The h Bar

Bohemian relativist

00:27

Are all partition functions with gauge freedom in Quantum Field Theory divergent?

bolbteppa

01:33

@ACuriousMind this Majorana existence thing is proven on page 10 here, I think your argument about duals and different spaces and it appearing unnatural is completely answered by their use of Schur's Lemma at the beginning, if you see some flaw with I'd love to see it

bolbteppa

01:52

@Charlie one natural way to get a Clifford algebra is by building up (a finite group of) Hermitian operators from the generators of certain finite-dimensional Hilbert spaces, but the Dirac equation idea is also (super!)-natural

@Bohemianrelativist you mean the qft version of a statistical mechanics partition function, or a path integral or something else

Bohemian relativist

02:38

@bolbteppa I mean the path integral $\int [dA]e^{iS[A]}$. The lecturer calls it a partition function.

bolbteppa

02:49

Well I don't know about all but it's probably good to think of e.g. Maxwell theory, the action can be written as $S = \int d^4 x A_{\mu} P^{\mu \nu} A_{\nu}$, this thing vanishes over a whole orbit of configurations so $\int [dA] e^{iS[A]} = \int [dA] = \infty$ integrating over these configurations alone

ACuriousMind

10:02

@bolbteppa The problem is not that I think the "unnatural" arguments are not proofs, it's that they appear to usually be not sure about what they're actually proving - that's why I cited three different texts that all claim to be showing "when Majoranas exist", but they all disagree with each other! What you cited there is equivalent to Fecko.

the answer I accepted explains these discrepancies satisfactorily - which is why I accepted it, after all

the paper by mike stone in the comments is also great because it discusses Wick rotation much more carefully than most other things I know w.r.t. spinors

B. Brekke

12:42

@ACuriousMind Sorry, I am not sure I am able to follow. I will try to restate my initial question. Consider a group of symmetries $G$ commuting with the Hamiltonian $H$. Using projection functions, I can construct basis functions for the irreps of $G$. Are all these basis functions eigenfunctions of $H$?

ACuriousMind

@B.Brekke If each distinct irrep occurs only once, then yes. Otherwise, not necessarily.

Take the hydrogen atom - we have a $\mathrm{SO}(3)$ symmetry, and each irrep with angular momentum $\ell$ only occurs once. All the spherical harmonics (=basis functions of your irreps) are eigenfunctions of $H$. But take the trivial symmetry, and you have just one irrep (the trivial rep) occuring infinitely often, there was can't say anything.

B. Brekke

@ACuriousMind When you say "irrep occurs only once", where is it occuring? In the Hamiltonian?

ACuriousMind

@B.Brekke in your space of states

B. Brekke

@ACuriousMind Okay, but in the hydrogen atom I have many states with angular momentum $\ell$, but they have distinct radial quantum number $n$?

ACuriousMind

@B.Brekke ah, you're right - the hydrogen atom is not an example

it becomes one if we fix $n$

B. Brekke

12:52

@ACuriousMind Still, this is very helpful. And I realize I am in the situation where I cannot claim that my basis functions are energy eigenfunctions.

Bohemian relativist

13:03

@bolbteppa what does $P^{\mu\nu}$ here mean?

Bohemian relativist

13:35

I think the action for the Maxwell theory is $S=\int dx^4 F_{\mu\nu}F^{\mu\nu}$

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Bohemian relativist

@bolbteppa does functional integral $\int [dA]$ always diverge?

Nihar Karve

@Bohemianrelativist $P^{\mu\nu}$ is $\frac12(\partial^2 g^{\mu\nu}-\partial^\mu\partial^\nu)$

Just integrate $-\frac14 \int d^4x \ F_{\mu\nu}F^{\mu\nu}$ by parts

Bohemian relativist

@NiharKarve OK, thank you, I just made that calculation several days ago in a lecture note to derive the propagator of photon, though I don't know why the propagator of photon is derived that way.

Nihar Karve

I suppose the naïve functional integration will diverge if the gauge transformation has a parameter space that is not discrete and finite

is gauge theory with finite gauge groups a thing?

Slereah

13:52

I don't think "finite gauges" are usually a problem?

ie $\mathbb{Z}_2$ invariant lagrangians aren't usually considered gauge

At worst you're just overcounting fields twice

ACuriousMind

it depends on your notion of "gauge theory" whether or not there are finite gauge groups :P

Slereah

I think having finite gauge, the overcounting would just cancel out from the normalization?

ACuriousMind

but it is common to consider theories where a U(1) gauge theory gets the global symmetry broken into some discrete $\mathbb{Z}_n$ - these still have some of the character of a gauge theory

Nihar Karve

@Slereah seems plausible

Slereah

14:12

A dumb argument, but I assume that if you do subsets of the configuration space, the Z2 symmetry will be something like $$\int_{C} = \int_{+C} + \int_{-C}$$

Assuming the action invariant, this just means twice the value of one sector

So if the numerator is invariant under that symmetry, I assume the two cancel out

Bohemian relativist

15:02

@NiharKarve but when we do the space integral $\int dx^4$, is the integral $\int dx^4$ always finite?

ACuriousMind

@Bohemianrelativist of course not - $\int 1 \mathrm{d}^nx$ diverges. For some functions $f$ the integral $\int f(x)\mathrm{d}^nx$ is finite, for others it is not.

Bohemian relativist

the functional integral is new to me. I don't know its difference from space integral.

ACuriousMind

the problem is that functional integration is a different kind of integration entirely - it is not a well-defined Riemann or Lebesgue integral, in most cases that physicists look at, it is mathematically not well-defined at all

usually physicists think about it as the limit of some integral on a lattice, but this limiting procedure is also not really well-defined in general

you should start by trying to understand the path integral in quantum mechanics, where we just integrate over one-dimensional paths $q(t)$ instead of fields on a higher-dimensional spacetime

most texts will derive this path integral precisely by a limiting procedure of the kind I alluded to above, and then wave their hands to "hope" that this generalizes to arbitrary fields. It doesn't, really, but amazingly physicists have figured out how to manipulate these integrals nevertheless in a way that yields useful results

Slereah

You just have to combine the parts converging to zero and the parts converging to infinity in pleasant ways

Although there are proper ways to do them if you want

but then it is less pleasant

it's all Polish spaces and whatnot

Bohemian relativist

15:27

@ACuriousMind $\int 1d^nx$ converges when the integral scope is finite, doesn't it?

ACuriousMind

if by "scope" you mean the set/volume we're integrating over, sure, but usually if you don't write any limits there it is implied you're integrating over all of $\mathbb{R}^n$

Bohemian relativist

@ACuriousMind I studied path integral in the Quantum Mechanics course with the book authored by Sakurai, but I have the impression that topic was not taught seriously.

so I have no impression of how that is calculated.

so I may need to check that book again to see if there is that calculation.

but I guess in QFT that's a serious topic.

bolbteppa

20:17

@ACuriousMind right, there's (at least) two general points in your post, one at the beginning concerns about duals transposes and a canonical choice etc, and another (obviously related) about different claims regarding $(p,q)$'s, my only point is whether you agree the use of Schur's Lemma in the paper I linked addresses the first concern

ACuriousMind

20:32

@bolbteppa no, because they're doing exactly what I object to - they're using complex conjugation on the space the $\gamma$ are represented on, and in the end charge conjugation turns out to be a random product of this "$B$" with $\gamma^0$. Sure, you "can show" the result doesn't actually depend on the choice of representation or conjugation, but when I say "elegant" I'm looking for an argument that doesn't need such choices to begin with

bolbteppa

21:01

I'm just not sure this is a valid concern, lets see:

1: In even dimensions, the $2n$ $\gamma_{\mu}$'s generate a finite group $C(\gamma) = \{\pm I, \pm \gamma_{\mu},... \}$ of order $2^{2n+1}$.

2: This group has $2^{2n}$ one-dimensional irreps, and only one non-trivial irrep of dimension $2^{n}$.

3: This implies that an irreducible matrix representation $\Gamma$ of $C(\gamma)$ is given by $2^n \times 2^n$ matrices.

4: The matrices $\Gamma(\gamma_{\mu})$ must satisfy the Clifford algebra $\{\Gamma(\gamma_{\mu}),\Gamma(\gamma_{\nu}) \} = 2 \eta_{\mu \nu} I$, indeed any represenation of the group $C(\gamma)$ must satisfy this.

5: Another collection of matrices, $\Gamma(\gamma_{\mu})^*$, the literal complex conjugates of the entries of the matrices $\Gamma(\gamma_{\mu})$, also provides a representation of $C(\gamma)$.

6: We can be absolutely sure of this because we know $\{\Gamma(\gamma_{\mu})^*,\Gamma(\gamma_{\nu})^* \} = \{\Gamma(\gamma_{\mu}),\Gamma(\gamma_{\nu}) \}^* = (2 \eta_{\mu \nu} I)^* = 2 \et…

7: Since it's also given by $2^n \times 2^n$ matrices, it also provides an irrep.

8: By Schur's lemma, since we have two irreps of the same group, these must be equivalent irreps, thus an invertible matrix $B$ exists such that $\Gamma(\gamma_{\mu})^* = B \Gamma(\gamma_{\mu}) B^{-1}$.

9: From $\Gamma(\gamma_{\mu}) = [\Gamma(\gamma_{\mu})^*]^*$ we can again use Schur's Lemma to say $B^* B$ is a multiple of the identity, and basically can define Majorana's now (see the ref above).

10: The rest of the proof on existence in certain dimensions uses the same ideas with transposes and adjoints.

11: Nowhere is there any concern about duals or complex conjugates acting on different spaces because we're talking about finite group representation theory here.

(For the group stuff in 1 to 3, see the Hilbert space thing I linked to above, it's explained in the 3 pages or so around there)

ACuriousMind

@bolbteppa Once again, I'm not saying their argument is wrong. It's a valid way to show what they want to show. It just doesn't have the kind of elegance I'm looking for. The inelegant part is that even if there is only one isomorphism class of irreps, there are potentially many different concrete choices of $\gamma^\mu$ (and hence $\gamma^\ast$).

You need to fix one to make this argument about conjugation, then argue this choice doesn't matter - that's possible, but that's precisely what I don't want to do

bolbteppa

I don't see where I fixed a basis for the matrices $\Gamma(\gamma_{\mu})$ anywhere, I used the word matrices but one can just say the word operator, I don't see how taking complex conjugates in any way forces one to fix a basis here

ACuriousMind

@bolbteppa The moment you write $(\gamma^\mu)^\ast$ you implicitly pick a specific representation

for instance, for dimensions in which Majoranas actually exist, you have a representation where the $\gamma^\mu$ have real entries and $B=1$

but you also have representations where the $\gamma^\mu$ have complex entries, and $B\neq1$

It's not about picking a basis, it's that "there is only one irrep" means that there is only one irrep up to isomorphy - it doesn't mean that $\gamma^\mu$ "in that representation" denotes a unique operator

bolbteppa

If it's not about a basis, then maybe you mean: We began from an arbitrary irreducible representation $\Gamma$ of $C(\gamma)$, then argued that $\Gamma^*$ is an equivalent irreducible representation - are you saying that this 'implicitly' picks the arbitrary $\Gamma$ we started from as if this is some inelegant choice? Schur's Lemma obviously says $\Gamma^*$ is equivalent to the $\Gamma$ so it's unavoidable to relate $\Gamma^*$ to $\Gamma$.

ACuriousMind

21:20

@bolbteppa Abstractly, you have that conjugation is a map $V\to \bar{V}$. Schur's lemma gives you that there is an invertible map $B: V\to \bar{V}$. In order to pretend that $\gamma^\mu$, $(\gamma^\mu)^\ast$ and $B$ live all on the same vector space, you have to pick a different isomorphism $V\to \bar{V}$ on the underlying vector space that identifies the space with its conjugate.

This isomorphism is uncanonical - it always exists, but it is not unique, you need to pick a basis in $V$ to define it, and depending on how you pick it, the $\gamma$ e.g. end up with real entries or not, and so "$B$" (really $B$ concatenated with the uncanonical isomorphism) interpreted as a map $V\to V$ ends up being the identity or not

when the paper you cited says things like "one can show this is independent of the choice of representation", in my terms above this means one must show this is independent of the choice of the uncanonical isomorphism

bolbteppa

I'm not sure about that, are we not doing conjugation on the space of representations $\{ \Gamma \}$ of a finite-dimensional group $G$. Here $\Gamma : G \to \mathrm{GL}(V)$ is one element of this space, the complex conjugate is just another map $\Gamma^* : G \to \mathrm{GL}(V)$. This ensures all the maps $\Gamma(\gamma_{\mu})$, $\Gamma^*(\gamma_{\mu})$ etc... still all act on the same vector space $V$.

Schur's lemma then says operations like complex conjugation, transposition or taking an adjoint canonically associates two irreps to one another. Similarly the act of conjugation etc on $B$ is on the space of intertwiners, it seems to have nothing to do with the vector space $V$ in $\mathrm{GL}(V)$ that these operators act on, $\Gamma(\gamma_{\mu})$, $\Gamma(\gamma_{\mu})^*$, $\Gamma(\gamma_{\mu})^T$, $B$, $B^*$,... all live "on the same vector space".

bolbteppa

21:44

To me the ugly thing about all this is the sign in $\gamma_{\mu}^T = - C \gamma_{\mu} C^{-1}$, obviously the sign makes sense thinking of the Dirac equation, (I think a dimensionality argument might also make it plausible to take this choice rather than $+$ but maybe misremembering that), but in general both choices are possible

Anytime a minus sign arises, e.g. Faraday's law, it causes endless grief, whereas a $+$ sign in some random definition goes unnoticed...

ACuriousMind

@bolbteppa To define this kind "complex conjugation", you need to pick a basis on $V$. Maybe a simpler example helps to show why: Consider $\mathbb{C}^2$. I pick a basis and the matrix that is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ on this vector space. Let's call the abstract operator represented by this matrix $A$.

If you define conjugation w.r.t. this basis, obviously $A^\ast = A$. But now I switch to the basis $e_1 \pm \mathrm{i}e_2$. In this basis, the matrix is $\begin{pmatrix}0 & -\mathrm{i}\\\mathrm{i} & 0 \end{pmatrix}$, and so $A^\ast \neq A$

That is, there is no basis-independent notion of complex conjugation of operators. You can only conjugate matrices, or you conjugate operators, but then the operators are between two conceptually different spaces

(on a completely unrelated note, this is also why introducing the Hermitian conjugate ${}^\dagger$ as the concatenation of the transpose and the complex conjugate is bad style, because the latter two are basis-dependent, but the former has a manifestly basis-independent definition in terms of an inner product)

bolbteppa

23:02

Yeah I agree it's not trivial to bat away all concerns with this which is why it's interesting, I'm not even more confused so have to think about this more first...

I think I'm just not appreciating how 'strict' you're being in wanting a 'canonical' definition where canonical I guess means basis independent in the strictest sense

20

A: basic difference between canonical isomorphism and isomorphims

Great question. Canonical is more a term of art than a word with a strict mathematical definition. It's sometimes used as a synonym for “natural” or “obvious,” although natural is yet another idiom and obvious is in the eye of the beholder. You might think of it as meaning independent of any c...

At the end of the day the result is about one basis existing in which the matrices are real so in a sense is a 'canonical' result not impossible when this is your end-goal, especially when there's concerns even about the existence of a canonical complex conjugation in and of itself :p

ACuriousMind

@bolbteppa I give a basis-independent definition of Majorana spinors in the question - the existence of a $\mathfrak{so}(p,q)$-equivariant real form on the representation space.

bolbteppa

Right yeah, I'm not sure if that's a 'correct' definition :p

Mark Giraffe

I made something.

bolbteppa

The problem with this is it really boils down to those matrix elements being real and I can't see how that doesn't inherently force the use of a basis somewhere thus making things noncanonical

ACuriousMind

@bolbteppa it implies the existence of a basis in which the matrices are real since you can use the real form to get a real vector space to which you can restrict the representation as the eigenspace of the real form with eigenvalue 1

bolbteppa

23:18

Unless you try to say this is a special case of the more abstract definition in your post, but this is blocked by the fact there is no canonical complex conjugation that canonically reproduces this condition about the matrix elements being real in at least one basis

ACuriousMind

Also, I didn't make this definition up, it's what O'Farrill uses in his notes.

Mark Giraffe

What do you think of the header?

ACuriousMind

@MarkGiraffe I'm not sure what I'm looking at.

bolbteppa

So if one just fixes a basis from the get-go, even though the basis is arbitrary, and then considers all things like complex conjugates and transposes on this matrix, you get to just ignore any questions about the inner product, the transposed matrix for each matrix in the representation is the operator in the space of representations whose matrices are the transposes of the original matrix in that basis, and that's that.

user 85795

23:53

The image of the logo is a bit fuzzy @MarkGiraffe

The color scheme is cool :-)

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