Wednesday, February 17, 2021

Irreversibility of measurements is straightforward

For some reasons, the recent 24 hours brought an explosion of the confused talk about the irreversibility of quantum mechanics (and especially the measurements in it).

Well, people have posted lots of other crazy stuff about quantum mechanics. For example, the dog with many phase-shifted legs above, one created by the Czech 3D artist in London, Mr Ondřej Zunka, is surely totally cool. But the idea that this animation has anything to do with the way how our quantum mechanical Universe works is dumb beyond imagination. Quantum mechanics doesn't add many new legs to the dogs. Quantum mechanics just makes the uncertainty about the existing legs more unavoidable and adds some new causal relationships between the properties and their uncertainties.

If you think that the centipede dogs capture the wisdom of quantum mechanics, I really beg you, stop reading, this blog post and this website is not for you.

OK, the remaining 1% continued reading up to this moment. You know, some people make the irreversibility of the measurements (which I discussed e.g. in 2016, click) extremely mysterious, doubtful, deserving to be presented by people who are high.

In reality, it is straightforward enough so that people could actually discover these things. So how does the irreversibility work in classical physics in quantum mechanics?

Classical physics tells us to study some degrees of freedom such as \(x(t)\), \(p(t)\)... In a closed isolated system, they evolve according to equations that are reversible – and often downright symmetric in time. The values of \(x(t),p(t)\) may be measured by experiments, predicted, measured again, verified. In principle, we may assume that someone (God?) knows the values of \(x(t),p(t)\) without any uncertainties, error margins, and doubts.

Quantum mechanics also tells us to measure observables such as \(x(t),p(t)\). Those are the only things that can actually be measured by a single experiment, and that is why they are technically called "observables" (as a "noun") in quantum mechanics. It is because they are the things that are "observable" (an adjective). However, quantum mechanics says that \(x(t),p(t)\), even though they can be separately measured by a measurement, cannot have simultaneously well-defined values prior to the measurement. Mathematically, it boils to the fact that observables in quantum mechanics are represented by linear, overwhelmingly mutually non-commuting, operators, e.g.\[ xp - px = i\hbar. \] Almost all pairs of operators refuse to commute with one another in quantum mechanics. It implies that they can't have a well-defined, uncertainty-free value at the same moment. More generally, almost all observables (linear Hermitian operators) are uncertain at every moment, at all times. It follows that the single certain trajectory or history \(x(t),p(t)\) cannot exist. There cannot exist any observer and any God that would know such a trajectory because it would violate the nonzero value of the commutator \([x,p]\).

The wave function or the density matrix summarizes the knowledge about \(x,p\), or anything else that can be measured. But the wave function \(\ket\psi\) itself isn't observable; and it isn't an observable. It is not observable (an adjective) because whenever you measure something, you only detect a random state (e.g. a position eigenstate) that the original wave function had an overlap with, but this always brutally damages the wave function. The wave function is a generalization of the probability distribution \(\rho(x,p;t)\) on the phase space in a description of classical physics with uncertainties (which is especially relevant in classical statistical physics).

Exactly because it's impossible to have any counterpart of \(x(t),p(t)\) from classical physics – or, equivalently, a probaility distribution that is equal to 100% or 0% for all well-defined questions (\(\rho\) would be a delta-function on the phase space...), and this fact is guaranteed by the uncertainty principle, quantum mechanics demands that we generalize the description of physics in terms of \(\rho(x,p;t)\) that existed in classical physics, not the description without uncertainties, \(x(t),p(t)\). And the quantum generalization of \(\rho(x,p;t)\) is the density matrix \(\rho\). In general, \(\rho\) may always be written as a sum\[ \rho = \sum_i p_i \ket{\psi_i} \bra{\psi_i} \] and because all the future probabilities and future \(\rho\) is a linear function of the past \(\rho\), it is very useful to consider a special form of the density matrix\[ \rho = \ket\psi \bra\psi \] in terms of the state vector \(\ket\psi\). The situation in which \(\ket\psi\) is possible (it is determined from a suitable \(\rho\) up to the overall phase) is called the "pure state". That may be a name both for a \(\ket\psi\) as well as the \(\rho\) calculated from the \(\ket\psi\) through the equation above. A pure state is a special example of a density matrix that has the form enough. Conversely, a totally general density matrix, the mixed state, will behave in a way that may be totally reconstructed from the behavior of pure states e.g. \(\rho\) of the form \(\ket\psi\bra\psi\).

But we are still generalizing \(\rho(x,p;t)\) in classical physics, not \(x(t),p(t)\). The generalization of the uncertainty-free trajectories or histories \(x(t),p(t)\) from classical physics is impossible in quantum mechanics. This impossibility is a consequence of the uncertainty principle. Everything that is new about quantum mechanics really followed from the uncertainty principle. So quantum mechanics generalizes the probabilistic description on the phase space. This 2011 blog post was dedicated to this quantum generalization of the Liouville equations etc. (Thanks, Erwin.)

Fine, in the absence of measurements, or between two measurements, the density matrix or the state vector evolve unitarily (which also guarantees that the evolution is reversible)\[ \ket\psi \to U \ket\psi\quad {\rm or} \quad \rho \to U \cdot \rho \cdot U^\dagger. \] But the measurements are abrupt events in which the probabilities change because the observer has just learned something. The corresponding change of the wave function or density matrix is a complex, quantum mechanical generalization of the Bayesian inference which has existed before quantum mechanics, too. The probability of possibilities that were incompatible with the fresh experimental result drops to zero; others are adjusted according to the formula. The quantum mechanical formula for the reduction of the wave function is\[ \ket\psi \to P_\lambda \ket\psi \] where \(P_\lambda\) is the projection operator on the subspace of the Hilbert space that contains the eigenvalues of the freshly measured observable \(L\) with the eigenvalue \(\lambda\) that was actually obtained. For density matrices, you must place the (Hermitian) projection operators on both sides of \(\rho\). The post-measurement state (\(\ket\psi\) or \(\rho\)) obtained from this formula is not normalized. You may add a normalization constant if you prefer to work with normalized wave functions or density matrices (with the trace equal to one) at all times.

Which of the eigenvalues \(\lambda\) is obtained by the measurement of \(L\) is uncertain in general. This is what all the uncertainty means. It is always the uncertainty about the future measurements. The probability of any particular outcome \(\lambda\) is obtained by Born's rule. Now, the last displayed formula with the projection operator is clearly irreversible e.g. because it sets the wave function at "most places" (in a representation) equal to zero and eliminates the information about "what the wave function was" over there.

In fact, it is much more obvious that the measurement is irreversible. Mathematically, an irreversible operation must be a map (a function): a fixed rule that maps any value of the argument (or initial wave function) to a particular result (e.g. the final wave function). But the replacement of the wave function that involves the projection operator \(P_\lambda\) isn't even a map. It is not a map because it depends on the random generator which picks \(\lambda\). Because it is not a map, it clearly cannot be reversible.

There are lots of physical examples and interpretations that explain why the measurement is clearly irreversible. For example, the von Neumann entropy \(S = -{\rm Tr}\, \rho \log \rho \) generally changes. Also, when we measure an object in a nice state, we perturb it. When we measure the position of a harmonic oscillator that was in the ground state \(\ket 0\), we get a position eigenstate instead. The oscillator starts to oscillate in brutal ways, it is no longer an energy eigenstate. It is not easy to return to an energy eigenstate. Well, we may measure the energy after we measured position and we are guaranteed to get an energy eigenstate after that. But it doesn't have to be the ground state \(\ket 0\) anymore. It may be an excited state. This change is possible because the energy \(H\) doesn't commute with the position \(x\). When this transition to another level occurs, it is not really a sign of the violation of the energy conservation law. The energy including the observer is still conserved and the transition of the oscillator to another level may be interpreted as a flow of energy between the oscillator and the observer.

Fine. So the observation (the measurement) is clearly irreversible. It is also dependent on the choice of the observer because there can't be any objective definition of what an observation is and what it isn't. From a very accurate external observer's viewpoint, what happens is a constant evolution to superpositions of all possibilities (superpositions even for macroscopic objects such as cats, humans, and planets) which are only projected when the external, very rigorous observer makes his careful, precise experiments. On the other hand, people and animals live inside and they have less rigorous standards for what they consider measurements and knowledge. They make measurements and update their knowledge all the time.

There is nothing wrong about the observer dependence of the wave functions or density matrices – or, equivalently, with the observer dependence of all the information about the state of the world – in quantum mechanics. It has existed in classical physics, too. The nontrivial distributions \(\rho(x,p;t)\) were also nontrivial just because we considered an observer with an incomplete knowledge. So the nontrivial (non-delta-function) character of such a distribution was an artifact that depended on an observer – and on his imperfect knowledge.

This is exactly the case in quantum mechanics, too. The only difference is that classical physics was compatible with the assumption that an ideal observer who knows a delta-function-like distribution \(\rho(x,p;t)\), and therefore an actual sharp trajectory \(x(t),p(t)\), exists. Quantum mechanics is built on the uncertainty principle that doesn't allow such an observer. Any observer – any physically legitimate description of physical systems according to quantum mechanics – will be a complex, quantum generalization of the nontrivial distributions \(\rho(x,p;t)\) that look nothing like the wave functions. Instead of real values of the function \(\rho(x,p;t)\) at individual points of the phase space, quantum mechanics needs to store all the information about the probabilities in \(\ket\psi\) or \(\rho\) that mostly contain complex numbers. (Wigner's tricks may transform the density matrix to a quasi-probability distribution on the phase space but this is no longer positively definite in general.)

If you are intelligent and you are looking at physics rationally, you will understand these basic rules after reading this blog post and/or an equivalent one (or a section in a book). You are not supposed to demand 100 pages of gibberish to memorize. If you don't get these things quickly enough, you are either too stupid or too stubborn. In either case, you are incompatible with modern physics. Give it up and go away, you have no chance to understand it correctly.

I don't plan to proofread this text.

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