Ladder operators for fermionic Fock space

Question

To describe multiple fermionic particles, we introduce a Fock space $$\mathcal H_F=V_{\alpha=1}\otimes V_{\alpha=2}\otimes \ ...$$ such that each $V_\alpha$ is a two dimensional vector space labelled by an index $\alpha$ that refers to a complete set of quantum numbers. The corresponding ladder operators for each such space can be written as $$a_\alpha=\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}, \quad a_\alpha^\dagger=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$ acting on a basis $\{|0\rangle_\alpha=(1,0)_\alpha^T, \ |1\rangle_\alpha=(0,1)_\alpha^T\}$. This works fine as $a_\alpha^\dagger$ sends $|0\rangle_\alpha$ to $|1\rangle_\alpha$, $a_\alpha$ does the opposite, and $a_\alpha^\dagger|1\rangle_\alpha=a_\alpha|0\rangle_\alpha=0$, so that the number of particles for each $\alpha$ can only be $0$ or $1$. It can then be checked that the various anticommutation rules $\{a_\alpha,a_\alpha^\dagger\}=1$, $\{a_\alpha,a_\alpha\}=\{a_\alpha^\dagger, a_\alpha^\dagger\}=0$ hold.

The only remaining problem is that for $\alpha\ne \beta$, there is still symmetry under the transformation $\alpha\leftrightarrow \beta$ when creating particles, against the antisymmetric nature of fermions: we wish instead to have operators such that $$b_\alpha^\dagger b_\beta^\dagger|\text{state}\rangle=-b_\beta^\dagger b_\alpha^\dagger|\text{state}\rangle.$$ I've been told that this can be accomplished by defining $b_\alpha^\dagger=a_\alpha^\dagger \eta_\alpha$, where $$\eta_\alpha=\prod_{\gamma=1}^{\alpha-1}\begin{pmatrix}-1 & 0 \\ 0 & 1 \end{pmatrix}_\gamma.$$ However, I don't see how this works or how it was derived. Could someone provide an explanation?

Answer 1

So any state in the Fock space can be parametrized by \begin{equation} |\nu \rangle = | (i_1,n_1), \ldots, \rangle, \end{equation} where $n_\alpha$ is the number of particles that ocupy the state discribed by the quantum number (potentially more than one) $i_\alpha$. For fermions $n_\alpha = 0,1$. Let us now assume without loss of generality that $\alpha < \beta$. Let us also assume that $n_\alpha \neq 0$ and $n_\beta \neq 0$ or else the statement is trivial. Then \begin{equation} \begin{split} b_\alpha^\dagger b_\beta^\dagger |\nu \rangle &= (-1)^{\sigma_\beta} b_\alpha^\dagger |(i_1,n_1), \ldots , (i_\beta, n_\beta + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha + \sigma_\beta} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split} \end{equation} where $\sigma_\beta$ is the number of occupied states with $i < i_\beta, \sigma_\beta = \sum_{i = 1}^{\beta -1} n_i$. This is exactly what you get from $\eta_\beta |\nu\rangle = (-1)^\sigma |\nu \rangle$ and likewise $\eta_\alpha |\nu\rangle = (-1)^\sigma |\nu \rangle$. Now if we let the operators act the other way around, \begin{equation} \begin{split} b_\beta^\dagger b_\alpha^\dagger |\nu \rangle &= (-1)^{\sigma_\alpha} b_\beta^\dagger |(i_1,n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha + \sigma_\beta + 1} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split} \end{equation} The additional minus sign comes from the fact, that there is now one more state with $i < i_\beta$ that is occupied, namely the $\alpha$-state. So this proofs $\lbrace b_\alpha^\dagger, b_\beta^\dagger \rbrace = 0$.

Answer 2

Let's consider just 2 "particles". We define \begin{equation} c = \left ( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right ), ~~ c^\dagger = \left ( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right ). \end{equation} The Hilbert space is given by the tensor product of the three Hilbert spaces. Therefore, we are looking for creation/annihilation operators for all the particles in the form \begin{equation} \begin{array}{c} c^\dagger_1 = c^\dagger \otimes 1,\\ c^\dagger_2 = \sigma^\dagger \otimes c^\dagger, \end{array} \end{equation} where $\sigma$ is a $2\times 2$ matrix (to be fixed). Commutation relations \begin{equation} \{c_1,c_1^\dagger \}=1, ~~ \{c_1^\dagger,c_1^\dagger \} = \{c_2^\dagger,c_2^\dagger\}= \{c_1,c_1 \} = \{c_2,c_2\}=0, \end{equation} are obviously satisfied. At the same time \begin{equation} \{c_2,c_2^\dagger\}=1, ~~ \{c_1^\dagger,c_2^\dagger\}=\{c_1,c_2\}=\{c_1^\dagger,c_2\}=\{c_1,c_2^\dagger\}=0 \end{equation} impose non-trivial conditions on $\sigma$. Namely, \begin{equation} \sigma\sigma^\dagger=\sigma^\dagger \sigma=1, ~~ \{\sigma^\dagger,c^\dagger\}=\{\sigma^\dagger,c\}=0, \end{equation} whose solution is \begin{equation} \sigma= \left ( \begin{array}{cc} e^{i\varphi} & 0 \\ 0 & -e^{i\varphi} \end{array} \right ), \end{equation} in particular \begin{equation} \sigma= \left ( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right ), \end{equation} works as well. In general the construction is similar \begin{equation} \begin{array}{c} c^\dagger_1 = c^\dagger \otimes 1 \otimes 1\otimes 1 \otimes \dots,\\ c^\dagger_2 = \sigma \otimes c^\dagger \otimes 1\otimes 1 \otimes \dots,\\ c^\dagger_3 = \sigma \otimes \sigma \otimes c^\dagger \otimes 1 \otimes \dots,\\ \dots \end{array} \end{equation}

Answer 3

I think you must first think about what the ladder operators on $\mathcal{H}_F$ are. There are $\textbf{not}$ just $a^\dagger = 1 \otimes \ldots \otimes a^\dagger_{n}$ but also involve (anti-)symmetrysation. So you might define \begin{equation} a^\dagger (\varphi): \mathcal{H}_\mathcal{A}^{(n)} \longrightarrow \mathcal{H}_\mathcal{A}^{(n + 1)}, \quad \end{equation} \begin{equation} (a^\dagger (\varphi) \psi) = \sqrt{n + 1} \mathcal{A} (\varphi \otimes \psi). \end{equation} Where $\mathcal{A}$ is the operator that anti-symmetrizes your state and $\mathcal{H}_\mathcal{A}^{(n)}$ is the $n$-particle Fock space of antisymmetrized states.