# relative phase/sign in $\Psi$ after exchange of composite particles with angular momenta

I'm reading Quantum Liquids by A.J. Leggett and became confused by the following statement in the first chapter.

Consider now a pair of such identical atoms. In the absence of appreciable coupling between the hyperfine and (atomic center-of-mass) orbital degrees of freedom, the energy eigenfunctions can always be written as a product of a “spin” function (i.e. a function of the hyperfine degrees of freedom) and an orbital function (i.e. a function of the coordinates of the atomic nuclei). The symmetry of the orbital wave function under exchange of the two atoms is $(−1)^L$ where $L$ is the relative orbital angular momentum. As to the spin wave function, the combination of two atomic (intrinsic) angular momenta each equal to $F$ yields possible values of the total dimer intrinsic angular momentum $K \equiv \left| F_1 + F_2 \right|$ equal to $0, 1, \ldots , 2F$.

$F$ here is the total atomic spin.

I have trouble understanding how the relative phase of the wave function results from the difference between the magnitude of the angular momenta when their $z$-projections ($z$ being the quantization axis) are not known.

Does anyone know how $L$ and $K$ are determined?

Ok, so "relative" here had to do with "relative motion" and not "difference".
Now... does anyone know the reasoning behind the factor of $(-1)^L$ ?

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I find it very hard to understand what you're asking. What do you mean by "how $L,K$ are determined?" They're eigenvalues of the corresponding operators and if written as a sum of some other angular momenta, they obey the rules for the addition of the angular momenta. Does it answer the last question? Concerning the previous question, why would $j_z$ affect symmetry considerations? It's the whole point of the rotational symmetry that different components in the multiplet have the same properties such as the symmetries. –  Luboš Motl Jul 7 '13 at 6:13
Also, you want to understand why a relative phase comes from a difference but there was no difference in the previous text... So which difference? –  Luboš Motl Jul 7 '13 at 6:13
Ok, then I must have misunderstood what "relative angular momentum" meant. So "relative" here comes from the relative coordinates of the two atoms instead of having to do with the difference. ... I was going in the wrong direction from the beginning, trying to bring in z-projections and Clebsch-Gordan coefficients etc. Thanks for the feedback. I actually made an edit stating my real question, about the (-1)^L factor. I may be able to answer it myself given some more time to think about it, but I'll leave it hanging here. –  DCY Jul 8 '13 at 4:51