The laws of physics are reversible and quantum information is never destroyed.

Given this, how do I time reverse the $U_{235}$ fission reaction, n which

${}^1_0n + {}^{235}_{92}U \rightarrow {}^{141}_{56}Ba + {}^{92}_{36}Kr + 3 {}^1_0n + \gamma +$ 202.5 MeV (in kinetic energy plus gamma ray energy).

That is, would reversibility require that if we bring together the $Ba$, $Kr$, $3ns$, and the $\gamma$ all at the same instant, a $U$-$235$ and a neutron might just pop out with a certain probability?

(EDIT: Here is a fun video from PBS Space Time explaining why quantum information can never be created or destroyed.)

  • $\begingroup$ Various laws of physics are not reversible. Information can be destroyed. $\endgroup$
    – Jon Custer
    Jun 7, 2020 at 2:58
  • $\begingroup$ Hi @JonCuster, I added some references for you that explain why quantum information is never destroyed. Classically you are right! But at the quantum level, things are different. $\endgroup$
    – vy32
    Jun 14, 2020 at 0:48

2 Answers 2


The reverse reaction, in which six particles join to make two, is allowed by time-reversal symmetry but forbidden by entropy considerations arising from classical thermodynamics. You just plain can't get those six particles to meet each other "just so"; there are too many ways for them to miss.

The way that thermodynamic entropy breaks time reversal symmetry is a subject of some discussion. Blanket statements about classical and quantum information are a way to lead yourself astray.

A rare example of a process with a three-body initial state is the "triple alpha" process by which helium fuses to carbon. It's so improbable that Hoyle proposed there must be an excited state in beryllium-8 to allow the sequential two-body reactions

$$\rm 3\alpha \to \alpha + {}^8Be^* \to {}^{12}C +\gamma $$

and successfully predicted the energy of the excited state.

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Triple-alpha_process#Discovery says that Hoyle predicted the C-12 resonance near 7.65 MeV, and en.wikipedia.org/wiki/Beryllium-8 says that Hoyle "postulated the existence of a resonance in carbon-12 within the stellar energy region of the triple-alpha process, enhancing the creation of carbon-12 despite the extremely short half-life of beryllium-8". $\endgroup$
    – PM 2Ring
    Jun 14, 2020 at 5:51
  • $\begingroup$ Thanks for the correction and the links! $\endgroup$
    – rob
    Jun 14, 2020 at 13:48
  • $\begingroup$ That's very exciting. Thanks for providing more information and the example of the Triple-alpha process, which I didn't previously know about. $\endgroup$
    – vy32
    Jun 14, 2020 at 15:02

The reverse reaction really is

$${}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3\ {}^1_0n + \gamma + 202.5\ \mathrm{MeV} \rightarrow {}^1_0n + {}^{235}_{92}\mathrm{U} $$

as you'd expect. Yes, your deduction from time symmetry is absolutely correct (as far as we know) - it has to be. The above will do the job. Yes, the four elements on the left will combine, fuse to form a uranium-235 nucleus and it will pop out a neutron just as going in the forward direction causes all those other things to pop out. No magic; it's entirely possible.

And yes, you're right to suspect it's not easy to make it happen. You'd have to shoot all 4 of those things (think about like 4 particle guns all aiming into one point) into the same tiny little area of space (remember that nuclei are measured in fermis [fm], $10^{-15}\ \mathrm{m}$, and your Uranium nucleus may be on the order of 10 fm, as a proton is about 1 fm), and with just the right energy - if you shoot with too much, you might end up "breaking" the nucleus you are trying to create: per nucleon, the binding energy of $^{235}_{92}\mathrm{U}$ is around 7 or 8 MeV, so if you put in maybe 8 MeV more than that 202.5, you stand a chance to end up with something not 235U, because you had enough energy to kick at least one more nucleon (in addition to the neutron that inevitably pops out) out of it.


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