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When reading Abrikosov's book AGD, there is a statement that 'It is obvious only that a Bose system can not have excitations with half-integral spins' (page 5).

I don't understand why this is the case. For Fermi systems, there are bosonic excitations, such as magnons and cooper pairs (they are actually composed of a spinon and electron, respectively, but they are both spin $\frac{1}{2}$). Does it mean that the excitations of certain systems are only allowed to carry the same spin as the original particles?

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    $\begingroup$ Well, you can generate fermion-like states from bosons when there are strong interactions in low numbers of spatial dimensions: physicsworld.com/a/how-to-transform-bosons-into-fermions. So maybe there are caveats to that statement. $\endgroup$
    – Andrew
    Commented May 16 at 15:57
  • $\begingroup$ As Andrew mentioned this does not hold generally, however if you are in a situation where the excitations of your system can be treated via a fock space approach (i.e. multiple particle states from the vacuum) then you can think of spin as being either added or subtracted when you combine particles. i.e. from two spin 1 particles you can form a spin 0 or a spin 2 particle. Now from this point it is obvious that bosons can only ever give rise to bosons as the sum or difference of two natural numbers is again a natural number but given an arbitrary sum of half-integers the result can be both. $\endgroup$
    – Thomas Tappeiner
    Commented May 16 at 16:00
  • $\begingroup$ @ThomasTappeiner Is it means that in a Fock space of certain spin, to construct an excited state, we have to use the corresponding creation or annihilation operators which statisfy the corresponding spin statistics, e.g., for Fermi systems, if we want to creat a phonon, we have to $c^{\dagger}c^{\dagger}|0\rangle$? $\endgroup$
    – Houmin Du
    Commented May 17 at 10:52
  • $\begingroup$ @HouminDu Yeah you can see it in that way. If your Hilbert space is given by the fock space then all excitation you can generate are of the form $\prod_i a_i^\dagger \lvert 0 \rangle$ ( or linear combinations thereof). If you only have bosonic operators available any product e.g. $(a_1^\dagger a_2^\dagger)$ will still transform as a bosonic operator (e.g. commute with other such operators etc). However for fermions you can easily check that if $a_i^\dagger$ naturally anticommutes objects of the form $a_i^\dagger a_j^\dagger$ will rather commute. $\endgroup$
    – Thomas Tappeiner
    Commented May 17 at 12:07
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    $\begingroup$ You cannot get an odd integer from adding even integers, but you can get even integers from adding odd ones. It is not hard to rigorously prove this, but if this is not already obvious to you, the proof will not be all that illuminating. Same holds for bosons and fermions. A pair of fermions is a boson, but a pair of bosons is still a boson. You can prove this but, if not obvious already, the proof will not help all that much anyway... $\endgroup$
    – AccidentalFourierTransform
    Commented May 17 at 17:02

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When reading Abrikosov's book AGD, there is a statement that 'It is obvious only that a Bose system can not have excitations with half-integral spins' (page 5).

I don't understand why this is the case.

It is because you can not add whole integers and end up with a half-integer.

For Fermi systems, there are bosonic excitations,

Yes, this is because you can add half-integers and end up with a whole integer. E.g., $$ 1/2 + 1/2 = 1 $$

Or, as a slightly less trivial example, with respect to angular momenta: $$ \frac{1}{2}\otimes\frac{1}{2} = 0\oplus 1\;, $$ where the $\otimes$ indicates a direct product and the $\oplus$ indicates a direct sum. For example, the direct product of two spin-1/2 fermions can result in an overall spin state with total spin 0 or 1.

Does it mean that the excitations of certain systems are only allowed to carry the same spin as the original particles?

No, not necessarily. For example, I can combine two spin-1 particles and end up with total spin angular momentum of 2, 1, or 0.

All the examples I provided follow the usual rules for "addition" of angular momentum in quantum mechanics. E.g., for the case of adding two angular momentum $j_1$ and $j_2$, the result will be in the range from: $$ |j_1 - j_2| $$ to $$ j_1 + j_2\;. $$

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