Symmetric, antisymmetric or neither

[fog37]

Hello,

If a composite system is formed by particles that are all fermions, the overall wavefunction must be antisymmetric. If the particles are all bosons, the wavefunction must be symmetric.

• What if the particles are not all identical particles (all electrons) but are all fermions? Does the wavefunction still needs to be antisymmetric regardless of the fact that the particles are not identical?

• What if the system was composed of particles that are both fermions and bosons? I bet the wavefunction is neither symmetric or antisymmetric, correct?

• For a two particle system composed of particle $1$ and particle $2$ and the states $|a>$ and $|b>$, why are the two states $|a>_1 |b>_2$ or $|b>_1 |a>_2$ not valid states? Are they valid but not general? Their superposition $c_{1} a>_1 |b>_2+ c_{2} |b>_1 |a>_2$ is a valid state but I am not sure why. What is missing to the the two product states?

Thanks!

 

[mfb]

The wave function has to be antisymmetric under the exchange of identical particles. For different particle types, there is nothing to exchange. If you have two identical fermions and two identical bosons, then it has to be antisymmetric under a fermion exchange and symmetric under a boson exchange.

Concerning the last bullet point: Those states are not (anti)symmetric. They work only if particles 1 and 2 are distinguishable.

 

[fog37]

Ok, thanks a lot.
For example, a neutron and an electron are not identical hence distinguishable. Identical particles are always distinguishable. Would would be a situation/problem in QM where we have to deal with distinguishable particles?

 

[Nugatory]

Identical particles are always distinguishable.

You mean INdistinguishable?

Would would be a situation/problem in QM where we have to deal with distinguishable particles?

You've already provided an example: A system containing an electron and a neutron.

 

[mikeyork]

Actually the anti-symmetric wavefunctions (or state vectors, or anti-commuting operators) for pairs of identical fermions as usually stated in the "symmetrization postulate" can also be re-expressed as symmetric wavefunctions by careful specification of the spin frames of reference. This has been well established by numerous authors, but is not widely known. The reason is actually very simple to see.

Consider the CM frame. The polar angles are related by

$\theta_b = \pi - \theta_a$ or $\theta_a = \pi - \theta_b$

and

$\phi_b = \phi_a \pm \pi$ or $\phi_a = \phi_b \mp \pi$

which is clearly symmetric in $\theta$ but asymmetric in $\phi$ under $a\leftrightarrow b$. If "exchange" $1\leftrightarrow 2$ refers to interchanging the labels $a$ and $b$ then the effect is a rotation by $2\pi$ about the z-axis on one particle relative to the other. This will change the sign of the wavefunction for fermions. If however "exchange" $1\leftrightarrow 2$ simply means a re-ordering of the product space leaving the polar relations between $a$ and $b$ unchanged then there is no sign change. It just so happens that conventionally, the distinction between $a\leftrightarrow b$ and leaving the relation between the polar angles unchanged by "exchange" is not made explicit. This would leave the fermion exchange two-valued with either a sign change or not, but it so happens that, in that case, the exchange phase can still be made single-valued if an implicit anti-symmetric choice be made.

All of this ambiguity can be avoided if, instead of the "symmetrization postulate" the spin-statistics theorem is expressed in terms of its observable consequences such as the superselection rule that $L + S$ is always even in the CM frame, or the rule that $S$ must be even when all other properties are the same..

More generally, states of multiple particles, whether indistinguishable or not, can always be symmetrically specified regardless of spin, but implicit assumptions regarding the relationship between spin frames of reference can introduce sign changes if those relationships are reversed for fermions when particles are interchanged. So the answer to the OP's question regarding a symmetry of a state of a boson and a fermion is "it depends", but it doesn't really matter as there is no resulting superselection rule to affect the physical observables.

The first refereed paper to be published showing how the conventional fermion anti-symmetry comes about was by Broyles in 1975 (soon after my own unpublished pre-prints from the University of Rome that same year). Similar papers were published by Bacry and Michael Berry.

Of course the explanation I have just given is a simplified summary. The full and complete argument is very detailed, particularly for massless particles, and depends on the assumption that particle "ordering" is an artifice of the theorist and not a property of nature so that it is always possible to specify a unique state vector to represent any multi-particle physical state.. The simplest complete argument is given in my paper:

"Symmetrizing The Symmetrization Postulate", AIP Conference Proceedings 545, "Spin-Statistics Connection And Commutation Relations", 2000.

A comprehensive list of authors who have published similar (but incomplete) proofs prior to 1999 is contained in my eprint

arxiv.org/pdf/quant-ph/9908078v2.pdf

Other authors who cited my work on this issue include A P Balachandran et al, O W Greenberg and A Jabs.