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Hilbert spaces of Majorana fermions

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:

  1. Every finite-dimensional representation of $\mathcal{M}$ can be written as a direct sum of irreducible representations of $\mathcal{M}.$
  2. 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 $a_{j_1}^{\dagger} \dots a_{j_m}^{\dagger}.$
  3. 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

  1. Basic representation theory of associative algebras
  2. Representations of the Majorana algebra
    1. An inner product, and semisimplicity of the algebra
    2. Even N
    3. Odd N
  3. 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:

  1. In an algebra, a left ideal is called simple if it has no nontrivial sub-ideals.
  2. 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.
  3. We'll show in the next section that the Majorana algebra is semisimple.
  4. Any finite-dimensional representation of a semisimple algebra can be written as a direct sum of irreducible representations.
  5. 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.
  6. 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.
  7. 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.

Now we are ready to prove a wonderful fact: every nontrivial irreducible representation of $\mathcal{A}$ must be isomorphic to one of the left ideals $\mathcal{L}_{j}$ in the semisimple decomposition of $\mathcal{A}.$ To see this, suppose that $V$ is a nontrivial irreducible representation of $\mathcal{A}.$ The identity on $V$ decomposes as
$$\rho(1) = \sum_j \rho(P_j).$$
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

Recall from the introduction that the Majorana algebra is the algebra $\mathcal{M}$ generated by $N$ symbols $\psi_1, \dots, \psi_N$ subject to the relation
$$\{\psi_j, \psi_k\} = \delta_{j k}.$$
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.
The punchline is that when $N$ is even there is only one nontrivial irreducible representation of $\mathcal{M}$; when $N$ is odd there are two. This is one reason that $N$ is usually taken to be even in physics, along with the point made in the introduction that when $N$ is even, $N$ Majorana fermions can be rearranged into $N/2$ Dirac fermions.

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.
The Majorana algebra $\mathcal{M}$ has a standard basis given by the products of fermions with ascending index order, i.e. the elements
$$1, \{\psi_j\}, \{\psi_{j} \psi_{k} | j < k\}, \dots$$
We'll abbreviate these by using the symbol $\psi_{j_1, \dots, j_{\alpha}}$ for the product $\psi_{j_1} \dots \psi_{j_{\alpha}},$ and will insert less-than symbols $\psi_{j_1 < \dots < j_{\alpha}}$ when denoting a fermion product that is an element of the standard basis.
We can introduce a complex inner product on $\mathcal{M}$ by declaring that this is (almost) an orthonormal basis, i.e., by writing
$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{k_1 < \dots < k_{\beta}} \rangle = \begin{cases} \frac{1}{2^{\alpha}} & \alpha = \beta, j_{\mu} = k_{\mu} \text{ for all } \mu = 1, \dots, \alpha \\ 0 & \text{otherwise} \end{cases}$$
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.)

Recall from section 1 that we can think of $\mathcal{M}$ as acting on itself via left multiplication; so we can think of any $a \in \mathcal{M}$ as a linear operator $a : \mathcal{M} \rightarrow \mathcal{M}.$ The inner product we've chosen on $\mathcal{M}$ allows us to take the adjoint, $a^{\dagger}.$ The crucial feature of this inner product is that 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 $\mathcal{M}.$

Since the taking the adjoint is antilinear, it suffices to show this property for fermion products $\psi_{\ell_1, \dots, \ell_{\gamma}}$. Without loss of generality, we can assume that this product contains no repeated fermions. Consider arbitrary basis elements $\psi_{j_1 < \dots < j_{\alpha}}$ and $\psi_{k_1 < \dots < k_{\beta}}.$ Previously, we explained that the expression
$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle$$
is nonzero if and only if we have
$$\psi_{j_1, \dots, j_{\alpha}, \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}}$$
proportional to the identity, with the inner product being given by $(-1)^{\alpha(\alpha-1)/2}$ times the constant of proportionality. Rearranging the order of the product doesn't change the question of whether the total product is proportional to the identity, so
$$\langle \psi_{k_1 < \dots < k_{\beta}} | \psi_{\ell_{\gamma}, \dots, \ell_{1}} | \psi_{j_1 < \dots < j_{\alpha}} \rangle$$
is nonzero exactly when
$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle$$
is nonzero. We would like to show that when the inner products are nonzero, they are equal; this will imply the identity
$$\psi_{\ell_1, \dots, \ell_{\gamma}}^{\dagger} = \psi_{\ell_{\gamma}, \dots, \ell_{1}}.$$

Consider now the case where the inner products are nonzero. Because the product $\psi_{\ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}}$ must reduce to 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 past 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
$$\psi_{j_1, \dots, j_{\alpha}, \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}} = \psi_{\ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}, j_1, \dots, j_{\alpha}}.$$
Since we assumed that there are no repeated fermions in the product $\psi_{\ell_1, \dots, \ell_{\gamma}},$ the total cost of reversing the order of that product is $(-1)^{\gamma (\gamma-1)/2},$ yielding
$$\psi_{j_1, \dots, j_{\alpha}, \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}} = (-1)^{\gamma (\gamma-1)/2} \psi_{\ell_\gamma, \dots, \ell_{1}, k_1, \dots, k_{\beta}, j_1, \dots, j_{\alpha}}.$$
We must now consider the cost of commuting each $k$ fermion past the $\ell$ fermions. First recall that in order for the inner product to be nonzero, the number $\alpha$ must be equal to the total number of 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
$$R = \frac{\gamma + \beta - \alpha}{2}.$$
(It might be easier to parse this as $\alpha = \gamma + \beta - 2 R.$) If a $k$ fermion is not repeated within the $\ell$ fermions, then the total sign accumulated in commuting it through the $\ell$ fermions is $(-1)^{\gamma}.$ If it is repeated, however, then the total sign is $(-1)^{\gamma-1}.$ So the total sign accumulated in commuting all $k$ fermions past all $\ell$ fermions is
$$(-1)^{(\gamma - 1) R} (-1)^{\gamma \cdot (\beta - R)} = (-1)^{\beta \gamma - R}.$$
Now we apply the identities
$$\langle \psi_{k_1 < \dots < k_{\beta}} | \psi_{\ell_{\gamma}, \dots, \ell_{1}} | \psi_{j_1 < \dots < j_{\alpha}} \rangle = (-1)^{\frac{\beta(\beta-1)}{2}} \psi_{k_1, \dots, k_{\beta} \ell_{\gamma}, \dots, \ell_{1}, j_1, \dots, j_{\alpha}}$$
$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle = (-1)^{\frac{\alpha(\alpha-1)}{2}} \psi_{j_1, \dots, j_{\alpha} \ell_1, \dots, \ell_{\gamma}, k_1, \dots, k_{\beta}},$$
which gives us
$$\langle \psi_{j_1 < \dots < j_{\alpha}} | \psi_{\ell_1, \dots, \ell_{\gamma}} | \psi_{k_1 < \dots < k_{\beta}} \rangle = (-1)^{-\frac{\alpha(\alpha-1) + \beta(\beta-1) + \gamma(\gamma-1) + 2 R + 2 \beta \gamma}{2}} \langle \psi_{k_1 < \dots < k_{\beta}} | \psi_{\ell_{\gamma}, \dots, \ell_{1}} | \psi_{j_1 < \dots < j_{\alpha}} \rangle.$$
Plugging in $\alpha = \gamma + \beta - 2 R,$ we find that the total sign is $(-1)^{2 R (\beta + \gamma - R)} = 1.$ This validates our assertion that the adjoint of a product of fermion operators is obtained by reversing the order of those fermions.

This is enough to prove, quite readily, that the Majorana algebra $\mathcal{M}$ is semisimple. If $\mathcal{L} \subset \mathcal{M}$ is a left ideal, then the orthocomplement with respect to our inner product is also a left ideal. This is because for any $\ell \in \mathcal{L}, \ell' \in \mathcal{L}^{\perp},$ and $a \in \mathcal{M},$ we have
$$\langle \ell | a \ell' \rangle = \langle a^{\dagger} \ell | \ell' \rangle.$$
Because $a^{\dagger}$ is in $\mathcal{M}$ and $\mathcal{L}$ is a left ideal, we have $a^{\dagger} \ell \in \mathcal{L},$ and thus $\langle a^{\dagger} \ell | \ell' \rangle = 0.$ So, since $a \ell'$ is orthogonal to $\mathcal{L},$ it follows that it is in $\mathcal{L}',$ and therefore that $\mathcal{L}'$ is a left ideal.

The statement that every left ideal has a complementary left ideal is what we took as the definition of a semisimple algebra, so it follows that $\mathcal{M}$ is semisimple.

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.
We'll take our hint for how to study the Majorana algebra at even $N$ from the physical observation that $N$ Majorana fermions can be combined into $N/2$ Dirac fermions. By grouping the Majorana operators into pairs $\{\psi_{2j-1}, \psi_{2j}\}$ for $j = 1, \dots, N/2,$ we define the "annihilation" and "creation" operators
$$a_j = \frac{\psi_{2j-1} + i \psi_{2j}}{\sqrt{2}},$$
$$a_j^{\dagger} = \frac{\psi_{2j-1} - i \psi_{2j}}{\sqrt{2}}.$$
One can then easily check the Dirac anticommutation relations $$\{a_j, a_k\} = \{a_j^{\dagger}, a_k^{\dagger}\} = 0, \{a_j, a_k^{\dagger}\} = \delta_{jk}.$$
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 $a_{j_1}^{\dagger} \dots a_{j_{\alpha}}^{\dagger} |0\rangle.$

But there's a bit of an ambiguity in this discussion: who are we to say that $a_{1}$ is an annihilation operator rather than a creation operator? Or $a_{2},$ or $a_{3}$? Why not consider Fock spaces where $a_{1}, \dots, a_{k}, a_{k+1}^{\dagger}, \dots, a_{N/2}^{\dagger}$ are the annihilation operators, with their adjoints being the creation operators? Your answer might be "well, those are obviously isomorphic representations," and you're right; but we'll find that all these representations live naturally inside $\mathcal{M}$ as mutually isomorphic direct summands, and by chasing them down we'll construct the decomposition of $\mathcal{M}$ into simple left ideals.
Now, a choice of which set of the $a_j$ operators counts as "annihilation" and which as "creation" can be labeled by a bit string of length $N/2,$ i.e., a string $\epsilon \in \{\pm 1\}^{N}.$ If $\epsilon_j$ is $+1$ then we'll say $a_j$ is the annihilation operator; if it's $-1$ then we'll say $a_j^{\dagger}$ is the annihilation operator. To construct a corresponding Fock space 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
$$(a_j^{\dagger} a_j)^2 = a_j^{\dagger} a_j,$$
$$(a_j a_j^{\dagger})^2 = (a_j a_j^{\dagger}),$$
$$(a_j a_j^{\dagger}) (a_j^{\dagger} a_j) = (a_j^{\dagger} a_j) (a_j a_j^{\dagger}) = 0.$$
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
$$P_{\epsilon} = \prod_{j=1}^{N/2} \Pi_{\epsilon_j}^{j}$$
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
$$P_{\epsilon} P_{\epsilon'} = \delta_{\epsilon, \epsilon'} P_{\epsilon},$$
so these form a set of orthogonal projectors as well.

It will be easiest to study the $P_{\epsilon}$ operators by writing out each $\Pi_{\epsilon_j}^{j}$ explicitly, which gives
$$P_{\epsilon} = \prod_{j=1}^{N/2} \frac{1 - 2 i \epsilon_j \psi_{2j-1} \psi_{2j}}{2}.$$
We can also expand this product out by grouping all of the terms where $(N/2-k)$ factors of 1 contribute, yielding
$$P_{\epsilon} = \frac{1}{2^{N/2}} \sum_{k=0}^{N/2} (- 2 i)^{k} \sum_{j_1 < \dots < j_k} \epsilon_{j_1} \dots \epsilon_{j_k} \psi_{2j_1 - 1} \psi_{2j_1} \dots \psi_{2 j_k - 1} \psi_{2 j_k}.$$
If we were to sum over all possible strings $\epsilon,$ we would see that every term that contains even a single $\epsilon_{j}$ vanishes; so we would be left with
$$\sum_{\epsilon \in \{\pm 1\}^{N/2}} P_{\epsilon} = \sum_{\epsilon \in \{\pm 1\}^{N/2}} \frac{1}{2^{N/2}} = 1.$$
This is wonderful — the operators $P_{\epsilon}$ are all mutually orthogonal projectors that sum to the identity. As such, the direct sum
$$\mathcal{M} = \oplus_{\epsilon \in \{\pm 1\}^{N/2}} \mathcal{M} P_{\epsilon}$$
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.

We'll now introduce the notation $\mathcal{L}_{\epsilon} \equiv \mathcal{M} P_{\epsilon}.$ We commented in section 1 that $\mathcal{L}_{\epsilon}$ is simple if and only if its projector $P_{\epsilon}$ cannot be decomposed further into two nontrivial orthogonal projectors. The first step toward showing this is to observe that for any Majorana operator $\psi_{k},$ we have
$$\psi_{k} P_{\epsilon} = P_{\epsilon'} \psi_{k},$$
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 $P_{\epsilon}.$

Now, consider an arbitrary product of Majorana operators, which we will denote by
$$\chi = \psi_{1}^{\eta_1} \dots \psi_{N}^{\eta_N},$$
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
$$P_{\epsilon} \chi P_{\epsilon} = \begin{cases} \chi P_{\epsilon} & \text{if $\eta_{2j-1} = \eta_{2j}$ for all $j=1, \dots, N/2$} \\ 0 & \text{otherwise}\end{cases}.$$
But we may easily verify the identity
$$\psi_{2j-1} \psi_{2j} \Pi_{\epsilon_j}^{j} = \frac{i \epsilon_{j}}{2} \Pi_{\epsilon_j}^{j}.$$
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
$$P_{\epsilon} a P_{\epsilon} \propto P_{\epsilon}. \label{eq:projector-proportionality} \tag{1}$$

Now, suppose that $\mathcal{M} P_{\epsilon}$ were not a simple left ideal. Then there would exist nonzero projection operators $\alpha, \beta$ satisfying $\alpha \beta = \beta \alpha = 0$ and $\alpha + \beta = P_{\epsilon}.$ But then we would have
$$(\alpha + \beta) \alpha (\alpha + \beta) = P_{\epsilon} \alpha P_{\epsilon} = \lambda P_{\epsilon},$$
where $\lambda$ is a constant of proportionality coming from equation (1). Expanding out the left side of this this equation, we obtain the identity
$$\alpha = \lambda P_{\epsilon}.$$
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.
Finally, we can show that the left ideals $\mathcal{M} P_{\epsilon}$ are isomorphic for $\epsilon \neq \epsilon'.$ This isn't hard; we already showed that by for any $\epsilon, \epsilon',$ there exists a Majorana operator $\chi$ satisfying
$$P_{\epsilon} \chi = \chi P_{\epsilon'}.$$
So right-multiplication by $\chi$ is a map from $\mathcal{M} P_{\epsilon}$ to $\mathcal{M} P_{\epsilon'},$ and it is easily checked to be a homomorphism. Because any product of Majorana operators is an invertible element of the algebra (its inverse being the product of the Majorana operators in reverse order, with an appropriate coefficient), this homomorphism is in fact an isomorphism.

So, we have learned that for $N$ even, the Majorana algebra can be decomposed as
$$\mathcal{M} = \oplus_{\epsilon \in \{-1, 1\}^{N/2}} \mathcal{M} P_{\epsilon},$$
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.

In particular, this tells us why the "Dirac fermion Fock space" usually used to study even numbers of Majorana fermions is the "right" Hilbert space to study. This Fock space is $2^{N/2}$ dimensional, since $N/2$ is the number of Dirac creation operators and for each creation operator we have a binary choice of whether to include the corresponding fermion in our state. It must, therefore, be the unique irreducible representation of the Majorana algebra. The standard inner product on this Fock space (the one that makes creation operators the adjoints of annihilation operators) is also isomorphic to the one on $\mathcal{M}$ that we constructed in section 2.1.

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 $-1/\sqrt{2}.$
Our starting observation will be that when $N$ is odd, the fermion product $\psi_{1} \dots \psi_{N}$ commutes with every element of the Majorana algebra $\mathcal{M}.$ It is also independent of the order in which we multiply the fermions. In terms of the inner product we defined in section 2.1, we have
$$(\psi_1 \dots \psi_N)^{\dagger} \psi_1 \dots \psi_N = \frac{1}{2^N}.$$
So $2^{N/2} \psi_{1} \dots \psi_N$ is unitary. By multiplying it by an appropriate phase, we can also make it Hermitian; the operator
$$\eta = i^{\frac{N(N-1)}{2}} 2^{N/2} \psi_1 \dots \psi_N$$
is easily checked to satisfy $\eta^{\dagger} = \eta, \eta^2 = 1.$

Because $\eta$ is Hermitian, it is diagonalizable. Because it squares to the identity, its eigenvalues must all be $\pm 1.$ Finally, because $\eta$ commutes with every element of $\mathcal{M},$ the eigenspaces of $\eta$ are left ideals. So the decomposition of $\mathcal{M}$ into the $+1$ and $-1$ eigenspaces of $\eta$ as
$$\mathcal{M} = \mathcal{M}_{+} \oplus \mathcal{M}_{-}$$
is a decomposition of $\mathcal{M}$ into complementary left ideals. Furthermore, because $\eta$ has different eigenvalues within these two ideals, we know that no nontrivial sub-ideal of $\mathcal{M}_{+}$ can be isomorphic as an $\mathcal{M}$-representation to any nontrivial sub-ideal of $\mathcal{M}_{-}$; so unless one of these ideals is empty, we know already that there will be at least two different irreducible representations of $\mathcal{M}.$

Within the first $N-1$ fermions, we can still construct creation and annihilation pairings as in the even-$N$ case, and construct the projectors
$$P_{\epsilon \in \{\pm 1\}^{(N-1)/2}} = \prod_{j=1}^{(N-1)/2} \frac{1 - 2 i \epsilon_j \psi_{2j-1} \psi_{2j}}{2}.$$
These projectors are still orthogonal, but they no longer sum to the identity element. In order to minimize our effort, it would be nice if we could make some small change to the projectors $P_{\epsilon}$ so that they lie in the ideals $\mathcal{M}_{+}$ or $\mathcal{M}_{-}.$ In fact, this is almost already true; one can easily check the identity
$$\eta P_{\epsilon} = i^{\frac{N(N-1)}{2}} i^{\frac{N-1}{2}} \sqrt{2} (\prod_{j} \epsilon_j) \psi_N P_{\epsilon}.$$
The product $\prod_j \epsilon_j$ is more convenient expressed as $(-1)^{|\epsilon_-|},$ where $\epsilon_-$ is the total number of values of $j$ for which $\epsilon_j$ is $-1.$ Simplifying a bit, we obtain
$$\eta P_{\epsilon} = i^{\frac{N^2 - 1}{2}} \sqrt{2} (-1)^{|\epsilon_-|} \psi_N P_{\epsilon}.$$
So if we add a factor to $P_{\epsilon}$ that is an eigenvector of $\psi_N$ with an appropriate eigenvalue, we'll obtain a genuine eigenvector of $\eta.$ One easy way to do this is via the identity
$$\psi_N \frac{1 \pm \sqrt{2} \psi_N}{2} = \pm \frac{1}{\sqrt{2}} \frac{1 \pm \sqrt{2} \psi_N}{2}.$$
The operators $(1 \pm \sqrt{2} \psi_N)/2$ commute with all other factors in $P_{\epsilon},$ and are easily checked to be orthogonal projectors. So, if we extend our string $\epsilon$ to the range $\epsilon \in \{-1, 1\}^{(N+1)/2},$ we can redefine $P_{\epsilon}$ as
$$P_{\epsilon} = \left(\prod_{j=1}^{(N-1)/2} \frac{1 - 2 \epsilon_j \psi_{2j-1} \psi_{2j}}{2}\right) \frac{1 \pm \epsilon_{(N+1)/2} \sqrt{2} \psi_{N}}{2},$$
then one can check quite straightforwardly the identities $P_{\epsilon} P_{\epsilon'} = \delta_{\epsilon \epsilon'} P_{\epsilon}$ and $\sum_{\epsilon} P_{\epsilon} = 1,$ along with
$$\eta P_{\epsilon} = i^{\frac{N^2 - 1}{2}} (-1)^{|\epsilon_-|} P_{\epsilon},$$
where $|\epsilon_-|$ now counts the term $\epsilon_{(N+1)/2}$ as well. Finally, because any odd integer $N$ has the property that either $(N-1)$ or $(N+1)$ is divisible by four, $(N^2 - 1)$ is divisible by eight and so $i^{\frac{N^2 - 1}{2}}$ is just $1.$

So we obtain the direct sum decomposition
$$\mathcal{M} = \oplus_{\epsilon} \mathcal{M} P_{\epsilon},$$
where each $\mathcal{M} P_{\epsilon}$ is a subideal of one of the ideals $\mathcal{M}_{+}$ or $\mathcal{M}_{-}$; the ones with $|\epsilon_-|$ even lie in $\mathcal{M}_{+},$ and the ones with $|\epsilon_-|$ odd lie in $\mathcal{M}_{-}.$ Now we just need to show that all of the $\mathcal{M} P_{\epsilon}$ ideals are simple, and that the ones in the same $\eta$-eigenspace are isomorphic; we will then have classified the two irreducible representations of $\mathcal{M}$ for $N$ odd.

Showing they are simple is more or less the same as in the even case. One can show, along the same lines as the argument in section 2.2, that $P_{\epsilon} a P_{\epsilon}$ is proportional to $P_{\epsilon}$ for any $a \in \mathcal{M},$ and then the argument given in section 2.2 shows that there can be no nontrivial orthogonal projectors summing to $P_{\epsilon}.$

Constructing isomorphisms is also more or less the same. Commuting $\psi_N$ through $P_{\epsilon}$ has no effect, while commuting $\psi_{2j-1}$ through $P_{\epsilon}$ changes the sign of both $\epsilon_{j}$ and $\epsilon_{(N+1)/2}.$ Since we always change two signs at once, we can never change the value of $|\epsilon_-|$ mod 2 by this procedure. But this is the only constraint; for any $\epsilon, \epsilon'$ with $|\epsilon_-| = |\epsilon'_-| \mod 2,$ we can always construct a fermion product that maps $\mathcal{M} P_{\epsilon}$ to $\mathcal{M} P_{\epsilon'}$ under right multiplication; since all fermion products are invertible, this is an isomorphism.
All that remains is to determine the dimensions of these simple ideals, and to give a simple model for each of the two inequivalent irreducible representations. First, we observe that $\mathcal{M}_{+}$ and $\mathcal{M}_{-}$ have the same dimension; to see this, let $I = \{j_1 < \dots < j_k\}$ be a set of ordered indices, and denote by $\psi_{I}$ the fermion product $\psi_{j_1} \dots \psi_{j_k}.$ Then the bases
$$\{ \frac{\psi_I + \eta \psi_I}{2} | |I| < N/2\},$$
$$\{ \frac{\psi_I - \eta \psi_I}{2} | |I| < N/2\}$$
are easily checked to be bases for $\mathcal{M}_{+}$ and $\mathcal{M}_{-},$ respectively. So we immediately see that the two spaces have the same dimension, which is $2^{N-1}$. Furthermore, we can show that the number of simple left ideals within each $\mathcal{M}_{+}$ is the same. The total number of simple ideals with $|\epsilon_-|$ odd must be the same as the total number of simple ideals with $|\epsilon_-|$ even, because the map $\epsilon_1 \mapsto -\epsilon_1$ is a bijection on the set of all $\epsilon$ strings that changes the parity of $|\epsilon_-|.$ Because the total number of $\epsilon$ strings is $2^{(N+1)/2},$ we conclude that there are $2^{(N-1)/2}$ strings for each parity of $|\epsilon_-|.$ (The first draft of this post had a more complicated proof of this fact — thanks to Brandon Rayhaun for helping me simplify it!)
So each of $\mathcal{M}_{+}$ and $\mathcal{M}_{-}$ contains $2^{(N-1)/2}$ simple left ideals, and since each of $\mathcal{M}_{+}$ and $\mathcal{M}_{-}$ is $2^{N-1}$ dimensional, we conclude that the simple left ideals of $\mathcal{M}$ have dimension $2^{(N-1)/2}.$ The two inequivalent irreducible representations of $\mathcal{M}$ are distinguished by whether the operator $\eta$ acts as plus or minus the identity.

Just as the unique irreducible representation of the even-$N$ Majorana algebra has a simple model as the Fock space of $N/2$ Dirac fermions, the two irreducible representations of the odd-$N$ Majorana algebra have simple models as well. As before, we group the first $(N-1)$ Majorana fermions into creation and annihilation operators $a_{1}, \dots, a_{(N-1)/2}.$ We then define a Fock space by starting with a base state $|0\rangle$ and asserting that it is sent to zero by the annihilation operators, and that the various strings of creation operators acting on this state are linearly independent. This is a $2^{(N-1)/2}$-dimensional vector space, and it almost completely specifies a representation of $\mathcal{M},$ except that we don't know how the operator $\psi_N$ should act. But via the creation and annihilation operators we already know how the Majoranas $\psi_1, \dots, \psi_{N-1}$ act on $|0\rangle$; specifying an irreducible representation amounts to specifying how $\eta$ acts on the base state; but because we know how all of the factors of $\eta$ other than $\psi_N$ act on the base state, specifying an eigenspace of $\eta$ is equivalent to specifying how $\psi_N$ acts on the base state. One can check quite trivially that for $\eta$ to act as $\pm 1,$ the operator $\psi_N$ must act on the base state as $\pm 1/\sqrt{2}.$ So for odd $N$, the two irreducible representations of $\mathcal{M}$ are modeled by Dirac-fermion Fock spaces together with the specification of whether $\psi_N$ acts on the base state as $+1/\sqrt{2}$ or $-1/\sqrt{2}.$

3. Consequences for physical fermions

The reason we've done all of this is to ask: if I have some Majorana algebra $\mathcal{M},$ and a Hamiltonian expressed in terms of the Majorana operators as
$$H = \sum_{k=1}^{N} \sum_{j_1 < \dots < j_k} c_{j_1, \dots, j_k} \psi_{j_1} \dots \psi_{j_k} ,$$
then what is its spectrum? First, it's always implicitly assumed that $H$ acts on a Hilbert space whose inner product matches that on the Majorana algebra, so $H$ is Hermitian if and only if, for all $c,$ we have
$$c_{j_1, \dots, j_k}^* = (-1)^{\frac{k(k-1)}{2}} c_{j_1, \dots, j_k}.$$
So $c$ must be real for $k(k-1) = 0 \mod 4,$ and pure imaginary for $k(k-1) = 2 \mod 4.$

The problem with the abstract question, "what is its spectrum?", is that we haven't specified the Hilbert space that $H$ acts on, or how it acts! We've only given it as an algebraic expression. But what we've learned in the previous section ameliorates this problem. Since $H$ is constructed entirely from $\mathcal{M},$ its spectrum on the irreducible representations of $\mathcal{M}$ is determined solely by the structure of those irreducible representations.

So, in the case that $N$ is even, there is only one irreducible representation of $\mathcal{M},$ which is the Fock space of $N/2$ Dirac fermions, and we can just compute the spectrum of $H$ on this space. Any other representation of $\mathcal{M}$ must be an integer number of copies of this representation, so the spectrum of $H$ on such a space will just have an extra integer degeneracy at every energy level.

In the case that $N$ is odd, we have two non-isomorphic representations; in general, the spectrum of $H$ will be different on these two representations. There's just no way around it: for a general Hamiltonian $H$ made up of an odd number of Dirac fermions, its spectrum is only well defined if you also specify a representation of $\mathcal{M}.$ But that's not so bad — to find the spectrum of $H$ in a general representation, you only need to compute its spectrum in the two irreducible ones and then extrapolate from there.

As an example, one can check that for three Majorana fermions, the Hamiltonian $\psi_1 + i \psi_2 \psi_3$ has spectrum $\pm (1 + \sqrt{2})/2$ in one of the irreducible representations, and $\pm (1 - \sqrt{2})/2$ in the other. This isn't too hard of an exercise; $\psi_1, \psi_2,$ and $\psi_3$ will be 2 by 2 matrices, and using the simple models for the two irreducible representations given at the end of section 2, $H$ can be written down and diagonalized explicitly.


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