Electron Indistinguishability and Repulsion

[Dario56]

If we had a system of $N$ non – interacting electrons than a wavefunction of such a system is a product of one-electron wavefunctions otherwise known as a Hartree product: $$ \Psi(x_1,x_2,...,x_N) = \prod_{n=1}^N \psi(x_n) $$

This means that in such a hypothetical system , it is possible to distinguish between electrons as the wavefunction of the system isn't antisymmetric to electron exchange (we can say which electron is in which orbital). However, I am not sure I understand how is it possible to distinguish between electrons even in such systems? As far as my understanding goes, electron indistinguishability is a consequence of electrons being quantum particles which means that they lack exact position prior to measurement.

What does such a fact have to do with electron repulsion?

 

[dextercioby]

Even if by absurd they do not interact, they are still fermions, so the "total wavefunction" must implement that in an antisymmetrization. Just think of the simplest two-electron system, the Helium atom. The electronic part must be a certain combination of one-electron wavefunctions.

 

[PeterDonis]

If we had a system of $N$ non – interacting electrons than a wavefunction of such a system is a product of one-electron wavefunctions

No, it isn't. Electrons are fermions, so the wave function of a multi-electron system must be antisymmetric under electron exchange.

There may be cases where the wave function you wrote down might be useful as an approximation, but as soon as you start asking about particle exchange, any such approximation must break down.

 

[Dario56]

Even if by absurd they do not interact, they are still fermions, so the "total wavefunction" must implement that in an antisymmetrization. Just think of the simplest two-electron system, the Helium atom. The electronic part must be a certain combination of one-electron wavefunctions.

In any real system, like helium atom you mentioned, Hartree product fails because electrons do interact in nature. If we take some hypothetical, non – interacting, system of electrons (not found in nature) than Hartree product should be the exact wavefunction of such a system. This can be proven.

If we take two electron system of non – interacting electrons, than we can define Hartree product as: $$ \Psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2) $$
Since electrons don't interact we can write Schrödinger equation for each of them separately: $$ H_1\psi_1(x_1) = E_1\psi_1(x_1) $$ $$ H_2\psi_2(x_2) = E_2\psi_2(x_2) $$
Also, because electrons don't interact, both Hamiltonian and the energy of the system is the sum of Hamiltonians and energies for each electron: $$ H = H_1 + H_2 $$ $$ E = E_1 + E_2 $$
Schrödinger equation for the system is: $$ H\Psi = E\Psi $$
Plugging previous equations into the left hand side of equation and assuming that Hartree product is the solution to Schrödinger equation of the system, we get: $$ (H_1 + H_2)\psi_1(x_1)\psi_2(x_2) = E\psi_1(x_1)\psi_2(x_2) $$
Taking into account that both Hamiltonian operators are one-electron and linear, we can pull $\psi_2(x_2)$ out of $H_1$ and $\psi_1(x_1)$ out of $H_2$ which yields: $$ \psi_2(x_2)H_1\psi_1(x_1) + \psi_1(x_1)H_2\psi_2(x_2) = E\psi_1(x_1)\psi_2(x_2) $$

Taking into account Schrödinger equation for each of electrons separately: $$ \psi_2(x_2)E_1\psi_1(x_1) + \psi_1(x_1)E_2\psi_2(x_2) $$

Hartree product can be factored which gives: $$ \psi_2(x_2)\psi_1(x_1) (E_1 + E_2) = E\psi_1(x_1)\psi_2(x_2) $$

Taking into account that for non - interacting electrons, energy of the system is the sum of energies of each electron, we proved that Hartree product is the exact wavefunction for the system of non - interacting electrons since both sides of the last equation are equal.

 

[PeroK]

Apart from ignoring the electron interaction, you also seem to be ignoring the Pauli exclusion principle. It follows that you can't call these particles "electrons". Instead, you have two distinguishable, non-interacting particles. But, these are not electrons in any shape or form.

 

[Dario56]

Apart from ignoring the electron interaction, you also seem to be ignoring the Pauli exclusion principle. It follows that you can't call these particles "electrons". Instead, you have two distinguishable, non-interacting particles. But, these are not electrons in any shape or form.

Pauli exclusion principle follows when we take into account that electrons are indistinguishable or when we take into account that wavefunction of any real system of electrons must be antisymmetric to electron exchange. If we than try to put two electrons in the same orbital, we get that the wavefunction of the system becomes equal to zero or in other words that it such state is impossible.

To explain my question more concisely: According to my understanding, the fact that electrons are indistinguishable follows from the fact that they are quantum particles which means they lack exact position prior to measurement. Whether electrons interact or not isn't important for their indistinguishability because it follows from basic principles of quantum mechanics not interaction.

However, this seem not to be correct in systems where there is no interaction between electrons because according to the proof I've written before, wavefunction of the non – interacting system isn't antisymmetric and yet it is still correct for such (non – realistic) system.

According to my understanding, even in such cases, wavefunctions should be antisymmetric, but it isn't.

If we measure positions of electrons in such a system, we should be able to tell which electron is which (which electron is in which orbital), but how is that possible if we don't know where both of electrons will be before the measurement is done?

 

[PeroK]

However, this seem not to be correct in systems where there is no interaction between electrons because according to the proof I've written before, wavefunction of the non – interacting system isn't antisymmetric and yet it is still correct for such (non – realistic) system.

You've assumed that the wavefunction is not antisymmetric. That doesn't prove anything except that in basic QM the exclusion principle is an additional axiom.

 

[TeethWhitener]

Today, the Hartree product is mainly a historical curiosity. I suppose It’s ok as a starting point for a many boson wavefunction, but it was very quickly superseded by the Slater determinant for approximating many-electron wavefunctions.

The Slater determinant is still used today in quantum chemistry, namely in the guise of Hartree-Fock and post Hartree-Fock theories, which approximate the full multielectron wavefunction as a single Slater determinant and a linear combination of Slater determinants, respectively.

Edit: the reason the Slater determinant is so much better is, as others have pointed out, that it ensures that all particles are antisymmetric to exchange at the outset by construction, which is the defining characteristic of fermions.

 

[PeroK]

but it was very quickly superseded by the Slater determinant for approximating many-electron wavefunctions.

And, hey, the Slater determinant is anti-symmetric!

 

[vanhees71]

Pauli exclusion principle follows when we take into account that electrons are indistinguishable or when we take into account that wavefunction of any real system of electrons must be antisymmetric to electron exchange. If we than try to put two electrons in the same orbital, we get that the wavefunction of the system becomes equal to zero or in other words that it such state is impossible.

To explain my question more concisely: According to my understanding, the fact that electrons are indistinguishable follows from the fact that they are quantum particles which means they lack exact position prior to measurement. Whether electrons interact or not isn't important for their indistinguishability because it follows from basic principles of quantum mechanics not interaction.

However, this seem not to be correct in systems where there is no interaction between electrons because according to the proof I've written before, wavefunction of the non – interacting system isn't antisymmetric and yet it is still correct for such (non – realistic) system.

According to my understanding, even in such cases, wavefunctions should be antisymmetric, but it isn't.

If we measure positions of electrons in such a system, we should be able to tell which electron is which (which electron is in which orbital), but how is that possible if we don't know where both of electrons will be before the measurement is done?

That doesn't make sense. Either you have two electrons. Then they are indistinguishable and are described by the two-particle fermionic Fock state, no matter whether the electrons are considered interacting or not, i.e., if you have a pure state describing it with a wave function this wave function must be antisymmetric under exchange of the two particles. In the position-spin basis that means $\psi(\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2)=-\psi(\vec{x}_2,\sigma_2;\vec{x}_1,\sigma_1)$.

 

[vanhees71]

Today, the Hartree product is mainly a historical curiosity. I suppose It’s ok as a starting point for a many boson wavefunction, but it was very quickly superseded by the Slater determinant for approximating many-electron wavefunctions.

The Slater determinant is still used today in quantum chemistry, namely in the guise of Hartree-Fock and post Hartree-Fock theories, which approximate the full multielectron wavefunction as a single Slater determinant and a linear combination of Slater determinants, respectively.

Edit: the reason the Slater determinant is so much better is, as others have pointed out, that it ensures that all particles are antisymmetric to exchange at the outset by construction, which is the defining characteristic of fermions.

If you self-consistently describe a fermionic many-body system with a Slater determinant of single-particle functions, you get the Hartree-Fock approximation, which often is a pretty good approximation. The most general wave function is, however, in the full fermionic Hilbert space, i.e., given by superpositions of Slater determinants.

 

[Dario56]

You've assumed that the wavefunction is not antisymmetric. That doesn't prove anything except that in basic QM the exclusion principle is an additional axiom.

Yes, and my assumption, was shown to be correct when derivation was carried out to its end (it isn't of course mine, this proof can be found in many quantum physics and chemistry textbooks, for example Atkins, Physical Chemistry (5. edition), chapter 13.4: Atomic Structure and Spectra, The Orbital Approximation, Justification).

If the wavefunction should be antisymmetric (and proof shows that it isn't) for such a system than this derivation must lead to incorrect result and you didn't mention what is wrong in it.

 

[PeroK]

Yes, and my assumption, was shown to be correct when derivation was carried out to its end (it isn't of course mine, this proof can be found in many quantum physics and chemistry textbooks, for example Atkins, Physical Chemistry (5. edition), chapter 13.4: Atomic Structure and Spectra, The Orbital Approximation, Justification).

If the wavefunction should be antisymmetric (and proof shows that it isn't) for such a system than this derivation must lead to incorrect result and you didn't mention what is wrong in it.

If you assume that electrons are distinguishable, non-interacting, spinless particles, then you may still have something useful in terms of an approximation. But, that's all you have. That model must break down sooner rather than later.

And it certainly doesn't mean that electrons are, therefore, distinguishable, non-interacting, spinless particles.

 

[PeterDonis]

If the wavefunction should be antisymmetric (and proof shows that it isn't)

You can't show anything about actual physical objects by doing a mathematical proof (and your "proof" proves nothing anyway beause it assumes that the wave function is not antisymmetric to begin with). You need to test your mathematical model by experiment. What experimental results show that the wave function of a multi-electron system is not antisymmetric? (Hint: there aren't any.)

 

[Dario56]

You can't show anything about actual physical objects by doing a mathematical proof (and your "proof" proves nothing anyway beause it assumes that the wave function is not antisymmetric to begin with). You need to test your mathematical model by experiment. What experimental results show that the wave function of a multi-electron system is not antisymmetric? (Hint: there aren't any.)

Firstly, you are missing the point because I am not talking about actual physical objects (electrons), I am talking about HYPOTHETICAL system of non – interacting electrons.

Secondly, proof isn't mine, it can be found in quantum physics and chemistry textbooks.

Thirdly, if assumption (assumption is that wave function of the system is Hartee product) leads to valid result than how can it be wrong? My point is, if Hartree product isn't valid solution (for system of non – interacting electrons) than carrying out this proof would lead to meaningless result and that would prove that Hartree product isn't solution to this problem. However, result we get isn't meaningless at least according to my understanding.

 

[Dario56]

If you assume that electrons are distinguishable, non-interacting, spinless particles, then you may still have something useful in terms of an approximation. But, that's all you have. That model must break down sooner rather than later.

And it certainly doesn't mean that electrons are, therefore, distinguishable, non-interacting, spinless particles.

Yes, this model is valid for some kind of hypothetical electrons which don't interact and it was a starting point for developing better models in quantum chemistry.

My question is how is it possible that electrons can be distinguished in non – interacting system since ,according to my understanding, indistinguishability of electrons has nothing to do with electron interaction, but with basic principles of quantum mechanics?

Proof above showed that if electrons don't interact, Hartree product is correct solution of Schrödinger equation. If this proof is wrong, why does it than give meaningful result when it is carried to its end?

 

[PeroK]

My question is how is it possible that electrons can be distinguished in non – interacting system since ,according to my understanding, indistinguishability of electrons has nothing to do with electron interaction, but with basic principles of quantum mechanics?

We've already established that you are not using the basic principles of QM, but a simplified model (with no QM justification) where electrons are what you assume them to be. Eventually that model will produce predictions that are odds with experimental data. But, until that happens you are free to assume what you like about electrons and see how far it takes you.

 

[hutchphd]

Proof above showed that if electrons don't interact, Hartree product is correct solution of Schrödinger equation. If this proof is wrong, why does it give meaningful result when it is carried to its end?

What does "meaningful" mean? The particle you describe is not an electron.

 

[PeroK]

What does "meaningful" mean? The particle you desribe is not an electron.

Moreover, it's not the SDE that is the problem. The PEP explains why all electrons do not reduce to the ground state.

 

[Dario56]

What does "meaningful" mean? The particle you describe is not an electron.

Not one you'll find in nature, but that is not the point. My point is that even if electrons didn't interact, they should nevertheless be indistinguishable because that is a consequence of electrons being quantum particles which means they lack exact position before the measurement and so there is no way to keep track of which electron is which. Such a fact holds no matter if electrons interact or not. According to my understanding, interaction should be irrelavant.

However, what is shown in the proof above is that for HYPOTHETICAL system of NON – INTERACTING ELECTRONS, wavefunction which satisfies the Schrödinger equation isn't antisymmetric (if non - interacting electrons are indeed indistinguishable, as I think they should be, wavefunction must be antisymmetric and yet it isn't). I don't understand how is that possible. Is it now clear what I am asking?

 

[PeroK]

However, what is shown in the proof above is that for HYPOTHETICAL system of NON – INTERACTING ELECTRONS, wavefunction which satisfies the Schrödinger equation isn't antisymmetric (if non - interacting electrons are indeed indistinguishable, as I think they should be, wavefunction must be antisymmetric and yet it isn't). I don't understand how is that possible. Is it now clear what I am asking?

You have a model of an atom surrounded by distinguishable particles. Why should that model fail mathematically? Even a classical model of the atom doesn't fail mathematically. Where it fails is in predicting atomic phenonema.

But, also, you have something implicity prevented all electrons falling into the ground state. In QM the reason is the axiomatic PEP. You (and Atkins) have simply decided you don't need a reason. You are simply going to postulate that your distingishable electrons keep to the required orbitals: 2, 8, 18 etc.

The SDE does not and cannot enforce antisymmetry. Anti-symmetry is enforced by the PEP.

 

[Dario56]

You have a model of an atom surrounded by distinguishable particles. Why should that model fail mathematically? Even a classical model of the atom doesn't fail mathematically. Where it fails is in predicting atomic phenonema.

But, also, you have something implicity prevented all electrons falling into the ground state. In QM the reason is the axiomatic PEP. You (and Atkins) have simply decided you don't need a reason. You are simply going to postulate that your distingishable electrons keep to the required orbitals: 2, 8, 18 etc.

The SDE does not and cannot enforce antisymmetry. Anti-symmetry is enforced by the PEP.

No one is talking about model failling mathematically. What does that even mean? Point is, if the wavefunction satisfies Schrödinger equation than this wavefunction must correctly describe our system since Schrödinger equation is a fundamental equation describing quantum systems.

We saw that Hartree product satisfies Schrödinger equation for system of non - interacting electrons (not something which yields good results for real electrons, but nevertheless true for non – interacting once). If Hartree product doesn't accurately describe our system than it wouldn't satisfy Schrödinger equation. That is my point.

Pauli exclusion principle (PEP) isn't an axiom or based only on observation (like Schrödinger equation), it follows from electron indistinguishability which requires antisymmetric wavefunction of the system, so if we try putting two electrons in the same orbital, wavefunction of the system becomes equal to zero or in other words such state is impossible. Electron indistinguishability in turn follows from basic principles of QM, as far as I understand.

 

[PeroK]

Pauli exclusion principle (PEP) isn't an axiom or based only on observation (like Schrödinger equation), it follows from electron indistinguishability which requires antisymmetric wavefunction of the system

The PEP is an axiom of QM. More generally, it's the anti-symmetry of the total wavefunction. It doesn't follow from indistinguishability. Likewise, for bosons the symmetry of the wavefunction is an axiom.

These axioms of QM can be proved for fermions and bosons using QFT - but that's a different ball game.

 

[PeterDonis]

you are missing the point

No, you are missing the point because you don't understand what you are modeling. See below.

I am talking about HYPOTHETICAL system of non – interacting electrons.

No, you aren't, because there are no such things. You are talking about some kind of hypothetical particles that are neither bosons nor fermions, but "distinguishable" particles that have a different kind of statistics. You can of course mathematically model such things, but to then ask questions about why your model does not correctly describe electrons is pointless, since your model is not a model of electrons to begin with. It is also pointless to ask why particles are not indistinguishable in your model, since your model assumes them to be distinguishable to begin with.

 

[PeterDonis]

The PEP is an axiom of QM.

More specifically, it's an axiom for fermions in QM. But the OP's hypothetical particles are not fermions. That's obvious from the model being described in the OP.

 

[TeethWhitener]

We saw that Hartree product satisfies Schrödinger equation for system of non - interacting electrons (not something which yields good results for real electrons, but nevertheless true for non – interacting once).

No, it doesn’t. As others have mentioned several times, the Hartree product satisfies the Schrodinger equation for a system of non-interacting particles. The fact that the wavefunction is not antisymmetric under particle exchange tells you immediately that those particles can’t be electrons.

 

[PeterDonis]

even if electrons didn't interact, they should nevertheless be indistinguishable

And that is correct. But your model is not a model of electrons, so you should not expect your model to describe this property of electrons. If you want to model indistinguishable electrons, you need to model indistinguishable electrons, not the hypothetical distinguishable non-fermions you are modeling.

 

[vanhees71]

The PEP is an axiom of QM. More generally, it's the anti-symmetry of the total wavefunction. It doesn't follow from indistinguishability. Likewise, for bosons the symmetry of the wavefunction is an axiom.

These axioms of QM can be proved for fermions and bosons using QFT - but that's a different ball game.

It can be proved in QFTs with spatial dimensions $\geq 3$. In 2D you can have anyons, and there are some quasiparticle anyons found in effective 2D structures in condensed-matter physics (e.g., graphene).

 

[Dario56]

No, you are missing the point because you don't understand what you are modeling. See below.No, you aren't, because there are no such things. You are talking about some kind of hypothetical particles that are neither bosons nor fermions, but "distinguishable" particles that have a different kind of statistics. You can of course mathematically model such things, but to then ask questions about why your model does not correctly describe electrons is pointless, since your model is not a model of electrons to begin with. It is also pointless to ask why particles are not indistinguishable in your model, since your model assumes them to be distinguishable to begin with.

Ahh, yes. Problem is the fact that indistinguishability should be taken from the start or in another words that Hartree product doesn't really describe non – interacting electrons, but distinguishable electrons which we know doesn't make sense in QM.

That holds, but it is strange that such an inaccurate model (which has no ground in reality) can even satisfy Schrödinger equation since Schrödinger equation is based on experiment and not on fictitious systems.

I think I've got it, thanks.

 

[vanhees71]

One should also stress that indistinguishability has nothing to do with "interacting" or "non-interacting" particles. It's just the Hilbert space of the many-body system, where the Hamiltonian (including the interactions) is defined.

 

[Dario56]

The PEP is an axiom of QM. More generally, it's the anti-symmetry of the total wavefunction. It doesn't follow from indistinguishability. Likewise, for bosons the symmetry of the wavefunction is an axiom.

These axioms of QM can be proved for fermions and bosons using QFT - but that's a different ball game.

When I was learning about PEP, it was actually described in a manner I said here. As far as I understand it, since fermions (and bosons) are indistinguishable, probability density function and expectation values of observables should be independent of any exchange between identical particles. This can only be done if wavefunction stays the same under exchange (bosons) or flips the sign (fermions).

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

 

[Dario56]

One should also stress that indistinguishability has nothing to do with "interacting" or "non-interacting" particles. It's just the Hilbert space of the many-body system, where the Hamiltonian (including the interactions) is defined.

Yes, this is what I claimed in the main post and also in comments. But, I was somewhat confused with how can Hartree product even satisfy the Schrödinger equation. We've gotten that out of the way here.

 

[PeroK]

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

An axiom is something you assume to be true as a fundamental postulate of the theory. You cannot prove the PEP from the SDE. Symmetric, anti-symmetric and wavefunctions that are neither can all be solutions to the SDE. You need another axiom (PEP) to force antisymmetry on fermionic wavefunctions.

 

[TeethWhitener]

That holds, but it is strange that such an inaccurate model (which has no ground in reality) can even satisfy Schrödinger equation since Schrödinger equation is based on experiment and not on fictitious systems.

The time evolution of any quantum mechanical system must satisfy the Schrodinger equation. There’s nothing about the Schrodinger equation itself that describes electrons specifically. To specify electrons, you have to make sure the wavefunction is antisymmetric (among other things).

 

[vanhees71]

When I was learning about PEP, it was actually described in a manner I said here. As far as I understand it, since fermions (and bosons) are indistinguishable, probability density function and expectation values of observables should be independent of any exchange between identical particles. This can only be done if wavefunction stays the same under exchange (bosons) or flips the sign (fermions).

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

That's not true. The indistinguishability of particles only implies that the arbitrary exchanges of two particles form a symmetry group of the system, i.e., the quantum theory of a system of $N$ indistinguishable particles must admit a unitary representation of the symmetric group $S(N)$. That at the end you end up only with the trivial (bosons) and the "alternating" (fermions) representation in 3 or more spatial dimensions is due to additional topological properties concerning the exclusion of the case that two (or more) particles cannot be at the same place. In two spatial dimensions you can have arbitrary realizations of that symmetry:

M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional
Integrals for Systems of Indistinguishable Particles, Phys.
Rev. D 3, 1375 (1970),
https://link.aps.org/abstract/PRD/v3/i6/p1375