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S Jul 15, 2022 at 9:30 history bounty ended Quantumwhisp
S Jul 15, 2022 at 9:30 history notice removed Quantumwhisp
Jul 9, 2022 at 9:30 comment added Blind Miner Have you read Appendix A.2 of Polchinski's String Theory, Volume 1? It addresses fermionic field configurations and path integrals, but I don't know if it meets your standard of rigor. Maybe give it a try.
Jul 8, 2022 at 13:05 answer added Qmechanic timeline score: 3
Jul 8, 2022 at 8:25 comment added Qmechanic Links to abstract pages for Jackiw 89, Symanzik 81 and Hatfield 92? Which pages?
S Jul 8, 2022 at 8:13 history bounty started Quantumwhisp
S Jul 8, 2022 at 8:13 history notice added Quantumwhisp Draw attention
Aug 24, 2019 at 5:25 history edited alexchandel CC BY-SA 4.0
sign error, lowercase fields
Aug 21, 2019 at 5:02 history edited alexchandel CC BY-SA 4.0
add ref
Aug 17, 2019 at 3:00 history tweeted twitter.com/StackPhysics/status/1162559765958090752
Aug 14, 2019 at 6:21 history edited alexchandel CC BY-SA 4.0
cite Jackiw for fermionic wave-functional
Aug 1, 2019 at 15:37 history edited alexchandel CC BY-SA 4.0
more examples
Aug 1, 2019 at 8:44 history edited Qmechanic
edited tags; edited tags
Aug 1, 2019 at 8:43 comment added alexchandel WP, PlanetMath, Wolfram, & nLab and their sources all disagree with you. Jackiw literally defines a fermionic wave-functional (as does Symanzik). And saying "$\theta_{x,0}$ takes no values" is as vacuous as saying "the Euclidean basis vector $\mathbf{e}_1$ takes no values;" no one ever suggested otherwise.
Aug 1, 2019 at 8:14 history edited alexchandel CC BY-SA 4.0
organize
Aug 1, 2019 at 7:57 history edited alexchandel CC BY-SA 4.0
cite Jackiw for fermionic wave-functional
Aug 1, 2019 at 6:31 history edited alexchandel CC BY-SA 4.0
minor
Aug 1, 2019 at 6:19 history edited alexchandel CC BY-SA 4.0
reword
Aug 1, 2019 at 0:37 comment added octonion I said Grassmann 'numbers' have no values. Something like $\theta_{x,0}$ appears in integrals much like a bosonic $\phi_x$ would. The difference is $\phi_x$ will take different values depending on the state so a functional makes sense. $\theta_{x,0}$ is already a member of the Grassmann algebra, it does not take new values. This is my last reply, good luck with your painstaking rigor
Jul 31, 2019 at 9:02 comment added alexchandel Is $\chi(\mathbf x)$ a field configuration, as in $\mathrm X(\mathbf x)$? I'd need my question answered to say.
Jul 31, 2019 at 9:01 comment added alexchandel $\chi(\mathbf x)=e^{i \mathbf{k}\cdot\mathbf{x}} \theta_{\mathbf x}$ isn't in $\mathbb{R}^3\to\mathbb C$; it's in $\mathbb{R}^3\to\Lambda(\mathbb{C}^{\mathbb R^3})$, specifically $\mathbb R^3\to\Lambda_1(\mathbb{C}^{\mathbb R^3})$. There are many other possibilities, e.g. $\chi:\mathbf{R}^3\to\Lambda(\mathbb{C}^{2\mathbb R^3})$ where $\chi(\mathbf x)=e^{i \mathbf{k}\cdot\mathbf{x}} \theta_{\mathbf x,0} + e^{-i \mathbf{k}\cdot\mathbf{x}} \theta_{\mathbf x,1}$. Painstaking rigor is needed in part because of all these superstitions about Grassmanns, like "Grassmann algebras have no values."
Jul 31, 2019 at 8:15 history edited alexchandel CC BY-SA 4.0
extend notation for the struggling
Jul 31, 2019 at 7:35 history edited alexchandel CC BY-SA 4.0
note emphasizing question
Jul 30, 2019 at 22:57 comment added octonion Is your $\chi$ ever a field configuration in the sense that it is like an eigenvalue associated with a state? You essentially turned the field configurations into a map from $R^3\rightarrow C$ by multiplying each $\theta_x$ by a complex number. But by using only $\theta_x$ at each $x$ I think you're starting to see what I'm getting at. Please think about it a little, sorry if I was a little harsh in my comment.
Jul 30, 2019 at 22:30 comment added alexchandel 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.
Jul 30, 2019 at 22:28 comment added alexchandel 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?
Jul 30, 2019 at 22:04 history edited alexchandel CC BY-SA 4.0
cleanup
Jul 30, 2019 at 21:58 comment added octonion 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?
Jul 30, 2019 at 21:54 comment added alexchandel Nope. Grassmann "numbers" are elements of a complex exterior algebra, and take values. You might be confusing them with their generators.
Jul 30, 2019 at 21:48 history edited alexchandel CC BY-SA 4.0
cleanup
Jul 30, 2019 at 21:47 comment added octonion 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.
Jul 30, 2019 at 21:44 history edited alexchandel CC BY-SA 4.0
shorten title, cleanup
Jul 30, 2019 at 21:28 history edited alexchandel CC BY-SA 4.0
shorten title
Jul 30, 2019 at 19:41 history asked alexchandel CC BY-SA 4.0