# Bosons, Spin Wavefunctions and Exchange Symmetry

I’m wondering about bosons and anti-symmetric spin states. (Throughout this post I’ll be using symmetric and anti-symmetric with respect to exchange of particle labels.)

It is well known that bosons have symmetric overall wavefunctions while fermions have anti-symmetric overall wavefunctions. For fermions, this overall anti-symmetry can be (and is!) created either by having an anti-symmetric spin wavefunction paired with a symmetric non-spin wavefunction or by a symmetric spin wavefunction paired with an anti-symmetric non-spin wavefunction.

Intuitively, one would suppose that bosons could equally produce a wavefunction with overall symmetry either by having two symmetric components or two anti-symmetric components. I’m currently tutoring a course on quantum mechanics which states without proof that bosons always occupy symmetric spin states, and thus need symmetric spatial states to produce an overall symmetry if those are the only degrees of freedom available. I’m perfectly willing to accept this, but I’d like to have something more convincing to tell my students than ‘the book says so’.

I tried to google appropriate terms with no luck and opened the question up to my fellow graduate students at tea, but there is no joy. So I turn to the physics minds of the Straight Dope: assuming that it is true that bosons never occupy antisymmetric spin states, why is this so? Why does the same reasoning not prevent fermions from occupying symmetric spin states?

Are you sure it’s not just talking about spin zero bosons?

If you have, say, a spin 1/2 fermion, you have two possible z-components of spin (call them |up> and |down>), and can form an antisymmetric state like:
|up> x |down> - |down> x |up>

Replacing the minus with a plus of course makes it symmetric.

However, if you have a spin 0 boson, you only have one possible z-component of spin (call it |0>).

You can still have the symmetric state:
|0> x |0> + |0> x |0>
which is really just |0> x |0> since we take our total state to be normalized.

But the antisymmetric state doesn’t exist:
|0> x |0> - |0> x |0> = 0

However, for higher-spin bosons, I don’t see why you couldn’t have an antisymmetric spin state, such as:
|1> x |3> - |3> x |1>

Clarification: When I wrote |1> and |3> above, a really had in mind what would normally be called |1> and |-1>, but any two of the three values (0, 1, and -1) would work.

The point is you need multiple states to form an antisymetric combination. If two particles can only occupy the same spin state (as is the case for spin=0), then the total spin state is necessarily symmetric.

Thanks for the answer - after further looking around, I’ve found an erratum on the course website. The statement is indeed not true and should not have been in the text to begin with. I thought something was fishy about it! Whew.

Your intuition is exactly right, and in fact that’s one of the motivations for “color” in quantum chromodynamics. Spin and isospin alone make the two up quarks in a proton symmetric, but they should be antisymmetric as fermions. Enter color and antisymmetric color-states to fix things up.