Jonginn Yun

Guiding picture

The formalism has three layers:

$$\boxed{\text{fermionic Fock space } \mathcal F_-} \qquad \boxed{\text{Grassmann algebra } \mathcal G} \qquad \boxed{\text{extended graded space } \mathcal G\mathbin{\widehat\otimes}\mathcal F_-}.$$

The physical fermionic Fock space $\mathcal F_-$ contains ordinary fermion states: the vacuum, one-particle states, two-particle antisymmetric states, and so on. The Grassmann algebra $\mathcal G$ contains formal anticommuting generators such as $\eta_i$ and $\bar\eta_i$. A fermionic coherent state is not an ordinary vector in $\mathcal F_-$. It is a Grassmann-valued vector, i.e. an element of something like

$$\mathcal G\mathbin{\widehat\otimes}\mathcal F_-.$$

The hat on $\mathbin{\widehat\otimes}$ means that both factors carry parity, and whenever two odd objects cross, one inserts a minus sign.

The most important formula in the whole note is

$$(A\mathbin{\widehat\otimes}B)(C\mathbin{\widehat\otimes}D) = (-1)^{\lvert B\rvert\lvert C\rvert} (AC\mathbin{\widehat\otimes}BD),$$

for homogeneous operators or algebra elements. This is the Koszul sign rule.

Ordinary vector spaces and tensor products

We work over $\mathbb C$, although almost everything below works over any field of characteristic not equal to $2$.

Definition 1 (Vector space). A complex vector space $V$ is a set equipped with addition and scalar multiplication by complex numbers, satisfying the usual linear axioms.

Definition 2 (Linear map). A map $T:V\to W$ is linear if

$$T(\alpha v+\beta w)=\alpha T(v)+\beta T(w)$$

for all $v,w\in V$ and $\alpha,\beta\in \mathbb C$.

Definition 3 (Ordinary tensor product). The tensor product $V\otimes W$ is the vector space generated by symbols $v\otimes w$, subject to the bilinearity relations

$$\begin{align*} (v_1+v_2)\otimes w &= v_1\otimes w+v_2\otimes w,\\ v\otimes(w_1+w_2) &= v\otimes w_1+v\otimes w_2,\\ (\lambda v)\otimes w &= v\otimes(\lambda w)=\lambda(v\otimes w). \end{align*}$$

If $A:V\to V'$ and $B:W\to W'$ are ordinary linear maps, the ordinary tensor product map

$$A\otimes B:V\otimes W\to V'\otimes W'$$

is defined by

$$(A\otimes B)(v\otimes w)=Av\otimes Bw.$$

No signs occur here. Signs enter only after we add parity.

Parity and super vector spaces

The group Z2

The symbol $\mathbb Z_2$ means the two-element additive group

$$\mathbb Z_2=\{\bar 0,\bar 1\},$$

with addition modulo $2$. In formulas we usually write $0$ and $1$ instead of $\bar 0$ and $\bar 1$. Thus

$$1+1=0 \pmod 2.$$

What parity is

Definition 4 ($\mathbb Z_2$-graded vector space, or super vector space). A $\mathbb Z_2$-graded vector space is a vector space $V$ together with a direct-sum decomposition

$$V=V_{\bar 0}\oplus V_{\bar 1}.$$

Elements of $V_{\bar 0}$ are called even. Elements of $V_{\bar 1}$ are called odd.

Definition 5 (Homogeneous element and parity value). A vector $v\in V$ is called homogeneous if $v\in V_{\bar 0}$ or $v\in V_{\bar 1}$. For a nonzero homogeneous vector, its parity is the value

$$\lvert v\rvert\in \mathbb Z_2$$

defined by

$$\lvert v\rvert=0 \quad \text{if } v\in V_{\bar 0}, \qquad \lvert v\rvert=1 \quad \text{if } v\in V_{\bar 1}.$$

Thus the parity function is not a linear map on all of $V$. More precisely, it is a function

$$\lvert \cdot\rvert: V_{\mathrm{hom}}\setminus\{0\}\longrightarrow \mathbb Z_2,$$

where $V_{\mathrm{hom}}=(V_{\bar 0}\cup V_{\bar 1})$ is the set of homogeneous vectors.

Warning 6 (Not every vector has a parity). If $v_0\in V_{\bar 0}$ and $v_1\in V_{\bar 1}$ are both nonzero, then

$$v=v_0+v_1$$

is generally not homogeneous. It is not correct to assign a single parity $\lvert v\rvert$ to such a sum. There is, however, a genuine linear operator associated with parity.

Definition 7 (Parity operator). The parity operator on a super vector space $V=V_{\bar 0}\oplus V_{\bar 1}$ is the linear map

$$\Pi_V:V\to V$$

defined by

$$\Pi_V(v_0+v_1)=v_0-v_1, \qquad v_0\in V_{\bar 0},\; v_1\in V_{\bar 1}.$$

Equivalently, $\Pi_V$ has eigenvalue $+1$ on $V_{\bar 0}$ and eigenvalue $-1$ on $V_{\bar 1}$. Thus there are two related but different notions:

object type meaning

---

$\lvert v\rvert$ function on homogeneous nonzero vectors to $\mathbb Z_2$ even or odd label $\Pi_V$ linear map $V\to V$ parity operator with eigenvalues $\pm 1$

Graded linear maps and operator parity

Definition 8 (Homogeneous linear map). Let $V=V_{\bar 0}\oplus V_{\bar 1}$ and $W=W_{\bar 0}\oplus W_{\bar 1}$. A linear map $T:V\to W$ is homogeneous of degree $\alpha\in\mathbb Z_2$ if

$$T(V_i)\subseteq W_{i+\alpha} \qquad \text{for } i\in\mathbb Z_2.$$

If $\alpha=0$, $T$ is even. If $\alpha=1$, $T$ is odd. So an even operator preserves parity, and an odd operator flips parity:

$$\begin{align*} T\text{ even} &: V_{\bar 0}\to W_{\bar 0},\quad V_{\bar 1}\to W_{\bar 1},\\ T\text{ odd} &: V_{\bar 0}\to W_{\bar 1},\quad V_{\bar 1}\to W_{\bar 0}. \end{align*}$$

For a homogeneous linear map, we write $\lvert T\rvert\in\mathbb Z_2$ for its degree.

Example 9. Let $V=\mathrm{span}\{e_0,e_1\}$, with $e_0$ even and $e_1$ odd. The operator

$$T e_0=e_1, \qquad T e_1=e_0$$

is odd. The operator

$$S e_0=2e_0, \qquad S e_1=-e_1$$

is even. The space $\mathrm{Hom}(V,W)$ itself becomes graded:

$$\mathrm{Hom}(V,W)_{\bar\alpha} = \{T:V\to W\mid T(V_i)\subseteq W_{i+\alpha}\}.$$

In particular, $\mathrm{End}(V)=\mathrm{Hom}(V,V)$ is a graded algebra under composition.

The graded tensor product of spaces

Definition 10 (Graded tensor product of super vector spaces). Let

$$V=V_{\bar 0}\oplus V_{\bar 1}, \qquad W=W_{\bar 0}\oplus W_{\bar 1}$$

be super vector spaces. Their graded tensor product

$$V\mathbin{\widehat\otimes}W$$

is the ordinary tensor product vector space $V\otimes W$, equipped with the grading

$$(V\mathbin{\widehat\otimes}W)_{\bar k} = \bigoplus_{i+j=k\;\mathrm{mod}\;2} V_{\bar i}\otimes W_{\bar j}.$$

Equivalently, for homogeneous pure tensors,

$$\lvert v\mathbin{\widehat\otimes}w\rvert=\lvert v\rvert+\lvert w\rvert\pmod 2.$$

This definition is important because it says: as a vector space,

$$V\mathbin{\widehat\otimes}W = V\otimes W.$$

The new structure is the grading. The hat reminds us that this tensor product lives in the category of super vector spaces, where signs appear in exchanges and operator products.

Example 11. Suppose $v$ is odd and $w$ is odd. Then

$$\lvert v\mathbin{\widehat\otimes}w\rvert=1+1=0\pmod 2.$$

So an odd tensor an odd vector is even.

The source of the minus sign: graded symmetry

In ordinary vector spaces, the flip map is

$$v\otimes w\longmapsto w\otimes v.$$

In super vector spaces, the natural flip map is not the ordinary flip. It is the graded flip.

Definition 12 (Graded flip, or braiding). For homogeneous $v\in V$, $w\in W$, define

$$\tau_{V,W}(v\mathbin{\widehat\otimes}w) = (-1)^{\lvert v\rvert\lvert w\rvert} w\mathbin{\widehat\otimes}v.$$

Extend linearly to all of $V\mathbin{\widehat\otimes}W$. Therefore:

$$\begin{align*} \text{even crosses anything} &\quad \Rightarrow \quad \text{no sign},\\ \text{odd crosses odd} &\quad \Rightarrow \quad \text{minus sign}. \end{align*}$$

This is the basic Koszul principle.

Graded tensor product of operators

The graded tensor product of spaces tells us the parity of $v\mathbin{\widehat\otimes}w$. To do calculations, we also need to know how tensor products of operators act.

Definition 13 (Action of a tensor product of homogeneous operators). Let $A:V\to V'$ and $B:W\to W'$ be homogeneous linear maps. Their graded tensor product operator

$$A\mathbin{\widehat\otimes}B:V\mathbin{\widehat\otimes}W\to V'\mathbin{\widehat\otimes}W'$$

is defined on homogeneous pure tensors by

$$(A\mathbin{\widehat\otimes}B)(v\mathbin{\widehat\otimes}w) = (-1)^{\lvert B\rvert\lvert v\rvert} Av\mathbin{\widehat\otimes}Bw.$$

Then extend linearly. Why does $\lvert B\rvert\lvert v\rvert$ appear? Intuitively, the operator $B$, which acts on the right factor, must pass the left vector $v$ before reaching $w$. If both are odd, a minus sign appears.

Warning 14. Some books choose a slightly different convention for tensor products of maps. The convention in [eq:operator-action] is the standard one compatible with the Koszul product rule [eq:koszul-algebra]. Once a convention is chosen, it must be used consistently.

Proposition 15 (Composition rule). *For homogeneous maps $A,B,C,D$ with compatible domains and codomains,

$$(A\mathbin{\widehat\otimes}B)(C\mathbin{\widehat\otimes}D) = (-1)^{\lvert B\rvert\lvert C\rvert} (AC\mathbin{\widehat\otimes}BD).$$

Proof. Apply the left-hand side to a homogeneous tensor $v\mathbin{\widehat\otimes}w$:

$$\begin{align*} (C\mathbin{\widehat\otimes}D)(v\mathbin{\widehat\otimes}w) &= (-1)^{\lvert D\rvert\lvert v\rvert} Cv\mathbin{\widehat\otimes}Dw. \end{align*}$$

Now apply $A\mathbin{\widehat\otimes}B$:

$$\begin{align*} (A\mathbin{\widehat\otimes}B)(Cv\mathbin{\widehat\otimes}Dw) &= (-1)^{\lvert B\rvert\lvert Cv\rvert} ACv\mathbin{\widehat\otimes}BDw \\ &= (-1)^{\lvert B\rvert(\lvert C\rvert+\lvert v\rvert)} ACv\mathbin{\widehat\otimes}BDw. \end{align*}$$

The total sign is

$$(-1)^{\lvert D\rvert\lvert v\rvert+\lvert B\rvert\lvert C\rvert+\lvert B\rvert\lvert v\rvert}.$$

On the other hand, $AC\mathbin{\widehat\otimes}BD$ acting on $v\mathbin{\widehat\otimes}w$ gives the sign

$$(-1)^{(\lvert B\rvert+\lvert D\rvert)\lvert v\rvert}.$$

Thus the left-hand side equals

$$(-1)^{\lvert B\rvert\lvert C\rvert}(AC\mathbin{\widehat\otimes}BD)(v\mathbin{\widehat\otimes}w).$$

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Graded tensor product of algebras

A super algebra is an algebra $A=A_{\bar 0}\oplus A_{\bar 1}$ whose product respects parity:

$$A_iA_j\subseteq A_{i+j}.$$

Definition 16 (Graded tensor product of super algebras). If $A$ and $B$ are super algebras, their graded tensor product algebra $A\mathbin{\widehat\otimes}B$ has underlying vector space $A\otimes B$, grading as above, and multiplication

$$(a\mathbin{\widehat\otimes}b)(a'\mathbin{\widehat\otimes}b') = (-1)^{\lvert b\rvert\lvert a'\rvert} aa'\mathbin{\widehat\otimes}bb'$$

for homogeneous $a,a'\in A$, $b,b'\in B$. This is the same sign rule as for operators. The middle objects $b$ and $a'$ must be exchanged to collect the $A$-parts together and the $B$-parts together.

Exterior algebra

Before defining fermionic Fock space, we need exterior powers.

Definition 17 (Exterior power). Let $\mathfrak h$ be a complex vector space. The $n$-th exterior power $\Lambda^n\mathfrak h$ is the vector space spanned by formal symbols

$$f_1\wedge\cdots\wedge f_n, \qquad f_i\in\mathfrak h,$$

subject to multilinearity and antisymmetry:

$$f_1\wedge\cdots\wedge f_i\wedge f_{i+1}\wedge\cdots\wedge f_n = - f_1\wedge\cdots\wedge f_{i+1}\wedge f_i\wedge\cdots\wedge f_n.$$

Also $\Lambda^0\mathfrak h=\mathbb C$. The antisymmetry implies

$$f\wedge f=0.$$

This is the algebraic origin of the Pauli exclusion principle.

Definition 18 (Exterior algebra). The algebraic exterior algebra of $\mathfrak h$ is

$$\Lambda \mathfrak h = \bigoplus_{n\ge 0}^{\mathrm{alg}}\Lambda^n\mathfrak h.$$

The superscript $\mathrm{alg}$ means algebraic direct sum: each element has only finitely many nonzero degree components. If $\dim \mathfrak h=N<\infty$, then

$$\Lambda^n\mathfrak h=0 \quad \text{for } n>N,$$

so

$$\Lambda\mathfrak h=\bigoplus_{n=0}^{N}\Lambda^n\mathfrak h.$$

If $\mathfrak h$ is infinite-dimensional, then $\Lambda^n\mathfrak h$ is nonzero for every finite $n$. However, the algebraic exterior algebra still contains only finite-degree sums. It does not contain actual infinite-degree wedge monomials such as

$$e_1\wedge e_2\wedge e_3\wedge\cdots$$

unless one introduces a different construction, such as an infinite wedge space. That is not the standard fermionic Fock space used for finite-particle sectors.

Fermionic Fock space

One-particle space

The starting input is a one-particle Hilbert space $\mathfrak h$. For example:

• for $N$ discrete modes, $\mathfrak h=\mathbb C^N$;

• for particles on a spatial domain, $\mathfrak h$ may be an $L^2$-space with spin labels.

Algebraic fermionic Fock space

Definition 19 (Algebraic fermionic Fock space). The algebraic fermionic Fock space over $\mathfrak h$ is

$$\mathcal F_-^{\mathrm{alg}}(\mathfrak h) = \bigoplus_{n=0}^{\infty,\mathrm{alg}}\Lambda^n\mathfrak h.$$

The vector $1\in \Lambda^0\mathfrak h=\mathbb C$ is called the vacuum and is usually denoted by

$$\lvert 0\rangle\quad\text{or}\quad \Omega.$$

An element of $\Lambda^n\mathfrak h$ is an $n$-fermion vector. The word “fermionic” means the $n$-particle wavefunction is antisymmetric.

Hilbert fermionic Fock space

If $\mathfrak h$ is a Hilbert space, each $\Lambda^n\mathfrak h$ inherits a Hilbert-space inner product. The physical Hilbert Fock space is the Hilbert direct sum

$$\mathcal F_-(\mathfrak h)=\bigoplus_{n=0}^{\infty}\Lambda^n\mathfrak h,$$

where $\oplus$ denotes the Hilbert direct sum. Concretely, a vector $\Psi\in\mathcal F_-(\mathfrak h)$ is a sequence

$$\Psi=(\psi_0,\psi_1,\psi_2,\ldots), \qquad \psi_n\in\Lambda^n\mathfrak h,$$

with

$$\sum_{n=0}^{\infty}\|\psi_n\|^2<\infty.$$

Warning 20 (No infinite wedge degree in standard Fock space). Even when $\mathfrak h$ is infinite-dimensional, standard Fock space does not contain one component of degree $\infty$. It contains components of degree $n=0,1,2,\ldots$, each finite. The Hilbert completion allows infinitely many finite-particle components to appear in a square-summable sequence.

Finite number of modes

If $\mathfrak h=\mathbb C^N$, then

$$\mathcal F_-(\mathbb C^N)=\bigoplus_{n=0}^N\Lambda^n\mathbb C^N.$$

This has dimension $2^N$. If $e_1,\\ldots,e_N$ is an orthonormal basis of $\mathbb C^N$, a basis of $\mathcal F_-(\mathbb C^N)$ is

$$e_{i_1}\wedge\cdots\wedge e_{i_k}, \qquad 1\le i_1<\cdots<i_k\le N.$$

In occupation-number notation, this is written as

$$\lvert n_1,\ldots,n_N\rangle, \qquad n_i\in\{0,1\}.$$

The fact that $n_i$ is either $0$ or $1$ is exactly the fact that $e_i\wedge e_i=0$.

Creation and annihilation operators

Let $\mathfrak h$ be a one-particle Hilbert space. We use the physics convention that the inner product $\langle f,g\rangle$ is conjugate-linear in $f$ and linear in $g$.

Definition 21 (Creation operator). For $f\in\mathfrak h$, the fermionic creation operator $a^\dagger(f)$ is defined on wedge monomials by

$$a^\dagger(f)(g_1\wedge\cdots\wedge g_n) = f\wedge g_1\wedge\cdots\wedge g_n.$$

In particular,

$$a^\dagger(f)\Omega=f.$$

Definition 22 (Annihilation operator). The annihilation operator $a(f)$ is the adjoint of $a^\dagger(f)$. On wedge monomials it is given by

$$a(f)(g_1\wedge\cdots\wedge g_n) = \sum_{k=1}^n (-1)^{k-1}\langle f,g_k\rangle g_1\wedge\cdots\wedge \widehat{g_k}\wedge\cdots\wedge g_n,$$

where the hat over $g_k$ means that $g_k$ is omitted. Also

$$a(f)\Omega=0.$$

For an orthonormal basis $e_i$, write

$$a_i=a(e_i), \qquad a_i^\dagger=a^\dagger(e_i).$$

Then

$$a_i^\dagger\Omega=e_i, \qquad a_i\Omega=0.$$

Proposition 23 (Canonical anticommutation relations). *The creation and annihilation operators satisfy

$$\{a(f),a^\dagger(g)\}=\langle f,g\rangle I,$$ $$\{a(f),a(g)\}=0, \qquad \{a^\dagger(f),a^\dagger(g)\}=0.$$

In mode notation,

$$\{a_i,a_j^\dagger\}=\delta_{ij}I, \qquad \{a_i,a_j\}=0, \qquad \{a_i^\dagger,a_j^\dagger\}=0.$$

Here $\{X,Y\}=XY+YX$.

Parity on fermionic Fock space

The fermionic Fock space is automatically graded by particle-number parity.

Definition 24 (Number operator). The number operator $\mathcal N$ is defined by

$$\mathcal N\psi=n\psi \qquad \text{for } \psi\in\Lambda^n\mathfrak h.$$

Definition 25 (Fermion parity operator). The fermion parity operator is

$$\Pi_{\mathcal F}=(-1)^{\mathcal N}.$$

Thus

$$\Pi_{\mathcal F}\psi=(-1)^n\psi \qquad \text{if } \psi\in\Lambda^n\mathfrak h.$$

Therefore

$$\mathcal F_{-,\bar 0}=\bigoplus_{\substack{n\ge0\\ n\text{ even}}}\Lambda^n\mathfrak h, \qquad \mathcal F_{-,\bar 1}=\bigoplus_{\substack{n\ge0\\ n\text{ odd}}}\Lambda^n\mathfrak h,$$

where both direct sums are Hilbert direct sums.

The vacuum is even:

$$\lvert 0\rangle\in\mathcal F_{-,\bar 0}.$$

A one-particle state is odd:

$$f\in\Lambda^1\mathfrak h\subset\mathcal F_{-,\bar 1}.$$

A two-particle state is even:

$$f\wedge g\in\Lambda^2\mathfrak h\subset\mathcal F_{-,\bar 0}.$$

Creation and annihilation operators are odd:

$$\lvert a_i\rvert=1, \qquad \lvert a_i^\dagger\rvert=1.$$

Indeed, they change particle number by $\pm 1$, so they flip even states to odd states and odd states to even states.

Grassmann algebras

Definition

Let $E$ be a finite-dimensional complex vector space with basis

$$\theta_1,\ldots,\theta_m.$$

The Grassmann algebra generated by these symbols is the exterior algebra

$$\mathcal G=\Lambda E.$$

Thus $\mathcal G$ is spanned by monomials

$$\theta_{i_1}\theta_{i_2}\cdots\theta_{i_k}, \qquad i_1<\cdots<i_k,$$

with multiplication determined by

$$\theta_i\theta_j=-\theta_j\theta_i, \qquad \theta_i^2=0.$$

Remark 26. The multiplication in $\mathcal G$ is usually written by juxtaposition rather than $\wedge$. Algebraically, it is the same exterior product.

Even and odd parts of a Grassmann algebra

The Grassmann algebra is graded by degree modulo $2$:

$$\mathcal G_{\bar 0} = \bigoplus_{k\text{ even}}\Lambda^k E, \qquad \mathcal G_{\bar 1} = \bigoplus_{k\text{ odd}}\Lambda^k E.$$

Thus

$$1,\; \theta_i\theta_j,\; \theta_i\theta_j\theta_k\theta_l$$

are even, while

$$\theta_i,\; \theta_i\theta_j\theta_k$$

are odd.

Eta and eta-bar variables

In fermionic coherent-state calculus, one often uses pairs of generators

$$\eta_1,\ldots,\eta_N, \qquad \bar\eta_1,\ldots,\bar\eta_N.$$

Formally, these are independent basis vectors of an auxiliary vector space

$$E=\mathrm{span}_{\mathbb C}\{\eta_1,\ldots,\eta_N,\bar\eta_1,\ldots,\bar\eta_N\}.$$

Then

$$\mathcal G=\Lambda E.$$

The bar in $\bar\eta_i$ is a label indicating the generator used for the bra side of coherent-state notation. In the purely algebraic path-integral formalism, $\bar\eta_i$ is not forced to be the complex conjugate of $\eta_i$. The two are independent Grassmann generators.

Berezin differentiation

Grassmann differentiation is not defined by a limit. There is no topology or order relation on a Grassmann algebra that would make such a limit meaningful. Instead, differentiation is defined algebraically.

Definition 27 (Left Berezin derivative). Let $\mathcal G=\Lambda(\theta_1,\ldots,\theta_m)$. The left derivative $\partial^L_i=\frac{\partial^L}{\partial\theta_i}$ is the unique odd linear operator $\mathcal G\to\mathcal G$ such that

$$\partial^L_i(\theta_j)=\delta_{ij}$$

and satisfying the graded Leibniz rule

$$\partial^L_i(uv) = (\partial^L_i u)v + (-1)^{\lvert u\rvert}u(\partial^L_i v)$$

for homogeneous $u\in\mathcal G$. The operator $\partial_i^L$ is odd because it lowers Grassmann degree by one.

Example 28. For two generators $\theta_1,\theta_2$,

$$\partial^L_1(\theta_1\theta_2)=\theta_2,$$

while

$$\partial^L_2(\theta_1\theta_2) = -\theta_1.$$

The second minus sign comes from the graded Leibniz rule because $\theta_1$ is odd. For a monomial with distinct generators,

$$\partial^L_j(\theta_{i_1}\cdots\theta_{i_k}) = \sum_{r: i_r=j}(-1)^{r-1} \theta_{i_1}\cdots \widehat{\theta_{i_r}}\cdots\theta_{i_k}.$$

Berezin integration

Berezin integration is also algebraic. It is coefficient extraction.

Definition 29 (One Grassmann variable). For $\mathcal G=\Lambda(\theta)$, define

$$\int d\theta\;1=0, \qquad \int d\theta\;\theta=1.$$

Thus, for $f(\theta)=a+b\theta$,

$$\int d\theta\;f(\theta)=b.$$

So, for one variable,

$$\int d\theta = \frac{\partial^L}{\partial\theta}$$

as linear functionals.

Definition 30 (Several variables, ordered convention). Fix an ordered list of generators

$$\theta_1,\ldots,\theta_m.$$

Define

$$\int d\theta_m\cdots d\theta_1\; \theta_1\cdots\theta_m=1,$$

and define the integral to be zero on all monomials not containing the full product $\theta_1\cdots\theta_m$. Equivalently,

$$\int d\theta_m\cdots d\theta_1 = \partial_m^L\cdots\partial_1^L.$$

The order of the differentials is part of the definition. Changing the order can change signs.

Example 31. With ordered variables $\theta_1,\theta_2$,

$$\int d\theta_2d\theta_1\;\theta_1\theta_2=1,$$

but

$$\int d\theta_2d\theta_1\;\theta_2\theta_1=-1.$$

The extended Grassmann-Fock space

Now we combine the Grassmann algebra and the fermionic Fock space.

Definition 32 (Grassmann-extended Fock space). Let $\mathcal G$ be the Grassmann algebra generated by $\eta_i,\bar\eta_i$, and let $\mathcal F_-(\mathfrak h)$ be the fermionic Fock space. The Grassmann-extended Fock space is

$$\mathcal F_\mathcal G = \mathcal G\mathbin{\widehat\otimes}\mathcal F_-(\mathfrak h).$$

As a vector space, this is $\mathcal G\otimes\mathcal F_-(\mathfrak h)$. As a super vector space, its parity is

$$\lvert g\mathbin{\widehat\otimes}\psi\rvert = \lvert g\rvert+\lvert \psi\rvert\pmod 2$$

for homogeneous $g\in\mathcal G$, $\psi\in\mathcal F_-(\mathfrak h)$. An element of $\mathcal F_\mathcal G$ has the form

$$\Psi=\sum_\alpha g_\alpha\mathbin{\widehat\otimes}\psi_\alpha,$$

where $g_\alpha\in\mathcal G$ and $\psi_\alpha\in\mathcal F_-(\mathfrak h)$.

Warning 33 (Not an ordinary physical Hilbert space). $\mathcal F_\mathcal G$ is not the physical Hilbert space of the fermionic system. It is a Grassmann-valued extension used as a calculational device. Fermionic coherent states live naturally in this extended space, not in the original physical Fock space $\mathcal F_-(\mathfrak h)$.

How operators act on the Grassmann-extended Fock space

There are two kinds of basic operators on the extended space.

Grassmann multiplication operators

For $\theta\in\mathcal G$, define left multiplication

$$M_\theta:\mathcal G\to\mathcal G, \qquad M_\theta(g)=\theta g.$$

If $\theta$ is odd, then $M_\theta$ is an odd operator. On $\mathcal G\mathbin{\widehat\otimes}\mathcal F_-$, it is represented as

$$M_\theta\mathbin{\widehat\otimes}I_{\mathcal F}.$$

Thus

$$(M_\theta\mathbin{\widehat\otimes}I)(g\mathbin{\widehat\otimes}\psi) = \theta g\mathbin{\widehat\otimes}\psi.$$

No sign appears here because the right operator is the identity, which is even.

Fock operators

A Fock operator such as $a_i$ acts on the Fock factor. On $\mathcal G\mathbin{\widehat\otimes}\mathcal F_-$, it is represented as

$$I_\mathcal G\mathbin{\widehat\otimes}a_i.$$

Since $a_i$ is odd, its action on a homogeneous tensor is

$$(I\mathbin{\widehat\otimes}a_i)(g\mathbin{\widehat\otimes}\psi) = (-1)^{\lvert g\rvert} g\mathbin{\widehat\otimes}a_i\psi.$$

This is just the general operator rule [eq:operator-action].

Example 34. Let $\eta$ be odd and let $\lvert 1\rangle$ be a one-particle state. Then

$$(I\mathbin{\widehat\otimes}a)(\eta\mathbin{\widehat\otimes}\lvert 1\rangle) = -\eta\mathbin{\widehat\otimes}a\lvert 1\rangle.$$

The minus sign appears because $a$ is odd and $\eta$ is odd.

Why Grassmann variables anticommute with fermionic operators

In the extended space, write

$$\eta := M_\eta\mathbin{\widehat\otimes}I, \qquad a := I\mathbin{\widehat\otimes}a.$$

Both are odd operators. Then

$$\begin{align*} \eta a &=(M_\eta\mathbin{\widehat\otimes}I)(I\mathbin{\widehat\otimes}a) \\ &=M_\eta\mathbin{\widehat\otimes}a, \end{align*}$$

while

$$\begin{align*} a\eta &=(I\mathbin{\widehat\otimes}a)(M_\eta\mathbin{\widehat\otimes}I) \\ &=(-1)^{\lvert a\rvert\lvert M_\eta\rvert} M_\eta\mathbin{\widehat\otimes}a \\ &=-M_\eta\mathbin{\widehat\otimes}a. \end{align*}$$

Therefore

$$a\eta=-\eta a.$$

This is not an additional arbitrary rule. It follows from the graded tensor product.

Berezin calculus on Grassmann-valued Fock vectors

Let

$$\Psi=\sum_\alpha g_\alpha\mathbin{\widehat\otimes}\psi_\alpha \in \mathcal G\mathbin{\widehat\otimes}\mathcal F_-.$$

A Berezin derivative acts on the Grassmann coefficient and leaves the Fock vector alone:

$$\frac{\partial^L}{\partial\theta_i}\Psi = \sum_\alpha \left(\frac{\partial^L g_\alpha}{\partial\theta_i}\right) \mathbin{\widehat\otimes}\psi_\alpha.$$

Equivalently, this is

$$\partial_i^L\mathbin{\widehat\otimes}I_\mathcal F.$$

Since the operator on the right factor is $I_\mathcal F$, no extra Koszul sign appears in this coefficientwise action.

Similarly,

$$\int d\theta_i\;\Psi = \sum_\alpha \left(\int d\theta_i\;g_\alpha\right) \mathbin{\widehat\otimes}\psi_\alpha.$$

Thus one should not say that a Fock vector itself is differentiated with respect to a Grassmann variable. The precise statement is: the Grassmann-valued coefficient is differentiated, while the Fock vector is carried along unchanged.

One-mode fermionic coherent states

Consider one fermionic mode. The Fock space is

$$\mathcal F=\mathrm{span}\{\lvert 0\rangle,\lvert 1\rangle\}, \qquad \lvert 1\rangle=a^\dagger\lvert 0\rangle.$$

The parities are

$$\lvert \lvert 0\rangle\rvert=0, \qquad \lvert \lvert 1\rangle\rvert=1, \qquad \lvert a\rvert=\lvert a^\dagger\rvert=1.$$

Let $\mathcal G=\Lambda(\eta,\bar\eta)$, with $\eta$ and $\bar\eta$ independent odd generators.

Definition 35 (One-mode ket coherent state). Define

$$\lvert \eta\rangle = 1\mathbin{\widehat\otimes}\lvert 0\rangle - \eta\mathbin{\widehat\otimes}\lvert 1\rangle \in \mathcal G\mathbin{\widehat\otimes}\mathcal F.$$

This vector is even: the first term is even plus even, and the second term is odd plus odd.

Now compute $a\lvert \eta\rangle$, where $a$ means $I_\mathcal G\mathbin{\widehat\otimes}a$:

$$\begin{align*} a\lvert \eta\rangle &= (I\mathbin{\widehat\otimes}a)(1\mathbin{\widehat\otimes}\lvert 0\rangle) - (I\mathbin{\widehat\otimes}a)(\eta\mathbin{\widehat\otimes}\lvert 1\rangle) \\ &= 1\mathbin{\widehat\otimes}a\lvert 0\rangle - \left((-1)^{\lvert \eta\rvert}\eta\mathbin{\widehat\otimes}a\lvert 1\rangle\right) \\ &= 0- \left(-\eta\mathbin{\widehat\otimes}\lvert 0\rangle\right) \\ &= \eta\mathbin{\widehat\otimes}\lvert 0\rangle. \end{align*}$$

On the other hand, multiplication by $\eta$ gives

$$\begin{align*} \eta\lvert \eta\rangle &= (M_\eta\mathbin{\widehat\otimes}I) \left(1\mathbin{\widehat\otimes}\lvert 0\rangle-\eta\mathbin{\widehat\otimes}\lvert 1\rangle\right) \\ &= \eta\mathbin{\widehat\otimes}\lvert 0\rangle-\eta^2\mathbin{\widehat\otimes}\lvert 1\rangle \\ &= \eta\mathbin{\widehat\otimes}\lvert 0\rangle. \end{align*}$$

Therefore

$$a\lvert \eta\rangle=\eta\lvert \eta\rangle.$$

This is an algebraic identity in $\mathcal G\mathbin{\widehat\otimes}\mathcal F$. It is not a statement that $a$ has an ordinary complex eigenvector in the physical Fock space.

Exponential notation for coherent states

The same one-mode ket can be written as

$$\lvert \eta\rangle=e^{-\eta a^\dagger}\lvert 0\rangle.$$

This notation is finite because $(\eta a^\dagger)^2=0$. Indeed,

$$e^{-\eta a^\dagger}=1-\eta a^\dagger.$$

Acting on the vacuum gives

$$(1-\eta a^\dagger)\lvert 0\rangle = \lvert 0\rangle-\eta\lvert 1\rangle.$$

The product $\eta a^\dagger$ is even because it is odd times odd.

For $N$ modes, define

$$\lvert \eta\rangle = \exp\left(-\sum_{i=1}^N \eta_i a_i^\dagger\right)\lvert 0\rangle.$$

The operator $\sum_i\eta_i a_i^\dagger$ is even. With the graded sign conventions above, one obtains

$$a_i\lvert \eta\rangle=\eta_i\lvert \eta\rangle.$$

Bras, overlaps, and a one-mode resolution of identity

A fully precise treatment of bras requires a convention for graded duals. This note records the common physics convention for one mode, enough to see how the Berezin integral works.

Take

$$\langle \bar\eta\rvert = \langle 0\rvert-\langle 1\rvert\bar\eta,$$

where $\bar\eta$ is an independent Grassmann generator placed on the right of the bra coefficient. Then

$$\langle \bar\eta \mid \eta'\rangle = 1+\bar\eta\eta' = e^{\bar\eta\eta'}.$$

For one pair $(\eta,\bar\eta)$, choose the integration convention

$$\int d\bar\eta\,d\eta\;\eta\bar\eta=1.$$

Equivalently,

$$\int d\bar\eta\,d\eta\;\bar\eta\eta=-1.$$

Then

$$e^{-\bar\eta\eta}=1-\bar\eta\eta=1+\eta\bar\eta.$$

With these conventions,

$$I_ \mathcal F = \int d\bar\eta\,d\eta\; e^{-\bar\eta\eta} \lvert \eta\rangle\langle \bar\eta\rvert.$$

Let us verify this directly. First

$$\lvert \eta\rangle\langle \bar\eta\rvert = \lvert 0\rangle\langle 0\rvert -\lvert 0\rangle\langle 1\rvert\bar\eta -\eta\lvert 1\rangle\langle 0\rvert +\eta\lvert 1\rangle\langle 1\rvert\bar\eta.$$

Multiplying by $e^{-\bar\eta\eta}=1+\eta\bar\eta$, the only terms contributing to the Berezin integral are those proportional to $\eta\bar\eta$. The off-diagonal terms do not contain both variables after multiplication, because an extra repeated variable gives zero. Hence

$$\begin{align*} \int d\bar\eta d\eta\;e^{-\bar\eta\eta}\lvert \eta\rangle\langle \bar\eta\rvert &= \lvert 0\rangle\langle 0\rvert + \lvert 1\rangle\langle 1\rvert \\ &=I_\mathcal F. \end{align*}$$

Warning 36 (Conventions matter). If one changes the order of $d\eta$ and $d\bar\eta$, or changes signs in the definitions of $\lvert \eta\rangle$ and $\langle \bar\eta\rvert$, corresponding signs in the overlap and identity resolution change. The mathematics is not ambiguous once all conventions are fixed.

What is essential and what is notation

The essential structures are these:

1. A one-particle Hilbert space $\mathfrak h$.

2. A fermionic Fock space

$$\mathcal F_-(\mathfrak h)=\bigoplus_{n=0}^\infty\Lambda^n\mathfrak h,$$

where $\oplus$ denotes the Hilbert direct sum. For $\dim\mathfrak h=N<\infty$, this stops at $n=N$.

3. A parity decomposition

$$\mathcal F_-=(\mathcal F_-)_{\bar 0}\oplus(\mathcal F_-)_{\bar 1}$$

by even or odd particle number.

4. An auxiliary Grassmann algebra

$$\mathcal G=\Lambda\mathrm{span}\{\eta_i,\bar\eta_i\}.$$

5. The extended super vector space

$$\mathcal G\mathbin{\widehat\otimes}\mathcal F_-.$$

6. The Koszul rule

$$(A\mathbin{\widehat\otimes}B)(C\mathbin{\widehat\otimes}D) = (-1)^{\lvert B\rvert\lvert C\rvert} AC\mathbin{\widehat\otimes}BD.$$

7. Berezin differentiation and integration, defined as algebraic operations on $\mathcal G$, acting coefficientwise on $\mathcal G\mathbin{\widehat\otimes}\mathcal F_-$.

Common confusions resolved

Is fermionic Fock space an exterior algebra?

For finitely many modes, yes: the fermionic Fock space is naturally

$$\mathcal F_-(\mathbb C^N)=\Lambda\mathbb C^N.$$

For an infinite-dimensional one-particle Hilbert space, the physical Hilbert Fock space is the Hilbert direct-sum completion

$$\mathcal F_-(\mathfrak h)=\widehat{\bigoplus}_{n=0}^\infty\Lambda^n\mathfrak h.$$

It is best thought of as the completed antisymmetric Fock space, built from exterior powers. The algebraic finite-particle subspace is the algebraic exterior algebra-like object

$$\mathcal F_-^{\mathrm{alg}}(\mathfrak h)=\bigoplus_{n=0}^{\infty,\mathrm{alg}}\Lambda^n\mathfrak h.$$

Does standard Fock space include infinite forms?

No. It includes finite $n$-particle sectors for all $n\ge 0$. A general Hilbert Fock vector may have infinitely many nonzero finite-particle components, but it is not a single infinite wedge monomial.

What exactly is parity?

There are two precise meanings:

• the parity value $\lvert x\rvert\in\mathbb Z_2$, defined only for homogeneous nonzero elements or homogeneous operators;

• the parity operator $\Pi:V\to V$, a linear map with eigenvalue $+1$ on even vectors and $-1$ on odd vectors.

What does the hat on the tensor product mean?

For spaces, $V\mathbin{\widehat\otimes}W$ is the ordinary tensor product vector space with the graded parity rule

$$\lvert v\mathbin{\widehat\otimes}w\rvert=\lvert v\rvert+\lvert w\rvert.$$

For operators or algebras, the hat also reminds us to use the Koszul multiplication rule

$$(A\mathbin{\widehat\otimes}B)(C\mathbin{\widehat\otimes}D) = (-1)^{\lvert B\rvert\lvert C\rvert}AC\mathbin{\widehat\otimes}BD.$$

What is being differentiated in Grassmann calculus?

Only the Grassmann coefficient is differentiated. If

$$\Psi=\sum_\alpha g_\alpha\mathbin{\widehat\otimes}\psi_\alpha,$$

then

$$\partial_i\Psi=\sum_\alpha (\partial_i g_\alpha)\mathbin{\widehat\otimes}\psi_\alpha.$$

The Fock vector $\psi_\alpha$ is not itself differentiated with respect to $\eta_i$.

Why does a eta equal minus eta a?

Because $a=I\mathbin{\widehat\otimes}a$ is an odd operator on the Fock factor, while $\eta=M_\eta\mathbin{\widehat\otimes}I$ is an odd multiplication operator on the Grassmann factor. The graded tensor product says odd operators from separate factors anticommute.

Minimal dictionary

Symbol	Meaning
$\mathfrak h$	One-particle Hilbert space.
$\Lambda^n\mathfrak h$	Antisymmetric $n$-particle sector.
$\mathcal F_-(\mathfrak h)$	Fermionic Fock space $\bigoplus_{n\ge0}\Lambda^n\mathfrak h$ as a Hilbert direct sum.
$\lvert 0\rangle$ or $\Omega$	Vacuum vector in $\Lambda^0\mathfrak h\cong\mathbb C$.
$a_i^\dagger$	Creation operator for mode $i$, odd.
$a_i$	Annihilation operator for mode $i$, odd.
$\Pi_\mathcal F=(-1)^{\mathcal N}$	Fermion parity operator.
$\mathcal G$	Grassmann algebra $\Lambda\mathrm{span}\{\eta_i,\bar\eta_i\}$.
$\eta_i,\bar\eta_i$	Independent odd Grassmann generators.
$\mathcal G\mathbin{\widehat\otimes}\mathcal F_-$	Grassmann-extended Fock space.
$\partial^L/\partial\eta_i$	Left Berezin derivative, an odd derivation.
$\int d\eta_i$	Berezin integral, coefficient extraction.

Final summary

Fermionic coherent-state calculus is not mysterious once the bookkeeping is made explicit:

$$\boxed{\text{fermionic Fock space is built from exterior powers,}}$$ $$\boxed{\text{Grassmann variables are generators of another exterior algebra,}}$$ $$\boxed{\text{the two are combined by a graded tensor product,}}$$ $$\boxed{\text{and Berezin calculus is coefficient extraction on the Grassmann algebra.}}$$

The signs are not arbitrary. They are exactly the signs forced by the graded flip

$$v\mathbin{\widehat\otimes}w\longmapsto (-1)^{\lvert v\rvert\lvert w\rvert}w\mathbin{\widehat\otimes}v$$

and by the corresponding Koszul multiplication rule.