# Trying to find QED symbol my professor uses to use in Latex?

I apologize for not knowing the name of this symbol, otherwise it might make it easier for me to track down! I also just haven't been able to find it on lists of symbols people use to represent QED, so I was hoping this community might be able to help me.

I have this one professor that, whenever he gets through a proof on the black board, when he finishes, he draws this symbol: and I was hoping to recreate it in Latex.

I, for the life of me, cannot seem to find this online though - did he pull this out of a hat? Is it just a unique symbol he made up? He always finished a proof with this, and since he scans his handwritten lecture notes and homework solutions, it shows up all over them as well.

Using websites such as Detexify, I have just been unsuccessful in finding it.

tl;dr I'm trying to recreate this symbol in Latex my professor uses to represent QED, but cannot find it.

edit: Here's an example where I use \blacksquare:

$$[\vec{L}^2,H] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(\vec{Q})] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(|\vec{Q}|) ]$$
$$= [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{Q}|)] + [L_y^2,\frac{\vec{P}^2}{2m} + V(|\vec{Q}|)] + [L_z^2,\frac{\vec{P}^2}{2m} + V(|\vec{Q}|)]$$
First looking at the $L_x^2$ component:
$$\rightarrow [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{Q}|)] = \frac{1}{2m}[L_x^2,P^2] + [L_x^2,V(|\vec{Q}|)]$$
$$= \frac{1}{2m}[L_x^2,P_x^2+P_y^2+P_z^2] + [L_x^2,V(|\vec{Q}|)]$$
$$= \frac{1}{2m} \bigg( [L_x^2,P_x^2]+[L_x^2,P_y^2]+[L_x^2,P_z^2] \bigg) + [L_x^2,V(|\vec{Q}|)]$$

\begin{flalign*}
[L_x^2,P_x^2] & = 0 & \\
[L_x^2,P_y^2] & = L_x \underbrace{[L_x,P_y]}_{=P_z} P_y + P_y[L_x,P_y]L_x + L_xP_y[L_x,P_y] + [L_x,P_y]L_xP_y &\\
& = i\hbar L_xP_z P_y + i\hbar P_yP_zL_x + \underbrace{i \hbar L_xP_yP_z}_{=-i\hbar L_xP_zP_y} + i\hbar P_zL_xP_y &\\
& =  i\hbar L_xP_zP_y - i\hbar L_xP_zP_y - i\hbar P_zP_yL_x + i\hbar P_zL_xP_y &\\
& = 0 &\\
[L_x^2,P_z^2] & = L_x [L_x,P_z] P_z + P_z[L_x,P_z]L_x + L_xP_z[L_x,P_z] + [L_x,P_z]L_xP_z  &\\
& = 0 &\\
\end{flalign*}
$$\rightarrow [L_x^2,P^2] = 0 \: \blacksquare$$
Similarly
$$[L_y^2,P^2] = [L_z^2,P^2] = 0 \: \blacksquare$$
We also know that $[L_x^2,V(|\vec{Q}|)] = 0$ because the angular momentum operators are generators of rotation about their respective axes, however the statement $V(\vec{Q}) = V(|\vec{Q}|)$ means that the potential is invariant under rotations, and commutes with the angular-momentum operators.
$$[L_x^2,V(|\vec{Q}|)] = [\frac{1}{2}(L_+ L_- + L_- L_+) + L_z^2,V(r) ] = \frac{1}{2} \bigg( [L_+ L_-,V(r)] + [L_- L_+,V(r)] \bigg) + [L_z^2,V(r)]$$
where
\begin{flalign*}
L_+ & = \hbar e^{i \phi} \bigg( \frac{\partial}{\partial \theta} + i \cot\theta \frac{\partial}{\partial \phi} \bigg) & \\
L_- & = \hbar e^{-i\phi} \bigg( - \frac{\partial}{\partial \theta} + i \cot\theta \frac{\partial}{\partial \phi} \bigg) &\\
L_z & = \frac{\hbar}{i} \frac{\partial}{\partial \phi} &\\
\end{flalign*}
However, none of these operators have a $\frac{\partial}{\partial r}$ term, meaning that they commute with $V(r)$, thus $[L^2,V(|\vec{Q}|)] = 0$ and since $\frac{1}{2m}[L_x^2,P^2] = 0$, $[\vec{L}^2,H] = 0$. $\blacksquare$


and what it looks like compiled: • Is \rotatebox{45}{\#} enough? – Phelype Oleinik Mar 10 '18 at 18:23
• Wow, that certainly looks like it to me! And to think the solution to all my googling was so simple... Thanks a lot for the comment. :D – Illari Mar 10 '18 at 18:25
• Good resources for finding symbols are symbols-a4 and detexify. However, I do not think the symbol you are looking for exists there. In my experience, the end of a proof (aka qed) is usually represented by \square or \blacksquare. Maybe those are good options for you, too. If you want to create the symbol your professor used yourself, have a look at this question. – schtandard Mar 10 '18 at 18:28
• My good friend, Wikipedia en.wikipedia.org/wiki/Q.E.D., says (under "Typographical forms used symbolically") that "Other authors have adopted two forward slashes (//) or four forward slashes (////)" and cites Rudin, Walter (1987). Real and Complex Analysis. McGraw-Hill. ISBN 0-07-100276-6 as an example of the latter. Interesting that // is half of what your professor used... – sgmoye Mar 10 '18 at 20:18
• @sgmoye oh yeah, I noticed that on the wiki page as well. He's never referenced Walter in class, but he's certainly of the age where he might have used/referenced that text. – Illari Mar 11 '18 at 1:33

## 1 Answer

You can remove the pencil-like behavior by removing pencildraw' from each line

\documentclass{article}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{tikz}

\usetikzlibrary{decorations.pathmorphing}

\newcommand*\myqed{%
\begin{tikzpicture}[scale = 0.3,
pencildraw/.style={
thick,
black!75,
decorate,
decoration={random steps, segment length = 0.8pt, amplitude=0.3pt}
},
]
\clip (0, 0) rectangle (1, 2);
\draw[pencildraw] (0, 0) -- ++ (45 : 3);
\draw[pencildraw, yshift = 0.6cm] (0, 0) -- ++ (45 : 3);
\draw[pencildraw] (1, 0) -- ++ (135 : 3);
\draw[pencildraw , yshift = 0.6cm] (1, 0) -- ++ (135 : 3);
\end{tikzpicture}
}

\renewcommand\qedsymbol{\myqed}

\begin{document}

\begin{proof}
A test text
\end{proof}

\end{document}
` • That's really gosh darn amazing, but I have to admit that I ran into some problems with \usepackage{amssymb} since various commands were already defined, and I'm not nearly familiar enough with Latex or knowledgeable enough to figure out which packages I would get rid of to get rid of the various "Command already defined" errors that popped up. – Illari Mar 10 '18 at 20:54
• @JomyBlue Which command are you using to write your proofs? – caverac Mar 10 '18 at 20:57
• Well, I'm using several packages that I've just been accumulating as necessary, and to be completely honest I never know exactly what each one has/does except for what I need it for, so I'm using and having not encountered problems with: \usepackage{graphicx} \usepackage{amsmath} \usepackage{MnSymbol} \usepackage{wasysym} \usepackage{subfig} \usepackage{float} \usepackage{ dsfont } \usepackage{ mathrsfs } \usepackage[makeroom]{cancel} \usepackage[margin=1in]{geometry} – Illari Mar 10 '18 at 21:00
• Sorry with how illegible that is – Illari Mar 10 '18 at 21:01
• @JomyBlue Can you show me an example of the code you plan to use to create a theorem and its proof? – caverac Mar 10 '18 at 21:02