Math for identifying bosonic atoms

[itssilva]

(And fermionic atoms, as well). Knowing the weird statement that rubidium atoms and such are bosons and reading some of the threads, I've convinced myself that it is a perfectly acceptable possibility; however, I'm not so sure if I learned how to point my finger and say atom X is a boson/fermion.
From what I see, the math is something like this: Wikipedia (BEC article) says ⁷Li is a boson; it has 3 e, 3 p and 4 n, right? All of which are fermions (s=1/2 for e; for p and n as well?). Now, to obtain the total spin of the ground state, divide each 'class' of fermion by 2; the remainder is the number of unpaired fermions in each class (1 e, 1 p, 0 n). This is the ground state, so the unpaired electron oughta be paired up with the proton (semiclassically: two magnets configured like N-S|N-S have less energy than N-S|S-N). End of the day: no leftover unpaired fermions, S=0. Boson. This is the rule of thumb I've adopted so far.
I say 'rule of thumb' because I'm highly suspicious of this method, since I've learned in my QM that spin addition is a quite fastidious procedure (determining Clebsch-Gordon coefficients and whatnot). Is this the correct math?
Also, I've heard in my QFT class (if I understood correctly) that, if you consider an energy scale less than that which would take to break apart a bound state, it 'looks like a fundamental particle' propagating away; is the same principle at work here? I mean, if the first excitation energy of ⁷Li is 5 eV and first ionization 11 eV (numbers I made up), does it mean that I oughta keep my lithium at energies lesser than 5 eV (in order to avoid spin-flipping) and/or 11 eV (to avoid ionization) to have it work with B-E statistics?

 

[fzero]

Orbital angular momentum is always an integer, so the only way to get a half-integer total angular momentum is to have an odd number of fermions. The rule you have is fine; an atom with an odd number of fermions must obey fermionic statistics.

To answer your second question. First, a spin flip changes the total angular momentum by $\pm 1$, so if the bound state started as a fermion, it will remain one after the spin flip. Second, the statistics of an ion will be determined by the same rule about odd/even number of fermions. If we remove a single electron from a fermionic atom we will have a bosonic ion and vice versa.

 

[DrDu]

From what I see, the math is something like this: Wikipedia (BEC article) says ⁷Li is a boson; it has 3 e, 3 p and 4 n, right? All of which are fermions (s=1/2 for e; for p and n as well?). Now, to obtain the total spin of the ground state, divide each 'class' of fermion by 2; the remainder is the number of unpaired fermions in each class (1 e, 1 p, 0 n). This is the ground state, so the unpaired electron oughta be paired up with the proton (semiclassically: two magnets configured like N-S|N-S have less energy than N-S|S-N). End of the day: no leftover unpaired fermions, S=0. Boson. This is the rule of thumb I've adopted so far.
I say 'rule of thumb' because I'm highly suspicious of this method, since I've learned in my QM that spin addition is a quite fastidious procedure (determining Clebsch-Gordon coefficients and whatnot). Is this the correct math?

You don't have to pair up the spins so that S=0 and normally, this won't be the case. However, the total spin will always be an integer number.
Arguing in terms of Clebsch Gordan coefficients: if s1 and s2 are half integer, so will be the z-components of spin. However, the z-components add up directly, so Sz will be integer. This is only possible if S is integer, too, as S_z ranges from S to -S in steps of 1.

 

[mathman]

A simple way to tell is by the number of neutrons in the nucleus. If odd it is a fermion, if even it is a boson. This is a result from the obvious fact that the number of protons equals the number of electrons.