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In this note$^1$ page 4 the author states that suppose $a_i$ acts on the whole vector space $V$ where $a_i$ is the annihilation operator of fermions. And since $[a_i^\dagger a_i,a^\dagger_j a_j]=0,\forall i,j$, and $a_i^\dagger a_i$ are hermitian operators, so we can construct a basis for them as $\left( {a_1}^{\dagger} \right) ^{\alpha _1}...\left( a_{n}^{\dagger} \right) ^{\alpha _n}|vac\rangle $ where $\alpha _i=0$ or $1$. Then the author states that the constructed vector space $W$ is a subspace of $V$, but I wonder that hermitian operators commuting implies that we have a common complete orthogonal basis, so how can $W$ be only a subspace of $V$?


1. "The Fermionic canonical commutation relations and the Jordan-Wigner transform" By Michael A. Nielsen, 2005

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    $\begingroup$ So the question is if $W$ is a proper subspace of $V$? $\endgroup$
    – Qmechanic
    Jul 16, 2022 at 14:05
  • $\begingroup$ @Qmechanic I've added more details to the link which I think is enough. It seems that $W$ is a proper subspace of $V$ from what the author wants to convey in the note and while I don't think so, so this is my question. $\endgroup$
    – narip
    Jul 16, 2022 at 14:24

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I think you are assuming that $\mathrm{dim}(V)=2^n$, which is not necessarily the case. As a somewhat trivial example, imagine you have an $8$-dimensional space $V$ spanned by orthogonal vectors $|\alpha_1,\alpha_2,\alpha_3\rangle$ with $\alpha_i \in \{0,1\}$, and consider the fermionic raising/lowering operators $a_1$ and $a_2$ where e.g. $$a_1|1,\alpha_2,\alpha_3\rangle = |0,\alpha_2,\alpha_3\rangle \qquad a_2|\alpha_1,1,\alpha_3\rangle = |\alpha_1,0,\alpha_3\rangle$$

By applying $a_1^\dagger$ and $a_2^\dagger$ to the vacuum state $|0,0,0\rangle$, we generate the 4-dimensional subspace $W$ which is spanned by $|\alpha_1,\alpha_2,0\rangle$. The orthocomplement $W^\perp$ is the subspace of $V$ which is spanned by $|\alpha_1,\alpha_2,1\rangle$. As a result, we may identify $V\simeq \mathbb C^4 \otimes \mathbb C^2$ (i.e. $|\alpha_1,\alpha_2,\alpha_3\rangle \simeq |\alpha_1,\alpha_2\rangle \otimes |\alpha_3\rangle$) where the set of operators $\{a_1,a_2\}$ and their adjoints act non-trivially only on $\mathbb C^4$.

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