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An electron pair as a composite system acts as a boson. So why can’t three/four electron “pairs” be present in a single 1 s Hydrogen atomic orbital?

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    $\begingroup$ Who told you that two electrons are the same thing as a boson? $\endgroup$
    – knzhou
    Apr 3, 2020 at 22:35
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    $\begingroup$ Since hydrogen has a single electron electron pairs are indeed quite rare in that system. $\endgroup$ Apr 3, 2020 at 23:09
  • $\begingroup$ Can you make an antisymmetric total spin function with three 1/2 spins? $\endgroup$
    – Socrates
    Apr 3, 2020 at 23:18
  • $\begingroup$ en.wikipedia.org/wiki/Pauli_exclusion_principle $\endgroup$
    – G. Smith
    Apr 3, 2020 at 23:58

2 Answers 2

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This is actually a good question, because there are composite particles that we treat as bosons. A great example is the atoms in magneto-optical traps that are cooled to produce Bose-Einstein condensates (BEC). All the constituent particles are fermions (electrons, neutrons, protons), but the composite particles are bosons.

The difference here is that the composite bosons (the atoms) are small compared to the size of the trap. The density of the resulting BEC is not actually higher than the density of an ordinary liquid, so we are not trying to squeeze the electrons themselves into a single state.

In the case of electrons in an atom, they are bound to the nucleus, not each other. The singlet of two electrons in the 1s orbital in Hydrogen aren't a single object $(e_{\uparrow} e_{\downarrow})$ orbiting the nucleus. If you were to try to add a new electron to that state it would see both other electrons and want to form an antisymmetric state with both of them, which is cannot, therefore it can't be in the orbital.

References:

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  • $\begingroup$ No you see, I am talking of two electron "pairs", i.e., two (e↑e↓) pairs. Each of this (e↑e↓) pair is a boson right (having a total spin of 1 (triplet) or 0 (singlet)) ? Both these spins spates are integer. Now when you bring 2 electron pairs, both of these pairs "individually" have integer spins. So, they can combine to form a symmetric spin state (say of total spin 2 which is a symmetric state) which means they can stay in the same state (which I claimed to be Hydrogen 1s orbital just because its a symmetric SPATIAL wavefunction). $\endgroup$ Apr 4, 2020 at 17:20
  • $\begingroup$ Is the same thing. The electron pair isn't a tightly-bound state of two electrons they aren't bound at all, so you can't treat these two pairs of electrons as two bosons. $\endgroup$ Apr 6, 2020 at 10:24
  • $\begingroup$ You would have to be able to antisymmetrize the whole multi-electron wavefucntion $\psi( \vec r_1, \vec r_2, \vec r_3, \vec r_4) $ with respect to the exchange of any two $r_i$, $r_j$. $\endgroup$ Apr 6, 2020 at 10:26
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    $\begingroup$ O okay I see. Probably my concept of treating 2 electron combination as like a single spin 1 particle (like a photon for example) is wrong in all cases. Thank You. $\endgroup$ Apr 8, 2020 at 13:15
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An electron pair as a composite system acts as a boson.

Hooold up.

If by "composite system" you mean a bound system, like a Cooper pair, then yes. In this case, two electrons are bound together so that you can stop talking about the individual electrons (two fermions) and instead describe this new state as a new "particle". Which has integer spin. So it's a boson.

But if by "composite" you just mean that I can draw a box around them and the two electron happen to fall within it, then no. If they still go on and about on their own accord then they will be described by their individual wavefunction and they all all behave like fermions. In an atom, for example, the electrons still follow the Pauli exclusion principle and pile up in ever higher energy levels. They actually do interact among each other (e.g. via Coulomb repulsion) but not in a way, like for the Cooper pair, to be described by a new excitation.

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  • $\begingroup$ Ahhh Great!! Yeah it seems I almost understood. However could you just tell me a few more things? 1) Does your reply mean that the composite electrons in a Cooper pair are NOT distinguishable from one another but ARE distinguishable in case of the electron pair in an atom? 2) If Question 1 is correct then how does one explain the SINGLET combination of 2 electrons in atoms (e↑e↓ - e↓e↑) / sqrt(2) ? Doesn't this state imply that the electron spins are entangled and we cannot tell apart the individual electron spin? $\endgroup$ Apr 8, 2020 at 13:27
  • $\begingroup$ 1) Electrons are indistinguishable always, in the true sense of the word: they are identical particles. In a Cooper pair, the two electrons can be described by a new quasiparticle, so that our theory can be written in terms of that, without ever requiring knowledge of the individual electrons and their properties. For instance, once the Cooper pair is formed, you don't care about the distance between the electrons. Whereas in a Helium atom the Coulomb repulsion between electrons depends on the electrons separation. $\endgroup$
    – SuperCiocia
    Apr 8, 2020 at 17:53
  • $\begingroup$ 2) Two electrons can have a singlet (anti-symmetric) or triplet (symmetric) spin wavefunction. But because they are fermions, their total wavefunction needs to be antisymmetric: so if you have a symmetric spatial wavefunction (e.g. two electrons in the 1s state of Heliu,), you need the spin part to be a singlet. [cont...] $\endgroup$
    – SuperCiocia
    Apr 8, 2020 at 17:56
  • $\begingroup$ [...] The two electrons are entagled, meaning they are described by a single wavefunction that cannot be decomposed into (electron 1) and (electron 2), but requires some crosstalk between the two. This does not mean you cannot know the electrons' individual spins, you know that their spin is $s=1/2$. You may not know if it's up or down until you measure it. $\endgroup$
    – SuperCiocia
    Apr 8, 2020 at 17:58
  • $\begingroup$ Your total spin is indeed $0$, because you are summing $1/2$ from one electron and $-1/2$ from the other electron (twice). Exactly for that reason, you are still considering them as separate. If you could replace them by a quasiparticle such as Cooper pair, you would not need to care about the individual electrons anymore. $\endgroup$
    – SuperCiocia
    Apr 8, 2020 at 18:01

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