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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?

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  • $\begingroup$ Why do you put brackets around equations? $\endgroup$ Sep 14, 2023 at 16:04
  • $\begingroup$ Is this really an important question @TobiasFünke? $\endgroup$
    – Mad
    Sep 14, 2023 at 16:05
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    $\begingroup$ Yes, at least for me it reduces readability. Also, use $$ for inline equations. $\endgroup$ Sep 14, 2023 at 16:07
  • $\begingroup$ What part of the explanation of the book is not clear? Have you computed an easy example to see what comes out? Say two or three particles for two or three single-particle states? $\endgroup$ Sep 14, 2023 at 16:07
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    $\begingroup$ My last comment. You write: "How is this calculation made? The argumentation in the book is not clear." and I asked what exactly is not clear. For me, the explanation (at least for the edition I have) is clear, and as I suggested, to do an easy example might help to understand these things better, i.e. to see how these factors indeed arise. But if you do not want to do this, fine. Moreover, yes, the inner product is well-defined and yes, the single-particle Hilbert space is the same for all particles - you couldn't introduce a permutation operator if this was not the case. $\endgroup$ Sep 14, 2023 at 16:16

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$$ \langle \phi_N^\pm \vert \phi_N^\pm \rangle = \frac{1}{N!} \sum_{\mathcal{P}} (\pm)^p \langle \phi_N \vert \mathcal{P} \vert \phi_N \rangle\tag{A} $$

I will assume you are OK with the above equation, which I have tagged as Eq. (A).

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.

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 $\frac{5!}{3!2!}$.

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    $\begingroup$ thank you for your answer, it is very illustrative and helpful. I will formulate a mathematical Proof and publish it here, it should not be difficult. If i was not clear in my question, i was asking for a mathematical proof, but your answer is just as good, as it illustrates why it is so, so i will be able to formulate this on my own now. $\endgroup$
    – Mad
    Sep 16, 2023 at 18:46
  • $\begingroup$ Sounds good to me. Have fun :) $\endgroup$
    – hft
    Sep 16, 2023 at 18:47
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    $\begingroup$ @Mad If this answered your question, consider to accept it. $\endgroup$ Sep 16, 2023 at 18:55

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