Dear friends,

Capture reactions are not the reverse of the decay. If one sends a beam of α particles upon 234Th (the daughter nucleus from the α decay of 238U), only part of the incident beam is absorbed (captured), the rest is reflected or passes freely. By reversing the decay I mean that the α wave emitted by the 238U source, be sent back and re-absorbed (captured) in entirety.

Every nuclear reaction has an inverse in the sense that we can bombard daughter nuclei and obtain parent nuclei, but not this is the question. The question is, as I said, if it is possible to reverse the process of emission obtaining total absorption.

If this total absorption is impossible, then we have to admit that we have to do with an irreversible process, and therefore, it is not correct to say that all the laws of the nature are time-symmetrical at the quantum level.

Dear Sofia,

the Mossbauer effect is based on gamma emission/absorption of a radioactive nucleus...it offers an interesting insight. I don't know if this emission can be defined  properly as a radioactive decay though. For sure the gamma photon which is emitted is exactly as the one absorbed.

The emission of gamma from Fe radioactive nuclei can be reversed: the emission from a Fe nucleus can  be followed by an absorption of a different Fe nucleus, if the system of both emitter and absorber is cooled at temperatures close to 0k. This is basically the nuclear recoilless resonance mechanism exploited also for the Harvard Tower experiments allowing very sharp resonance bandwidths.

Such process was originally performed by Mossbauer with Ir nuclei and is a very interesting one since it creates a link between reversibility and zero entropy if systems are brought close to zero kelvin.

My lovely Stefano,

The question is whether the absorption is total, i.e. when irradiating the nucleus with the γ rays, the incoming beam is totally absorbed, or saying otherwise, all the nuclei in the target jump to the γ-excited level.

Though, the rigorous indication that the decay is reversible would be, as I said, if the emitted wave could be returned in entirety to the emitting nucleus.

Kindest regards from me!

Yes Sophia the absorption is total, there is resonance and emitter and receivers are static. 

Dear Stefano,

It seems to me that you mix two things, the gamma energy and the gamma flux. Let me explain:

In the scattering theory we have a wave sent to a target, and a scattered wave. To make the case simple assume that the input beam is a plane wave, and that the scattered beam contains exactly the same type of particles as the incoming beam. Now, if the total outgoing flux Φout (i.e. integrated over all the directions of scattering), is equal to the total incoming flux Φin (i.e. integrated over the transversal section of the input beam), there is no absorption. However, if Φout < Φin, we have absorption.

My question is whether one can get Φout = 0.

What you say, so I understand, is that the energy ħω of the emitted photon is equal to that of the absorbed photon. Of course the energies are equal if the sources are so strongly cooled as the Mossbauer effect requires. But that does not guarantee total absorption, i.e. Φout = 0.

(I believe that the fact that the decay cannot be reversed should be a NO GO law, but I am not sure. It would be interesting to see which conclusions could be drawn if we would assume that the decay YES is reversible.)

My best wishes.

There is something called the Quantum Zeno Effect, whereby one, by observing an unstable state more and more often, eventually prevents it from ever decaying! It is quite a lot discussed in the literature, and also in Wikipedia articles.  There is also experimental evidence that it is a real effect. Not observable in stars, but in microscopic systems. It is not quite the same as what you asked for, but pretty close.

And yes, in microscopic systems, which couples very weakly (theoretically not at  all) to the environment, one may have considerable revival of an excited level, because an initial configuration

|exited state> |rest of coupled system>

is not an eigenstate of the full Hamiltonian, but must be expanded in a small number of true eigenstates. Then this configuration will change with time. If the eigenfrequencies are commensurate (not realistic) there will be an exact revival after some time. In practice this is difficult to obtain, but one may observe that the probability of finding the system in the excited state in periods will increase with time.

I think quantum error correcting schemes are methods to achieve similar effects.  

Sofia> a Hamiltonian for the decay process, comprising a part describing the nucleus, a part describing the environment, and a part describing the interaction between the two. If such a Hamiltonian exists, we can write the Schrodinger equation, solve it, and obtain a wave-function describing the decay process. Next, since the Schrodinger equation is symmetrical in time, the decay process should be reversible.

It is easy to simulate systems like that. If your environment is sufficiently large (a 100x100 matrix Hamiltonian can be sufficient), you can observe a decay process in beautiful accordance with the Fermi Golden rule up to a (rather) long time T. By then making a time reversal of the full wave function, and continue running for a time T, you will be back to the initial state (as long as numerical round-off errors can be ignored)! 

By reducing the Hilbert space dimension of the environment, the decay process starts to deviate from the Fermi Golden rule prediction, and on observes revival behavior. The whole process is not significantly different from the observed increase of entropy in classical Hamiltonian systems. The main point is that if the environment is large enough, it becomes essentially impossible to find some state which leads you back to a very distinct initial state after a not-too-long time.

Stefano> the Mossbauer effect

The main point of the Mössbauer effect is that the recoil due to emission or absorption is taken up by the whole lattice, not an individual nuclei. This is why emission and absorption frequencies can be equal to very high accuracy. Otherwise one would have to include the changed kinetic energy of the nucleus in the energy budget.

Dear Kåre,

I see that my question interests you, and I am glad and thankful.

Let me tell you some more things. I had a long talk about a year ago on this topic with Naomichi Hatano, a young (and very kind) researcher who deals with such things. He described to me a particular case in which he thought/believed that he can reverse a phenomenon analogous with the decay. But, for proving that the decay is a unitary transformation, i.e. time reversible, it's not enough to find a particular case in which the reversal can be done. NO! One single case in which the reversal cannot be done, and the unitarity was disproved.

Now, in the light of the above, I quote here some statements of yours:

=> "By reducing the Hilbert space dimension of the environment, the decay process starts to deviate from the Fermi Golden rule prediction, and on observes revival behavior."

NO!!! Don't go to particular and favorable cases as Naomichi tried to do!

=> "By then making a time reversal of the full wave function, and continue running for a time T, you will be back to the initial state (as long as numerical round-off errors can be ignored)!"

Practically, how do you implement the time-reversal? Mirrors? You'll get interference between the incoming and outgoing wave. You can't stop the so-called coupling with environment that causes emission.

I will refer to the Zeno effect too, but separately. There exists something simpler for stopping the emission from beginning. But, the problem is not to stop the emission from beginning, but to reverse it. This is the essential factor that impedes returning the emitted beam into the parent nucleus.

I heard once an expression that pleased me, "to play the devil's advocate". If I understand properly the meaning, this is what we have to do when checking an assumption, in our case the unitarity of the decay, attack it from all the sides.

Sofia> Practically, how do you implement the time-reversal?

For most cases it would be very difficult to implement in practice, in particular for nuclear processes. But that does not rule out that it can be done in principle.

There is one class of phenomena where it is possible, the spin echo technique.

And there is a debated case of whether unitary evolution really breaks down: evolution in the presence of a black hole.

And there are arguments that in big enough systems almost every pure state looks equivalent to a thermal state, as long as one can only make observations on a (relatively) small subsystem.

Dear Kåre,

I'd like to react to the idea of quantum Zeno effect. If we want to impede an atom from de-exciting we can place it between two metallic plates so that the distance between the plates be smaller than the wavelength of the photon to be emitted by the atom. Such an experiment was done.

The idea would be good for the disintegration of a nucleus too, however the distance between the plates has to be too much small. Let's take a typical energy of an emitted alpha particle, ~ 4MeV = 6.4 x 10-6erg. The linear momentum is therefore p ~ 9.21 x 10-15g cm/s, s.t. λ ~ 2πħ/p = 0.68 x 10-12cm. So, λ is of the order of magnitude of a nucleus. I don't know how can one bring two plates so closely that the distance between them be 104 times smaller than the atom radius. 

But, again, the problem here is not how to impede the decay to start, but how to reverse it.

APROPOS: this so-called "coupling with the environment" doesn't seem to you fantasy? Which environment? What is there in the environment, to be coupled with? People even write a Hamiltonian of the environment, and I don't know of what exactly is this Hamiltonian at all. Do you maybe know?

Dear Sofia!

You asked: Is it possible to reverse the decay of ... an unstable nucleus? Well, it's like following equation:

31-ga-71 ↔ 32-ge-71 + e–

The profile of binding energy of the nuclei 71 shows a small energy difference between stable nucleus 31-71 and the adjoining nuclei. The variabel internal energy leads sometime to reverse decays. There are some more nuclei which show the same behavior, for example the nucleus 49-in-113 and all Lead nuclei.

My Regards

Hans

To explore detailed quantum phenomena with current technology, it certainly make more sense to focus on the atomic and nanometer scale than on the nuclear scale.

The point about the 'environment' is that it must contain a very large number of accessible degrees of freedom, the details of which should be essentially irrelevant. For thermal equilibration it should be characterized by the standard thermodynamic parameters, for a measurement apparatus with perhaps additional details. It will naturally become entangled with the subsystem we want to analyze and observe. So 'what is there' could be almost anything.

For a measurement process this leads to a strong correlation between the state of the measured object and the state of the (large) measurement 'pointer'. However, it is a great mystery how and when the quantum mechanical AND description of the total system becomes a classical OR description; maybe that is where our 'free will' comes into play :-)

Hans,

Thanks for your comment, but this is not what I asked. I didn't ask whether a reaction can go in opposite direction, but whether the emitted wave can be sent back in such a way as to be re-encapsulated in entirety into the emitting nucleus. I mean, if the absorption can be total. For more details, see on the 1st page my answer to Stefano with Φout < Φin.

Dear Kåre,

You say "it certainly make more sense to focus on the atomic and nanometer scale than on the nuclear scale". I am not sure what you try to say. But, my reaction is that I focus on the issue and scale where I have a problem.

As to "For a measurement process this leads to a strong correlation between the state of the measured object and the state of the (large) measurement 'pointer' ", my problem is not the measurement problem and the limit between classical and quantum.

The question is WHAT is there inside the environment, with WHAT there is coupling. Degrees of freedom, as you mention, are PROPERTIES, not objects. Which are the objects characterized by these properties named degrees of freedom? You say "almost anything". Can you give an example?

Just take into consideration that at the start of the decay, the anything has to be void, vacuum. And still, it should have degrees of freedom. This is why this story with coupling with the environment seems to me fairy tales. To speak of degrees of freedom when no object is present is like putting clothes on a man who is not there.

Sofia,

I did quite understand your question.

But: We have not yet understand the reality of the microcosm resp. the world of particles and nuclei. All what we have are theories and mathematical equations. My point of view is to declare reality by reality, what means at first I need to understand the processes and then I can apply mathematics correctly. Mathematics is the end of knowledge but not the beginning.

Your interesting question is deduced from the mathematical world view in physics. Maybe the reality is much more different. We shall talk about this again.

Hans

Sofia> I am not sure what you try to say.

I was saying that it is better to focus on situations which are accessible experimentally. I.e., atoms and larger objects. The old-fashioned trade of Quantum Mechanics, the calculation of scattering and decay amplitudes, and converting this to cross-sections and lifetimes by use of the Fermi Golden rule, is extremely well tested. The deep issues of Quantum Mechanics involve situations which are not easily treated this way, and in particular how the transition from quantum to classical can occur. That means going to bigger and bigger quantum mechanical objects.

Dear Kåre,

> "it is better to focus on situations which are accessible experimentally. I.e., atoms and larger objects"

No problem. The de-excitation of an excited atom is absolutely analogous with the nuclear decay. It is treated in quantum mechanics beginning from a Hamiltonian containing three components, analogous to those in Feshbach's model of nuclear decay (in fact, Feshbach's theory describes nuclear reactions, the decay is a particular case). The components are: a Hamiltonian for the atom, one for the e.m. field alone, and one for the interaction between the electron and e.m. field.

So, the question is the same: can we make the photon return and be confined back to the atom (of course, by raising an electron to a higher level)?

Please don't be tempted to say "yes" quickly. Imagine an e.m. beam irradiating the atom; part of the beam would be reflected, not absorbed. As to the absorbed part, the so-called coupling with environment is present, s.t. the decay process would start on the spot, even if, eventually, the half-life is long.

So, is the decay reversible? Can we indeed re-confine the emitted wave to the atom/nucleus ?

Sofia> So, is the decay reversible? Can we indeed re-confine the emitted wave to the atom/nucleus ?

Let me think about this as a single exited atom in an otherwise empty universe (static, and of finite volume), decaying into a de-exited atom plus a single photon:

* If the volume of the universe is sufficiently small, the atom would not even decay, because there would be no room for a photon with the right energy (the difference between the atomic exited and ground state energies) in this universe.

* If we now very rapidly change the size of the universe a tiny bit (theorists can act like God :-) ), the exited state would no longer be an exact eigenstate of the full Hamiltonian, and we would obtain a state which a tiny bit is in an superposition of essentially two eigenstates. By considering the atomic part of the state only, it would seem like it has a large (slightly time dependent) probability of being excited, and a corresponding small probability of being de-excited.

* By increasing the universe further (at a time when the atom is excited with probability one), one would encounter a size where a resonance condition is fulfilled. I.e. there will be room for a photon with the right energy in the universe, and strong mixing between "the atom and that photon" will occur. The atom may start to oscillate between states where it is excited with probability one, and where it is de-excited with probability one.

* By increasing the universe even further, more and more photonic states comes into play, and more and more of them will approximately fulfill the resonance condition (and thus couple strongly to the atom). The time development will more and more look like the atom is decaying with the emission of a photon, as described by standard theory. But by looking long enough and close enough one would note that there is some revival behavior.

* In a rather big universe, one should realize that there is a tiny probability that an emitted photon travels (f.i.) around the universe in a time t, and re-excites the atom at that time. As the size of the universe becomes (practically) infinite, that time also becomes (practically) infinite, and we can never observe any re-excitation.

To get around that, one would have to introduce "stuff" which makes it look like the atom is in a small universe. Which I think becomes possible to an increasing degree with the development of nanotechnology.

So the short form of my answer is in principle not impossible.

Sofia> The question is WHAT is there inside the environment, with WHAT there is coupling.

In atomic physics, like the situation I discussed above, the environment is essentially the electromagnetic modes of the universe. The electromagnetic field may vanish (except for zero-point fluctuations), but its degrees of freedom is always there.

And the coupling is between the available modes of the electron and the electromagnetic fields (too a much smaller degree also including the atomic nucleus).

Your question seems very hard to answer. I try to expose my opinion. Let us take an example of the gamma transition of the nucleus, because it is much more precise than others in both theoretically and experimentally. It is very hard to say whether it is reversible or irreversible. However, it may be said to be basically irreversible. When the nucleus emits an gamma-ray through the transition from the first excited state to the ground state, this gamma-ray never return to the nucleus. Even if it is forced to return to the original nucleus, its energy is not exactly equal to that of the excited state because it gets inevitably recoil in the instant when the gamma-ray is emitted.Thereby, it cannot be reabsorbed by the nucleus to come back to the excited state. Even the Moessbauer scattering is not perfect but some ambiguity remain. These varieties are somewhat covered by the width of the excited state which is linked to the lifetime of the state. By setting a reflector (e.g. metallic plate) near the nucleus, the gamma-ray can be forced to return. However, the cross section for the elastic scattering of the gamma-ray is not 100% but a small portion. Furthermore, we should take into account the solid angle and that atomic vibration on the surface of the metallic plate is not negligible. Not only that, the probability of gamma absorption by the nucleus is small enough as well.

Therefore, even in this case, reversibility is extremely small.

I will give you the following references:

[1]Il-T. Cheon, Europhysics J. A33,213-221(2007).

[2]Il-T. Cheon, Int. J. phys. Sci. 5(1) 12-17(2015)

With best.  Il-Tong Cheon

Il-tong, Happy New Year!

I see, with pleasure, that you are professional in nuclear theory.

Let me just stress that the major problem in the re-absorption of the emitted ray, if we retro-reflect it with mirrors, it's not the difference in energies. First of all, the Mossbauer technique offers indeed a good amelioration of this difference, but we can't escape the low cross-section of absorption. That is a major impediment in reversibility.

Also, I claim that even if part of the retro-reflected ray is absorbed, the nucleus which absorbed it will immediately become excited and in principle, the emission process starts.

But there is more than that. You see, irreversibility implies non-unitarity. QM dislikes non-unitary processes, and I assume that you are acquainted with the Friedrichs' model for decay, and with Feshbach's theory for reactions. The decay is a particular case in this theory. Both Friedrichs and Feshbach described a Hamiltonian for the decay. Well, if there exists a Hamiltonian, the decay should be unitary. But, if the decay is not unitary, there can't exist a Hamiltonian for it.

 Bottom line, please tell me, can there be a proof that the re-absorption probability can never be equal to 1?

I am no expert in the scattering theory, I hope that you are. Does the scattering theory say something in this direction?

Best regards,

Sofia

Dear Henk,

Please, let's leave aside the Universe. It is a much more complex issue. Also, cosmology is not a field in which I am competent, s.t. I can't express a wise opinion.

You say: "You need to set-up the wavefunction at some time after the decay, reverse all necessary momenta and wave-vectors . . ."

We can't do this reversion. A very good classical analogue is a trial to reverse the evolution of a gas (Loschmidt's paradox). To reverse all the velocities we need to put mirrors behind every molecule, and immediately withdraw the mirrors for not stopping the movement of the other molecules. Is such a thing realizable?

With the emitted quantum wave the things are similar, because the emitted wave doesn't have a sharp energy, but some line width.

However, we have a much more severe problem. Please have a look at the answer of Il-tong Cheon, and my comment to it, they are just above your post. If we send back the emitted wave upon the nucleus, most of the beam is scattered, not absorbed.

I am no specialist in the scattering theory, but as far as can see there, we cannot reduce to zero the scattered part of the wave (i.e. the part which is not absorbed), unless we set to zero the scattering potential. But, in this way, the wave sent upon the nucleus would simply not "see" the nucleus, would just pass by it. Though, again, I am no specialist in scattering, it seems that Il-tong Cheon is better than I. 

Best regards,

Sofia

Dear Sofia,

This time, I will discuss on two key questions and others can be referred to some reference books.

It should be reminded that all processes of gamma emission and absorption have to proceed in recoilless way, because the energy of gamma-ray should not be changed for the case discussed in the previous communication. In more precisely, the emitted gamma-ray has flexibility as much as the width of the excited state. Accordingly, all processes should take place at least within this energy flexibility, even if these recoil effects cannot be perfectly killed.

[1] Mirrors: Any material of mirrors consists of atoms and molecules. Therefore, we should analyze the processes of light reflection in the level of atoms and molecules. These atoms and molecules bound on the material surface (e.g. metals) are always vibrating. This vibration makes a recoil effect on gamma-ray and, then, the reflected gamma-ray cannot keep the same amount energy as the incident gamma-ray. Such effect of vibration is usually described by the Debye-Waller factor. In order to maximize this factor, one should choose material of mirror with high Debye temperature.

Moreover, the gamma-ray reflected backward is not concentrated on a line but rather spread out because of the Quantum Mechanical properties. That is, even if the beam is scattered backward, it forms a solid angle which is very small for a single nucleus (or atom) as a target. Remind that there are various processes for the collision of the gamma-ray with the nucleus (atom), i.e. coherent, incoherent, photo-electric and so on. As you have seen, the situation is not so simple as that in classical picture.

[2]The probability of the gamma-ray absorption by a nucleus is not unity. A pure nucleus cannot exist practically in nature. It exists in a form of atom. Not only that, the nucleus consists of nucleons. Accordingly, when the gamma-ray approaches the nucleus, various reactions take place. For example, as mentioned above, aside from the absorption, coherent and incoherent scattering, photo-electric processes occurs in nuclear level and atomic level. Thus, the probability of gamma-ray absorption by a nucleus is given as a fraction of the gamma absorption cross section over the total cross section (which is a sum of all these cross sections). It is, of course, smaller than one.

Good luck !

With best regards.

Il-Tong Cheon

Dear Il-tong,

Thank you very much for your illuminating explanations.

Now, I suppose that you are aware of Feshbach's reaction theory and of Friedrichs' model for decay. In Feshbach's theory the decay is only a particular case of reaction. Both these people assume a Hamiltonian for decay, consisting in 3 terms: the nucleus Hamiltonian, the Hamiltonian of the emitted type of radiation, and the interaction Hamiltonian. Anyway, if there is a Hamiltonian for the decay, then the decay must be a unitary process, and therefore must be reversible.

People of competence in this domain, and whom I asked frankly whether the Friedrichs and Feshbach decay models can be wrong, gave me confuse answers. So, I ask you: are these models WRONG?

On the other side, in QM we similarly have a Hamiltonian for the de-excitation of an excited atom. The Hamiltonian is similar, a part for the atom, a part for the e.m. field, and a part describing the interaction between the e.m. field and the electron. Is something wrong with this model? 

With thanks in advance,

Sofia

Henk Smid ... agree, I have had a lot of success in reversibility in mathematical models. Although related to quantum compression, the issue is in the same pitchers square. It is possible to create a Hamiltonian as described by Sofia, and work it both ways in the vast majority of cases. Though it isn't a simple ableian relationship between the values. Good luck achieving this is practice though.

Dear Panagiota,

Tanks for your suggestion. It would be useful indeed.

In classical mechanics, an elastic ball falling on the floor performs a reversible movement, the ball hits the floor, it returns up to (almost) the initial height, again falls, etc. But, when hitting the floor, part of the energy is lost by heat. So, the movement is not perfectly reversible. 

The same in QM, to find a perfectly reversible movement, it seems to me that it is impossible. Let's take for example the movement of a plane wave in the direction z, see figure.

In the figure a there is a segment from a plane wave. The movement will be reversed by reflection when encountering the mirror M.

Now, an ideal plane wave is infinite in all directions, it occupies all the space, even if its movement is in one defined direction, z in our case. Therefore, we have all the time a direct wave and the reflected wave. The two waves are identical, except for the direction of movement - so, we have reversibility.

 But, an ideal plane wave does not exist. A real wave-packet is limited in volume, see fig. b. It has a front border and a rear border. Due to the limitation in length, there is, by the uncertainty principle, a spread in the velocities. This fact has as a consequence the continuous lengthening of the front and rear border. Thus, the reflected wave-packet won't be identical with the direct one.

We can get only approximate reversibility.

By the way, your question is very good, and I suggest you to ask it in a separate question, e.g. "can someone offer an example of reversible process in QM?"

Strongest than that, I would add to the above question the continuation: "And if there is no perfectly reversible process, is that because of the 2nd law of the thermodynamics (increasing entropy?"

My kindest regards, and I regret that I am limited to 1 upvote per comment, I would give you more upvotes.

Sofia

Henk> Unfortunately, this last option seems (technically at least) impossible.

Certainly! As I mentioned in a earlier post, it is much better to start with some more accessible systems. There is one (semiclassical) class of systems where (good) time reversal can be achieved, instruments using the spin echo technique. Something similar might be possible in suitable atomic systems.

Hi, Kåre,

How was the celebration of the New Year? I assume that in Norway you have a lot of snow at this time of the year. We hardly have a few centimeters of snow on mount Hermon, and people come with their children to enjoy by throwing at one another balls of snow.

Well, I subscribe to what says Henk. You can't reverse all the momenta. Imagine a classical gas. Placing a mirror behind a particle would stop the movement of other particles. Placing a mirror behind each particle in the gas, besides being technically non realizable, would stop the movement in the gas, unless you can place and withdraw the mirrors instantly. It's not possible, nothing goes instantly in the nature.

Restoring the movement of a quantum wave puts the same problem, but harder. You see, a resonant level - only such a level emits - has a width. That means there is a range of velocities. Where can you put mirrors in the emitted wave, for reversing all the velocities? A wave is not discrete as a gas, it is a continuum.

Shall we surround a nucleus with a spherical mirror? The first part of the wave that would touch the mirror would be the fastest velocities. They would be the first returned to the nucleus. Slower velocities would touch the mirror later, and return to the nucleus, ever with a bigger delay. So, the wave-packet returning from the mirror is not the same as the emitted one, just with reversed velocities. No, it is wider.

Could this be the way how the 2ND PRINCIPLE OF THERMODYNAMICS manifests itself in QM?

Dear Sofia!

It's fundamentally possible to reverse any process or reaction. It's merely a question of the way to reach the initial state. But your question lead to a perpetuum mobile on particle level.

Otherwise, in case of a beta decay how should the "neutrinos" return to the starting point?

The comparisons between particle reactions and mechanical processes are not useful.

Hans

Dear Panagiota,

Good day to you too!

Yes, this is the question, if one can create only incoming waves, and if they could be absorbed totally in the nucleus, i.e. no more waves outside.

The quantum process is not completely equivalent to the classical one, in the example with waves in a kitchen bowl we cannot create an excited compound nucleus that absorbs - there is no absorption there, only reflection.

My question aims indeed in the direction that the decay is a NON-unitary process. It's not a trivial fact, there are people that claim that non-unitarity does not exist at the microscopic (quantum) level.

I wish you a good new year!

Sofia

Dear Hans,

==> It's fundamentally possible to reverse any process or reaction. It's merely a question of the way to reach the initial state.

This mere question on the way to do the reversal, is THE question. The Schrodinger equation seems to be indeed time-symmetrical, but in the lab, can you return the emitted wave and re-confine it, in entirety, to the emitting nucleus? Please see the explanations of Il-tong Cheon on page 3. They are very professional. The re-absorption by the emitting nucleus, has a very low probability.

Also, don't forget that for the decay process we use complex energies. That makes the process not time-symmetrical. WHY do we use complex energies instead of real ones?

Best regards,

Sofia

Dear Henk, and all the other friends,

There is among us a person who is a specialist in these matters. His name is Il-tong Cheon, and one can see his explanations and indication of further references on page 3.  As he explained, even if we put mirrors and retro-reflect the beam emitted from unstable nuclei, the absorption probability is not 1. That means, the emitted particle won't be re-absorbed with certainty, part of the wave would be reflected by the nuclei.

I also add that even the part of the wave which is absorbed, would be re-emitted because the nucleus created by re-absorption is unstable and begins immediately to output a wave.

But, the best thing is to read the explanations of Il-tong Cheon, page 3.

Dear Henk,

Can you describe more specifically what you mean by "Reversal of time of the wavefunction of the universe" ?

 I suppose that you don't mean the reverse of the evolution of the far galaxies. Then, what should be reversed? Can you name it?

You see, we have two early theoretical models of the decay: one due to Friedrichs and one due to Feshbach. Both models suppose that the decay happens because a certain coupling with the vacuum. I posted in the past a question about this coupling, i.e. what is there in the vacuum to allow the supposed coupling? A coupling is between two objects. But vacuum is nothing, emptiness.

This coupling with the vacuum seems to me a trial to describe as unitary, a process which isn't unitary. In practice, when we write the Schrodinger equation for decay, we use complex energies, i.e. we admit that the decay is non-unitary.

Similarly I ask you, when you say to reverse the wave-function of the universe what exactly you have in mind? Just recall that in general we do our experiments in deep vacuum, i.e. high isolation.

Best regards,

Sofia

Hello Panagiota, good day!

=> It makes no sense to demand that "the same" alpha particle be absorbed by "the same" nucleus. I do not see the need for mirrors. Take any daughter nucleus and shoot it with any alpha particle with the desired energy. 

The wave emitted by a nucleus has a certain line-width, a spectrum of energies. To answer the question of reversibility you should first create a wave with the same spectrum. Also the target of daughter nuclei have to be kept at the same very low temperature, otherwise you will have a problem with the recoil of these nuclei, and the recoil takes part of the energy of the incoming particles, making them unfit for the absorption.

=> I can sit and monitor one parent nucleus 

How do you think to monitor one single nucleus? What you think that you can measure on it? These nuclei have to be in the form of a very cool solid target, to avoid recoil.

=> Why should I expect the daughter nucleus to absorb an alpha particle every time I shoot one against it? A sample of daughter nuclei, a flow of alpha particles and an absorption rate are more relevant entities I think.

It seems to me that you miss the essence of the question. Let me elaborate a bit. The quantum theory dislikes non-unitary processes. There are people that claim that the microscopic (quantum) world behaves only unitarily, i.e. all the processes are unitary. My question is, in essence: "is that true? What about the decay, is it a unitary process? If it is, then let's try to reverse it. Can we?"

This is why I question whether, reflecting back the emitted wave, it will be re-absorbed with probability 1.

Well, Il-tong Cheon says: NO! There is no such thing as re-absorption probability equal to 1, because if we send a beam of particles on a target there is also scattering, not only absorption.

But, in fact, I am not so sure. I don't know experiments in which the emitted beam is sent back to a target which is the identical to the emitting source. I also doubt whether the experiment is possible, because of the dispersion of the emitted wave - faster Fourier components  run quicker, slower components run more slowly, and on the way from source to target (or back to the source) the wave is distorted.

About the entanglement with vacuum I'll tell you in another answer, the present one is already quite long.

Best wishes

Dear Panagiota,

I'd like first to be sure that you read what I said about unitarity of the decay. If the decay is unitary, then it is reversible.

The question here is whether the decay is non-unitary i.e. whether it provides an example of non-unitary quantum process. As I said, if it is unitary then it should be reversible. In that case, indeed, sending the emitted wave from a probe of alpha-unstable nuclei, to a probe of daughter nuclei identical to the source probe as it remains after emission, all the alpha particles should be absorbed. Practically, we shouldn't get scattered alphas from the target probe.

Well, there are a couple of additional details, but this is the problem in main lines.  

Dear Panagiota,

I believe that you know what is unitarity, though, the problem is probably not entirely clear to you.

A unitary transformation Û has the following properties: given an initial state |ψ1> and a final state |ψ2>

(1) Û|ψ1> = |ψ2> , and is the pair: daughter nucleus and alpha outside. Since |ψ1> and |ψ2> are just functions, not vectors, it is sufficient to write

(1') Ûψ1 = ψ2 , and  Û†ψ2* = ψ1*

The unitary transformation Û is the decay, of course if the decay is unitary. The equality (2) says that beginning with the state ψ2* one should end with the state ψ1*. Now, the state ψ2*  is the complex conjugate of ψ2. In the simplest case it may be obtained by reversing the linear momenta in the Fourier expansion of ψ2. That means, we send the wave back. As to the state ψ1, let's assume, just for simplicity, that it is real.

What is important for your question is that the state to which we should return by reversing the process, is ψ1, the alpha inside the nucleus.  

Dear Panagiota,

Which decoherence? Our experiments are done in deep vacuum. And with detector we have interaction only after the reversal, i.e. beyond the target, by testing if there are alpha particles which were not absorbed by the target.

We don't disturb the alpha wave on the way from source to target.

Dear Henk,

Indeed I don't understand you. Why should I reconstruct the wave-function when I can take a probe of alpha-emitting nuclei? The probe produces the wave-function.

Now, what you say about equations, "reversal of time is very easy in equations", please don't be sure when the issue is the decay. Don't forget that for the decay we write the Schrodinger equation with complex energies, not real. A transformation of this type is not unitary, as a theoretician you know that exp(iHt) is not unitary when H has complex eigenvalues.

You probably would ask: if so is the situation, then the decay is non-unitary. What's the problem?

The problem is that there exists two theoretical works that represent the decay as a unitary process. What these work do, is to construct a Hamiltonian with three parts: a part representing the nucleus, one representing the environment, and one representing a so-called coupling between the nucleus and the environment. These works belong to two theoreticians Friedrichs and Feshbach. On Feshbach's work are based today many studies. However, ultimately, people write a so-called "effective Hamiltonian" with complex energies.

So, the question is, the decay-model of Friedrichs and that of Feshbach are eluding? Which coupling with the environment? Such experiments are done in vacuum, and vacuum is nothing, emptiness.

This comments is a reaction to Panagiota's post, immediately above.

The question if the decay should be treated as a unitary process under coupling with the environment, or as a non-unitary process, i.e. leaving aside the idea of coupling with the environment, has noting to do with Schrodinger's cat. The problem here is not the difference between the quantum world, in which there exists coherent superposition of states, and the classical world where there is no such thing.

Let me stress the background of the question. In order to construct a Hamiltonian with real values for the decay process, Friedrichs and Feshbach introduced the idea of coupling with the environment.

Therefore, to my question about the reversibility of the decay, some users answered: "for reversing the decay one should also reverse the evolution of the environment".

Here is the problem: which environment? What exactly is this environment with which a coupling is supposed?

In QM we know many examples of coupling, e.g. the spin-orbit coupling, the coupling between the magnetic moment of a particle and a magnetic field, etc. These couplings are well understood, they appear because of superposition of fields. But in the coupling with the environment, it is not obvious which fields come into play. To be specific, in the case of alpha decay, the question is which field is there in the environment, so that the parent nucleus, or the alpha particle to be emitted, interacts with.

Thus, to reply to the answer proposed by some users, which environment has to be reversed, what is there in the environment? The only thing that clearly has to be done for reversing the decay, is to send back the emitted wave to the nucleus.

I have an objection here to what Panagiota said, "You don't even know that the nucleus, or any particle, has decayed until you have detected the products." The nucleus to which we send the wave back is not in the state of either decayed, or not yet decayed - indeed it is not a cat (a classical object) but a quantum object. So it is entangled with the emitted particle and their joint state is a superposition of 1) alpha out + daughter nucleus, and 2) the parent nucleus with the alpha inside. Also, we don't aim at detecting the decay products, our task is to reverse the process, so, we send the emitted wave back.

Dear Henk and other friends,

Here is an additional argument against the so-called coupling with the environment.

Let's assume that indeed one can construct a Hamiltonian with real eigenvalues for describing the decay process. And let's assume, like in the models of Friedrichs and Feshbach, that it contains three parts: 1) one describing the emitting nucleus, one describing the environment, and one describing the interaction between them.

Let's now consider one of the eigenvalues of this total Hamiltonian, Etotal. As it is well-known the energy levels of an unstable nucleus are not sharp, they have a width. In other words, the α particle has a line-width, a range of energies, most of them (or, maybe, all of them), different from Etotal. Then, when we detect an α particle, with energy, say, E, belonging to this range and differing from Etotal, where goes the difference Etotal - E?

Worse than that, could it be that part of the energies in the range exceed Etotal? If the answer is positive, and E > Etotal, where from is taken the difference E - Etotal ? From the so-called environment? But, what if the experiments are done in vacuum?