We have antisymmetric wavefunctions for a fermionic systems and symmetric wavefunctions for bosonic systems that give us a hint that quantum states can be occupied by a single fermion(or none) while multiple bosons can occupy single quantum state(including zero bosons in a state). This shows how fermions behave in themselves i.e. in a pure fermionic gas and how bosons behave in themselves i.e. in pure bosonic gas. Let us take a system of one fermion and one boson. I want to know how these two behave in presence of each other. What can we say about this system?
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1$\begingroup$ I think that the title may be misleading: as I understand, the question is "which are the properties of a many-body wave function for a system comprised of both fermionic and bosonic particles?". E.g. a gas of Helium (bosonic) and Lithium (fermion). Is my interpretation correct? $\endgroup$– QuilloOct 22, 2022 at 21:03
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They will interact through their Hamiltonian. For example, if the fermions are electrons and the bosons are electromagnetic radiation, the interaction Hamiltonian could be $$ \bar \psi \gamma^\mu \psi A_\mu,\qquad ({\rm relativistic\; case})$$ or $$ \frac{e}{m}\vec{p}\cdot \vec{A},\qquad ({\rm non-relativistic\;QM})\;.$$
If the interaction Hamiltonian is zero, they will both evolve independently, each in their own world, without interacting.
But I think you are asking for specifically if their interactions are affected by the boson/fermion statistics. I guess this would affect the form of the interaction Hamiltonian. The spin-statistics theorem says that fermions have to be half-integer spin and bosons integer. This severely limits the choices of Lorentz-invariant/rotation,translation-invariant interaction Hamiltonians. Of course, zero is always allowed.
answered Aug 31, 2020 at 10:20
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$\begingroup$ Am I correct that there must be an even number of fermions but can be any number of bosons in the interaction? For example in $\bar \psi \gamma^\mu \psi A_\mu$ there are two fermions ($\bar\psi$ and $\psi$) and one boson $A_\mu$. $\endgroup$– md2perpeAug 31, 2020 at 10:33
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$\begingroup$ I even think that the total charge should be zero which also is the case in that term. $\endgroup$– md2perpeAug 31, 2020 at 11:07
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$\begingroup$ Yes, otherwise charge will not be conserved, in which case it does not make sense to think of it as charge anymore. $\endgroup$ Aug 31, 2020 at 11:10