Let $\mathcal{H}$ be a Hilbert space with orthonormal basis $\{\varphi\_{k}\}\_{k\in I}$. Take $\mathcal{H}^{\otimes n} := \overbrace{\mathcal{H}\otimes\cdots\otimes \mathcal{H}}^{\mbox{$n$ times}}$. An element of $\mathcal{H}^{\otimes n}$ can be expressed as:
$$\psi = \sum\_{\{k\_{1},...,k\_{n}\}\subset I}\alpha\_{k\_{1},...,k\_{n}}(\varphi\_{k\_{1}}\otimes \cdots \otimes \varphi\_{k\_{n}})$$
with $\alpha\_{k\_{1},...,k\_{n}} = \langle \varphi\_{k\_{1}}\otimes\cdots\otimes\varphi\_{k\_{n}},\psi\rangle$. Let us define $\sigma^{\*}$ as an operator on $\mathcal{H}^{\otimes n}$ which acts on the basis elements as:
$$\sigma^{\*}(\varphi\_{k\_{1}}\otimes \cdots \otimes \varphi\_{k\_{n}}) := \varphi\_{k\_{\sigma(1)}}\otimes\cdots \otimes \varphi\_{k\_{\sigma(n)}}$$
where $\sigma$ is a permutation of the set $\{1,...,n\}$. We extend $\sigma^{\*}$ to all $\mathcal{H}^{\otimes n}$ by linearity. Now, one can define:
$$A\_{n}:= \frac{1}{n!}\sum\_{\sigma}\epsilon\_{\sigma}\sigma^{\*}$$
an antisymmetrization operator on $\mathcal{H}^{\otimes n}$. Here $\epsilon\_{\sigma}$ is the sign of the associate permutation $\sigma$. Then $A\_{n}$ is an orthogonal projection and, if $A\_{n}\mathcal{H}^{\otimes n}$ denotes its range, the fermionic Fock space is defined to be:
$$\mathcal{F}\_{f}(\mathcal{H}) := \bigoplus\_{n=0}^{\infty}A\_{n}\mathcal{H}^{\otimes n}$$
with $A\_{0}\mathcal{H}^{0} := \mathbb{C}$.


Alternatively, let us say that a tensor $\psi \in \mathcal{H}^{\otimes n}$ is antisymmetric if $\sigma^{\*}\psi = \epsilon\_{\sigma}\psi$ for every permutation $\sigma$. Take $\wedge^{n}\mathcal{H}$ to be the subspace of all antisymmetric tensors of $\mathcal{H}^{\otimes n}$ and $\wedge^{0}\mathcal{H} := \mathbb{C}$.


**Question:** Can I use the second approach to define fermionic Fock spaces in an equivalent way, as before? In other words, if I set $\mathcal{F}'\_{f}(\mathcal{H}) := \bigoplus\_{n=0}^{\infty}\wedge^{n}\mathcal{H}$, does it follow that $\mathcal{F}\_{f}(\mathcal{H}) = \mathcal{F}'\_{f}(\mathcal{H})$? Equivalently: is it possible to prove that every $\psi \in \wedge^{n}\mathcal{H}$ can be expressed as $\psi = \frac{1}{n!}\sum\_{\sigma}\epsilon\_{\sigma}\sigma^{\*}\varphi$ for some $\varphi \in \mathcal{H}^{\otimes n}$?



Representation of an arbitrary element on a fermionic Fock Space
[This is not research level, so probably does not belong on MO, but I think the question is well-asked.]


If $\psi\in\wedge^n\mathcal{H}$ then by definition $\sigma^\*\psi = \epsilon\_\sigma \psi$ for each permutation $\sigma$ and so as $\epsilon\_\sigma \in \{\pm 1\}$ we have
$$ A\_n\psi = \frac{1}{n!} \sum\_\sigma \epsilon\_\sigma \sigma^\*\psi
= \frac{1}{n!} \sum\_\sigma \epsilon\_\sigma \epsilon\_\sigma \psi
= \psi. $$


Actually, the converse is much more interesting, as it really involves a little bit of representation theory. By the converse, I mean: show that if a tensor is in the range of $A\_n$ then it is anti-symmetric. So, $\psi = A\_n\varphi$ for some arbitrary $\varphi$. Then for a permutation $\sigma$,
$$ \sigma^\*\psi = \frac{1}{n!} \sum\_\tau \epsilon\_\tau \sigma^\*\tau^\*\varphi. $$
Set $\rho = \tau\sigma$ and notice that $\sigma^\*\tau^\*(\otimes\_i \varphi\_{k\_i}) = \sigma^\*(\otimes\_i \varphi\_{k\_{\tau(i)}}) = \sigma^\*(\otimes\_i \varphi\_{l\_i})$ say, where thus $l\_i = k\_{\tau(i)}$. Then $l\_{\sigma(i)} = k\_{\tau(\sigma(i))}$ and so $\sigma^\*\tau^\*(\otimes\_i \varphi\_{k\_i}) = \otimes\_i \varphi\_{k\_{\tau\sigma(i)}} = (\tau\sigma)^\*(\otimes\_i \varphi\_{k\_i})$. Thus we have an *anti-*representation of the symmetric group. As $\epsilon$ is a group homomorphism, $\epsilon\_\tau = \epsilon\_{\rho\sigma^{-1}} = \epsilon\_\rho\epsilon\_\sigma$. Thus
$$ \sigma^\*\psi = \frac{1}{n!} \sum\_\rho \epsilon\_\rho\epsilon\_\sigma \rho^\*\varphi
= \epsilon\_\sigma A\_n\varphi = \epsilon\_\sigma \psi. $$
So $\psi\in\wedge^n\mathcal{H}$.


