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I've been introduced to multiple particle systems in quantum mechanics, and in the case of the $2$-electron system, I'm facing this massive confusion.

In the ground state of a $2$-electron system, you have both electrons occupying the same exact energy level, one is spin up, the other is spin down. My question is, why does this wave function have to have a singlet spin part?

I've read many answers here, and it claims that, since the spatial wavefunction for two electrons in the same energy level, is symmetric under exchange, the spin part must be antisymmetric. This is because, the particle is a fermion, and must always have an antisymmetric wavefunction.

However, according to the books that I've read, ( Griffiths, Zettili ), the concept of the symmetric and antisymmetric wave function is important only when the two particles are indistinguishable. However, that is not the case here. Due to the spin projection, the two particles here are distinguishable. In that case, why do we care so much about creating an antisymmetric wave function ?

Is the book wrong or misleading here, or is there a better reason for the singlet state? The book and the answers over here, clearly seem to contradict each other, unless I'm missing something. Any explanation would be highly appreciated.

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    $\begingroup$ You can't label them; you can not tell which one is in up state and which one is in down state. All you can tell is that one is like this and the other is like this. In other words: It is impossible to label them, even in principle. $\endgroup$ Aug 20, 2021 at 10:56
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    $\begingroup$ For the first part: Yes you are correct. That is the reason why we consider them indistinguishable. For the second question: First of all a good question. Think about it for a while. Suppose you conduct an experiment in which you know that you have an electron and proton. You take a cathode ray tube and record scintillation on the cathode. This tells you that the particle hitting it was obviously the proton. Replace this by 2 electron situation: Suppose you see a single scintillation on the anode. Do you know which electron hit it? The only thing you know... $\endgroup$ Aug 20, 2021 at 12:24
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    $\begingroup$ ... is that one of them hit it. You can't question which one hit it. The electrons have same everything: charge, mass, spin etc. Don't make the mistake of thinking that obviously it is you who doesn't know which electron hit the anode, but deep down there is some objective reality to which one hit it. As per the standard Copenhagen interpretation,there is literally no way to tell that the particles even existed before the scintillation,let alone the question of which one hit the screen. In other words, not only can you not label them; there is no label on them even outside your knowledge $\endgroup$ Aug 20, 2021 at 12:28
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    $\begingroup$ Answering "Why is mass a good label but spin orientation not?" Because both electrons can have either value of $L_z$. How about mass? Can an electron ever weigh $1836 m_e$ $\endgroup$ Aug 20, 2021 at 12:31
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    $\begingroup$ @Sarthak I think I finally understand this. I think my biggest problem was thinking, that if two particles exist in different states, they must be distinguishable. that is clearly not the case. In order to be distinguishable, one or more intrinsic properties like mass, spin etc should differ irrespective of what particular state they are in. $\endgroup$ Aug 20, 2021 at 12:33

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Yes but which one has its spin up, and which one its spin down?

Surely calling the first electron "one" and the second electron "the other" is arbitrary so the trick we have to deal with this arbitrariness and make the state independent of this arbitrary labelling is to make the wavefunction either symmetric (for bosons) or antisymmetric (for fermions). This way, there is no such thing as electron "one" and electron "other". That's why we make a state symmetric or antisymmetric.

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    $\begingroup$ Okay, so basically, the $z$ component of spin cannot be used as a label, so these particles are indistinguishable. However, what if instead of two electrons, we had two 'distinguishable fermions like an electron and a proton for example, in that case, are the particles distinguishable? Do we need symmetric or antisymmetric wavefunction then? Or can spin take any of the 4 states, 1 singlet and 3 triplets, as the symmetry doesn't matter $\endgroup$ Aug 20, 2021 at 12:08
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    $\begingroup$ Why is the mass of a particle a good label that can distinguish between the two, but the z component of spin is not ? Is it because of the hamiltonian? Also, if both had spin up, the exclusion principle wouldn't have allowed that. Doesn't the exclusion principle, treat these as different distinguishable particles ? $\endgroup$ Aug 20, 2021 at 12:11
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    $\begingroup$ I like your comments. This is a subtle point. The component $s_z$ depends on a choice of basis (or here, quantization axis) in the sense that you could have chosen $s_x$ rather than $s_z$. The spin quantum number $s$ or mass or electric charge do not depend on a choice of basis, so these are the ones to be used to referee indistinguishability. If I sound unconvincing, it is because I do not like the vague concept of "indistinguishability": I prefer to make a symmetry argument along my answer to this question: physics.stackexchange.com/a/652546/36194 and references therein. $\endgroup$ Aug 20, 2021 at 12:46
  • $\begingroup$ Thank you so much. I think I finally understand this. I think my biggest problem was thinking, that if two particles exist in different states, they must be distinguishable. that is clearly not the case. One electron in state 1, and the other in state 2, is exactly as indistinguishable as one up electron and one down. I failed to notice this. In order to be distinguishable, one or more intrinsic properties like mass, spin etc should differ irrespective of what particular state they are in. Is this valid ? $\endgroup$ Aug 20, 2021 at 12:54
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    $\begingroup$ yes this is an acceptable way of thinking about this. $\endgroup$ Aug 20, 2021 at 13:12

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