# What's the space of eigenvalues/field configurations for a fermion?

In the Schrödinger picture of quantum field theory, the field eigenstates of a real scalar field $$\hat\phi(\mathbf x)$$ with $$\mathbf x \in\mathbb R^3$$ are the states $$\hat\phi(\mathbf x)|\phi\rangle=\phi(\mathbf x)|\phi\rangle$$, with field configurations $$\phi$$ in the space $$\mathbb R^3\to\mathbb R\;$$.

This field eigenbasis defines the wave-functional $$\Psi[\phi]=\langle\phi|\Psi\rangle$$, and expresses the vacuum state, free field wave-functional $$\Psi_0[\phi]=\langle\phi|\Psi_0\rangle$$ as (Jackiw 89)

$$\begin{array}{cl} \Psi_0[\phi] &= C\prod_k e^{-\omega(\mathbf k)\frac{\tilde\phi(\mathbf k)^2}{2}\epsilon^3} \\ &\to C e^{-\frac{1}{2}\int\frac{d^3k}{(2\pi)^3} \omega_{\mathbf k}|\tilde\phi(\mathbf k)|^2} \\ &= \operatorname{det}^{\frac{1}{4}}\left(\frac{K}{\pi}\right)\; e^{-\frac{1}{2}\int d\mathbf{x} \int d\mathbf{y}\, \phi(\mathbf{x}) K(\mathbf{x},\mathbf{y}) \phi(\mathbf{y}) } = \operatorname{det}^{\frac{1}{4}}\left(\frac{K}{\pi}\right)\; e^{-\frac{1}{2}\phi\cdot K\cdot\phi}.\\ \end{array}$$

Jackiw (and Symanzik 81, and Hatfield 92) also discusses the wave-functional $$\Psi[\chi]=\langle\chi|\Psi\rangle$$ of a fermionic field $$\hat\chi(\mathbf x)$$ with field eigenstates $$\hat\chi(\mathbf x)|\chi\rangle = \chi(\mathbf x)|\chi\rangle$$, but in less detail. (Though he does derive a form of the fermionic vacuum state wave-functional, $$\Psi_0[\chi]\propto\operatorname{det}^{-\frac{1}{4}}\left(\Omega\right)\; e^{\frac{1}{2}\chi\cdot\Omega\cdot \chi}$$.)

Question: In what space are the field configurations $$\chi(\mathbf x)$$?

In other words, what fills in $$\chi:\mathbb R^3\to\mathrm (\,\_\,)\,$$? (For both spin-½ and spin-3/2, and for the Majorana and Dirac cases?)

Assumptions:

Ignore UV/IR issues and assume Minkowski space is the well-defined limit of the cyclic lattice $$\mathbf x\equiv a\mathbf n, \mathbf n\in \mathbb{Z}^4_N$$ as $$(a,N)\to(0,\infty)$$.

Assume the Grassmann algebra over vector space V with basis $$\{\mathbf{e}_i\}$$ is denoted $$\Lambda\,\mathrm{V}$$, with generating elements $$\{\theta_{\mathbf{e}_i}\}$$.

• for example, the dual numbers are $$\Lambda\,\mathbb R$$, with generator $$\theta_1$$ and general element $$z=c_0+c_1\theta_1$$

• the infinite-dimensional Grassmann algebra over the complex free vector space of $$\mathbb R^3$$ is $$\Lambda\,\mathbb{C}^{\mathbb R^3}$$, with generators $$\theta_{\mathbf x}|\mathbf x\in\mathbb R^3$$ and a general element $$z=\sum_{k=0}^\infty \sum_{\{\mathbf{x}_1,\cdots,\mathbf{x}_k\}\,\subset\, \mathbb R^3} \frac{1}{k!}c_{\mathbf{x}_1\cdots\mathbf{x}_k} \theta_{\mathbf{x}_1}\cdots\theta_{\mathbf{x}_k}$$. Example functions $$f:\mathbb R^3\to\Lambda\,\mathbb{C}^{\mathbb R^3}$$ are $$f(\mathbf{x})=e^{i\mathbf{k}\cdot\mathbf{x}}\theta_{\mathbf{x}}$$ and $$f(\mathbf{x})=e^{i\mathbf{k}\cdot\mathbf{x}}\theta_{\mathbf{k}}$$ for some $$\mathbf{k}\in\mathbb{R}^3$$.

• two generators per point in $$\mathbb R^3$$ would be $$\Lambda\,\mathbb{C}^{2\mathbb R^3}$$, with generating elements $$\theta_{\mathbf x,a}|\mathbf x\in\mathbb R^3,a\in\{0,1\}$$. An example function $$f:\mathbb R^3\to\Lambda\,\mathbb{C}^{2\mathbb R^3}$$ is $$f(\mathbf{x})=e^{i\mathbf{k}\cdot\mathbf{x}}\theta_{\mathbf{x},0}+e^{-i\mathbf{k}\cdot\mathbf{x}}\theta_{\mathbf{x},1}$$.

• one generator per function $$\psi:\mathbb R^3\to\mathbb C$$ would be $$\Lambda\,\mathbb{C}^{\mathbb{C}^{\mathbb{R}^3}}$$

• I think you're on the wrong track. Grassmann 'numbers' don't take values, so there is no way to define the distinct field configurations that would appear in a wave functional. – octonion Jul 30 '19 at 21:47
• Nope. Grassmann "numbers" are elements of a complex exterior algebra, and take values. You might be confusing them with their generators. – alexchandel Jul 30 '19 at 21:54
• I suspect you are the one confusing something. Back up a little. Take a single component of a fermion field at a given point of spacetime and define its eigenvalue as Grassmann number. You can form an exterior algebra with other components and other points of spacetime, but that eigenvalue itself doesn't take distinct values. This is not like the bosonic case. How are you going to define a wave functional like this? – octonion Jul 30 '19 at 21:58
• I might be, I mirror the definitions that Wolfram, PlanetMath, and WP (and its sources) use to define the Grassmann algebra, with values $z\in\Lambda(V)$. Are they all wrong? – alexchandel Jul 30 '19 at 22:28
• Using the above notation, a simple counterexample to the notion there are no distinct functions yielding Grassmanns is $\chi:\mathbf{R}^3\to\Lambda(\mathbb{C}^{\mathbb R^3})$ where $\chi(\mathbf x)=e^{i \mathbf{k}\cdot\mathbf{x}} \theta_{\mathbf x}$ with $\mathbf{k}\in\mathbb R^3$, but there are infinitely more. – alexchandel Jul 30 '19 at 22:30