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?

  • $\begingroup$ I am not totally convinced by what you are proving here as you did not distinguish 2 different representations. Normally whether a particle is boson or fermion is determined by its $su(2,C)$ representation's dimension. A boson is a representation of odd dimensional representation and a fermion is a representation of even dimensions. A bunch of identical particles is merely tensor product of representations. The symmetric groups acting on a chosen set of basis elements of the representation to induce representation. Your result is the joint spectrum of $S_n$ with cartan element of $su(2,C)$. $\endgroup$ – user45765 Jun 2 at 17:41
  • $\begingroup$ Thanks. Since I haven't studies lie groups yet (I think it was predictable from the question) I understood almost nothing from your comment. Hence I am left with the question of what I have actually proved. $\endgroup$ – Lorenzo Jun 3 at 7:34
  • $\begingroup$ I would recommend reading Brian Hall's Lie group, Lie algebra and representation's section on $su(2,C)$ or $sl_2(C)$ representations. That book has very low prerequisite. Then you can read either shankar or sakurai's quantum mechanics for its boson fermion part. After that you can go back to brian hall's book for clebsch gordan coefficients, then you can compared to quantum mechanics book for identical particles or addition of angular momentum. $\endgroup$ – user45765 Jun 3 at 13:30
  • $\begingroup$ @Lorenzo I really think you should learn a bit of Lie Groups and Algebras before embarking in this kind of problems, stay humble $\endgroup$ – Francesco Bilotta Jun 6 at 17:46

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