The representation theory of the Clifford algebra you are asking about can be understood using the theory of associative superalgebras. In this context, the irreducible $\mathbb{Z}_2$-graded representations are described in section 2 of this paper by Brundan and Kleshchev (see example 2.10).

The upshot is that the Clifford algebra $C(n)\cong C(1)\otimes\cdots\otimes C(1)$ as associative superalgebras (see the introductory paragraphs from section 2 and example 2.2 in loc. cit.). This makes the construction of irreducible representations fairly easy.

The superalgebra $C(1)=\mathbb{C}[c]/(c^2-1)$ is self associate meaning that the left regular module is simple as a supermodule, but decomposes into two non-isomorphic (and ungradable) submodules after forgetting the grading (the generator $c$ acts by $\pm\sqrt{-1}$ on the submodule of the left regular module spanned by $(\sqrt{-1}\pm c)$). Let $U(1)$ denote the self associate representation of $C(1)$ and $U_\pm$ the two irreducible ungradable subrepresentations.

Now, using the isomorphism $C(2)\cong C(1)\otimes C(1)$ and Lemma 2.9, the 4-dimensional $C(2)$-supermodule $U(1)\otimes U(1)$ decomposes as two 2-dimensional isomorphic absolutely irreducible modules (meaning they stay irreducible after forgetting the grading). Let $U(2)$ denote one such copy. Fixing a basis of homogeneous vectors for $U(2)$, we can represent the generators $c_1$ and $c_2$ of $C(2)$ as matrices by $$ c_1\mapsto\begin{bmatrix} 0&1\\1&0\end{bmatrix}\;\;\;\mbox{and}\;\;\;c_2\mapsto\begin{bmatrix} 0&\sqrt{-1}\\-\sqrt{-1}&0\end{bmatrix}.$$

Now we can complete the construction. If $n=2k$ is even, then using Lemma 2.9 and the isomorphism $C(n)\cong C(2)^{\otimes k}$ the module $U(n)=U(2)^{\otimes k}$ is the unique irreducible module of dimension $2^k$. By the same reasoning, if $n=2k+1$, there are two nonisomorphic irreducible modules of dimension $2^k$, namely, $U_+(n)=U(2)^{\otimes k}\otimes U_+$ and $U_-(n)=U(2)^{\otimes k}\otimes U_-$.

It should be emphasized that for associative superalgebras $A$ and $B$, the superalgebra $A\otimes B$ has multiplication given by $$(a_1\otimes a_2)(b_1\otimes b_2)=(-1)^{p(a_2)p(b_1)}(a_1b_1)\otimes(a_2b_2)$$ so care needs to be taken when computing the action of $C(n)$ on these irreducible modules. I prefer to think of them sitting inside the left regular representation: \begin{align*} U(2k)&\cong C(n).(\sqrt{-1}-c_1c_2)(\sqrt{-1}-c_3c_4)\cdots(\sqrt{-1}-c_{n-1}c_n)\\ U_\pm(2k+1)&\cong C(n).(\sqrt{-1}-c_1c_2)\cdots(\sqrt{-1}-c_{n-2}c_{n-1})(\sqrt{-1}\pm c_n) \end{align*}

The representation theory of the Clifford algebra you are asking about can be understood using the theory of associative superalgebras. In this context, the irreducible $\mathbb{Z}_2$-graded representations are described in section 2 of this paper by Brundan and Kleshchev (see example 2.10).

The upshot is that the Clifford algebra $C(n)\cong C(1)\otimes\cdots\otimes C(1)$ as associative superalgebras (see the introductory paragraphs from section 2 and example 2.2 in loc. cit.). This makes the construction of irreducible representations fairly easy.

The superalgebra $C(1)=\mathbb{C}[c]/(c^2-1)$ is self associate meaning that the left regular module is simple as a supermodule, but decomposes into two non-isomorphic (and ungradable) submodules after forgetting the grading (the generator $c$ acts by $\pm 1$ on the submodule of the left regular module spanned by $(1\pm c)$). Let $U(1)$ denote the self associate representation of $C(1)$ and $U_\pm$ the two irreducible ungradable subrepresentations.

Now, using the isomorphism $C(2)\cong C(1)\otimes C(1)$ and Lemma 2.9, the 4-dimensional $C(2)$-supermodule $U(1)\otimes U(1)$ decomposes as two 2-dimensional isomorphic absolutely irreducible modules (meaning they stay irreducible after forgetting the grading). Let $U(2)$ denote one such copy. Fixing a basis of homogeneous vectors for $U(2)$, we can represent the generators $c_1$ and $c_2$ of $C(2)$ as matrices by $$ c_1\mapsto\begin{bmatrix} 0&1\\1&0\end{bmatrix}\;\;\;\mbox{and}\;\;\;c_2\mapsto\begin{bmatrix} 0&\sqrt{-1}\\-\sqrt{-1}&0\end{bmatrix}.$$

Now we can complete the construction. If $n=2k$ is even, then using Lemma 2.9 and the isomorphism $C(n)\cong C(2)^{\otimes k}$ the module $U(n)=U(2)^{\otimes k}$ is the unique irreducible module of dimension $2^k$. By the same reasoning, if $n=2k+1$, there are two nonisomorphic irreducible modules of dimension $2^k$, namely, $U_+(n)=U(2)^{\otimes k}\otimes U_+$ and $U_-(n)=U(2)^{\otimes k}\otimes U_-$.

It should be emphasized that for associative superalgebras $A$ and $B$, the superalgebra $A\otimes B$ has multiplication given by $$(a_1\otimes a_2)(b_1\otimes b_2)=(-1)^{p(a_2)p(b_1)}(a_1b_1)\otimes(a_2b_2)$$ so care needs to be taken when computing the action of $C(n)$ on these irreducible modules. I prefer to think of them sitting inside the left regular representation: \begin{align*} U(2k)&\cong C(n).(\sqrt{-1}-c_1c_2)(\sqrt{-1}-c_3c_4)\cdots(\sqrt{-1}-c_{n-1}c_n)\\ U_\pm(2k+1)&\cong C(n).(\sqrt{-1}-c_1c_2)\cdots(\sqrt{-1}-c_{n-2}c_{n-1})(1\pm c_n) \end{align*}

The representation theory of the Clifford algebra you are asking about can be understood using the theory of associative superalgebras. In this context, the irreducible $\mathbb{Z}_2$-graded representations are described in section 2 of this paper by Brundan and Kleshchev (see example 2.10).

The upshot is that the Clifford algebra $C(n)\cong C(1)\otimes\cdots\otimes C(1)$ as associative superalgebras (see the introductory paragraphs from section 2 and example 2.2 in loc. cit.). This makes the construction of irreducible representations fairly easy.

The superalgebra $C(1)=\mathbb{C}[c]/(c^2-1)$ is self associate meaning that the left regular module is simple as a supermodule, but decomposes into two non-isomorphic (and ungradable) submodules after forgetting the grading (the generator $c$ acts by $\pm\sqrt{-1}$ on the submodule of the left regular module spanned by $(\sqrt{-1}\pm c)$). Let $U(1)$ denote the self associate representation of $C(1)$ and $U_\pm$ the two irreducible ungradable subrepresentations.

Now, using the isomorphism $C(2)\cong C(1)\otimes C(1)$ and Lemma 2.9, the 4-dimensional $C(2)$-supermodule $U(1)\otimes U(1)$ decomposes as two 2-dimensional isomorphic absolutely irreducible modules (meaning they stay irreducible after forgetting the grading). Let $U(2)$ denote one such copy. Fixing a basis of homogeneous vectors for $U(2)$, we can represent the generators $c_1$ and $c_2$ of $C(2)$ as matrices by $$ c_1\mapsto\begin{bmatrix} 0&1\\1&0\end{bmatrix}\;\;\;\mbox{and}\;\;\;c_2\mapsto\begin{bmatrix} 0&\sqrt{-1}\\-\sqrt{-1}&0\end{bmatrix}.$$

Now we can complete the construction. If $n=2k$ is even, then using Lemma 2.9 and the isomorphism $C(n)\cong C(2)^{\otimes k}$ the module $U(n)=U(2)^{\otimes k}$ is the unique irreducible module of dimension $2^k$. By the same reasoning, if $n=2k+1$, there are two nonisomorphic irreducible modules of dimension $2^k$, namely, $U_+(n)=U(2)^{\otimes k}\otimes U_-$ and $U_-(n)=U(2)^{\otimes k}\otimes U_-$.

It should be emphasized that for associative superalgebras $A$ and $B$, the superalgebra $A\otimes B$ has multiplication given by $$(a_1\otimes a_2)(b_1\otimes b_2)=(-1)^{p(a_2)p(b_1)}(a_1b_1)\otimes(a_2b_2)$$ so care needs to be taken when computing the action of $C(n)$ on these irreducible modules. I prefer to think of them sitting inside the left regular representation: \begin{align*} U(2k)&\cong C(n).(\sqrt{-1}-c_1c_2)(\sqrt{-1}-c_3c_4)\cdots(\sqrt{-1}-c_{n-1}c_n)\\ U_\pm(2k+1)&\cong C(n).(\sqrt{-1}-c_1c_2)\cdots(\sqrt{-1}-c_{n-2}c_{n-1})(\sqrt{-1}\pm c_n) \end{align*}

The representation theory of the Clifford algebra you are asking about can be understood using the theory of associative superalgebras. In this context, the irreducible $\mathbb{Z}_2$-graded representations are described in section 2 of this paper by Brundan and Kleshchev (see example 2.10).

The upshot is that the Clifford algebra $C(n)\cong C(1)\otimes\cdots\otimes C(1)$ as associative superalgebras (see the introductory paragraphs from section 2 and example 2.2 in loc. cit.). This makes the construction of irreducible representations fairly easy.

The superalgebra $C(1)=\mathbb{C}[c]/(c^2-1)$ is self associate meaning that the left regular module is simple as a supermodule, but decomposes into two non-isomorphic (and ungradable) submodules after forgetting the grading (the generator $c$ acts by $\pm\sqrt{-1}$ on the submodule of the left regular module spanned by $(\sqrt{-1}\pm c)$). Let $U(1)$ denote the self associate representation of $C(1)$ and $U_\pm$ the two irreducible ungradable subrepresentations.

Now, using the isomorphism $C(2)\cong C(1)\otimes C(1)$ and Lemma 2.9, the 4-dimensional $C(2)$-supermodule $U(1)\otimes U(1)$ decomposes as two 2-dimensional isomorphic absolutely irreducible modules (meaning they stay irreducible after forgetting the grading). Let $U(2)$ denote one such copy. Fixing a basis of homogeneous vectors for $U(2)$, we can represent the generators $c_1$ and $c_2$ of $C(2)$ as matrices by $$ c_1\mapsto\begin{bmatrix} 0&1\\1&0\end{bmatrix}\;\;\;\mbox{and}\;\;\;c_2\mapsto\begin{bmatrix} 0&\sqrt{-1}\\-\sqrt{-1}&0\end{bmatrix}.$$

Now we can complete the construction. If $n=2k$ is even, then using Lemma 2.9 and the isomorphism $C(n)\cong C(2)^{\otimes k}$ the module $U(n)=U(2)^{\otimes k}$ is the unique irreducible module of dimension $2^k$. By the same reasoning, if $n=2k+1$, there are two nonisomorphic irreducible modules of dimension $2^k$, namely, $U_+(n)=U(2)^{\otimes k}\otimes U_+$ and $U_-(n)=U(2)^{\otimes k}\otimes U_-$.

It should be emphasized that for associative superalgebras $A$ and $B$, the superalgebra $A\otimes B$ has multiplication given by $$(a_1\otimes a_2)(b_1\otimes b_2)=(-1)^{p(a_2)p(b_1)}(a_1b_1)\otimes(a_2b_2)$$ so care needs to be taken when computing the action of $C(n)$ on these irreducible modules. I prefer to think of them sitting inside the left regular representation: \begin{align*} U(2k)&\cong C(n).(\sqrt{-1}-c_1c_2)(\sqrt{-1}-c_3c_4)\cdots(\sqrt{-1}-c_{n-1}c_n)\\ U_\pm(2k+1)&\cong C(n).(\sqrt{-1}-c_1c_2)\cdots(\sqrt{-1}-c_{n-2}c_{n-1})(\sqrt{-1}\pm c_n) \end{align*}