Does someone know of a clear (pedagogical) example where one can really see(with the math) where interaction and measurement are not synonymous in quantum mechanics?

  • I know that every measurement involves a certain interaction with the outside world (e.g. momentum gain from a photon), which results in the system collapsing into one of its eigenstates, meaning a pure state.

  • On the other hand, it is also known that not all interactions result in collapsing into eigenstates of the system, so they are in principle very different from what we call "measurement".

It would be definitely nice to also see a bit of the math behind, maybe for simplicity just restricting it to operator algebra and showing how measurement and interaction are defined, shedding a clear light on their difference. I must admit, from a purely physical point of view, I don't know their difference either.


6 Answers 6


This is not a settled question. Just as it is still debated whether or not there is wavefunction collapse, so is it debated what exactly we should understand by a measurement. In the following, we will go through the ideas behind the von Neumann measurement scheme, which is one way to try and talk about measurement in quantum mechanics.

An interaction happens every time when two quantum systems cannot be neatly separated anymore. Given two systems with Hilbert spaces of states $\mathcal{H}_1,\mathcal{H_2}$, they are interacting if entangled states between them are allowed, that is, if the physical Hilbert space of states of the whole system is not $\mathcal{H}_1 \times \mathcal{H}_2$, but $\mathcal{H}_1 \otimes \mathcal{H}_2$. A measurement is a special kind of interaction.

Let $\mathcal{H}_q$ be the Hilbert space of the object of which we want to measure some property, and $\mathcal{H}_m$ the Hilbert space of the apparatus which we will use to perform the measurement. Note that the von Neumann scheme treats the apparatus as a quantum object, just like everything else. The distinguishing quality of a suitable measurement apparatus is that the unitary time evolution of the combined system (perhaps + the environment, too) acts as

$$ \lvert \psi \rangle \otimes \lvert \phi \rangle \overset{\mathcal{U(\tau)}}{\mapsto} \sum_n c_n \lvert \psi_n \rangle \otimes \lvert \phi_n \rangle$$

where $\psi_n$ are some uniquely determined states of the measured system and $\phi_n$ are orthogonal states of the apparatus w.r.t. to the operator we want to measure that can be "read off macroscopically" (they are "pointer states"). Now, the apparatus is supposed to be "macroscopic", so we can determine the pointer state just by looking at it. So, each pointer state corresponds uniquely to a state $\psi_n$ of the system. This process is called measurement of the second kind, and of the first kind if the $\psi_n$ are eigenstates of an observable.

Note that the form the unitary evolution is supposed to take here is special. In general, it looks like $$\lvert \psi \rangle \otimes \lvert \phi \rangle \mapsto \sum_{i,j} c_{ij} \lvert \psi_i \rangle \otimes \lvert \phi_j \rangle$$

where we would not have such a correspondence between pointer states and object states. The $\lvert \psi_n \rangle \otimes \lvert \phi_n \rangle$ occurring in the time evolution of the measuring/decohering system are not a basis of the (tensor) space of states, while, in general, the time evolution should entail all basis vectors $\lvert \psi_i \rangle \otimes \lvert \phi_j \rangle$, of which the basis for the "special" time evolution we suppose for a measurement apparatus is just a subset (namely that where $i = j$). One can devise many arguments why it should take such a form when going to "classical sizes", this is the study of decoherence and einselection.

  • $\begingroup$ Thanks for this. Very interesting. I hope you don't mind if I ask some small question as the answer is rather compact. 1) If I understand correctly, $\otimes$ stands for a tensor product, which means that the $\mathcal{H}_1 \otimes \mathcal{H}_2$ results in a Hilbert space dimension of $dim{1}*dim{2}$, but what kind of product is $\times$? 2) Could explain a bit more how one should understand "if entangled states between them are allowed"? 3) from your answer, to me the distinction between measurement/interaction remains very foggy, what interactions aren't measurements then? Thanks $\endgroup$
    – user929304
    Commented Dec 12, 2014 at 15:19
  • 1
    $\begingroup$ @user929304: 1. $\times$ is the Cartesian product, which is the "classical notion" of a combined space. 2. States which are in $H_1 \otimes H_2$, but not in $H_1 \times H_2$ are called entangled. Formally, any state not in the image of the Segre embedding is entangled. 3. Measurements are characterized by the special form of the time evolution (for small times $\tau$, even!). Any interaction that does not have such a time evolution is not a measurement in this scheme. $\endgroup$
    – ACuriousMind
    Commented Dec 12, 2014 at 15:39
  • 3
    $\begingroup$ @PhotonicBoom Be careful, you shouldn't think of $\mathcal{H}_1 \otimes \mathcal{H}_2$ as a requirement for entangled systems, it is rather a requirement for all composite systems in quantum mechanics! In Classical Mechanics we use the Cartesian product to compose state spaces of systems, which if you will, is a composition that doesn't allow one to represent correlated or interacting systems in it. Another way of composing vector spaces, is by Tensor Products, which preserves all the required properties (e.g. distributivity...), this non-Cartesian way of composing vector spaces is (...) $\endgroup$
    – Ellie
    Commented Dec 22, 2014 at 0:33
  • 3
    $\begingroup$ @PhotonicBoom considerably larger than $\mathcal{H}_1 \times \mathcal{H}_2.$ (dim($n \times m $)$=n+m,$ dim($n \otimes m $)$=n*m,$) Now $\otimes$ may not be the only way of combining vector spaces with inner product, but it's one that has all the needed properties and is capable to include all sorts of composite systems, be it interacting in an entangled manner or not! Meaning any state $\sum_{i,j} c_{ij} \lvert \psi_i \rangle \otimes \lvert \phi_j \rangle \in \mathcal{H}_1 \otimes \mathcal{H}_2.$ Now entanglement arises for special coefficients $c_i^{1}$ and $c_j^{2}$ in the superposition. $\endgroup$
    – Ellie
    Commented Dec 22, 2014 at 0:33
  • 1
    $\begingroup$ This doesn't actually answer the question... $\endgroup$
    – DanielSank
    Commented Dec 24, 2014 at 14:49

When does an interaction drop the system into an eigenstate? (i.e. when is a measurement=)

This is an ill-posed question because, first of all, the system $S$ doesn't drop into any state but each observer $O$ has a state about it, as a state $\rho$ is nothing but the coding of past measurements (so it should be named with reference to being dependent on $O$ and $S$ and not only the latter, that is why 'the state lives on the observer-system link' and is not intrinsic to just the system, that would be its algebra of observables). Interactions create correlations during the isolated evolution $\hat{U}(t_1,t_0)$ of the closed system $O\otimes S$ and only those which are stable and robust in time can be regarded as measurements from the point of view of $O$, where the process of decoherence has a major role to play. In this sense, your question is of a phenomenological nature regarding when and how different interactions between observers, systems, aparata... create stable, robust and reliable correlations between the observables of $O$ and $S$ as seen by another observer $O'$, something which depends completely on the nature of the hamiltonian $\hat{H}_{O-S}$, the strength of the couplings and the time scales involved.

I believe the answer by @ACuriousMind is quite to the point. This topic easily turns into philosophical discussions even by experts (that's why he says it is "unsettled"), but I believe some conservative but objective conclusions are unavoidable. You ask for some mathematical description but I believe the difficulty is precisely in epistemological confusions and not formalism. For a more mathematically detailed exposition please refer to the articles and the end of this answer:


The world is made of interacting physical systems $S_i$ characterized by their degrees of freedom, i.e. observable properties $\mathcal{A}^{(S_i)}_j$ subject to algebraic relations, in general noncommutative, which characterize their spectra of measurable values and complementarity (in particular, commutator relations are equivalent to specifying Heisenberg's uncertainty relations to pairs of observables and they are also usually enough to deduce their spectral representations). General operator algebras with physically motivated analytic properties are represented by operators $\hat{A}^{(S_i)}_j$ acting on Hilbert spaces. Any physical description must be done from within the world by some system studying other systems, thus fixing a system of reference $\mathcal{O}:=S_0$ specifies an observer which/who will describe the rest (or the subsystem of interest $S$ disregarding the rest as environment) by the correlations it gets by measuring their observables. Observables are then what characterize the connections, the links, between systems since they are 'channels of information' through which systems may affect/know each other via interaction. Unknowing yet how this works, $\mathcal{O}$ interacts with $S$ obtaining values of the spectra of $\hat{A}^{(S)}_j$, but if any two of them do not commute, $[\hat{A},\hat{B}]\neq 0$ the eigenvalues measured are incompatible to be well-defined at the same time

Example: measuring spin $\hat{S}_z$ in a Stern-Gerlach apparatus gives a definite eigenvalue $\pm 1/2$ with $50\%$ chance each, say $+$, this is preserved if another $\hat{S}_z$ is measured in succession by the same or other observer, $100\%$ is again $+$, but if $\hat{S}_x$ is measured afterwards then a new measurement $\hat{S}_z$ reveals that the original information $+$ was lost and again $50\%$ is obtained, therefore incompatible observables do not have common eigenstates and represent properties not well-defined at the same time.

Therefore $O$ measures from $S$ at the same time only at most maximal sets of compatible observables because these ideally have definite sharp eigenvalues at the same time, thus any system $S$ is specified by 'complete windows of interaction' given by all the possible maximally compatible subsets $\{\hat{A}, \hat{B},...\},\{\hat{X},\hat{Y}...\}...$, of its observables algebra. Now, at every interaction time between $O$ and $S$, $O$ obtains through its senses/apparata at most a collection $|a,b...\rangle$ of simultaneously well-defined eigenvalues from a complete set of observables (which is selected depends on the senses/apparata intervening at that and each interaction). So $O$ 'observes' system $S$ at $t_0$ in the state $|\Psi(t_0)\rangle =|a,b,..\rangle$, because the maximally compatible properties defining $S$ have those values at that time. Now, at a later time $O$ interacts again with $S$ but through another compatible set of observables giving a state of measured values $|\Psi(t_1)\rangle =|x,y,..\rangle$. The collection of eigenvalues of the compatible operators form an eigenstate in the Hilbert space representation.

Everything that non-relativistic mechanics is about is to study how to predict the probability that having measured a system in state $|\psi\rangle$ it will be observed at state $|\chi \rangle$ at a later time. That is given by a probability distribution on the space of possible future states for each current state, which in the general noncommutative case is the same as giving a normalized positive linear functional on the observable algebra, and by Gleason's theorem every such functional can be represented by a density operator $\rho$. Indeed, complete eigenstates $|\Psi\rangle$ of some maximal set of compatible observables are in bijective correspondence to density operators which are rank-1 projectors, i.e. $\rho_\psi=|\psi\rangle\langle\psi|$, called pure states (general mixed states are convex combinations of pure states and represent statistical mixtures of our uncertainty which pure state was actually measured). In this way, a state $\rho$ of $S$ is not only the record of the values $O$ observed but a probability disposition for future observations, since Gleason's theorem guarantees that the probability of measuring a future eigenvalue $m$ of observable $\hat{M}$ is given by the expectation value of its projector $\langle \hat{P}_m\rangle =tr(\rho\cdot\hat{P}_m)$, and by spectral decomposition the probabilities and expectations of any observables are obtained, including then transition probabilities between eigenstates $\mathcal{P}(\psi\mapsto\chi)=|\langle\chi |\psi\rangle|^2 =tr(\rho_\chi\cdot\rho_\psi)$.

This is the kinematics of quantum mechanics because no dynamical evolution has been taken into account between observer-system interactions at $t_0$ and $t_1$. When $O$ is not interacting with $S$, the latter is considered isolated or a "closed system" and is left alone to evolve during $\Delta t=t_1-t_0$. Now if $S$ is to be considered the same system at different times, everything which characterizes it must remain invariant, i.e. the observable algebra must be the same at different times, which means the operators representing it must be related by an algebra-automorphism $\mathcal{U}(t_1,t_0)$ tending to the identity when $t_1\rightarrow t_0$, this by Stone-von Neumann theorem is given by a unitary operator $\hat{U}(t_1,t_0)$ transforming the operator representation of the algebra as $$\hat{A}(t_1)=\hat{U}^\dagger (t_1,t_0)\cdot\hat{A}(t_0)\cdot\hat{U}(t_1,t_0)$$ which is nothing else that Schrödinger's equation in Heisenberg's picture. Therefore, if $O$ measured a state $\rho_0$ initially, if at a time $t_1>t_0$ observer $O$ interacts with $S$ to measure observable $\hat{B}(t_1)$ and obtain eigenvalue $b$ with probability $tr(\rho\cdot\hat{P}_b(t_1))$ where the projector evolves by the same unitary transformation by the same reasoning. As soon as $O$ perceives/detects/measures a new value $m$ of any observable $\hat{M}$ of $S$, it must update the information it had stored in $\rho_0$ by the new state $$\rho_1 =\frac{\hat{P}_m(t_1)\cdot\rho_0\cdot\hat{P}_m(t_1)}{tr(\rho_0\cdot\hat{P}_m(t_1))}.$$ This is Lüder's rule and is the noncommutative generalization of Bayes' rule for updating probability distributions by conditioning upon new information, as is justified by Duvenhage's article linked above or in Busch-Lahti articles and their books.

Now, the nature of the evolution operator IS WHAT ENCLOSES THE INTERACTIONS of the subsystems within $S$ as seen by $O$. Why is so? because a priori the automorphism preserving the identity of the system may depend on other systems, and that is what is meant by 'interaction'. Indeed, by Stone's theorem the unitary evolution operator can be recast into its infinitesimal generator the hamiltonian operator: a hermitian operator $\hat{H}_S$, with bounded from below spectra, characteristic of the isolated evolution of the whole system $S$, satisfying $U(t_1,t_0)=\exp (-i\hat{H}_S\Delta t/\hbar)$. The operator $\hat{H}_S$ must depend on the observables of $S$ and those of the other systems interacting with it, for them to intervene in the time evolution. Reasons of symmetry (e.g. homogeneity and isotropy) motivate the terms of the hamiltonian responsible of free evolution (kinetic terms like $\hat{p}^2/2m$) which only involve the observables of the system, so interactions must appear in the form of couplings between different systems observables (like $\hat{\mathbf S}_1\cdot\hat{\mathbf S}_2$ for spin interactions). Given all this, any observer $O$ just needs to know the Hamiltonian of $S$ (or the free Hamiltonian of its subsystems and the interaction couplings) to be able to evolve the observables $\mathcal{A}^{(S)}_j$ in an operator representation via Heisenberg's equation using $\hat{U}(t_1,t_0)$, wich along with the information it already has in a previous measured state $\rho_0$ allows it to establish all the probability distributions of future measurements via the laws established above. But time evolution can be mathematically mapped into state-space instead which just amounts to evolving $\rho(t)$ and fix $\mathcal{A}^S$ giving the usual Schrödinger's equation.

The categorical error in ontology is bestowing reality to intermediate states $\rho(t)$ while $S$ is isolated between measuring times $t_0<t<t_1$: $\rho(t)$ evolves $\rho_0$ into a superposition of eigenstates which generically do not correspond to a pure state of measurements/information of any observer, i.e. Schrödinger's picture intermediate states are not justified to represent physical sates from an operational empiricist interpretation of mechanics. We should suspend judgment about the realist asserting a superposition of all histories and such, because they are not seen by any physical observer and come by from a mathematical convenience outside the original physically motivated meaning of observables and states we started with.

It is clear that the description any observer can make of any system, is intrinsically incomplete because it does not include the $O-S$ interaction. This is unavoidable as observables are meaningful only as "communication channels" between systems. However, quantum mechanics is complete because another observer $O'$ can model by the same theory the measurements made by $O$ by considering the coupled system $O\otimes S$, since now $O'$ has access to both the observables of $O$ and those of $S$ and can study their couplings. Hence, a measurement of $S$ by observer $O$ is nothing but an interaction in the time evolution of the $O\otimes S$ system as seen by another observer $O'$. What $O'$ sees is degrees of freedom of $O$ getting correlated with degrees of freedom of $S$ (e.g. the apparata of $O$ is always measured by $O'$ to be in the same position when $S$ is measured by $O'$ to be in eigenvalue $a$), and those correlations of observables are measurements. The dynamics used by $O'$ through $\hat{U}$ uses $\hat{H}_{O-S}$, something which $O$ could not use, that is why it was not able to describe its own interaction and measurement process. Besides, quantum mechanics is consistent because dynamics guarantees that $O$ and $O'$ get the same values when comparing upon measuring the eigenvalues of the same observable of $S$ if this was left to evolve isolated.

Read Smerlak, Rovelli or Englert for discussions of how to apply the quantum formalism consistently and check out that different observers may have different information on the system but neverthelesss agree. (In short, they agree because either they are decohered and measure compatible observables so they get the same eigenvalues, or they are observers out of contact who are regarded as part of the closed system from the point of view of each other; however if they are to compare measurements they must interact, thus decohering and each observer will see <an eigenvalue of the system> and <the other observer seen that eigenvalue> both with the same probability, grating consistency; cf. Rovelli's articles).

van Kampen’s moral: If you endow the mathematical symbols with more meaning than they [operationally] have, you yourself are responsible for the consequences, and you must not blame quantum mechanics when you get into dire straits…

SUMMARY: the answer to what makes the system project into an eigenstate is precisely the tricky part: the system doesn't drop into anything, only the observer's information about the system "drops" into an eigenstate, because eigenstates are the results of measuring observables of the system by the observer. Your "dropping" is the "collapse of the wavefunction" which is just the noncommutative analogue of the classical Bayesian update of probability distributions after new knowledge is acquired. There is nothing "collapsing" or "dropping" at each measurement because the state is an informational device, a book-keeping of the observer-system past interaction history. The whole point of my answer is that physical interpretation is justified only for Heisenberg's picture: states do not evolve and then collapse, system's observables evolve and you update the info you have at each measurement. Quantum Mechanics just gives the conditional probabilities of questions "having measured a,b.. what is the probability of measuring x,y.. later?". Thus the "state is on the observer-system link" because different observers may have different information of the system due to different past interaction history, so the state is not something intrinsic of the system but relative/relational. However as soon as observers get into contact, dynamics guarantees they will see the same measurements as long as they are of compatible observables and the system was isolated. This is because in the quantum case measuring an observable destroys previous information on a complementary observable. In classical mechanics a common state of the system among observers is possible because it is the commutative limit and all observables have well defined values at the same time for a pure state.

Therefore a measurement is any interaction of the system $S$ with the observer $O$ which establishes a robust, stable correlation between some of their degrees of freedom, but this is a process that can only be described by another observer $O'$ who will see it depend on the couplings between the observables of the compound system $S\otimes O$. So only those interactions in the coupling of the hamiltonian $\hat{H}_{O-S}$ which create correlations robust enough to appear at the experimental sensitivity of $O$ will qualify $O'$ to say that the interaction of $O$ and $S$ was a measurement.

Einstein taught us time and length are relative, Heisenberg taught us that quantum entities do not have absolute definite eigenstates in-between measurements (because if insisting on Schrödinger's picture, the general intermediate superposed state given by unitary evolution, is not justified to be empirically physical in the sense that there is no observer for which it is an eigenstate, as generically there are no physical observables which diagonalize it).

  • $\begingroup$ Thanks for follow up comments, feel free to add them at the end of your answer, as an edit like "In response to questions: ...". So we can free up the comments section for further questions. If the interaction exists only on the observer-system link as you say, then how can two different observers ever obtain the same measurement if not for the collapse of the system into an stationary state after the first measurement? $\endgroup$
    – user929304
    Commented Dec 18, 2014 at 15:39
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    $\begingroup$ @user929304 following your suggestion I deleted my comments and added them partially along with a new long more formal justification of my answer explaining every essential concept of the formalism from the beginning to frame your question in context. I hope it is more clear now. $\endgroup$ Commented Dec 18, 2014 at 19:51
  • $\begingroup$ Thank you for the write up. It will be a long read for me, and there are a lot of things that will need some digesting. I will get back to you with questions once I have roughly understood your answer. thanks $\endgroup$
    – user929304
    Commented Dec 19, 2014 at 13:53
  • $\begingroup$ @user929304 thanks, you should also download the referenced articles and try to read them, as they elaborate on the same idea and its subtleties/details on and on. Basically the answer is the first paragraph and summarized in the last three, and the explanation tries to make clear that your question is a subtle one because it is actually the hardcore confusion of the theory, namely, that the quantum state is not a 'material substance' of the system, but 'informational knowledge' of the observer. Wittgenstein: 'do not confuse substance [the system] with substantive [the relative state]'. $\endgroup$ Commented Dec 19, 2014 at 17:55

All measurements are based on interactions , but not all interactions are measurable. The "set of measurements" is a subset of the "set of interactions".

The simplest mathematical way to see this is by Feynman diagrams. Feynman diagrams have one to one correspondence with integrals, and when they describe a measurement, the crossection or lifetime can be easily set up for calculation.

feynman diagrams

Feynman diagrams for first order interactions

Top left the diagram describes a number of interactions: if time goes at +y it is electron electron scattering into electron electron by exchanging a virtual photon . It can also be interpreted with the time going at +x , then it is electron positron scattering into a virtual photon which turns again into electron positron.

top right: time on +y, a neutron decaying into a proton an electron antineutrino and an electron.

botom left: time +y , green quark scatters off/turns into a blue quark by exchanging a green-antiblue gluon , and out goes a blue quark on the left and a green quark on the right. Colors are a label for specific quantum numbers, a type of charge.

botom right: time along +y, a proton scatters of a neutron by exchanging a pion: it is an effective diagram from the time before quarks were discovered , and the calculation worked !

Here is a write up which shows how Feynman diagrams turn into calculations.

Three of the four diagrams describe first order calculations of a measurement, the ones in red. What one can measure in an experiment are on mass shell particles , i.e. the fourvector describing the particle gives the mass square of the particle. Off mass shell, virtual particles are not measurable in an experiment, their existence is validated by the fact that the mathematics works and one gets a fit of data and calculations.

The fourth blue green diagram describes mathematically an interaction , where the incoming and outgoing are quark lines and the exchanged boson is a gluon. All three are off mass shell, as quarks and gluons cannot be isolated in beams and their crossection of interaction measured. For the calculation to have a meaning, the external lines must be dressed with at least another quark (describing a meson). This is a clear instance where there is an interaction not accessible to measurement without a higher level intervention/approximation: incoming are nucleons or mesons, outgoing quark jets or more nucleons and mesons.

Now all of the above is true as long as we are talking of electrons and protons with some energy that can be measured on individual electrons and protons.

Electromagnetic interactions of very low energy are the interactions that hold the world we observe together, from the aggregate of atoms and molecules emerges the classical physics of mechanics electrodynamics and thermodynamics. In principle the collective "interaction" could be written in terms of Feynman diagrams , but the input and output lines would be mostly virtual, off mass shell even though slightly. The experimenter can measure only the collective set, temperature, pressure, classical trajectories,.... The zillions of interactions going on with the ~10^23 of atoms/molecules are unmeasurable.

  • 1
    $\begingroup$ Take the simple crystal which is in a quantum mechanical state. X ray studies do not disturb those state, the elastic scattering gives information, but the x rays interacted with the crystal and were measured on the film. So the measurement can be non destructive as far as the isolated wave function goes but still there are real particles measured to get the information. Collapse means that the system interacted , the old wavefunction is no longer valid, an instant on the probability distribution( psi**2) was taken for the particles in the game, and there is a new solution. $\endgroup$
    – anna v
    Commented Dec 12, 2014 at 16:20
  • $\begingroup$ In the crystal example, the system x-ray+crystal psi "collapsed" each time to give us the information $\endgroup$
    – anna v
    Commented Dec 12, 2014 at 16:24
  • $\begingroup$ But if the measurement is done on the film, then we're studying the x-ray interaction with the film and completely exclude the crystal from the picture, no? I always thought the interaction with crystal resulted in collapsing the x-ray photons into some eigenstates, which upon further measurement (e.g. on a film as you say) will not change anymore. $\endgroup$
    – user929304
    Commented Dec 12, 2014 at 17:11
  • $\begingroup$ There is a psi**2 of the system crystal+xray, gives the probability to find the xray at some (x,y) on the film. The spot is the "collapse", the accumulation is the probability distribution. The x-ray photons are scattered ( Feynman diagram can be written) by the atoms/molecules in the crystal . $\endgroup$
    – anna v
    Commented Dec 12, 2014 at 19:13

I don't intend to repeat the points already made in the other answers, consider this as a small addition to those, with the intention to give a more practical description (reminding some of the basic ideas) without getting into observer-philosophies.

Quite clearly the act of observing, i.e. measuring a quantum system can be done via many different non-equivalent ways. E.g. one can measure a photon's position using a silvered screen plate, which results in destroying the photon afterwards. One can use polarization filters to measure polarization of a photon (just an example), relying on momentum exchange with electrons and inducing a current. Photon scattering is another possibility, where the system then interacts with a photon...you get the idea! This should clarify that a measurement is by definition nothing but an interaction between an external physical system and the quantum system under study, where the original system does not necessarily comply a unitary evolution anymore, as was made clear in the other answers. (we'll get back to this point).

Now you may ask what happens to the postulate in QM, which claims that quantum systems evolve according to unitary evolution? Well this postulate holds for closed quantum systems, i.e. quantum systems without an interaction with an external system. But upon a measurement, nothing is undermined, as one can always define a larger system, containing the measurement quantum system together with the original system under study, which would then again correspond to an isolated system evolving unitarily.

From the other answers you must already know that in QM one talks about different kinds of measurements. The two more practical and easily understandable ones are:

  1. General quantum measurement: Evolution described by a unitary transformation.

  2. Projective measurement

A measurement is performed in order to obtain definite information about a certain observable describing our system at hand. In QM observables (physical quantities you can measure) are described by self-adjoint Hermitian operators $M$, which have a lot of nice properties, some of them are:

  • Self-adjoint: $M^{\dagger}=M$, where $\dagger \leftrightarrow (^*)^T:$ Transpose of complex conjugated operator.

  • They are normal: $MM^{\dagger}=M^{\dagger}M$

  • They're diagonalizable and have real eigenvalues, $m \in \mathbb{R}$.

  • They can be represented by a spectral decomposition (if positive definite), i.e. given the eigenvalues $m_i$ and corresponding eigenstates $|i\rangle: \rightarrow$ $M=\sum_i m_i \left|i\rangle \right. \left.\langle i\right|$

Now we describe general quantum measurements by the set of such Hermitian operators $M$, which act on the state space of the system at hand. The eigenvalues $m$ correspond to the possible outcomes of the measurement, more precisely upon measurement of a system initially in state $|\psi\rangle$, the result $m$ can occur with the probability: ($M_m$: index $m$ denotes the measurement outcome that can occur, associated with the projector $\left|m\rangle \right. \left.\langle m\right|$) $$prob(m)=tr\left(M_m^{\dagger}M_m \left|\psi\rangle \right. \left.\langle \psi\right|\right)=\langle \psi \left|\right.M_m^{\dagger}M_m \left|\right. \psi \rangle$$

The state of the system after measurement is then: $$\left| \psi_m \rangle \right.= \frac{M_m \left|\right. \psi \rangle}{\sqrt{\langle \psi \left|\right.M_m^{\dagger}M_m \left|\right. \psi \rangle}}$$

Now if we limit the measurement operators just to the ones that fulfill the completeness relation, then we speak of general quantum measurements. Completeness relation (a necessity for the sum of probabilities to be one.): $$\sum_m M_m^{\dagger} M_m=I.$$

Whereas in projective measurements in addition to fulfilling the completeness relation, the operators also satisfy the condition of orthogonal projectors (orthogonal eigenvectors), i.e. $M_mM_{m'}=\delta_{m,m'}M_m.$ Projective measurements have a lot of other nice features which I will not get into here, but the main point to retain is that, due to the orthogonality condition, they allow for a better distinction between quantum states of a system. As was explained in the answer given by ACuriousMind, such measurements are not necessarily equivalent to the general quantum measurements, as they do not follow a unitary time evolution. But they can be made equivalent into one another if one considers an augmented isolated system (external device+quantum system), in which unitary transformations occur. An important property of projective measurements is that they are repeatable, meaning if after the first measurement (done by Alice e.g.) $m$ was obtained, a second measurement (done by Bob e.g.) would give the same outcome, and the repetition doesn't change the state of the system. This repeatability of projective measurements does not imply that all measurements in QM are repeatable.

Main references:


Quantum optics demonstrates the existence of interaction-free measurements: the detection of objects without light—or anything else—ever hitting them.

Paul Kwiat, Harald Weinfurter and Anton Zeilinger SciAm November 1996


  • $\begingroup$ What is your question then? $\endgroup$
    – dolun
    Commented Dec 13, 2014 at 13:03
  • 2
    $\begingroup$ Please refrain from link-only-answers. Try writing a bit on what you have to share, and leave a link for further reference! Thanks. $\endgroup$
    – Ellie
    Commented Dec 13, 2014 at 17:22

QM says that interactions are not always measurements. The canonical example is Schrödinger's cat. We have countless trillions of interactions per microsecond for hours - the state of the system is a very complex huge wave function which lives in a hilbert space one can only describe as awesome. Then the box is opened, and the cat is dead or alive - but its state is complex.

This absurd construction (for instance that the individual smell molecules both exist and don't exist for hours) is ridiculous. QM is almost certainly wrong, and there exists some complexity limit where interactions and measurement are the same thing.


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