# Ladder operators for fermionic Fock space

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?

So any state in the Fock space can be parametrized by $$$$|\nu \rangle = | (i_1,n_1), \ldots, \rangle,$$$$ where $$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$$ or else the statement is trivial. Then $$$$\begin{split} b_\alpha^\dagger b_\beta^\dagger |\nu \rangle &= (-1)^{\sigma_\beta} b_\alpha^\dagger |(i_1,n_1), \ldots , (i_\beta, n_\beta + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha + \sigma_\beta} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split}$$$$ where $$\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$$. Now if we let the operators act the other way around, $$$$\begin{split} b_\beta^\dagger b_\alpha^\dagger |\nu \rangle &= (-1)^{\sigma_\alpha} b_\beta^\dagger |(i_1,n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots \rangle \\ &= (-1)^{\sigma_\alpha + \sigma_\beta + 1} |(i_1, n_1), \ldots , (i_\alpha, n_\alpha + 1), \ldots (i_\beta, n_\beta + 1), \ldots \rangle , \end{split}$$$$ The additional minus sign comes from the fact, that there is now one more state with $$i < i_\beta$$ that is occupied, namely the $$\alpha$$-state. So this proofs $$\lbrace b_\alpha^\dagger, b_\beta^\dagger \rbrace = 0$$.
• Sorry for replying this late. The answer is clear but I think there might be a small mistake: when $b_\alpha^\dagger$ acts on $n_\alpha=1$ it should reduce by one unit the number of occupied fermionic states with $\gamma<\beta$, so $\sigma_\beta'=\sigma_\beta-1$. The end result is still the same as the two operators still differ by a $-1$ factor. In short, wouldn't it be more precise to write $(-1)^{\sigma_\alpha+\sigma\pm 1}$ for the second expression? Jul 10, 2021 at 22:39
• If $b_\alpha^\dagger$ acts on $n_\alpha = 1$ you get zero. That's why I mainly focused on the case $n_{\alpha} = n_\beta = 0$. Therefor I never apply $b_\alpha^\dagger$ to an $n_\alpha = 1$ state. If you apply $b_\beta^\dagger$ you only increase the number in the $\beta$ state by one and you get this $(-1)^{\sigma_\beta}$ factor. Because I get an additional minus sign for every occupied state with $\gamma < \beta$ we should get $\sigma_\beta^\prime$ = \sigma_\beta + 1$because there is one more state occupied then previously. But you're write of course in that it doesn't really matter. Jul 11, 2021 at 8:38 • Btw, I made a small edit and in the second equation from$b_\alpha^\dagger$to$b_\beta^\dagger$maybe that's where the confusion came from. Sry about that. Jul 11, 2021 at 8:39 Let's consider just 2 "particles". We define $$$$c = \left ( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right ), ~~ c^\dagger = \left ( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right ).$$$$ The Hilbert space is given by the tensor product of the three Hilbert spaces. Therefore, we are looking for creation/annihilation operators for all the particles in the form $$$$\begin{array}{c} c^\dagger_1 = c^\dagger \otimes 1,\\ c^\dagger_2 = \sigma^\dagger \otimes c^\dagger, \end{array}$$$$ where $$\sigma$$ is a $$2\times 2$$ matrix (to be fixed). Commutation relations $$$$\{c_1,c_1^\dagger \}=1, ~~ \{c_1^\dagger,c_1^\dagger \} = \{c_2^\dagger,c_2^\dagger\}= \{c_1,c_1 \} = \{c_2,c_2\}=0,$$$$ are obviously satisfied. At the same time $$$$\{c_2,c_2^\dagger\}=1, ~~ \{c_1^\dagger,c_2^\dagger\}=\{c_1,c_2\}=\{c_1^\dagger,c_2\}=\{c_1,c_2^\dagger\}=0$$$$ impose non-trivial conditions on $$\sigma$$. Namely, $$$$\sigma\sigma^\dagger=\sigma^\dagger \sigma=1, ~~ \{\sigma^\dagger,c^\dagger\}=\{\sigma^\dagger,c\}=0,$$$$ whose solution is $$$$\sigma= \left ( \begin{array}{cc} e^{i\varphi} & 0 \\ 0 & -e^{i\varphi} \end{array} \right ),$$$$ in particular $$$$\sigma= \left ( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right ),$$$$ works as well. In general the construction is similar $$$$\begin{array}{c} c^\dagger_1 = c^\dagger \otimes 1 \otimes 1\otimes 1 \otimes \dots,\\ c^\dagger_2 = \sigma \otimes c^\dagger \otimes 1\otimes 1 \otimes \dots,\\ c^\dagger_3 = \sigma \otimes \sigma \otimes c^\dagger \otimes 1 \otimes \dots,\\ \dots \end{array}$$$$ • Thank you for your reply. I am struggling a bit to understand the role of the$\sigma$operators. In a bosonic space of course we can define$c_2^\dagger=1\otimes c^\dagger$. Why is the presence of$\sigma$required in this case? To me, it seems like it imposes a sort of ordering on the Fock state, which is counter-intuitive. Jun 16, 2021 at 15:53 I think you must first think about what the ladder operators on $$\mathcal{H}_F$$ are. There are $$\textbf{not}$$ just $$a^\dagger = 1 \otimes \ldots \otimes a^\dagger_{n}$$ but also involve (anti-)symmetrysation. So you might define $$$$a^\dagger (\varphi): \mathcal{H}_\mathcal{A}^{(n)} \longrightarrow \mathcal{H}_\mathcal{A}^{(n + 1)}, \quad$$$$ $$$$(a^\dagger (\varphi) \psi) = \sqrt{n + 1} \mathcal{A} (\varphi \otimes \psi).$$$$ Where $$\mathcal{A}$$ is the operator that anti-symmetrizes your state and $$\mathcal{H}_\mathcal{A}^{(n)}$$ is the $$n$$-particle Fock space of antisymmetrized states. • Hi. Honestly, I don't see how this answers my question. Jun 16, 2021 at 14:44 • Maybe, I was a little fast. I think I used a different basis for the Fock space. The index$\alpha$labels states of the system not particles, right? I just used the tensor-product of the single-particle states. So sry for the confusion. The idea remains the same though. You have to think about what the ladder operators on the Fock space are. You only defined them on$V_\alpha$. And you construct the ladder operators by antisymmetrized tensorproducts of$a^{(\dagger)}$. I'll edit my answer with more details. Jun 16, 2021 at 14:59 • In my understanding, when we write for instance something like $$|1_\alpha,0_\beta, 1_\gamma\rangle$$ in fermionic space we just mean that we have two particles, one in the state$\alpha$, one in the state$\gamma$(and zero everywhere else). We don't have to worry about anti-symmetrizing the state itself as long as the operators$a_\alpha^\dagger, a_\beta^\dagger$that create these particles obey the correct anticommutation rules. Isn't this correct? Jun 16, 2021 at 15:02 • Just read your comment. I understand what's going on now: you are trying to build a single ladder operator defined on the whole space for each possible quantum state$\alpha$. Instead what I have defined is a collection of operators acting on local vector spaces. The 'drawback' to your approach is that you now have to impose a symmetry/antisymmetry constraint on the state instead of relying only on (anti-)commutation relations between the local operators, isn't that right? But that is not what I am trying to do in my question. Jun 16, 2021 at 15:08 • Yes, sry did not read your original question propperly. In your basis, should't$a_\alpha^\dagger a_\beta^\dagger = a_\beta^\dagger a_\alpha^\dagger$for$\alpha \neq \beta\$, because there are different states and so the operators act on different parts of the Hilbert space. Jun 16, 2021 at 15:11