I'm studying mathematics and I began a course in quantum statistics, in which I got to the discussion related to indistinguishibility of particles. My professor's notes are not very clear and rigourous, but I understood from them that:
- Two quantum states are physically equivalent iff they differ by a complex phase factor $$|\Psi \rangle \equiv |\Psi' \rangle \iff |\Psi'\rangle = e^{i \phi} |\Psi\rangle$$
- Particles $|\psi_i\rangle$ within a state $|\Psi\rangle \in \bigotimes_{i=1}^n H_i$ are indistinguishable iff $$P|\Psi\rangle \equiv |\Psi\rangle \quad \forall P \in S_n$$ that is if given a generic permutation of the variables ($S_n$ is the symmetric group), the corresponding state is physically equivalent to the original.
- In a quantum system of identical particles, these are indistinguishible (which is postulate?)
Then my professor notes jump and say that there are two kinds of particles: bosons and fermions, and that their representation in $\bigotimes_{i=1}^n H_i$ is given by: $$|\Psi\rangle_{Bosons} = \sqrt{\frac{\prod_n m_n!}{N!}}\sum_{\pi \in S_n} |\psi_{\pi(1)}\rangle|\psi_{\pi(2)}\rangle\dots|\psi_{\pi(n)}\rangle$$ with $m_n$ number of particles in the $|\psi_n\rangle$ state, while $$|\Psi\rangle_{Fermions} = \sqrt{\frac{1}{N!}}\sum_{\pi \in S_n} \text{sign}(\pi)|\psi_{\pi(1)}\rangle|\psi_{\pi(2)}\rangle\dots|\psi_{\pi(n)}\rangle$$ with no further explanation. So I proceeded, trying to derive the latter expressions. I started with noticing that by imposing that $$\pi|\Psi\rangle \equiv |\Psi\rangle$$ we are saying that $|\Psi\rangle$ is an eigenstate of $\pi \in S_n$, and we need to impose this for all possible permutations $S_n$. So, since the permutations have a spectrum $\sigma(\pi) \subseteq \{\pm 1\}$ then we need that $$|\Psi\rangle \in \bigcap_{\pi \in S_n}(E_\pi(+1) \cup E_\pi(-1))$$ where by $E_\pi(\pm 1)$ I mean the eigenspace of $\pi$ associated to the eigenvalue $\pm 1$. Now we have that in general $|\Psi\rangle$ will be a linear combination (up to normalization) of the permuted elements: $$|\Psi\rangle = \sum_{\pi \in S_n} f(\pi) |\psi_{\pi(1)}\rangle|\psi_{\pi(2)}\rangle\dots|\psi_{\pi(n)}\rangle$$ By imposing $$ \text{for all transpositions } \tau \in S_n (\tau |\Psi\rangle = |\Psi\rangle \text{ or } \tau |\Psi\rangle = -|\Psi\rangle)$$ We get that $f:S_n \rightarrow \mathbb{Z}^*$ is a homomorphism, and we have that there are only two homomorphisms from $S_n$ into $\mathbb{Z}^*$:
- $f_1 = 1$
- $f_2 =$ sign
So we can catalogue particles according to this result, that is the combination's coefficients are given by $f_1$ in case of Bosons, and by $f_2$ in case of Fermions. In particular we get that $$|\Psi\rangle_{Bosons} \in \bigcap_{\tau \in S_n} E_\tau(+1)$$ and $$|\Psi\rangle_{Fermions} \in \bigcap_{\tau \in S_n} E_\tau(-1)$$ Is this line of reasoning acceptable and\or reasonable?