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When I am learning about constructing wavefunctions for identical particles, I am taught to write down the wavefunctions of well defined energy then symmetrise them according to exchange symmetry.

Why does this method guarantee to give us a complete set of eigenstates? In essence, why can we be sure that states with well defined energy provides a complete set of eigenstates?

I am wondering is the reason behind it is because hamiltonians for identical particles always commute with the exchange operator, i.e. $[H,P]=0$.

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There are two parts two this question. Let's first think if the state space. For distinguishable particles this would just be the tensor product (in the Hilbert space sense, i.e. the completion of the algebraic tensor product) of the parts. For nice enough function spaces these tensor products can be seen as function spaces again, with a domain that is the Cartesian product.

Since our particles are indistinguishable, many of these states are equivalent (or equal), so that the actual state space is a quotient of this space, in which $v\otimes w$ is equal to $w\otimes v$, and likewise for more factors. It is a basic fact of linear algebra that this quotient is isomorphic to a subspace of the same tensor product space, consisting of elements of the form $v\otimes w + w\otimes v$, and the general isomorphism maps a representative of an equivalence class to its symmetrization

$$v_1\otimes\cdots\otimes v_n \mapsto \frac1{n!}\sum_{\sigma\in\mathfrak {S}_n}v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)},$$

where $\mathfrak{S}_n$ is the group of permutations of the set $\{1,\ldots,n\}$ (and extended by linearity and continuity). When working with function spaces, this is equally explicit, because the right hand space consists of symmetric functions of its arguments, and a general function on the same domain is sent to its symmetrization.

For the second part of your question, your multi-particle Hamiltonian is usually defined in terms of the multi-particle space (before symmetrization). The fact that it commutes with the symmetrization operator $P$ means that $P$ maps energy eigenstates to energy eigenstates:

$$HP|\lambda\rangle = PH|\lambda\rangle = P\lambda|\lambda\rangle = \lambda P|\lambda\rangle,$$

which was intuitively clear as well.

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