As per Wikipedia,

The above discussion generalizes readily to the case of $N$ particles. Suppose there are $N$ particles with quantum numbers $n_1, n_2, ..., n_N$. If the particles are bosons, they occupy a totally symmetric state, which is symmetric under the exchange of any two-particle labels: $$|n_1,n_2,...,n_N;S\rangle =\sqrt{\frac{\prod_nm_n!}{N!}}\sum_p|n_{p(1)}\rangle|n_{p(2)}\rangle\cdots |n_{p(N)}\rangle $$ The quantity $m_n$ stands for the number of times each of the single-particle states $n$ appears in the N-particle state.

But Kardar's book, Statistical Physics of Particles, says

The bosonic subspace is constructed as $$|\vec k_1,\cdots \vec k_N\rangle _+=\frac{1}{\sqrt{N_+}}\sum_P P |\vec k_1,\cdots \vec k_N\rangle_\otimes $$ Proper normalization requires $N_+=N! \prod_{\vec k}n_{\vec k}!$.

A particular one-particle state may be repeated $n_{\vec k}$ times in the list.

But both of them are quite different. Why it is so?


1 Answer 1


I am also looking for an answer to this, but I believe it is the latter. I think about it like this:

There are definitely $\frac{N!}{\prod_{n}(m_n!)}$ distinct permutations that change $|n_1 \rangle \cdots |n_N \rangle$ which are orthonormal and for each of those distinct permutations, there are $\prod_{n}(m_n!)$ permutations that dont change $|n_1 \rangle \cdots |n_N \rangle$

For notational convenience define $k = \frac{N!}{\prod_{n}(m_n!)}$ and $l = \prod_{n}(m_n!)$

So we can group all permutations of the $|n_1 \rangle \cdots |n_N \rangle$ into $k$ groups, each group containing the same state $l$ times and picking a representative state from each group, say states $\psi_1, \cdots, \psi_k$ gives us $k$ orthonormal states

So we can write \begin{align*} \bigg \| \sum_{\sigma \in S_N} |n_{\sigma(1)} \rangle \cdots |n_{\sigma(N)} \rangle \bigg\|^2 &= \sum_{i=1}^k \| l \psi_i \|^2 \\ &= l^2 \sum_{i=1}^k \| \psi_i\|^2 \\ &= l^2 k \\ &= N! \prod_{n}(m_n!) \end{align*}

Let me know what you think.


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