If we attach labels, say "1" and "2", to the electrons (as we do when we write down a wavefunction) then the theory must be constructed in such a way that a measurement cannot decide whether the electron being observed is number 1 or number 2. Thus, in the question we do see two different electrons, one free (incident) and one bound but we should not be able to decide whether the free electron is number 1 or number 2, similarly the bound electron, ie, we should (for fermions) antisymmetrize the wave function.

Dear Joul, I'm not an expert on electron-atom collisions but I believe that the part of the wavefunction that is antisymmetric with respect of the exchange of the electrons is the spin-part, but a collision involving only Coulomb potentials is spin-insensitive, thus we only care for the spatial part of the wavefunction and it should be symmetric given that the detection takes place after the collision, and one doesn't know which of the electrons reached the detector.

I hope this could be useful, either way I will transmit your question to Omar Fojón and Carlos Stia, they are our experts on electron-atom collisions.

Cheers!

Dear Boghos: Within a time-independet approach, the wf of incident particle is not localized, it represents a stationary flow (the same the scattered wave), then identical particles necessarily are entangled.

But if one adopts the operational interpretation that "particles are indistinguishable if they can not be distinguished," then entanglement can be avoided in a time-dependent approach. While the wave packet which describes the incident electron has not overlapping with the bound one, they are distinguishable. An evolving wave function must respect causality: Electron of atom can not be in a unbound state before the incident electron approaches sufficiently.

Do you agree with this?

Regards.

My guess is that you always need to antisymmetrise. In particular, if you detect the unbound electron, you cannot really be sure which electron it was. More to the point, if the electrons are well separated, that is, have small spatial overlap, then, at least with respect to position measurements, you will get nearly the same results whether you antisymmetrise or not. That is why, in practice, if you have two distant H atoms, you need not really antisymmetrise the electron wave functions. However, as they come closer to each other, the antisymmetrisation becomes essential for the chemical bond.

Yes. Two-electron wave-function have to be antisymmetric, so electrons are indistinguishable.

The wave function of any multi-fermionic system is expressed as a linear combination of anti-symmetric Slater determinants and hence is anti-symmetric. This may not be a trivial exercise. Please consult a text for a recipe of how this is done.

Salut Bogos,

Les deux électrons sont indiscernables donc il te faut antisymétriser ta f.o.  à 2 électrons.

The short answer to the full statement of the question-which is different from the leading statement-is No. The statement about the two electrons being in the same field is misleading-and irrelevant.  And the reason is that the physical situation is e + atom -> e + atom. The atom and the electron are distinguishable particles. So there aren't any indistinguishable particles in the same state here. The fact that the atom is a bound state of electrons and the nucleus is irrelevant for the question. There isn't a state of two or more electrons, in this problem,  but the transition described is that of  |electron> x |atom> -> |electron> x |atom>, so the transition amplitude is  

,

where U is the evolution operator of the system |electron>|atom>. That's all.  U acts on the (sub)space of one electron states, not on the- here non-existent- (sub)space of multi-electron states-that can't be resolved.  The latter must be antisymmetrized, were it relevant, not the former-where the operation doesn't even make sense. One doesn't (anti)symmetrize initial and final states, since there doesn't exist any operator that acts on the tensor product of initial and final states-that's the confusion here. 

So the wavefunction is a sum of products of the wavefunction of the electron and that of the atom. 

If, however, one considers the situation where one can resolve the constituents of the atom, then, of course, things change: Then one does have a configuration of more than one electrons as  initial state and another configuration of more than one electrons in the  final state and one must take into account that the electrons are indistinguishable particles and project out the configurations one is interested in accordingly. The evolution operator is completely different, of course. One does know that one of the electrons in the initial state is physical-it's the probe; and that one of the electrons in the final state will be, similarly projected as a physical state. 

If no collision takes place, there isn't any interaction, so the initial electron is the final electron-everything factorizes throughout.  However, even if a collision with the atom does takes place, if it isn't possible to resolve the constituents, the statement that the final electron is the ``same'' or isn't is meaningless, since the final electron will be in a superposition of states, given by the electrons of the atom-however the interaction explicitly projects onto the one-electron subspace. So throughout only one-electron states are treated.

If the electrons can be resolved, then the superposition described above must first be calculated, that's the only difference. Once more, the statement that the final electron   corresponds to the initial electron is meaningless. The projection selects a one-electron state, that's all.

Even if no interaction, in the sense of electroweak interaction, takes place, the fact that the electrons are indiistinguishable does result in an interaction, namely entanglement: Precisely because the fact that it is possible to resolve the electrons of the atom, means that the initial state is a multi-electron state, that projects one single-particle superposition on the mass shell (that's the probe electron state) and the final state is another multi-electron state, that projects one single-particle superposition on the mass shell. 

The difference with the first case is that there can't be any entanglement defined with one-particle states.

The relation between the states of the  initial particles and the  states of the final particles is, precisely, given by the asymptotic form of the evolution operator-this asymptotic expression is known as the S-matrix element, relevant for the process.  One can say a lot about what happens. One shouldn't confuse issues (mostly historical)  that are relevant for an indefinite number of degrees of freedom, i.e. quantum field theory, with issues that are relevant for a finite number of degrees of freedom, i.e. non-relativistic quantum mechanics. Standard material in any textbook on quantum mechanics or quantum field theory. 

The statement that the electron has interacted with the atom is equivalent with the statement that its wavefunction acquires a phase, relative to the ``free'' state.  This phase can, in principle, and, also, perhaps, indirectly, in practice, be measured by interference experiments. Standard consequence of the unitary evolution. 

So, to return to the topic of the discussion, if one can express the problem by an effective interaction of an electron and an atom, in the non-relatvistic approximation, then one is dealing with one particle states and there isn't any issue. If one must express the problem by an effective interaction of many electrons-the probe electron and the electrons of the atom, then one, must, indeed, write the appropriate Slater determinant(s) for the initial and final states and sum over the possible states that describe the electrons that one isn't observing, in a way that's consistent with the assumptions about the model of the atom one is probing. That gives the transition amplitude unambiguously. 

Whether the electrons are identical or not is a statement that, as has been recalled, is decided by experiment (and theory) through the measurement of the relative phase and its calculation. For free electrons, the relative phase of any two is π. For interacting electrons there is a contribution to the relative phase, due to the interaction. Classical particles, that can be distinguished, don't have such a relative phase-it's not defined at all. All standard calculations in quantum mechanics, but that don't have much to do anymore, with the question discussed here. 

The statement that two electrons are identical *means* different things in quantum physics and in classical physics. In the former it implies statements about relative phases-but, also, that they can't be individually labeled in a multiparticle state. In the latter it implies that a relative phase can't be assigned any meaning and that they can be individually labeled in a multiparticle state. These statements are postulates in non-relatvistic quantum mechanics-of relevance here-and theorems in relativistic quantum mechanics, cf. http://ejde.math.txstate.edu/conf-proc/04/w1/wightman.pdf

But the electron in the initial state can't be compared to the electron in the final state, as such-the statement is meaningless. What is a meaningful statement is, however, irrelevant for this discussion, where such a situation doesn't arise, anyway. One is discussing electrons in the initial state separately from electrons in the final state. This makes sense, inasmuch as it's possible to define free electron states, that describe electrons that *haven't* interacted with the atom in question.

That expression, as given, is incorrect, precisely, because there isn't any tensor product between initial and final electrons-that's what's meaningless, as has been mentioned in previous messages.  Once more: if the physical model is the interaction of one electron and one atom, then the interaction Hamiltonian is V(...)|electron>|atom>=(Π_k(|electron_k>))|nucleus>, then the interaction Hamiltonian of the probe electron with such an atom is the operator that acts on the antisymmetrized product of the state of the probe electron with that of the other electrons. So the wavefunction of the complete system would be what's called a Slater determinant of the appropriate size. Its evolution is then described by the corresponding many-body Hamiltonian and the result is a many-body wavefunction, that describes the state of the Z +1 electrons. These are indistinguishable; This is the transition amplitude and one must then project on the one-electron subspace, that describes physical electrons, in order to obtain the corresponding one-electron final state.

Of course it's easier to start with, say, the hydrogen atom as a toy model and go on from there. 

In fact, the confusion that's apparent in the question is due to the following misunderstanding: The total Hamiltonian, H, is a sum of three terms: H_e + H_a + V

H_e is the Hamiltonian of the probe electron: it acts on one-electron states, call them |e>, that include spatial and spin degrees of freedom. H_a is the Hamiltonian of the atom and acts on the states of the atom, |a>, but is a one particle operator, since the atom is considered as a whole.  V is the interaction and acts on the tensor product, |e>|a>. Of course the H_e = H_e x I in the |e>|a> basis and H_a = I x H_a in the |e>|a> basis. 

The evolution is with the full Hamiltonian, H, one obtains |e>|a>(t)=exp(-iHt)|e>|a>(0). One is only interested in |e>(t), so the measurement operator, that prepares the electron state involves a sum over all possible states of the atom. One thus obtains a one-electron state. 

In the second case, H_a is Z-body Hamiltonian and V is a (Z+1)-body operator. The space of states (ignoring the nucleus) is the antisymmetric tensor product over all the electrons. They can't be distinguished, between probe and ``atomic'', within this subspace. The information that Z of them are bound just means that H_a acting on a Z-dimensional subspace of the Z+1-dimensional space has discrete spectrum,  that's all. The interaction will mix the electron states, therefore the Z-dimensional subspace, at any instant, that describes the atom will not be preserved by the evolution operator, since this can't be block diagonal, since the electrons are indistinguishable.

One solves the Z+1-body Schrödinger equation, sums over the Z-dimensional subspace that corresponds to the atom and ends up with a one-electron state that's the final state. 

Of course, in practice, it's useful to consider approximations, where the probe electron, in fact, interacts with a ``small'' number among all possible electrons of the atom. In principle, it should be possible to obtain the effective Hamiltonian for the atom, as a whole, by a ``coarse graining procedure'', whereby one traces over more and more electrons of the atom.

That the ``atom'' remains  a meaningful concept throughout is described by the statement that the full Hamiltonian, H, that acts on the Z+1-body wavefunction, has a Z-body subspace, where its spectrum is discrete, thus describes bound states, and a one-body subspace, with continuous spectrum, that describes the probe; were the Z+1-body spectrum discrete, it would mean the probe gets captured (this is in the reference frame where the atom itself isn't moving, else, of course, there's the Hamiltonian that represents the motion of the atom itself as a whole). If the subspace that has continuous spectrum is of higher dimension, then there are more particles in the final state, that don't have bounded motion.

In principal it's right, but for further calculations it is actually indifferent

Dear Prof. Joulakian,

                                       I guess you have the answer but let me provide one nevertheless because I have done research on Brownian Motion of charged particles with mass say electrons in gases in the Econophysics systems model developed by me. You can refer to my Political Environment and String Theory of Stock markets paper which are published papers on my RG page. You will see that I have derived four Theorems and tested them using experimental data and simulation with respect to Systems Integration of the random waves (non chaotic) as they can give rise to vectorial meanfields and drive the system to equilibrium preserving the Arrow of Time and D-Branes String Duality preserving Information Meanfield which can Kinematically describe the Dynamic Equilibrium and the resulting stock markets with Lagrangean Actions and Market potential Actions. I hope it will be helpful. SKM QC

this is a two-electron Coulomb problem, actually a H- ion.

not essentially different to the Helium atom problem

a standard textbook  problem.

note: two (or more) electron ststes are separated by a superselection rule, this means, there exits no superposition of the two electron states.

The specific Hamiltonian is not required for the application of the general principles required to answer the question.  One certainly does not gain a one electron state from the Hamiltonian, which is applied as an operator on Fock space according to the usual rules. Nor can one, ignore the amplitude for the bound and free electrons to switch in scattering.

Should two electrons, in the same field, be considered as indistinguishable, if one is in a bound state and the other unbound (thus detectable)

IMHO, particles in orthogonal states are distinguishable so far as the said orthogonality is exact. This implies closed systems, and no measurement. In real open systems, orthogonality, and hence indistinguishability, are always an approximation. Electrons in me and my wife are in principle indistinguishable. However, unless we once collide with ultrarelativistic velocity, this does not seem to affect our relationship in any discernible way.

Electrons are not tiny machine screws. If they were, a tiny engraving tool could mark each with a distinct serial number and they would be distinguishable. No thing is smaller than an electron and no word play can change that truth. I wish the moderator had flagged this question as unfriendly.