# Invariant subspaces of multi-particle Hamiltonians

I haven't found anywhere that discusses multi-particle quantum mechanics exactly like this, and I'm not sure the correct thing to Google. If anyone has any pointers to a reference that might help, I'd appreciate it!

In first-quantization, the Hamiltonian operator $$\hat{H}$$ acts on wavefunctions of $$N$$ variables $$\psi(x_1,...,x_N)$$. For example, a standard $$N$$-body Hamiltonian might look like $$\hat{H}=-\sum_i\frac{1}{2m}\frac{\partial^2}{\partial x_i^2}+\frac{1}{2}\sum_{i\neq j}V(x_i-x_j)$$. If this is a Hamiltonian describing identical particles, it must commute with the permutation operators defined by $$\hat{P}_\sigma\psi(x_1,...,x_N)\equiv\psi(x_{\sigma(1)},...,x_{\sigma(N)})$$.

If we consider the set of all $$\hat{H}$$ that commute with the permutation operators, these $$\hat{H}$$ form a semigroup. It's obvious that this semigroup has at least two invariant subspaces. The antisymmetric subspace (wavefunctions for Fermions) is preserved, and the symmetric subspace (wavefunctions for Bosons) is also preserved under this semigroup.

My question is: How many invariant subspaces are there for general $$N$$? For $$N=2$$, the symmetric and antisymmetric subspaces span the space of all functions, so there are only two invariant subspaces. For $$N>2$$, there's at least a third subspace, the set of functions orthogonal to all antisymmetric and symmetric functions, that is clearly preserved by the semigroup. But it's not clear to me if this space breaks into smaller invariant subspaces, or if the semigroup mixes all the states orthogonal to the symmetric and antisymmetric subspaces.

I've asked a similar question on the math stackexchange, where I've phrased it in terms of finite-dimensional representation theory, if anyone finds that simpler to think about than symmetries of wavefunctions.

The spaces of defnite symmetry correspond to the representation spaces of the symmetric group are labelled by Young diagrams. The number of diagrams for $$N$$ particles is given by the number of partitions of $$N$$.
• The permutations acting on wavefunctions definitely form a representation of the symmetric group, but dividing this representation into its irreps doesn't seem to be what I am looking for. For example, the wavefunction $\psi(x_1,...,x_N)=\phi(x_1)...\phi(x_N)$ spans a 1D trivial subrepresentation, but it does not span an invariant subspace of the semigroup, since the semigroup contains uniform rotations $\phi(x_1)...\phi(x_N)\rightarrow\bar\phi(x_1)...\bar\phi(x_N)$ for any function $\bar\phi$. Jun 22, 2020 at 18:32
• I don't understand your $\phi\to \bar\phi$ example. Decomposable wavefunctions don't even form a subspace (sum of decomposables is not usually decomposable). Since all your $H$'s commmute with the primitive idempotents (young symmetrizers) of $S_N$, I'd have said that the inmages of those projectors on the multiparticle space are all invariant under the set of $H$'s that compute with permutations. So for example the set of all symmetric functions form a subspace corresponding to the Young diagram with a single row of squares. Jun 22, 2020 at 21:48
• Ok, after Googling around a bit I think I understand what you're saying. What I was missing was the tensor product structure of the decomposition into an irrep of $S_N$ and an irrep of the Hamiltonian operators. So, e.g., the state $\phi(x_1)\cdots\phi(x_N)$ spans a 1D trivial irrep of $S_N$, but the structure I care about is the irreps of $S_N$ that are tensored with the irreps of $H$. Jun 23, 2020 at 16:10