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

Question

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:

Answer 1

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}