In the book "Nolting, Theoretical Physics Part 5/2" (German), on Page 264, Formula 8.80, the author introduces second quantization in the case of identical particles. One considers the product space $(\mathcal{H}_N = \bigotimes_i \mathcal{H}_i)$, where the kets $(\vert \phi^i_j \rangle)$ represent the j-th base vector of the i-th particle. A ket of the product space of identical particles is then given by:
\begin{align}
\vert \phi_N \rangle &= \vert \phi_{\alpha_1}, \phi_{\alpha_2}, \cdots, \phi_{\alpha_N} \rangle \\
&= \vert \phi_{\alpha_1}^1, \phi_{\alpha_2}^{2}, \cdots, \phi_{\alpha_N}^{N} \rangle \\
&= \vert \phi_{\alpha_1}^1 \rangle \vert \phi_{\alpha_2}^{2} \rangle \cdots \vert \phi_{\alpha_N}^{N} \rangle \\
&= \vert \phi_{\alpha_1}^1 \rangle \otimes \vert \phi_{\alpha_2}^{2} \rangle \otimes \cdots \otimes \vert \phi_{\alpha_N}^{N} \rangle
\end{align}
To find symmetrical and antisymmetrical states, one introduces an operator (S) and considers its action on kets as (anti)symmetrizing, resulting in:
$$
\vert \phi_N \rangle ^\pm = S \vert \phi_N \rangle
$$
Here, the plus sign represents Bosonic states, and the minus sign represents Fermionic states.
In addition, occupation numbers are introduced such that $(\sum_i n_i = N).$
In a small calculation , it is shown that
$$
\langle \phi_N^\pm \vert \phi_N^\pm \rangle = \frac{1}{N!} \sum_{\mathcal{P}} (\pm)^p \langle \phi_N \vert \mathcal{P} \vert \phi_N \rangle
$$
where $p$ is the number of permutations and $(\mathcal{P})$ is a permutation operator that acts on $(\phi_N)$ such that $(\mathcal{P} \phi_N = \vert \phi_1^{i_1}, \phi_2^{i_2}, \cdots \phi_N^{i_N} \rangle).$
The book then makes the claim, which I do not understand nor have been able to derive, that:
$$
\langle \phi_N^\pm \vert \phi_N^\pm \rangle = \frac{1}{N!} \prod_{i=1}^{N} n_i!
$$
How is this calculation made? The argumentation in the book is not clear.
Additionally, according to a discussion on Math Stack Exchange, vectors need to be ordered for the scalar product to be defined. If we are permuting the upper indices, how can we evaluate such a product, especially if, for example, a state from space 1 is to be taken with a state from space 3? Does "identical particles" mean identical Hilbert spaces? Is it set to be zero?
Best Answer
I will assume you are OK with the above equation, which I have tagged as Eq. (A).
Suppose that every orbital (or "base vector") was different, in this case the occupation numbers would all be $n_i=1$ and the result would be: $$ \langle \phi_N^\pm \vert \phi_N^\pm \rangle = \frac{1}{N!}\;. $$
This is consistent with Eq. (A) (tagged above), since for this case only the identity permutation can contribute. This is true since if there is any non-trivial permutation then you end up with a orbital product like $\langle\phi_i|\phi_j\rangle$, where $i\neq j$.
OK, now suppose that one of the orbitals (e.g., the orbital $\phi_{15}$) is occupied by two particles. For this orbital $n_{15}=2$ and all others are $n_i=1$. In this case you only get a single $2!$ in the numerator since there are only $2!$ permutations that give a non-zero result (the two that swap the particles in orbital 15).
In general, think of a state of the product space as being denoted by an ordered sequence of orbital numbers like, for example, if there are 12 particles and 5 orbital states: $$ [1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 5]\;. $$ In the above example, there are $12!$ possible permutations, but all $3!$ of the permutations where I only permute "1" orbitals will end up looking the same. Similarly, all $2!$ permutations where I only permute "2" orbitals will look the same. Similarly, all $4!$ permutations where I only permute "3" orbitals will look the same. And finally, all $2!$ permutations where I only permute "5" orbitals will look the same. In fact, I will have a total of $3!\times 2! \times 4! \times 1 \times 2!$ of states that "look the same." Since I can permute around all the same-named orbitals as much as I want.
It is only the states that "look the same" that can contribute to the sum, since otherwise I will have a inner product of two orthogonal orbitals giving zero.
The above example about which states "look the same" can be generalized to say that $\Pi_i n_i!$ states will "look the same."
You will encounter similar calculations when you have to "choose" items out of a set. E.g., "5 choose 3" is equal to 10, since you can "choose" things by forcing them to be in certain positions, but then you have to "mod out" by the number chosen (3) and the number unchosen (2), so the result is $\frac{5!}{3!2!}$.