When we talk about fermions in quantum mechanics, we talk about two kinds: Dirac and Majorana. Both of these are supposed to have the property that if I create a fermion of type $1$ and then a fermion of type $2,$ the resulting state will be related by a minus sign to the state where I create a fermion of type $2$ and then a fermion of type $1.$ But there is a decision to make as to what should happen if we try to create two type-$1$ fermions. Dirac fermions are defined by the property that if you try to create two type-$1$ fermions, the state is completely annihilated to the zero vector. Majorana fermions are defined by the property that if you try to create two type-$1$ fermions, they annihilate one another and leave the total state unchanged.

These properties are summarized by saying that the algebras of operators that create and annihilate the two different types of fermions should be different. The Dirac fermions have creation and annihilation operators $a_j, a_j^{\dagger}$ satisfying

$$\{a_j, a_k\} = \{a_j^{\dagger}, a_k^{\dagger}\} = \{a_j, a_k^{\dagger}\} = \delta_{j k}.$$

The Majorana fermions, on the other hand, have Hermitian creation operators $\psi_j$ that satisfy

$$\{\psi_j, \psi_k\} = 2 \delta_{j k}.$$

It is actually more common among field theorists to define the algebra by

$$\{\psi_j, \psi_k\} = \delta_{j k},$$

so that the creation operator for a type-$j$ fermion is $\sqrt{2} \psi_j.$ This is because, as noted in my post about canonical anticommutation relations, the second anticommutation relation is the one that follows from Lagrangian quantization with a standard kinetic term. We'll use this second convention going forward.

Majorana fermions can be thought of as fundamental constituents of Dirac fermions, because an algebra of $n$ Dirac fermions can always be rewritten as an algebra of $2n$ Majorana fermions via the linear transformation

$$\psi_{2j-1} = \frac{a_{j} + a_{j}^{\dagger}}{\sqrt{2}},$$

$$\psi_{2j} = \frac{a_{j} - a_j^{\dagger}}{\sqrt{2} i}.$$

Conversely, an even number $N$ of Majorana fermions can be grouped into $N/2$ Dirac fermions via

$$a_j = \frac{\psi_{2j-1} + i \psi_{2j}}{\sqrt{2}}.$$

On the other hand, an odd number of Majorana fermions can never be grouped exactly into Dirac fermions.

If we want to study a system of fermions of either kind, we need to specify a Hamiltonian made up of the fermion operators; for example, if we want to arrange $N$ Majorana fermions in a line and couple them locally, we might choose the Hamiltonian

$$H = i \sum_{j=1}^{N-1} \psi_{j} \psi_{j+1}.$$

But there's more to a system than just an operator algebra and a Hamiltonian: to actually ask questions like "what is the spectrum of the $H$?", we need to specify a Hilbert space and how the algebra acts on that Hilbert space. In other words, we need to specify a **representation** of the fermion algebra. Often you'll see people skip over this issue and just ask about the spectrum of a fermionic Hamiltonian without specifying a representation; but without knowing anything about the representation, that question isn't necessarily well defined.

The purpose of this post is to classify the finite-dimensional representations of the Majorana algebra $\mathcal{M},$ defined as the algebra generated by $N$ fermions with anticommutation relations $\{\psi_{j}, \psi_{k}\} = \delta_{jk} $. This algebra is more traditionally called a Clifford algebra, and is well studied by mathematicians under that name, but because I want to emphasize the connection to fermions I'll use the term "Majorana algebra" throughout.

The main statements we'll prove are:

- Every finite-dimensional representation of $\mathcal{M}$ can be written as a direct sum of irreducible representations of $\mathcal{M}.$
- When $N$ is even, there is only one irreducible representation of $\mathcal{M}.$ It has dimension $2^{N/2},$ and is isomorphic to the Fock space representation of $N/2$ Dirac fermions. This is the representation generated by taking a base vector $|0\rangle$ and requiring it to be annihilated by the operators $a_{j},$ and taken to orthogonal vectors by any string of creation operators $a_{j_1}^{\dagger} \dots a_{j_m}^{\dagger}.$
- When $N$ is odd, there are two irreducible representations of $\mathcal{M},$ each of which has dimension $2^{(N-1)/2}.$ These are isomorphic to the Fock space representation of a Dirac-fermion grouping of the first $(N-1)$ fermions, plus a decision of whether $\psi_N$ should act on the base state as $\psi_N |0\rangle = \frac{1}{\sqrt{2}} |0\rangle$ or $\psi_N |0\rangle = - \frac{1}{\sqrt{2}} | 0 \rangle.$

So the odd and even cases are quite different. If $N$ is even, then we're in the clear. The only thing we could really mean by the "spectrum of $H$" is the spectrum of $H$ within the unique irreducible representation. In any other representation of $\mathcal{M},$ because it must be a direct sum of some integer $m$ copies of the unique irreducible representation, the spectrum of $H$ will just be the irreducible spectrum with an $m$-fold degeneracy at each level. Incidentally, because every Dirac fermion algebra is isomorphic to an even-$N$ Majorana fermion algebra, this implies that the Dirac Fock space is the only irreducbile representation of Dirac fermions.

If $N$ is odd, however, talking about the "spectrum of $H$" requires asking two questions: what is the spectrum of $H$ in each of the two irreducible representations? For general Hamiltonians, they will be different. And in a general representation, the spectrum will be $m$ copies of the spectrum in the first irreducible representation, and $n$ copies in the second, where $m$ and $n$ are independent integers.

In practice, for many analytical calculations, people work in the "fundamental representation" of $\mathcal{M},$ which is the representation of $\mathcal{M}$ on itself by left multiplication, with an inner product specified by declaring that the basis

$$\{\psi_{j_1} \dots \psi_{j_k} | j_1 < \dots < j_k\}$$

is orthogonal, with the norm-squared of each basis element given by $2^{-\# \text{fermions}}$. This can be thought of as a Fock space built off of a base vector $|0\rangle$ corresponding to the identity element of $\mathcal{M},$ where every ordered string of Majorana fermion operators takes $|0\rangle$ to an orthogonal state. We'll see in section 2 that when you do this for even $N$, you end up with $2^{N/2}$ copies of the unique irreducible spectrum, and when you do it for odd $N$, you end up with $2^{(N-1)/2}$ copies of each of the two unique irreducible spectrums.

But there are many analytical calculations where working in the irreducible representations is more convenient, and it is **always** more convenient for doing numerical simulations of fermionic systems. Diagonalizing a matrix on a space of dimension $2^{(N-1)/2}$ or $2^{N/2}$ is much, much easier than on a space of dimension $2^{N}.$

So let's get to it.

In **section 1**, I give an overview of some basic lemmas and tools for studying representations of associative algebras.

In **section 2**, I prove the above-stated facts about the irreducible representations of the Majorana algebra $\mathcal{M}.$

In **section 3**, I give some further comments on the physical consequences of the representation-theoretic properties of $\mathcal{M}$ explored in this post.

Because this post is quite long, I've added "punchline" summaries at the start of each section/subsection except for the last one (which is very short).

I learned pretty much everything I know about Majorana representation theory from chapters 4 and 14 of Ambar Sengupta's book "Representations of Algebras and Finite Groups," though the pedagogy of this post is very different from the pedagogy in that book.

**Prerequisites:** Basic comfort with abstract algebra. I'll say things like "direct sum" and "first isomorphism theorem" without explaining what they mean.

__Table of Contents__

- Basic representation theory of associative algebras
- Representations of the Majorana algebra
- Consequences for physical fermions

__1. Basic representation theory of associative algebras__

Our goal in the next section will be to classify all irreducible representations of the Majorana algebra $\{\psi_j, \psi_k\} = \delta_{jk}.$ This is an associative algebra, and it'll be easiest to proceed if we first compile a few general observations about irreducible representations of associative algebras.

Since this section is rather mathy, it might help to summarize the main points for readers who would like to skip it (or have a road map going ahead). The essential ideas are:

- In an algebra, a left ideal is called simple if it has no nontrivial sub-ideals.
- A finite-dimensional algebra is called "semisimple" if for every left ideal $\mathcal{L},$ there is a
*complementary*left ideal $\mathcal{L}',$ i.e. one satisfying $\mathcal{A} = \mathcal{L} \oplus \mathcal{L}'.$ This is equivalent to the statement that $\mathcal{A}$ can be written as a direct sum of simple left ideals. - We'll show in the next section that the Majorana algebra is semisimple.
- Any finite-dimensional representation of a semisimple algebra can be written as a direct sum of irreducible representations.
- Every irreducible representation of a semisimple algebra is isomorphic to one of the simple left ideals in its semisimple decomposition; in this statement, a left ideal is thought of as a representation of the algebra on itself via left multiplication.
- By point 4, we can classify finite-dimensional representations of the Majorana algebra by classifying its irreducible representations. By point 5, we can do this by classifying the simple left ideals in its semisimple decomposition.
- In any semisimple algebra $\mathcal{A},$ for any pair of complementary left ideals $\mathcal{L} \oplus \mathcal{L}',$ there exist
*right*projection operators $P$ and $P'$ satisfying $P + P' = 1, P^2 = P, (P')^2 = P', P P' = P' P = 0,$ and $\mathcal{L} = \mathcal{A} P, \mathcal{L}' = \mathcal{A} P'.$ These will be useful technical tools in section 2.

An **associative algebra** is a vector space with a linear, associative multiplication map. So if $\mathcal{A}$ is an associative algebra, and we have some elements $a, b, c \in \mathcal{A},$ then there are also elements of $\mathcal{A}$ corresponding to the expressions $ab,$ $ba,$ $abc,$ etc. (Note: the multiplication is not necessarily commutative.) Furthermore, this multiplication rule satisfies

$$a (b + c) = a b + a c.$$ We will consider mainly associative algebras where the underlying field is the complex numbers. We will also only consider algebras that have an identity element, which we'll call $1.$ Finally, we will only consider algebras where the underlying vector space is finite-dimensional.

A **representation **of $\mathcal{A}$ is a vector space $V$ together with a map $\rho$ from $\mathcal{A}$ to the space of linear operators on $V,$ such that the map $\rho$ respects the algebraic structure. I.e., given $a, b, c \in \mathcal{A}$ and $\alpha \in \mathbb{C},$ we have

$$\rho(\alpha a (b + c)) = \alpha (\rho(a) \rho(b) + \rho(a) \rho(c)).$$

We also require that $\rho(1)$ is the identity operator.

A subspace $W \subset V$ is said to be a **subrepresentation **if the action of $\mathcal{A}$ leaves that subspace invariant, i.e., if we have

$$\rho(\mathcal{A}) W = W.$$

The subspace $V$ is obviously a subrepresentation of $V,$ as is the trivial subspace $\{0\}.$ The representation $V$ is said to be **irreducible** — sometimes **simple** — if the only subrepresentations are $V$ and $\{0\}.$

A representation of $\mathcal{A}$ is said to be **semisimple** if every subrepresentation $W \subset V$ has a complement, i.e., if there exists another subrepresentation $W'$ satisfying $V = W \oplus W'.$ Any subrepresentation of a semisimple representation is itself semisimple; observe that for $U \subset W$ a representation, there is a complement $U'$ in $V$ with

$$U \oplus U' = W \oplus W'.$$

But then we have

$$U \oplus \Pi_W(U') = W,$$

where $\Pi_W$ is the projection onto $W$ within the direct sum $W \oplus W'$; this equality follows from $U \subset W.$ Since $\Pi_W(U')$ is itself a representation, we conclude that $W$ can be decomposed as a direct sum of subrepresentations. Proceeding by induction, so long as $V$ is finite-dimensional it follows that $V$ can be written as a direct sum of irreducible (simple) representations. The converse is also true; if $V$ can be written as a direct sum of irreducible representations, e.g. $V = \oplus_j V_j,$ then every subrepresentation $W \subset V$ can be written as

$$W = \oplus_j \Pi_j(W),$$

where $\Pi_j$ projects onto $V_j.$ Since $\Pi_j(W)$ is itself a representation contained in $V_j,$ which is irreducible, we must have $\Pi_j(W) = V_j$ or $\Pi_j(W) = 0$. So there is some finite set of $V_j$'s with $W = \oplus_j V_j,$ and the direct sum of the $V_j$'s with $\Pi_j(W) = 0$ is a complementary subrepresentation.

Now, there is one natural representation of $\mathcal{A},$ which is its left action on itself:

$$\rho(a) \cdot b = a b.$$

The subrepresentations of this left-representation are exactly the **left ideals** of $\mathcal{A},$ i.e., subspaces $\mathcal{L} \subset \mathcal{A}$ satisfying

$$a \mathcal{L} \subset \mathcal{L} \quad \text{for all $a \in \mathcal{A}$}.$$

I think of this as the statement that "$\mathcal{L}$ is absorbing on the left." It can be written more compactly as $\mathcal{A} \mathcal{L} = \mathcal{L}.$ Note that this condition automatically implies that $\mathcal{L}$ is a subalgebra, not just a subspace. The algebra $\mathcal{A}$ is said to be **semisimple** if its left-representation is semisimple, and the left ideal $\mathcal{L}$ is said to be **simple** if it has no nontrivial sub-ideals. So per the previous paragraph, if $\mathcal{A}$ is semisimple and finite dimensional then it can be written as a direct sum of simple left-ideals, $\mathcal{A} = \oplus_j \mathcal{L}_{j}.$

We will see in the next section that the Majorana algebra is semisimple. In anticipation of that fact, let's learn some basic facts about semisimple algebras.

First, because we have a direct sum decomposition of $\mathcal{A},$ the identity element $1$ decomposes uniquely into the various summands as

$$1 = \sum_j P_j.$$

Since we have $P_j \in \mathcal{L}_{j},$ which is a left ideal, we have $a P_j \in \mathcal{L}_j$ for any $a \in \mathcal{A}.$ In particular, for any $\ell_k \in \mathcal{L}_{k},$ we have

$$\ell_k = \sum_j \ell_k P_j,$$

which, by uniqueness of the direct sum decomposition, gives

$$\ell_k P_j = \delta_{k j} \ell_k.$$

In particular, this implies $P_j P_k = \delta_{j k} P_k$; so the algebra elements $P_j$ are *orthogonal right projectors* onto the corresponding left ideals.

If $\mathcal{L}_j$ is not simple, then because it is semisimple it has a nontrivial direct sum decomposition $\mathcal{L}_{j} = \mathcal{M} \oplus \mathcal{N},$ with right projectors $M$ and $N$ that satisfy $M + N = P_j, M \neq 0, N \neq 0, M^2 = M, N^2 = M, M N = N M = 0.$ Conversely, if we had algebra elements $M$ and $N$ satisfying these relations, then the subalgebras $\mathcal{L} M$ and $\mathcal{L} N$ would be nontrivial left ideals satisfying $\mathcal{L} = \mathcal{L} M \oplus \mathcal{L} N.$ This identity obviously holds as a sum; it holds as a direct sum because any $\ell \in \mathcal{L}_j$ is uniquely decomposed as $\ell M + \ell N,$ with uniqueness coming identities $M N = N M = 0.$ For any decomposition $\ell = m + n,$ we have

$$m = (m + n) M = (\ell M + \ell N) M = \ell M,$$

and similarly $n = \ell N.$

So we have learned that in a semisimple algebra, left ideals can be classified by their right projectors, and simplicity of a left ideal is equivalent to its projector not being expressible as a sum of nontrivial, orthogonal projectors.

$$\rho(1) = \sum_j \rho(P_j).$$

Therefore, for any nonzero $|v\rangle \in V,$ at least one $\rho(P_j)$ must send $|v\rangle$ to a nonzero vector. So fix any nonzero $|v\rangle,$ and any corresponding $\rho(P_j)$ with $\rho(P_j) |v\rangle \neq 0.$ The set $\rho(\mathcal{L}_{j})|v\rangle$ is a nonzero subspace of $V$; furthermore, since $\mathcal{L}_{j}$ is a left ideal, it is also a subrepresentation. So since $V$ is irreducible, we must have

$$\rho(\mathcal{L}_{j}) |v\rangle = V.$$

This gives us a natural map from $\mathcal{L}_{j}$ to $V$ defined by

$$\ell \mapsto \rho(\ell)|v\rangle.$$

This map is easily seen to be a homomorphism of representations, and surjectivity is exactly the statement $$\rho(\mathcal{L}_{j}) |v\rangle = V.$$ Injectivity follows from the fact that the kernel of this map, i.e., the set of all $\ell$ with $\rho(\ell) | v \rangle = 0,$ is easily seen to be a left ideal contained in $\mathcal{L}_j.$ But since $\mathcal{L}_{j}$ was assumed to be simple, and since the kernel cannot be all of $\mathcal{L}_{j}$ by the assumption $\rho(P_j) |v\rangle \neq 0,$ we conclude that the kernel is trivial, and therefore that the map from $\mathcal{L}_{j}$ to $V$ is an isomorphism of representations.

So, we have shown that **the irreducible representations of a semisimple algebra are in one-to-one correspondence with its simple left ideals**. In the next section, we will show that the Majorana algebra is semisimple, and classify its simple left ideals to find its irreducible representations.

As a final comment, we show that any finite-dimensional representation of a finite-dimensional, semisimple algebra is semisimple, i.e., it can be decomposed into a direct sum of irreducible representations. Let $V$ be an arbitrary representation of $\mathcal{A}$, and let $|v_1\rangle, \dots, |v_D\rangle$ be a basis for $V.$ The algebra $\oplus_{k=1}^D \mathcal{A}$ is semisimple because each copy of $\mathcal{A}$ is semisimple. We can define a map from $\mathcal{A}$ to $V$ by

$$\phi : a_1 \oplus \dots \oplus a_D \mapsto a_1 v_1 + \dots + a_1 v_D.$$

This is clearly surjective, since we could take $a_k$ to be the unit element of $\mathcal{A}$ times an arbtirary complex coefficient, which lets us construct arbitrary linear combinations of basis elements. This is also a homomorphism for the representation of $\mathcal{A}$ on $\oplus_{k=1}^{D} \mathcal{A}$ given by

$$a (a_1 \oplus \dots \oplus a_D) = a a_1 \oplus \dots \oplus a a_D,$$

which is easily verified. So, since we have a surjective homomorphism of representations, we obtain the representation isomorphism

$$(\oplus_{k=1}^{D} \mathcal{A}_{k}) / \operatorname{ker}\phi \cong V$$

via the first isomorphism theorem. Since $\operatorname{\ker}\phi$ is a subrepresentation, and since the quotient $(\oplus_{k=1}^{D} \mathcal{A}_k) / \operatorname{\ker} \phi$ is isomorphic to any complement of $\operatorname{\ker}\phi$ in $\oplus_{k=1}^{D} \mathcal{A},$ we conclude that $V$ is isomorphic to a subrepresentation of a semisimple representation, and is therefore semisimple itself.

__2. Representations of the Majorana algebra__

$$\{\psi_j, \psi_k\} = \delta_{j k}.$$

In

**subsection 2.1**, we will construct a useful inner product on $\mathcal{M},$ then use that inner product to show that $\mathcal{M}$ is semisimple. In

**subsections 2.2 and 2.3**, we will then construct the semisimple decomposition of $\mathcal{M}$ explicitly, and therefore classify the irreducible representations of $\mathcal{M}.$ It turns out that the cases $N$ even and $N$ odd are different, so we will treat them separately in subsections 2.2 and 2.3.

__2.1 An inner product, and semisimplicity of the algebra__

**Punchline: The purpose of this subsection is to show that in the inner product where the standard basis elements of the Majorana algebra are orthogonal with norm-squared given by $2^{-\# \text{fermions}}$, the adjoint of a Majorana operator is given by $(\psi_{j_1} \dots \psi_{j_{\alpha}})^{\dagger} = \psi_{j_{\alpha}} \dots \psi_{j_1}.$ Then, we show that this implies the Majorana algebra is semisimple.**

$$1, \{\psi_j\}, \{\psi_{j} \psi_{k} | j < k\}, \dots$$

$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{k_1 < \dots < k_{\beta}} \rangle = \begin{cases} \frac{1}{2^{\alpha}} & \alpha = \beta, j_{\mu} = k_{\mu} \text{ for all } \mu = 1, \dots, \alpha \\ 0 & \text{otherwise} \end{cases}$$

and extending the inner product by linearity. For general, non-index-ordered products $\psi_{j_1, \dots, j_{\alpha}}$ and $\psi_{k_1, \dots, k_{\beta}},$ the inner product is nonzero if and only if the product $\psi_{j_1, \dots, j_{\alpha}, k_1, \dots, k_{\beta}}$ is proportional to the identity element of $\mathcal{M},$ and in that case is given by $(-1)^{\alpha (\alpha-1)/2}$ times the constant of proportionality. (This comes from the fact that $\frac{(-1)^{\alpha (\alpha-1)/2}}{2^{\alpha}}$ is the constant of proportionality in the case where both fermion products are properly ordered.)

**it closes in $\mathcal{M}$ under adjoints**. A priori, the adjoint $a^{\dagger}$ is a linear transformation acting on $\mathcal{M},$ but it need not be one of the linear transformations that can be realized via left multiplication by another element of $\mathcal{M}.$ (Indeed, there will be many more linear transformations acting on $\mathcal{M}$ than there are elements of $\mathcal{M}$; the space of transformations is $2^{N} \times 2^{N}$ dimensional, while $\mathcal{M}$ itself is $2^N$ dimensional.) In the Majorana algebra, however, we will now show that the adjoint of any element $a \in \mathcal{M}$ is also in $\mathcal{M}.$

$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle$$

$$\langle \psi_{k_1 < \dots < k_{\beta}} | \psi_{\ell_{\gamma}, \dots, \ell_{1}} | \psi_{j_1 < \dots < j_{\alpha}} \rangle$$

$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle$$

$$\psi_{\ell_1, \dots, \ell_{\gamma}}^{\dagger} = \psi_{\ell_{\gamma}, \dots, \ell_{1}}.$$

*exactly*the same set of fermion operators in $\psi_{j_1, \dots, j_{\alpha}},$ the total cost of commuting a single $j$-labeled fermion through $\psi_{\ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}}$ is $(-1)^{\alpha - 1},$ since we pick up a minus sign for every fermion we commute pa

**s**t

*except*the one that matches the one we're moving. Moving every $j$ fermion to the end of the product incurs a total sign $(-1)^{\alpha (\alpha-1)} = 1.$ As such, we have

$$\psi_{j_1, \dots, j_{\alpha}, \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}} = \psi_{\ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}, j_1, \dots, j_{\alpha}}.$$

*distinct*fermions in the product $\psi_{\ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}}.$ From this, we may conclude that the number of

*repeated*fermions is

$$R = \frac{\gamma + \beta - \alpha}{2}.$$

and

$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle = (-1)^{\frac{\alpha(\alpha-1)}{2}} \psi_{j_1, \dots, j_{\alpha} \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}},$$

$$\langle \ell | a \ell' \rangle = \langle a^{\dagger} \ell | \ell' \rangle.$$

__2.2 Even N__

**Punchline: Up to isomorphism, there is only one finite-dimensional irreducible representation of an even number of Majorana fermions; it has dimension $2^{N/2},$ and is isomorphic to the standard Fock space of $N/2$ Dirac fermions.**

$$a_j = \frac{\psi_{2j-1} + i \psi_{2j}}{\sqrt{2}},$$

$$a_j^{\dagger} = \frac{\psi_{2j-1} - i \psi_{2j}}{\sqrt{2}}.$$

The standard Fock space representation of the Majorana algebra is then constructed by fixing some "vacuum state" $|0\rangle$ and asserting that this state is annihilated by all $a_j$ operators but not by the $a_j^{\dagger}$ operators, constructing our representation as the span of all states of the form $a_{j_1}^{\dagger} \dots a_{j_{\alpha}}^{\dagger} |0\rangle.$

*inside of*$\mathcal{M},$ we need to find an element of $\mathcal{M}$ to act as our base state, i.e., an element $P_{\epsilon}$ in $\mathcal{M}$ satisfying $a_{j} P_{\epsilon} = 0$ for $\epsilon_j = +1$ and $a_{j}^{\dagger} P_{\epsilon} = 0$ for $\epsilon_{j} = -1.$ We'd also like these elements to have the properties of mutually orthogonal projectors, so we can study them as right projectors onto the left ideals corresponding to each Fock space. We can guess what these elements might be by observing that the Dirac anticommutation relations imply

$$(a_j a_j^{\dagger}) (a_j^{\dagger} a_j) = (a_j^{\dagger} a_j) (a_j a_j^{\dagger}) = 0.$$

So the operators $a_j a_j^{\dagger}$ and $a_j^{\dagger} a_j$ have all the right properties to be mutually orthogonal projection operators. Furthermore, we clearly have

$$a_j (a_j a_j^{\dagger}) = a_j^{\dagger} (a_j^{\dagger} a_j) = 0.$$

Let's denote by $\Pi_{+1}^{j}$ the operator $a_j a_j^{\dagger}$ and by $\Pi_{-1}^{j}$ the operator $a_j^{\dagger} a_j.$ Then for any string $\epsilon,$ the operator

is annihilated by all the correct operators; it is annihilated by $a_j$ for $\epsilon_j = +1,$ and by $a_j^{\dagger}$ for $\epsilon_{j} = -1.$ One can easily check this by noting that the $\Pi_{\epsilon_j}^{j}$ operators commute with one another at different values of $j.$ This statement, coupled with the "orthogonal projector" properties of each $\Pi_{\epsilon_j}^{j}$, implies that we have

so these form a set of orthogonal projectors as well.

$$P_{\epsilon} = \prod_{j=1}^{N/2} \frac{1 - 2 i \epsilon_j \psi_{2j-1} \psi_{2j}}{2}.$$

is a decomposition of $\mathcal{M}$ into left ideals. It only remains to show that these are simple, and then that they are all mutually isomorphic.

where $\epsilon$ differs from $\epsilon'$ on the index $j$ that contains the $\psi_k$ operator; i.e., the index satisfying $k=2j$ or $k=2j-1.$ This is easily verified by inspection of the explicit expression for $P_{\epsilon}.$

$$\chi = \psi_{1}^{\eta_1} \dots \psi_{N}^{\eta_N},$$

where each $\eta_{k}$ is either $0$ or $1.$ When we commute $\chi$ through the projector $P_{\epsilon},$ the string $\epsilon$ is unchanged if and only if, for each $j = 1, \dots, N/2$, we have $\eta_{2j-1} = \eta_{2j},$ since this is the condition under which $\psi_{2j-1}^{\eta_{2j-1}} \psi_{2j}^{\eta_{2j-1}}$ commutes with $\Pi_{\epsilon_j}^{j}$ (and since it automatically commutes with all of the other $\Pi$ operators). So, since we have $P_{\epsilon} P_{\epsilon'} = 0$ for $\epsilon \neq \epsilon'$ and $P_{\epsilon}^2 = P_{\epsilon},$ we may conclude

$$P_{\epsilon} \chi P_{\epsilon} = \begin{cases} \chi P_{\epsilon} & \text{if $\eta_{2j-1} = \eta_{2j}$ for all $j=1, \dots, N/2$} \\ 0 & \text{otherwise}\end{cases}.$$

$$\psi_{2j-1} \psi_{2j} \Pi_{\epsilon_j}^{j} = \frac{i \epsilon_{j}}{2} \Pi_{\epsilon_j}^{j}.$$

So in the case that $\chi$ has $\eta_{2j-1} = \eta_{2j}$ for all $j,$ we may conclude that $P_{\epsilon} \chi P_{\epsilon}$ is a complex multiple of $P_{\epsilon}.$ In fact, this is always true; if we have $\eta_{2j-1} \neq \eta_{2j}$ for some $j,$ then the complex multiple is just zero. Since this is true for an arbitrary product of fermions, we conclude that for

**any**operator $a$ in the Majorana algebra, we have

$$\alpha = \lambda P_{\epsilon}.$$

If we square both sides of this equation, then we obtain

$$\alpha = \lambda^2 P_{\epsilon} = \lambda \alpha.$$

So we either have $\lambda = 1,$ which gives $\alpha = P_{\epsilon},$ or $\alpha = 0.$ In either case, $\alpha$ is a trivial projector. We conclude, by contradiction, that $\mathcal{M} P_{\epsilon}$ must be a simple left ideal.

$$P_{\epsilon} \chi = \chi P_{\epsilon'}.$$

$$\mathcal{M} = \oplus_{\epsilon \in \{-1, 1\}^{N/2}} \mathcal{M} P_{\epsilon},$$

where each $\mathcal{M} P_{\epsilon}$ is a simple left ideal and all of the different left ideals are isomorphic. From the considerations of section 1, we immediately conclude that there is only one finite-dimensional irreducible representation of this algebra, and it is isomorphic to any of the left ideals in this decomposition. Because $\mathcal{M}$ has dimension $2^N,$ and all of the left ideals are isomorphic, and there are $2^{N/2}$ terms in the direct sum decomposition, we conclude that the dimension of each left ideal is

$$\dim(\mathcal{M} P) = 2^{N/2}.$$

So the upshot is that if we find

**any**representation of the even-N Majorana algebra with dimension $2^{N/2},$ it must be isomorphic to the unique irreducible representation.

__2.3 Odd N__

**Punchline: When $N$ is odd, there are two non-isomorphic irreducible representations of $\mathcal{M},$ each of which has dimension $2^{(N-1)/2}.$ They are equivalent to the Fock space of a Dirac grouping of the first $(N-1)$ fermions, plus a choice of whether $\psi_N$ acts on the base state as $+1/\sqrt{2}$ or $-1/\sqrt{2}.$**

$$(\psi_1 \dots \psi_N)^{\dagger} \psi_1 \dots \psi_N = \frac{1}{2^N}.$$

$$\mathcal{M} = \mathcal{M}_{+} \oplus \mathcal{M}_{-}$$

$$P_{\epsilon \in \{\pm 1\}^{(N-1)/2}} = \prod_{j=1}^{(N-1)/2} \frac{1 - 2 i \epsilon_j \psi_{2j-1} \psi_{2j}}{2}.$$

*(The first draft of this post had a more complicated proof of this fact — thanks to Brandon Rayhaun for helping me simplify it!)*

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