# Matrix representation of the CAR for the fermionic degrees of freedom

The canonical anticommutation relations (CAR) for a fermionic degree of freedom can be written as follows:

$$a^2 = \left( a^{\dagger} \right) ^2 = 0,$$ $$a a^{\dagger} + a^{\dagger} a = 1.$$

This admits a 2-dimensional matrix representation:

$$a=\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right),\;a^{\dagger}=\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right).$$

It is easy to verify that the CAR algebra holds. The dimensionality of this representation is 2, which shouldn't come as a surprise, because due to the Pauli exclusion principle the number of fermionic excitations occupying a degree of freedom could either be 0 or 1.

Now, consider a system of $$N$$ fermionic degrees of freedom. The CAR algebra reads:

$$a_i a_j = - a_j a_i, \; a^{\dagger}_i a^{\dagger}_j = -a^{\dagger}_j a^{\dagger}_i$$ $$a_i a^{\dagger}_j + a^{\dagger}_j a_i = \delta_{ij}.$$

I am interested in the generic formula for a matrix representation of such CAR. According to the Pauli principle, it should have the dimensionality of $$2^N$$.

Note that simply taking

$$a_i = 1 \otimes \dots \otimes a \otimes \dots \otimes 1$$

gives you a different algebra – for example, $$a_i$$ and $$a_j$$ commute for $$i \neq j$$ instead of anti-commuting. Hence, this isn't the sought matrix representation.

The answer is the so-called Jordan-Wigner transformation. This expresses the canonical ladder operators describing $$N$$ fermionic modes in terms of Pauli operators on a tensor product Hilbert space of $$N$$ spin-1/2 particles. Given a set of fermionic operators $$\{\hat{a}_j\}$$, for $$j = 1,2,\ldots,N$$, which satisfy the CAR $$\{\hat{a}_j,\hat{a}_k^\dagger\} = \delta_{jk}$$ and $$\{\hat{a}_j,\hat{a}_k\} = 0$$, the matrix representation is given by $$\hat{a}_k = \left( \prod_{j=1}^{k-1} \hat{\sigma}^z_j \right)\hat{\sigma}^-_k.$$ Here, I am using the standard Pauli operators $$\hat{\sigma}^{x,y,z}$$ and $$\hat{\sigma}^\pm = \tfrac{1}{2}(\hat{\sigma}^x \pm {\rm i}\hat{\sigma}^y)$$, which satisfy the usual algebra $$[\hat{\sigma}^x,\hat{\sigma}^y] = 2{\rm i}\hat{\sigma}^z$$ and cyclic permutations thereof, while $$\hat{\sigma}^{z,\pm}_j$$ acts on the Hilbert space of particle $$j$$, e.g. $$\hat{\sigma}^{z}_j = \underset{j-1}{\underbrace{\mathbb{1} \otimes\cdots \otimes \mathbb{1}}}\otimes \hat{\sigma}^{z} \otimes \underset{N-j}{\underbrace{\mathbb{1} \otimes\cdots \otimes \mathbb{1}}}.$$ By direct computation, and using the convenient property of the Pauli operators $$\{\hat{\sigma}^\pm,\hat{\sigma}^z\} = 0$$, you can prove that the CAR hold. Thus, any matrix representation of the Pauli operators will suffice.