23
$\begingroup$

I understand that the quantum no-deleting theorem dictates that it's impossible to delete quantum information, so what happens to the quantum information of a particle and an antiparticle when they annihilate each other?

$\endgroup$
26
$\begingroup$

Particle + antiparticle annihilation preserves quantum information. It is often said that annihilation creates a pair of photons, but that's a big simplification. It only applies to the electron + positron, and even in that case it may lead to more than 2 photons. If the electron's and positron's spins are parallel, an odd number of photons must be produced, so in that case at least 3 photons are produced, since production of a single photon is prohibited by momentum conservation. If the electron's and positron's spins are antiparallel, then an even number of photons are produced. In either case, producing more than the minimum number of photons has a low probability. If the electron and positron have very high kinetic energy, other particles may be produced, eg D or B mesons, or even the weak gauge bosons, if the KE is high enough.

The interaction cross-section for neutrino + antineutrino annihilation is very small, and the reaction is only probable for extremely energetic particles, since the result is a Z boson, which has a huge mass, as mentioned in this answer.

When more complex particles annihilate, things get messy. From Wikipedia:

[...] when a proton encounters an antiproton, one of its quarks, usually a constituent valence quark, may annihilate with an antiquark (which more rarely could be a sea quark) to produce a gluon, after which the gluon together with the remaining quarks, antiquarks, and gluons will undergo a complex process of rearrangement (called hadronization or fragmentation) into a number of mesons, (mostly pions and kaons), which will share the total energy and momentum.

The newly created mesons are unstable, and unless they encounter and interact with some other material, they will decay in a series of reactions that ultimately produce only gamma rays, electrons, positrons, and neutrinos. This type of reaction will occur between any baryon (particle consisting of three quarks) and any antibaryon consisting of three antiquarks, one of which corresponds to a quark in the baryon.

In summary, the results of matter + antimatter annihilation preserve quantum information because the results depend on the type(s) of particle, their spin, and their kinetic energy.

$\endgroup$
  • 1
    $\begingroup$ I must be missing something. When you say "In summary, the results of matter + antimatter annihilation preserve quantum information because the results depend on the type(s) of particle, their spin, and their kinetic energy," that seems equivalent to saying that if you know the sum of several numbers, you will somehow know what each of those numbers is. $\endgroup$ – D. Halsey Dec 31 '18 at 14:34
  • 2
    $\begingroup$ @D.Halsey Except that annihilation doesn't produce a single "sum", it produces multiple particles, and conserves energy and momentum (linear & angular). $\endgroup$ – PM 2Ring Dec 31 '18 at 14:42
  • 1
    $\begingroup$ @D.Halsey Not quite, because not all numbers are possible. A better analogy is factorization - when you multiply prime numbers together, you get one number out of N, but you lose no information. It works both ways, which is quite critical for physics. $\endgroup$ – Luaan Dec 31 '18 at 15:33
  • 1
    $\begingroup$ @PM2Ring: Any neutrino and antineutrino can annihilate. They can produce a virtual Z boson (which can have much more mass than either of the original two particles). The Z boson can then decay through virtual intermediate products into a pair of photons. This is extremely unlikely, both because the Z boson can't decay into photons directly and because the Z boson is so massive, but I believe it is theoretically possible. $\endgroup$ – Peter Shor Dec 31 '18 at 16:12
  • $\begingroup$ @Peter Fair enough. But it is very low probability. Eg, the antineutrinos released by a beta emitter don't have much chance of reacting with a solar neutrino. But I'll adjust my answer. $\endgroup$ – PM 2Ring Dec 31 '18 at 16:32
13
$\begingroup$

First of all, the particles don't disappear: they turn into other particles. But it's not as simple as just "electron + positron -> two photons". A state containing an electron and a positron is really only well defined as $t \to -\infty$, when we can think of the two particles as being infinitely far apart and hence having no probability of annihilation. As soon as time starts to move, the state begins to evolve into a massive superposition of all the possible decay products: two photons, three photons, $n$ photons, and, if the energies are high enough, other particle-antiparticle pairs (muons, quarks, whatever), hadrons, Higgs bosons, and whatever you want. And don't forget, given the particle content you still have a lot of room to maneuver by distributing energies, momenta and spins among the different particles in the possible final states.

This complicated state is the one that holds all the information of the initial state, because the time evolution was just dictated by Schrödinger's equation, and is thus unitary. As soon as you actually observe some decay product, you collapse the state, and hence lose information1. If you just observe that the result is two photons with, say, $3\, \mathrm{GeV}$ of center-of-mass energy and some given momenta, can you deduce that the initial state was an electron and a positron? No, there are lots of possible initial states that result in two photons with that energy. You really need that huge superposition if you want to run the clock backwards and recover the initial state. That's where the information is.

1 At least in the Copenhagen interpretation, but let's not get into that.

$\endgroup$
  • $\begingroup$ Nice answer. But got me thinking: if you started at t=-inf in that complicated particle superposition and ran Schrodinger's equation forward you would end up with the simple electron + positron pure state. So what you are saying if you start with a highly privileged particle eigenstate it will typically evolve into a complex superposition. At this point I feel like saying "entropy has increased" but of course entropy = 0 throughout. So my question is whether there a physical quantity which measures the complexity of a superposition? $\endgroup$ – Bruce Greetham Dec 31 '18 at 16:56
  • $\begingroup$ @BruceGreetham As far as I know, there is no such measure (but there are many things I don't know). As far as QM is concerned, it's just one pure state evolving into another; the entropy is zero all the time. $\endgroup$ – Javier Dec 31 '18 at 17:06
1
$\begingroup$

Your question is a very deep one and I fear there exists no simple answer in the year 2019.

You are asking for quantum information, not entropy. Classical information of a message is defined as

$ I = \log_2 N $

N is the number of possible alternatives, I the number of bits that make up the message.

A physical message is a part of the physical world. In QT this corresponds to a subspace of the total Hilbert space. The amount of quantum information is the number of qubits in the subspace. This means that the subspace must be separable into a finite number of qubits. [More general you could also allow qutrits and so on.] The important point is the "finite". The subspace must be a product space of a finite number N of qubits. Then you can retrieve a maximum of N classical bits out of it.

The process you are talking about is something like

$ \psi_{initial}(q_1,q_2) \rightarrow \psi_{final}(q_3,q_4) $

with $q_x$ (x = 1..4) being continuous 4-momentums of the 2 incoming and (example) 2 outgoing particles. The process is not

$ \psi_{initial,q_1,q_2} \rightarrow \psi_{final,q_3,q_4} $

where $q_1$ .. $q_4$ would be discrete indices of the vector components in the momentum basis. Dependency on a continuous momentum variable (or on continuous spacetime) means an infinite number of corresponding base vectors, and therefore the amount of information in the message "2 incoming particles" corresponds to an infinite number of qubits. And the final amount of information is again infinite. Now you could say, this infinite information is preserved, because it remains infinite.

If you look at this process leaving 3 particles

$ \psi_{initial}(q_1,q_2) \rightarrow \psi_{final}(q_3,q_4,q_5) $

If all $q_x$ were indices and bounded, you could always define an index mapping

$ q_1,q_2 \mapsto q_i $

$ q_3,q_4,q_5 \mapsto q_f $

so that the index sets {$q_i$} and {$q_f$} are as mighty as the products of {$q_1$},{$q_2$}, and of {$q_3$},{$q_4$},{$q_5$}, respectively. This means N can be preserved. It may not appear very intuive, but this even holds in the continuous case. The "infinite number of qubits" belonging to a state depending on a continous variable is the same as for 2 such states or the same as for a state depending on 2 continuous variables. This number is known as "cardinality of the continuum" in mathematics and denoted by $|\mathbb {R}|$.


What is often meant when talking about "quantum information is preserved" is the entropy, not the information defined above. Classically the entropy is the expectation value of the information you can gain from outcomes (of a certain alternative).

$ H = - \sum_{i=1}^{n} p_i \log_2 p_i $

$p_i$ are the probabilities of different alternatives. In QT you have the von Neumann entropy

$ H = -\mathrm {tr} ( \rho \ln \rho ) $

which again is the expectation value of information about a state you can gain from outcomes of "measurements". Because you usually calculate with pure initial and final states, the entropy remains zero during such processes, as Javier said already.

But even if the states in your particle subspace weren't pure but entangled with the environment, the entanglement entropy would be preserved as long as no interaction between particles and environment occurs during the process. This is because

$ \rho \mapsto U \rho U^{-1} $

with a unitary operator U that annihilates together with its inverse under the trace (where operators can be cyclically rotated). So as long as your process can be represented by a unitary operator acting only in the particle subspace, the entanglement with the environment is preserved.

But in a particle experiment you want to start without entanglement. At the end of the experiment you must entangle the final particles with the measurement apparatus and an outcome is chosen (again we don't know why which outcome is chosen, decoherence theory does not solve the problem of outcomes). The measurement apparatus determines how the entanglement is established and how many final states can be distinguished (by different states of the apparatus). At the end we get discrete floating point numbers (a computer has a finite set of FP numbers) in a bounded space as finite information. This is an example of how we establish finite information, though in the theoretical model the quantum system always represents an infinite amount of information. Of course in such a final step the information is not preserved.

$\endgroup$
  • $\begingroup$ Sorry, I made a mistake when saying "This means N can be preserved. But not in the continuous case. An area is more mighty than a line. " The last to sentences are not correct and I'll try to update my answer now... See for example cardinality of the continuum in Wikipedia. $\endgroup$ – Harald Rieder May 26 '19 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.