any particular set of springs is doing, or how any particular drum is vibrating, and also the above equation (the eigenvalue equation) does not describe the act of measurement.

However, and this is what motivated the question of course, we know that operators bear some sort of relationship to the physical operation of measurement. So what is that relation exactly? The relation is as follows.

The physical process of measurement can be mathematically modeled as a projection (like projecting a vector in 3 dimensions onto a plane or a line). The operator representing the observable being measured tells you what are the directions in Hilbert space the state vector of the system, i.e. $| \psi \rangle$, can possibly be projected on, and in each particular measurement just one of those directions is picked$^*$. Mathematically this is $$ | \psi \rangle \rightarrow N P_q | \psi \rangle $$ where $N$ is a normalization constant and $$ P_q = | u_q \rangle \langle u_q | $$ and the probability that this particular projection happens is $$ \mbox{Prob}(q) = | \langle u_q | \psi \rangle |^2 $$ Notice that these final equations do not involve $\hat{Q}$, and yet $\hat{Q}$ is in the background, providing the "menu" $\{ | u_q \rangle \}$ and $\{ q \}$.

As I said at the start, I realise the above is well known to the OP; I don't want to insult anyone's intelligence, just increase general understanding for others who may be learning QM.

$^*$ Final comment, for completeness: when there are several mutually orthogonal states all with the same eigenvalue, then the projection is not onto a single direction or eigenstate, but onto a plane or higher-dimensional space.

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Quantum observables, i. e. hermitian/selfadjoint operators are mathematical representations of measurable quantities. The spectrum of the operator, namely the set \begin{align*} \sigma(F) = \bigl \{ z \in \mathbb{C} \; \; \vert \; \; F - z \; \, \mbox{not invertible} \bigr \} \end{align*} of complex numbers where F - z is not invertible, is the set of possible outcomes of measurements. Of course, for hermitian operators, the spectrum is real.

The statistics of the outcome distribution is captured by the “projection-valued measure” associated to $F$, which quantifies the “density of states” in a given spectral region.

The measurement process itself has nothing to do with the observable. If you wish to model that, then you need to modify the hamtiltonian, i. e. dynamics. There is no collapse of wave functions to some fixed point.

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    $\begingroup$ How can the measurement process have "nothing to do" with the observable, when the possible measurement outcomes are determined by the observable? I understand that standard QM treats the relation axiomatically and instrumentally, but as I ask in my post, it seems like it would be a pretty strange coincidence for measurement to in reality have "nothing to do" with the very thing that determines the possible results of measurements. It seems like there is a conceptual issue going unaddressed. $\endgroup$ – user1247 Jan 31 at 6:38
  • $\begingroup$ The way to include the measurement is to add a model for the measurement apparatus, i. e. the total Hamiltonian $H_{\mathrm{tot}} = H_{\mathrm{sys}} \otimes \mathbf{1}_{\mathrm{meas}} + \mathbf{1}_{\mathrm{sys}} \otimes H_{\mathrm{meas}} + H_{\mathrm{int}}$ consists of the system, the measurement apparatus and their interaction. Any observable $F_{\mathrm{sys}}$ on the system described by $H_{\mathrm{sys}}$ is then promoted to $F_{\mathrm{sys}} \otimes \mathbf{1}_{\mathrm{meas}}$, i. e. the presence of the apparatus changes the dynamics, not the observable. $\endgroup$ – Max Lein Jan 31 at 7:11
  • $\begingroup$ Yes, you are just re-stating the standard definition, which is fine but it does not address my question. $\endgroup$ – user1247 Jan 31 at 16:53
  • $\begingroup$ Well, I think I have: you claimed there was some connection between the observable and the measurement process. And I have shown you that the observable is not modified by the measurement process. What changes is the time evolution, not the observable. Of course, in the Heisenberg picture, the time-evolved observable will no longer have a product structure due to the interaction, but again, this is due to the measurement process changing the time evolution. $\endgroup$ – Max Lein Feb 1 at 7:06
  • $\begingroup$ You haven't really shown anything; you've just re-stated the standard description: that the hamiltonian describes time evolution including system-apparatus interaction, and that observables determine possible measurement outcomes. This is fine, but it ignores my question, relating to the conceptual relationship between the operator-measurement connection, e.g. "why are possible momentum measurement outcomes given by the spectrum of the space translation generator?" I fully understand standard QM and the ordinarily factorized questions of time evolution vs state reduction. $\endgroup$ – user1247 Feb 1 at 17:00
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This is the essence of the decoherence-based understanding of measurement. The idea is that the operators in the Hamiltonian act upon the system's state, causing it to decohere into a probability distribution at a certain basis, called a "pointer basis". For example, because physical interactions are generally based on the position operator, they often cause (approximate) decoherence in the position eigenbasis, which is why things appear to be localized - particles have trajectories and positions.

The operators that represent measurement in QM can hence be considered to "work" because they are diagonal in the same basis as the pointer-basis that the measurement-process induces decoherence in.

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  • $\begingroup$ This makes perfect sense for position whose eigenfunctions are delta functions in the position basis, but what about, say, the momentum operator? Why should the spectrum of possible momenta be determined by finding the states (in the position basis) that are unchanged when acted on by infinitesimal spatial translations? Why can't one measure momenta for a state that is changed under spatial translations? $\endgroup$ – user1247 Jan 31 at 17:29
  • $\begingroup$ I think that's a mathematical property stemming from the Hilbert structure underlying QM, unrelated to the measurement itself. When you assume a position eigenbasis and its operator, it follows that its conjugate momentum would be the generator of spatial translation, and the uncertainty relation kicks in. $\endgroup$ – PhysicsTeacher Feb 3 at 3:49
  • $\begingroup$ I don't think that addresses why we can only measure the eigenvectors of that conjugate momentum. Stepping back, it vaguely makes sense that spatial translation operator would be related to p, and could be used as a mathematical tool to extract p from a given state, but left unexplained is why the measurement process happens to produce only eigenvectors of that particular operator. It seems these should be two different things: 1) some operator used as a math tool to extract info, and 2) the same op's spectrum are the results of the measurement process (no matter the precise interaction). $\endgroup$ – user1247 Feb 3 at 17:08
  • $\begingroup$ It is two different things. The physical measurement process diagonalizes the state in some pointer basis on one hand, and the choice of an operator to represent the measurement-process is just taking an operator that is diagonal in that basis on the other. That's why we can also choose arbitrarily the units in the measurement operator, it's just a mathematical tool used to represent in condensed form the effect of the physical measurement-process. $\endgroup$ – PhysicsTeacher Feb 6 at 4:34
  • $\begingroup$ So for example, if we measure spin 1/2, we find eigenvectors of the pauli matrices, but really we are measuring some pointer-basis that is correlated with the spin property, some position or momentum (say) of whether the electron took an "up" or "down" trajectory. So you are saying the spin operators (say) are clearly not fundamental, but are just a useful tool for describing a pointer-basis that has binary outcomes. OK, but the original question still deflates to asking, conceptually, why the operator that is diag when measuring p is the generator of space translations, and same for spin. $\endgroup$ – user1247 Feb 7 at 0:29
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I think that in the same sense as previous members have answered the hamiltonian has a certain role to play in the time evolution of the system.If viewed from the schrodinger picture it is the state of the system which evolves but in heisenberg picture it is the operator or rather the expectation value which evolves in time.These two completely equivalent to each other.I think it is an analogue of phase space volume in classical mechanics.And when any other operator commutes with the hamiltonian then it is called a constatnt of motion.This is what I have inferred from it.

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