Timeline for Why coupling constants with negative mass dimensions lead to non-renormalizable theories?
Current License: CC BY-SA 4.0
49 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2023 at 11:49 | comment | added | Qmechanic♦ | 30. Relation to resistor networks mathoverflow.net/q/364948 $\alpha_e=$resistance in an edge $e$. $\alpha^{-1}_e=$conductance in an edge $e$. Ohm's law $\quad-j_a={\cal A}^{-1}_{ab}{\cal E}_{b,v} U_v$; current conv $\quad {\cal E}_{u,a}j_a=\delta_{u,v}-\delta_{u,w}$; $\quad e_v={\cal L}^{\prime}U^{\prime}$; Resistance $\quad R_{vw}=\frac{\det({\cal L}^{v,w}_{v,w})}{\det({\cal L}^{w}_{w})}$ from Cramer's rule; | |
| Oct 4, 2023 at 11:49 | comment | added | Qmechanic♦ | $\quad G\backslash\gamma$ IR-irreducible subgraph of $G\Leftrightarrow G\backslash\gamma$ IR and $G/\gamma$ 1VI. $\quad\gamma$ motic$\Leftrightarrow\forall\mu$ proper subgraph that is $p$-connected and $mp$-cover of $\gamma: L(\mu)<L(\gamma)$. | |
| Oct 3, 2023 at 11:48 | comment | added | Qmechanic♦ | $G_m$=subgraph of all massive propagators in $G$ (with nec. vertices). $G_p$=subgraph of all external momentum vertices in $G$. Let $\gamma\subseteq G$ be a subgraph. $\quad\gamma$ $m$-cover of $G\Leftrightarrow\gamma\supseteq G_m$. $\quad\gamma$ $p$-cover of $G\Leftrightarrow\gamma\supseteq G_p$. $\quad\gamma$ $p$-connected$\Leftrightarrow\gamma_p\subseteq$ conn. comp. of $\gamma$. $\quad G\backslash\gamma$ IR subgraph of $G\Leftrightarrow\gamma$ $mp$-cover of $G$ and [each conn. comp. of] $\gamma/(G_m\cup G_p)$ is 1PI. (Hm. Speculate whether to replace $G\backslash\gamma$ with $G/\gamma$?) | |
| Oct 2, 2023 at 8:32 | comment | added | Qmechanic♦ | And then $t=T/\gamma$ is spanning $(n^"+1,\ell^")$-subgraph $G/\gamma$ with $V(t)=V(G/\gamma)$; $n^"+1=|\pi_0(t)|\leq|\pi_0(T)|=n+1$; $0\leq L(t)=\ell^"$; $E(\tau)=V(G/\gamma)+\ell^"-n^"-1$. $\quad V(T)=V(\tau)-|\pi_0(\gamma)|+V(t)$; $\quad E(T)=E(\tau)+E(t)$; $\quad0\leq\ell^"-\ell+\ell'=L(t)-L(T)+L(\tau)=|\pi_0(\tau)|-|\pi_0(\gamma)|-|\pi_0(T)|+|\pi_0(t)|=n^"-n+n'$; $\Box$ | |
| Oct 2, 2023 at 8:31 | comment | added | Qmechanic♦ | Proof: $\quad|\pi_0(G/\gamma)|=1$; $\quad V(G)=V(\gamma)-|\pi_0(\gamma)|+V(G/\gamma)$; $\quad E(G)=E(\gamma)+E(G/\gamma)$; $\quad L(G)=L(\gamma)+L(G/\gamma)$; Let $T\subseteq G$ be a spanning $(n+1,\ell)$-subgraph with $V(T)=V(G)$; $|\pi_0(T)|=n+1$; $L(T)=\ell$; $E(T)=V(G)+\ell-n-1$. Then $\tau=T|_{\gamma}$ is a spanning $(n'+|\pi_0(\gamma)|,\ell')$-subgraph in $\gamma$ [since $V(\tau)=V(\gamma)$ all vertices in $\gamma$ belongs to $\tau$] with $n'=|\pi_0(\tau)|-|\pi_0(\gamma)|\geq 0$; $0\leq L(\tau)=\ell'\leq\ell$; $E(\tau)=V(\gamma)+\ell'-n'-|\pi_0(\gamma)|$. | |
| Sep 24, 2023 at 15:13 | comment | added | Qmechanic♦ | ${\cal K}_{00}(G)-{\cal K}_{00}(\gamma){\cal K}_{00}(G/\gamma)={\cal O}\left((\alpha_{G/\gamma}^{-1})^{V(G/\gamma)}\right)$ only contain less powers of $\alpha_{\gamma}^{-1}$ /more powers of $\alpha_{G/\gamma}^{-1}$. $\quad{\cal U}_{00}(G)-{\cal U}_{00}(\gamma){\cal U}_{00}(G/\gamma)={\cal O}\left(\alpha_{\gamma}^{L(\gamma)+1}\right)$; $\quad{\cal F}_{00}(G)-{\cal U}_{00}(\gamma){\cal F}_{00}(G/\gamma)={\cal F}_{00}(\gamma){\cal U}_{00}(G/\gamma)+{\cal U}_{10}(\gamma){\cal F}_{01}(G/\gamma)+{\cal O}\left(\alpha_{\gamma}^{L(\gamma)+2}\right)={\cal O}\left(\alpha_{\gamma}^{L(\gamma)+1}\right)$; | |
| Sep 24, 2023 at 14:41 | comment | added | Qmechanic♦ | Assume $G$ connected $|\pi_0(G)|=1$ while $\gamma\subseteq G$ not nec. conn. and might have loops. Then $\quad {\cal K}_{n,0}(G)=\sum_{n',n^"\in\mathbb{N}_0}{\cal K}_{n',0}(\gamma){\cal K}_{n^",n'+n^"-n}(G/\gamma)$; $\quad\bar{\cal K}_{n,\ell}(G):=\sum_{m\in\mathbb{N}_0}{\cal K}_{n-m,\ell-m}(G)$ $=\sum_{n',n^",\ell',\ell^"\in\mathbb{N}_0}^{n-\ell=n'-\ell'+n^"-\ell^"}{\cal K}_{n',\ell'}(\gamma){\cal K}_{n^",\ell^"}(G/\gamma)$ $=\sum_{m\in\mathbb{N}_0}\bar{\cal K}_{m,\ell}(\gamma)\bar{\cal K}_{n,m}(G/\gamma)$ where $\ell'\leq\ell, n^"\leq n$; UV:$\alpha_{\gamma}\ll1$. IR:$\alpha_{G/\gamma}\gg1$. | |
| Sep 24, 2023 at 11:17 | comment | added | Qmechanic♦ | $\quad{\cal K}(G)=\alpha_e^{-1}\theta(e\!\in\!G){\cal K}(G/e)+{\cal K}(G\backslash e)$; $\quad{\cal U}(G)=\theta(e\!\in\!G){\cal U}(G/e)+\alpha_e{\cal U}(G\backslash e)$; $\quad{\cal F}_0(G)=\theta(e\!\in\!G){\cal F}_0(G/e)+\alpha_e{\cal F}_e(G\backslash e)$; $\quad\det({\cal L}^{\prime}(G))=\text{affine in }\alpha_{12}^{-1}=\alpha_{12}^{-1}\det({\cal L}^{\prime}(G/e_{12}))+\det({\cal L}^{\prime}(G\backslash e_{12}))$; The first term adds row 1 to row 2 and column 1 to column 2. For ${\cal F}_0(G)$ induction wrt. $\alpha_{12}$ in $\det({\cal L}^{\prime n-1}_{n-2}(G))$. | |
| Sep 24, 2023 at 11:17 | comment | added | Qmechanic♦ | Induction: $\quad{\cal K}_{n,\ell}(G)=\theta(e\!\in\!G)\left(\alpha_e^{-1}{\cal K}_{n,\ell}(G/e)+{\cal K}_{n,\ell}(G\backslash e)\right)$ $+\theta(e\!\notin\!G)\left\{\theta(v_1,v_2\!\in\!G)(\ldots{\cal K}_{n-1,\ell}(G/e)+\ldots{\cal K}_{n,\ell+1}(G/e)) +(1\!-\!\theta(v_1,v_2\!\in\!G)){\cal K}_{n,\ell}(G\backslash e)\right\}$ (spanning subgraph $\gamma\subseteq G$: cases $e\in\gamma$; $G\ni e\notin\gamma$; $e\notin\gamma,G$); | |
| Sep 22, 2023 at 10:17 | comment | added | Qmechanic♦ | If $G_1\cap G_2\subseteq v$ then ${\cal K}_n(G_1\cup G_2)=\sum_{r=0}^n{\cal K}_r(G_1){\cal K}_{n-r}(G_2)$; $\quad{\cal U}_n(G_1\cup G_2)=\sum_{r=0}^n{\cal U}_r(G_1){\cal U}_{n-r}(G_2)$; $\quad{\cal F}_n(G_1\cup G_2)=\sum_{r=0}^n{\cal F}_r(G_1){\cal U}_{n-r}(G_2)+\sum_{r=0}^n{\cal U}_r(G_1){\cal F}_{n-r}(G_2)$; $\det({\cal L}^{\prime v}_w(G))=0$ if $v$ and $w$ are not connected in $G$. | |
| Sep 22, 2023 at 10:17 | comment | added | Qmechanic♦ | Vanishing theorem: ${\cal K}_{n,\ell}(G)=0\Rightarrow{\cal K}_{\leq n,\geq\ell}(G)=0$. Similar for ${\cal U}_{n,\ell}(G)$ and ${\cal F}_{n,\ell}(G)$. If $G$ has no external momenta $p_v=0$ then ${\cal F}_{n,\ell}(G)=0$. If $G$ has no masses, similar story. Scaleless case. $\quad {\cal U}_{n=0,\ell}(\text{tree})=\delta_{\ell}^0$; | |
| Sep 21, 2023 at 6:24 | comment | added | Qmechanic♦ | $\quad{\cal K}_{n,\ell}(G)=\sum_{\text{span. $(|\pi_0(G)|+n,\ell)$-subgraph }\gamma}\prod_{e\in\gamma}\alpha_e^{-1}$; $\quad{\cal U}_{n,\ell}(G)=\sum_{\text{span. $(|\pi_0(G)|+n,\ell)$-subgraph }\gamma}\prod_{e\notin\gamma}\alpha_e$; $\quad{\cal F}_{n,\ell}(G)=\sum_{\text{vertex pair }i,j}p_i\cdot p_j$ $\sum_{\text{span. $(|\pi_0(G)|+1+n,\ell)$-subgraph }\gamma\text{ with terminal vertices }i,j}\prod_{e\notin\gamma}\alpha_e$; $\quad n,\ell\in\mathbb{Z}$. (If $n,\ell<0$ then it is zero.) Laplacian factorizes in conn. comp. | |
| Sep 21, 2023 at 6:24 | comment | added | Qmechanic♦ | $\quad\det(0\times0\text{-matrix})=1$; A single vertex is a 1-tree. $\quad{\cal K}_{n,\ell}(v)=\delta_n^0\delta_{\ell}^0={\cal U}_{n,\ell}(v)$; $\quad{\cal F}_{n,\ell}(v)=0$; Undirected graphs: $\quad{\cal K}_{0,0}(v_1\!\ni\!e\!\in\!v_2)=\alpha_e^{-1}$; $\quad{\cal U}_{0,0}(v_1\!\ni\!e\!\in\!v_2)=1$; $\quad{\cal F}_{0,0}(v_1\!\ni\!e\!\in\!v_2)=p_1\cdot p_2 \alpha_e$; (even for self-loop $v_1=v_2$ if we change loop number $\ell=0\to\ell=1$.) | |
| Sep 19, 2023 at 9:36 | comment | added | Qmechanic♦ | $\det(M^I_I)=\sum_{D\text{ with terminal vertices }I}r(b(D))\prod_{h=1}^{\nu(D)}\{r(c_h(D))-s(c_h(D))\}$ $=\sum_{|I|\text{-tree }F\text{ with terminal vertices }I}s(e(F))$; $\quad\Gamma\backslash\gamma$ ($\Gamma/\gamma$) is $\Gamma$ where each edge of $\gamma$ is removed (shrunk to a point and its vertices identified), respectively. The quotient can have self-loops. "Double" quotient $\Gamma//\gamma$ is $\emptyset$ if $L(\gamma)>0$. Set deletion may violate mom. cons. (Paradigm: Try to avoid set deletion $\Gamma\backslash\gamma$.) Remove pertinent row/column in Lagrangian. | |
| Sep 18, 2023 at 11:19 | comment | added | Qmechanic♦ | Thm 3.1 with $s=r$: The vertex corresponding to the $w$th deleted column must be hit. Spanning graph=each dynamical (=not spectator) vertex $v\notin I$ has an output but not nec. an input. $\quad w\in I\backslash J$ has input but no output. $\quad w\in I\cap J$ has no output; possibly input. $\quad w\in I$ can only sit in second place of $-s/r_{\Box w}$. $\quad v\in J\backslash I$ has output; possibly input. $\quad v\notin I\cup J$ has output; possibly input. (Ignoring conn. comp. consisting of a single vertex.) | |
| Sep 18, 2023 at 8:09 | comment | added | Qmechanic♦ | $\quad\det(M)=\sum_{\pi\in S_n}\sigma(\pi)m_{1\pi(1)}m_{2\pi(2)}\ldots m_{n\pi(n)}$ $=\sum_{\pi\in S_n}\sigma(\pi)m_{\pi(1)\pi(2)}m_{\pi(2)\pi(3)}\ldots m_{\pi(n)\pi(1)}$; $s$-variables comes in cycles only. Each $s$-loop has a minus. Now consider the remaining diag. $r$-variables. Each conn. comp. has 0 or 1 loops. Consider spanning graph $D=B\cup\bigcup_1^{\nu(D)} C$. [Old:Make disjoint partition $R\sqcup S=\{1,\ldots,n\}$ of rows.] $\quad\det(M)=\sum_{\text{directed graph }D}r(b(D))\prod_{h=1}^{\nu(D)}\{r(c_h(D))-s(c_h(D))\}$. Error: $\nu(D)=$# of cycles rather than conn. comp.?? | |
| Sep 18, 2023 at 7:26 | comment | added | Qmechanic♦ | 50. J.W. Moon doi.org/10.1016/0012-365X(92)00059-Z Digraph = directed graph. Their "root" is more correctly called terminal element. Thm 2.1: Spanning sub-graph=each dynamical (= not spectator) vertex $v\notin I$ has an output but not nec. an input, i.e. $-s/r_{v\Box}$ is present in each term. So $n$ edges. Each vertex has 0 or 1 output, possibly many inputs. Don't forget that cycles of length $\geq 3$ are directed. | |
| Sep 17, 2023 at 2:13 | comment | added | Qmechanic♦ | 40. sites.math.rutgers.edu/~zeilberg/mamarimY/DM85.pdf complete graph p.67. (Incompleting a graph can destroy but not create loops.) Each vertex in directed graph can only have 1 output (0 if we delete rows). Deleting $v$th row<=>$v\in I$<=>$v$th vertex terminal<=>$v$ has no output. Idea: Introduce $r_{vw}$ with $n=E<w\leq V=n+\pi_0$ for terminal vertices. (Recall that a single vertex is a 1-tree.) Case $I=J$: Each conn. comp. of a spanning tree must have precise 1 terminal vertex $v\in I$. OTOH, $w\in J\backslash I$ is never a terminal vertex, possibly an initial vertex. | |
| Sep 16, 2023 at 13:22 | comment | added | Qmechanic♦ | Lagrangian/determinants originally comes from Gaussian integration but polynomials such as Kirchhoff, Symanzik, Tutte, etc, seem to be the right generalization (incl. self-loops). 39. mathematicalgemstones.com/gemstones/… Blog with Zeilberger's idea. | |
| Sep 15, 2023 at 14:07 | comment | added | Qmechanic♦ | Matrix-Tree Theorem: 20. en.wikipedia.org/wiki/Laplacian_matrix 21. en.wikipedia.org/wiki/Kirchhoff%27s_theorem 22. en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula 23. julesjacobs.com/notes/kirchoff/kirchoff.pdf 24. math.brown.edu/reschwar/M1230/cauchy.pdf 25. personalpages.manchester.ac.uk/staff/mark.muldoon/Teaching/… No proof but mentions Tutte. 26. mathoverflow.net/q/73385 Induction. | |
| Sep 15, 2023 at 12:13 | comment | added | Qmechanic♦ | 4. Overlapping divergencies destroys the paradigm of only needing to renormalizing 1PI vertices and self-energies. Kreimer: "One of the achievements of renormalization theory that it disentangles ODs in terms of nested and disjoint ones." Idea: Does ODs go away in $\alpha$-parametrization? | |
| Sep 15, 2023 at 7:10 | comment | added | Qmechanic♦ | 3. Dirk Kreimer. UV $\overline{MS}$ Renormalization. arxiv.org/abs/q-alg/9607022 $<\ldots>=$div. part p.23. Knot/link p.30. Skein relation p.32. ${}_j\Delta$ p.33. $\zeta(3)$=trefoil p.68. arxiv.org/abs/q-alg/9707029 Hopf algebra. $Z=$CT. $\quad-R$=div. part. Forest formula is non-overlapping p.5. arxiv.org/abs/hep-th/9810022 Overlapping divergences (OD). arxiv.org/abs/1512.06409 Cosmic Galois group. Prop 2.2. Def 3.1 arxiv.org/abs/2003.04301 | |
| Sep 15, 2023 at 7:10 | comment | added | Qmechanic♦ | UV: Tropical $\alpha$-degree (of subgraph $\gamma$)$=\sum_{e\in\gamma}\nu_e\tau_e-\frac{D(G)+d}{2}L(\gamma)+\frac{D(G)}{2}(L(\gamma)(+1)_0)$ $=-\frac{D(\gamma)}{2}\left(+\frac{D(G)}{2}\right)_0>0$ for UV conv (Gen. case $\neq 0$); IR: (of subgraph $G/\gamma$)$=-\frac{D(G/\gamma)}{2}\left(+\frac{D(G)}{2}\right)_{\neq0}<0$ for IR conv (0 case most important); Test ${\cal F}_{00}(G/\gamma)=0$; (Ignoring gauge-fixing. Panzer seems to argue via the non-projective formula.) Note that SDOD grows if we destroy edges without destroying loops. Hence it is enough to check 1PI subgraphs. | |
| Sep 13, 2023 at 8:10 | comment | added | Qmechanic♦ | Assume $D(G)<0$ is UV conv.?? UV-div $k_e\to\infty\Rightarrow\alpha_e\to0$. IR/soft/collinear div. $k_e,m_e,p_v\to0\Rightarrow\alpha_e\to\infty$. Tropical formulas $\quad\alpha_e\equiv e^{\tau_e}$; Tropical $\tau$-function (of graph $G$)$=\sum_{e\in G}\nu_e\tau_e-\frac{D(G)+d}{2}\max{\cal U}_{00}(G)+\frac{D(G)}{2}\max{\cal F}_{00}(G)$; Momentum-IBPs are often more practical than $\alpha$-sector-decomposition. | |
| Sep 13, 2023 at 8:10 | comment | added | Qmechanic♦ | $\quad{\cal L}_{ii}=\sum_{e\in i}\alpha_e^{-1}$; $\quad{\cal L}_{i\neq j}=-\alpha_e^{-1}$ if $i\in e\ni j$ else $0$; $\quad {\cal E}\begin{pmatrix} 1\cr \vdots \cr 1 \end{pmatrix} =0$ $\quad\Rightarrow\quad$ $\det{\cal L}=0$; To show that $\det({\cal L}^1_1)=\det({\cal L}^n_n)$ (for conn. graph) compare ${\cal L}^1_1$, ${\cal L}^{1,n}_{1,n}$, ${\cal L}^n_n$. Primary $e_0$-sector: $\alpha_e\leq\alpha_{e_0}$. Hepp sectors: $\alpha_{\pi(1)}\leq\ldots\leq\alpha_{\pi(I)}$. | |
| Sep 12, 2023 at 13:32 | comment | added | Qmechanic♦ | Notes for later (old): $\quad \widetilde{S}=\sum_e \alpha_e(k_e^2+m_e^2) + i\sum_{v\in V}k_v\cdot x_v =k^T{\cal A}k +i x^T({\cal E}^Tk+\Delta p) +m^T{\cal A}m$ $=(k+i{\cal A}^{-1}{\cal E}\frac{x}{2})^T{\cal A}(k+i{\cal A}^{-1}{\cal E}\frac{x}{2}) +\frac{x^T}{2}{\cal L}\frac{x}{2} +ix^T\Delta p +m^T{\cal A}m$; | |
| Sep 11, 2023 at 7:33 | comment | added | Qmechanic♦ | Notes for later (old): $k_v=\sum_{e\in E}k_e{\cal E}_{e,v}+\Delta p_v$; $\quad \Delta p_v:=p_v-q$; $\quad I=\left[\prod_{e\in E}\int\frac{d^dk_e}{\pi^{d/2}} \frac{\Gamma(\nu_e)}{(k_e^2+m_e^2)^{\nu_e}}\right]\left[\prod^{\prime}_{v\in V\backslash \{v_0\}} \pi^{d/2} \delta^d (k_v)\right] $ $ =\left[ \prod_{e\in E} \int_{\mathbb{R}_+}d\alpha_e~\alpha_e^{\nu_e-1}\int\frac{d^dk_e}{\pi^{d/2}} \right]\left[\prod_{v\in V}\int\frac{d^dx_v}{(4\pi)^{d/2}} \right]\int\frac{d^dq}{(2\pi)^d} \exp(-\widetilde{S})$; $\quad {\cal L}:={\cal E}^T{\cal A}^{-1}{\cal E}$; | |
| Sep 9, 2023 at 14:43 | comment | added | Qmechanic♦ | $S=\ldots=(k+{\cal A}^{-1}{\cal E}^{\prime}\frac{x^{\prime}}{2})^T{\cal A}(k+{\cal A}^{-1}{\cal E}^{\prime}\frac{x^{\prime}}{2})$ $-(\frac{x^{\prime}}{2}-{\cal L}^{\prime -1} p^{\prime})^T{\cal L}^{\prime}(\frac{x^{\prime}}{2}-{\cal L}^{\prime -1}p^{\prime}) +p^{\prime T}{\cal L}^{\prime -1}p^{\prime}+m^T{\cal A}m-i\epsilon$; For $L=1$ one-loop, choose gauge fixing fct. $\sum_{e\in E}H_e\alpha_e={\cal U}$. Note that kin. ${\cal F}$ drops out if $D(G)=0$ log. div. Period = Residue of $1/\epsilon$ in log. div. graph with no kin. $\quad P(G)={\rm Res}_{D(G)=0}I(G)=\int\ldots {\cal U}(G)^{-d/2}$; | |
| Sep 8, 2023 at 12:35 | comment | added | Qmechanic♦ | $\quad S=\sum_e\alpha_e(k_e^2+m_e^2-i\epsilon) +\sum^{\prime}_{v\in V\backslash\{v_0\}}k_v\cdot x_v$ $=k^T{\cal A}k +x^{\prime T}({\cal E}^{\prime T}k+p^{\prime}) +m^T{\cal A}m-i\epsilon$ $=(k+{\cal A}^{-1}{\cal E}^{\prime}\frac{x^{\prime}}{2})^T{\cal A}(k+{\cal A}^{-1}{\cal E}^{\prime}\frac{x^{\prime}}{2}) -\frac{x^{\prime T}}{2}{\cal L}^{\prime}\frac{x^{\prime}}{2} +x^{\prime T}p^{\prime} +m^T{\cal A}m-i\epsilon$ | |
| Sep 6, 2023 at 10:30 | comment | added | Qmechanic♦ | $I=\ldots\stackrel{\alpha^{\prime}_e=\lambda\alpha_e}{=}$ $\hbar^L\left[\prod_{e\in E}\int_{\mathbb{R}_+}d\alpha_e~\alpha_e^{\nu_e-1}\right]\delta(1\!-\!\sum_{e\in E}H_e\alpha_e)\frac{\Gamma(-D(G)/2)}{{\cal U}^{d/2}}\left(\frac{\cal F}{\cal U}\right)^{D(G)/2}$ projective/scale invariant/Cheng-Wu theorem; (Compare with I&Z eq. (6-91) with $(+,-,-,-)$.) | |
| Sep 4, 2023 at 6:48 | comment | added | Qmechanic♦ | $I=\ldots=\left(\frac{\hbar}{i}\right)^Li^Li^{-dL/2}\left[\prod_{e\in E}i^{\nu_e}\int_{\mathbb{R}_+}d\alpha_e~\alpha_e^{\nu_e-1}\right]{\cal U}^{-d/2}\exp(-i({\cal F}/{\cal U}-i\epsilon))$ non-projective $=\hbar^Li^{-D(G)/2}\left[\prod_{e\in E}\int_{\mathbb{R}_+}d\alpha_e~\alpha_e^{\nu_e-1}\right]\int_{\mathbb{R}_+}\!d\lambda~\delta(\lambda\!-\!\sum_{e\in E}H_e\alpha_e){\cal U}^{-d/2}\exp(-i({\cal F}/{\cal U}-i\epsilon))$ linear gauge. | |
| Aug 30, 2023 at 15:26 | comment | added | Qmechanic♦ | $\quad I=\left[\prod_{e\in E}\frac{\hbar}{i}\int\frac{d^dk_e}{\pi^{d/2}}\frac{\Gamma(\nu_e)}{(k_e^2+m_e^2-i\epsilon)^{\nu_e}}\right]\left[\prod^{\prime}_{v\in V\backslash\{v_0\}}\frac{i}{\hbar}\pi^{d/2}\delta^d(k_v)\right]$ $=\left[ \prod_{e\in E} \frac{\hbar}{i}i^{\nu_e}\int_{\mathbb{R}_+}d\alpha_e~\alpha_e^{\nu_e-1}\int\frac{d^dk_e}{\pi^{d/2}}\right]\left[\prod^{\prime}_{v\in V\backslash\{v_0\}}\frac{i}{\hbar}\int\frac{d^dx_v}{(4\pi)^{d/2}}\right]\exp(-iS)$ | |
| Aug 29, 2023 at 14:24 | comment | added | Qmechanic♦ | Multiple parallel edges (bananas) are allowed. $\quad\frac{\Gamma(\nu)}{k^2+m^2-i\epsilon}=i^{\nu}\int_{\mathbb{R}_+}d\alpha~\alpha^{\nu-1}e^{-i(k^2+m^2-i\epsilon)\alpha}$; Schwinger parametrization (without coupling constant factor $\prod_v\tilde{g}_v $ and traditional physics Fourier factor $(4\pi)^{-dL/2}$). Minkowski signature $(-,+,+,+)$. physics.stackexchange.com/q/752398/2451 | |
| Aug 29, 2023 at 10:09 | comment | added | Qmechanic♦ | $\quad{\cal F}_0/\det{\cal A}=p^{\prime T}\text{Cofactor}({\cal L}^{\prime})p^{\prime}$ $=\sum^{\prime}_{i,j\in V\backslash\{v_0\}}p_i(-1)^{i+j}\det({\cal L}^{\prime}{}^i_j)p_j$; Homogeneity:$\quad[{\cal U}_{n,\ell}/\det{\cal A}]=[{\cal K}_{n,\ell}]=[\alpha^{-1}]^{V-\pi_0-n+\ell}$; $\quad[{\cal U}_{n,\ell}]=[\alpha]^{L+n-\ell}$; $\quad[{\cal F}_{n,\ell}/\det{\cal A}]=[\alpha^{-1}]^{V-\pi_0-1-n+\ell}$; $\quad[{\cal F}_{n,\ell}/{\cal U}_{n,\ell}]=[\alpha]$; $\quad[{\cal F}_{n,\ell}]=[\alpha]^{L+1+n-\ell}$; $\quad [\alpha]=\text{mass dim}^{-2}$; | |
| Aug 29, 2023 at 8:23 | comment | added | Qmechanic♦ | Amputated not nec. conn. graphs, i.e. each end of a line has a vertex. (Paradigm: Trees are graphs.) SDOD $\omega(G)\equiv D(G)=dL(G)-2\sum_{e\in G}\nu_e$. Future notation: $G\equiv\Gamma$. $\quad k_v=\sum_{e\in E}k_e{\cal E}_{e,v}+p_v$; $\quad{\cal L}^{\prime}:={\cal E}^{\prime T}{\cal A}^{-1}{\cal E}^{\prime}$; $\quad{\cal U}_{0,0}\equiv{\cal U}_0\equiv{\cal U}={\cal K}\det{\cal A}$; $\quad{\cal K}_{0,0}\equiv{\cal K}_0\equiv{\cal K}=\det{\cal L}^{\prime}$; $\quad{\cal F}/{\cal U}={\cal F}_0/{\cal U} +m^T{\cal A}m$; $\quad{\cal F}_0/{\cal U}=p^{\prime T}{\cal L}^{\prime -1}p^{\prime}$; | |
| Aug 27, 2023 at 11:32 | comment | added | Qmechanic♦ | $n$-tree=tree with $n$ conn. comp. Spanning subgraph=touches all vertices. (Touch by vertex is allowed, and is in fact assumed to be the case.) $(n,\ell)$-graph=graph with $n$ conn. comp. and $\ell$ loops. Componentwise 1PI=don't cut into more conn. comp. by removing 1 line. Componentwise 1VI=don't cut into more conn. comp. by removing 1 vertex. (I.e. the notions make sense even if the graph is not connected.) A self-loop $\int\frac{d^dk}{\pi^{d/2}}\frac{\hbar}{i}\frac{\Gamma(\nu)}{(k^2+m^2-i\epsilon)^{\nu}}=\hbar m^{d-2\nu}\Gamma(\nu\!-\!\frac{d}{2})$ factorizes. Assume no self-loops. | |
| Aug 17, 2023 at 8:51 | comment | added | Qmechanic♦ | 5. E. Panzer, PhD thesis, arxiv.org/abs/1506.07243 Be aware that Panzer's SDOD $\omega=-\frac{D}{2}$ in eq. (2.1.18) is normalized with a factor $-\frac{1}{2}$. Laplacian $(V\!-\!\pi_0)\times(V\!-\!\pi_0)$ matrix. 6. arxiv.org/abs/0910.0114 Lemma 50. 7. arxiv.org/abs/0910.5429 8. Bogner & Weinzierl arxiv.org/abs/1002.3458 Laplacian $V\times V$ matrix. 9. arxiv.org/abs/2202.12296 $(+,-,-,-)$. $\rm L$ seems to be missing in eq. (12). Eq. (B1) correct. 10. osti.gov/etdeweb/biblio/21380826 15. math.stackexchange.com/a/4764382/11127 | |
| Jun 15, 2023 at 7:33 | comment | added | Qmechanic♦ | Notes for later: $\quad \sum_i V_i ( n_i -2) =2(L-1) +\sum_f E_f $. So at a certain $\hbar$/loop-order, the number of vertices $V_i$ with $n_i\geq 3$ legs (in the amputated correlation function) is finite. 2-vertices (apart from the kinetic terms) contain derivatives. If we assume Lorentz symmetry such 2-vertices are irrelevant $[\lambda_i]<0$. If the kinetic term is unique (e.g. by homogeneity & isotropy), then all other 2-vertices has $[\lambda_i]\neq 0$. It is natural to assume that all relevant or marginal 2-vertices belong to the free (rather than the interaction) part of the action. | |
| Jul 6, 2022 at 11:31 | comment | added | Qmechanic♦ | Notes for later: 1. For a disconnected diagram with $\pi_0$ connected components, the first term $d\to \pi_0d$ is replaced in eq. (1). 2. In QM $d=1$ the scalar field $[\phi]=\frac{d-2}{2}=-\frac{1}{2}$, so $D$ increased with number of external legs. | |
| May 22, 2022 at 20:19 | comment | added | Qmechanic♦ | NB: Note that SDOD may underestimated the dependence of the UV cut-off: Toyexample in $d=1$: $\quad\int^{\Lambda_H}_{\Lambda_L}\!\frac{mdp}{p^2+m^2}$ $=\int^{\Lambda_H}_{\Lambda_L}\!\frac{mdp}{2i}\left(\frac{1}{p-im}-\frac{1}{p+im}\right)$ $=[\arctan\frac{p}{m}]^{\Lambda_H}_{\Lambda_L}$ $=[{\rm sgn}(p)\frac{\pi}{2}-\frac{m}{p}+{\cal O}((\frac{m}{p})^2)]^{\Lambda_H}_{\Lambda_L}$. Naively $D=-1$ but effectively $D=0$! Notes for later: userswww.pd.infn.it/%7Eferuglio/Falkowski.pdf p.6: eq. (1.7) blocking/dimless coupling constants. p. 6-7: $\hbar$. | |
| Feb 7, 2022 at 12:15 | comment | added | Qmechanic♦ | A field $\overline{\phi}_f=\frac{\phi_f}{\sqrt{\hbar}}$. A source $\overline{J}_f=\frac{J_f}{\sqrt{\hbar}}$. A coupling constant $\lambda_i$ should be replaced with $\overline{\lambda}_i=\hbar^{n_i/2-1}\lambda_i$, where $n_i=\sum_f n_{if}$ is the number of legs on a vertex of $i$th type. 2. Q: How about _not_ putting $c=1$ for non-relativistic physics? 3. Q: Could vertex numerator momentum factors conspire to become projection operators with an effective lower UV momentum dependence than their power suggests? A: No. | |
| Oct 13, 2021 at 18:27 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Minor formatting
|
| Mar 27, 2020 at 21:37 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
|
| Mar 27, 2020 at 9:29 | comment | added | Qmechanic♦ | Correction to answer (v7): Replace the word Feynman diagram with the word Feynman loop diagram. Notes for later: 1. Q: Can we generalize to not putting $\hbar=1$? A: All quantities $Q$ are of the form $[Q]=(\text{reciprocal length})^{\#}(\text{angular momentum})^{\#\#}=L^{-\#}J^{\#\#}$. The corresponding $\hbar$-reduced quantity $\overline{Q}:=\frac{Q}{\hbar^{\#\#}}$ with powers of $\hbar$, so that $[\overline{Q}]=L^{-\#}$ only. Now let $[\cdot]$ denotes inverse length dimension. Momentum $p=\hbar k$ should be replaced with wavevector $k=\overline{p}$. Action $S=\hbar\overline{S}$. | |
| Aug 21, 2019 at 21:25 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Minor
|
| Aug 21, 2019 at 17:21 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
|
| Aug 20, 2019 at 9:29 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
|
| May 23, 2019 at 10:40 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
|
| May 20, 2019 at 18:09 | history | answered | Qmechanic♦ | CC BY-SA 4.0 |