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Suppose I have an electron and a proton, and that the electron is in the spin-up state, and that the proton is in the spin-down state. The particles are distinguishable, so I should just be able to write the wavefunction as $$|\Psi_{1}\rangle=|\uparrow\rangle_{e}|\downarrow\rangle_{p}$$ Clearly, this wavefunction is neither symmetric nor antisymmetric, since $|\Psi_1\rangle$ is not at all proportional to $$|\Psi_{2}\rangle=|\uparrow\rangle_{p}|\downarrow\rangle_{e}$$ So, from that standpoint this would seem to be neither a fermion or a boson. However, the particle also wouldn't be an angular momentum / spin eigenfunction; it would be in an even superposition of one of the spin-1 triplet states and the spin-0 singlet state. So, if you measured it's total spin, it would be integral, which seems to indicate that it would still be a boson, even though the wavefunction is not symmetric under particle exchange. Which is it? Is there another name for it? What type of statistics would such particles follow then?

Along this line of thought, it seems strange to me that a larger system of particles, such as helium 4, or heavier elements, being large combinations of protons, neutrons, and fermions, should conspire to create a perfectly symmetric or anti-symmetric wavefunction as required for the behavior we see of such particles, since such combinations of wavefunctions form such an infinitely small fraction of all the other ways you could combine the wavefunction.

Can anyone help me?

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  • $\begingroup$ You're treating the combination as a single particle, so when you think about its symmetry under the exchange operator, you should not be interchanging the electron and proton as you have done above, but should look at the wavefunction for states containing two different instances of this combination. $\endgroup$ – ApproximatelyTrue May 31 '15 at 19:08
  • $\begingroup$ Thanks for the comment. Could you be more specific when you say I "should look at the wavefunction for states containing two different instances of this combination"? Are you saying that the wavefunction I gave is not actually a valid wavefunction? I understand for identical particles you'd have to consider combinations such as |up>|down>+-|down>|up>, but it seems like for distinguishable particles I should not be so limited. $\endgroup$ – aquirdturtle May 31 '15 at 19:17
  • $\begingroup$ Actually, I think I understand now. You can tell if e.g. an electron is a fermion or a boson by considering whether it is symmetric or anti-symmetric under particle exchange of two identical electrons. I should therefore consider the exchange of TWO identical hydrogen atoms, not the exchange of particles within the hydrogen atom. $\endgroup$ – aquirdturtle May 31 '15 at 19:29
  • $\begingroup$ Yes, that is what I meant. @Sebastian explains it in detail in his answer. $\endgroup$ – ApproximatelyTrue Jun 1 '15 at 8:02
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The question you have to ask, is what happens under swapping the entire sets of coordinates of two of the composite particles! You don't have to think of spin addition and the spin-statistics-theorem (especially since spin and orbital angular momentum cannot necessarly be kept apart for composite particles due to spin orbit coupling!).

Taking your hydrogen example, the wave function of two hydrogen atoms has a position and spin dependence for each electron: $$\Psi(\vec r_{e1}, \sigma_{e1}, \vec r_{p1}, \sigma_{p1}, \vec r_{e2}, \sigma_{e2}, \vec r_{p2}, \sigma_{p2}) = -\Psi(\vec r_{e2}, \sigma_{e2}, \vec r_{p1}, \sigma_{p1}, \vec r_{e1}, \sigma_{e1}, \vec r_{p2}, \sigma_{p2}) = \Psi(\vec r_{e2}, \sigma_{e2}, \vec r_{p2}, \sigma_{p2}, \vec r_{e1}, \sigma_{e1}, \vec r_{p1}, \sigma_{p1})$$ Where the sign flips in each step as the electron and protons are both fermions. The result of this procedure is the wave function where the position of both entire hydrogen atoms have been swapped. As there is no sign change, the wave function is symmetric in the coordinate sets for hydrogen, giving that hydrogen is a boson.

Following this procedure, you can easily see, that a system composed of an even number of fermions is a boson, and a system composed of an odd number of fermions must be a fermion.

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    $\begingroup$ I think I understand now. I was trying to consider the exchange of particles within the hydrogen atom, when really in considering whether the hydrogen atom is a fermion or a boson, when I really should have been considering symmetry under exchange of two hydrogen atoms. Thanks! $\endgroup$ – aquirdturtle May 31 '15 at 19:32
  • $\begingroup$ I still don't get the answer to the question: the $H_2$ molecule is a boson but is a Hydrogen atom a boson or a fermion? $\endgroup$ – hyportnex May 31 '15 at 22:57
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    $\begingroup$ What I show, is that the $H$ atom is a boson. The two hydrogen atoms in this many particle wavefunction are well separated (so they do not form a $H_2$ hydrogen molecule). $\endgroup$ – Sebastian Riese May 31 '15 at 23:06
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    $\begingroup$ @user31748, you show that the $H$ atom is a boson by considering the interchange of two identical $H$ atoms. You would show that the $H_2$ molecule is a boson by considering the interchange of two identical $H_2$ molecules. $\endgroup$ – aquirdturtle Jun 2 '15 at 17:48

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