AccidentalFourierTransform

so you cannot patch up your local choices consistently along the entire manifold

locally you can declare pretty much whatever you want, but generic manifolds will have global obstructions

@Jagerber48 you cant: not every manifold admits every structure. For example, CP1 x CP1 admits spinors but CP2 does not. In general it is a non-trivial question whether a given manifold admits spinors at all

(2/2) Here, your "natural" notion of vector (namely, a GL(7) vector bundle in the 7-dim rep) is fine but very incomplete: you can get much more information about the manifold if you work with G2 vector bundles instead. But there is no canonical notion of "vector" here, there are actually two fundamental representations. So which one would you call a "vector"? We have to be knee deep in rep theory to even ask this question!

@Jagerber48 Unfortunately, your definition of (physics) vector does require representation theory, it's just that the required rep theory is so basic that you are used to it, and it seems too obvious to be relevant. So to make things a little more interesting and less obvious, I invite you to think what it means to have a vector in a G2 (or other special holonomy) manifold. (1/2)

@Jagerber48 There is representation theory all over your two comments. You may choose not to use the words representation theory, but it is 100% there. If you want to use geometric language, you could say: a metric gives you a canonical restriction of the structure group $GL(n)\to O(n)$. And $O(n)$ is not simply connected so there is a canonical cover $Pin(n)$ which is simply connected. A spinor is an element of the $Pin(n)$ bundle, i.e., there is a 2-to-1 canonical map to $TM$. This is of course just a rephrasing of the rep theory statement about reps of O(n) vs Pin(n). It is equivalent.

thx

i forgot about the badge mechanic thing and insta-nuked this post, could someone please double-check? (its a new user, we should be extra careful...)

if only we could build string bombs

"how the strong force has changed civilization" i'm sure the japanese might have something to say about this :-)

@satan29 ah yes, lubos, the #NotLikeTheOtherGirls of white conservative men

sorry this last year i became a germanphobe

AAAAAAAAAAAAAAAAAAAAA

btw does anyone have any idea what's going on here:

arnold is just 10 answers from his, so he should be getting it in no time

whoo :-)

Thanks!

I fear reviewers will most likely miss the context and vote as looks-fine

should it be flagged as not-an-answer?

and the rest is just a list of (irrelevant) references

the first sentence is very much false (and it does not answer the issue in the OP)

hi all, I would appreciate some feedback on this answer: physics.stackexchange.com/a/608256/84967

at least those of he who shall not be named

it would be great if several 10k's go though the posts of, er... problematic users, and vote to delete other non-answers

this might be a good time to do some further clean up

i was going to point out that we probably should, given that the post made it into the arxiv

i see the terrible answer here was finally deleted, good!

done and done :-)

cheers

happy new year everyone!

@Xnero awesome, thanks :-)

(In case it is relevant: I am a Spanish citizen, I have two covid shots that I got in Spain, so it is EU certified)

would this be something I can ask on main? or is it off-topic?

basically, I find conflicting information online for whether I need a PCR test to fly from Canada to Spain

hi y'all, I've got a question but I'm not sure if it is appropriate for main

↑↑↑

@MadMax 1) Yes. In $d=1$ the minimal spinor is one-dimensional, so $\chi$ is just a single operator. In $d\ge2$ the minimal spinor has more components, so all formulas acquire a spinor index, so $\{\chi_{i,\alpha},\chi_{j,\beta}\}=i\delta_{ij}C_{\alpha\beta}$ where $C$ is a matrix in spinor space (that depends on the number of spacetime dimensions and the basis of gamma matrices). 2) Yes, it is a hermitian operator, the classical $\psi^*=\psi$ lifts to $\psi^\dagger=\psi$ (in the Majorana basis; in other bases you get a matrix in spinor space). 3) Yes, it is a fermionic operator.

@MadMax It means that, as an operator, one has $\chi_1^2=1/2$ (or $\chi_1^2=1$, again depending on conventions). This is in contrast to the classical $a$-number result $\chi_1^2=0$. The Majorana operator does not square to zero, rather to the identity. This property is behind every "weird" behaviour of Majorana modes, from cond-mat to string theory.

@MadMax Sure. The canonical definition [not MY definition] of the Dirac-Poisson bracket (see wikipedia link above) is $\{a,b\}_D=\{a,b\}-\sum_{ij}\{a,\ell_i\}\{\ell,\ell\}^{-1}_{ij}\{\ell_j,b\}$, where $\{\cdot,\cdot\}$ denotes the standard Poisson bracket $\{a,b\}=\sum_i \frac{\partial a}{\partial\psi_i}\frac{\partial b}{\partial \pi_i}+(-1)^{|a||b|}(a\leftrightarrow b)$. It is not equal to $\chi_1\chi_2-\chi_2\chi_1$. I already explained above that $\{\chi_i,\chi_j\}=i\delta_{ij}$, so if you want it spelled out, one has $\{\chi_1,\chi_1\}=\{\chi_2,\chi_2\}=i$ and $\{\chi_1,\chi_2\}=0$

@MadMax 1) Yes, as long as they are both Majorana and $\chi_1\neq\chi_2$. 2) That they anti-commute. So (as operators!) $\chi_1\chi_2=-\chi_2\chi_1$.

@MadMax I brought it up, but you wrote $\{\chi_1,\chi_2\}=\chi_1\chi_2-\chi_2\chi_1$. I never wrote this, this expression is a commutator, which is how it is defined in the quantum theory. In the classical theory the symbol $\{a,b\}$ does not denote $ab\pm ba$. [It denotes something like $\frac{\partial a}{\partial \psi}\frac{\partial b}{\partial \pi}+(a\leftrightarrow b)$ plus corrections due to the constraint]. In my answer, $\{\cdot,\cdot\}$ denotes a bracket, and as a bracket one has $\{\chi_i,\chi_j\}=i\delta_{ij}$ (or perhaps $2i\delta_{ij}$ depending on conventions).

@MadMax What is $\{\cdot,\cdot\}$ here? Classical Dirac bracket? Or anti-commutator? Are we doing classical field theory or quantum field theory?

@MadMax The bracket is really a Dirac-Poisson bracket because the system is constrained. But, long story short, the basic idea is that the conjugate momentum is $\pi=\frac{\partial L}{\partial\dot\psi}=i\psi^\dagger$. Therefore $\{\pi,\psi\}=1$ implies $\{\psi^\dagger,\psi\}=i$. The full story requires resolving constraints and so on, but this doesn't change the main conclusion. [Note that when you quantize, you replace the Dirac bracket by $i$ times the anti-commutator, so the $i$ effectively disappears, and you get the standard Clifford algebra]

@MadMax "Hermitian" is a notion defined on operators, and the first part of my answer covers that. "Real" is a notion defined on scalars, and the second part of my answer covers that. I wholeheartedly agree that the discussion is impossible if we disagree on basic definitions, but here the burden is on you. What is your definition of "hermitian" for a scalar? How is it different from "real"? I don't see this difference explained in the OP, did I miss it?

No, I am not talking about an expectation value. I am talking about an operator. More precisely, the matrix representation of the operator in a given basis. It just so happens that this basis is an eigenbasis so the diagonal elements coincide with the expectation values. But this is not true in other bases.

@Forge It is not a valid concept. Think of point-particle mechanics. Classically, you have a position $x(t)$ that satisfies $\ddot x=F$. Quantum-mechanically, you have an operator $\hat x(t)$ that satisfies $\ddot{\hat x}=\hat F$. It makes no sense to claim that "Quantum fluctuations have an effect on $x$, in that the it becomes higher at some instants in time and become lower at some other instants in time". It is just a meaningless statement. Now replace "position" with "field". The statement is just as meaningless when applied to $x(t)$ as it is when applied to $\phi(t,x)$.

@parker neither; see en.wikipedia.org/wiki/Distribution_(mathematics)