This question may be pretty naive. But I wanted to confirm it anyway. Also if people can tell me more about this fact, I will be grateful.

I am looking at a simple problem of n-particle free Hamiltonian:

\begin{align} \label{1} \tag{1} H_N=\sum\limits_{i=1}^NH_i \;with\; H_i=-\frac{\hbar^2}{2m}\nabla^2_{\vec{r}_i}+V(\vec{r}_i) \end{align}

Now, suppose the single particle energy eigenstates are $\{u_j(\vec{r}_i)\}_{j=0}^{n}$ such that \begin{align}\tag{2}\label{2}H_iu_j(\vec{r}_i)=e_ju_j(\vec{r}_i)\end{align}

We can very easily list a basis for the Hilbert space spanned by the eigenstates of $H_N$, viz. $$\tag{3}\label{3}\{\prod\limits_{i=1}^Nu_{j_{i}}(\vec{r}_i):j_i \text{ picks out an energy eigenstate of $H_i$ }\}$$ and each such eigenstate in the list given has an eigenvalue of $\sum\limits_{i=1}^{N}e_{j_i}$.

Now, I want to consider the bosonic state space only. Then, I take the basis elements from (\ref{3}) and construct all possible sum of permutations of the form: \begin{align} \tag{4}\label{4}\{\sum\limits_{\text{permutations of} (p_{1},p_{2},...,p_{k})}\prod\limits_{i=1}^Nu_{j_i}(\vec{r}_i):j_i\in \{p_1,p_2,...,p_k\}, \{p_1,p_2,...,p_k\} \text{ label a set of k distinct energy eigenstates of } H_i\} \end{align}

Thus the new basis,(\ref{4}), consisting of only symmetrized eigenstates is smaller in size than the one in (\ref{3}).

Is this correct? For then it would mean the bosonic sector of the starting Hilbert space is a proper subspace of it. Same would hold for the fermionic sector.

Is the combination of bosonic and fermionic sector also a proper subspace of the starting Hilbert space?

Soliciting comments and answers.


1 Answer 1


Consider the single particle Hilbert space, $\mathcal{H}$. For $N$ particles, the total Hilbert space is simply given by $\mathcal{H}_{total} = \otimes_{i=1}^{N} \mathcal{H}$, i.e., the tensor product of the single particle Hilbert spaces. In this language, your Hamiltonian above can be rewritten as $H = \sum_{i}^{N} \{(H_1\otimes I \otimes I...\otimes I) + (I \otimes H_2 \otimes I \otimes ... \otimes I) + ... + (I \otimes ... \otimes I \otimes H_N)\}$

Now, bosons or fermions are simply given by the symmetric and antisymmetric state vectors respectively. But symmetrization or antisymmetrization is still just about taking specific linear combinations. Antisymmetrization, for example, is given by

$ {\frac {1}{\sqrt {N!}}}\sum _{p}\operatorname {sgn} (p)|n_{p(1)}\rangle |n_{p(2)}\rangle \cdots |n_{p(N)}\rangle $ where $p$ are permutations.

Therefore, such states still live in the same Hilbert space, $\mathcal{H}_{total}$.

Next, it is easy to see that the space of symmetrized and antisymmetrized wavefunctions is not the total Hilbert space (since there are states that are neither symmetric not antisymmetric). Therefore, these states just form a subspace at best for the entire Hilbert space. Hence, bosonic and fermionic sectors are just subspaces of the total $N$-particle Hilbert space.

Regarding the last part on combinations of bosons and fermions, you cannot superpose a boson and a fermion $--$ this is restricted by something called superselection rules. Also see this arXiv article about superselection rules. States from the bosonic and fermionic sectors cannot be superposed with each other.


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