I would just add a \strut
to both characters
\documentclass[final,t]{beamer}
\mode<presentation>
\usepackage{tikz}
\usepackage{pgfplots}
\tikzstyle{gate2} = [circle,fill=white,draw=black,minimum size=1cm]
\begin{document}
\begin{tikzpicture}
\node[gate2] (mu_0) at (0, 0) {$\strut\mu_0$};
\node[gate2, baseline = (mu_0.base)] (Lambda_0) at (2, 0) {$\strut\Lambda_0$};
\end{tikzpicture}
\end{document}
![enter image description here](https://i.stack.imgur.com/Z4kO9.jpg)
Nodes on a regular polygon
It is possible to put circular nodes with various sizes on a circle and corners of a regular polygon and draw arrows between them, which are also on the circle.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{bending}
\begin{document}
\begin{tikzpicture}[
->,
thick,
main node/.style={circle, fill=blue!20, draw},
]
\newcommand*{\MainNum}{5}
\newcommand*{\MainRadius}{1.5cm}
\newcommand*{\MainStartAngle}{90}
% Print main nodes, node names: p1, p2, ...
\path
(0, 0) coordinate (M)
\foreach \t [count=\i] in {A, Hello\\World, 3, foobar, $\cdot$} {
+({\i-1)*360/\MainNum + \MainStartAngle}:\MainRadius)
node[main node, align=center] (p\i) {\t}
}
;
% Calculate the angle between the equal sides of the triangle
% with side length \MainRadius, \MainRadius and radius of circle node
% Result is stored in \p1-angle, \p2-angle, ...
\foreach \i in {1, ..., \MainNum} {
\pgfextracty{\dimen0 }{\pgfpointanchor{p\i}{north}}
\pgfextracty{\dimen2 }{\pgfpointanchor{p\i}{center}}
\dimen0=\dimexpr\dimen2 - \dimen0\relax
\ifdim\dimen0<0pt \dimen0 = -\dimen0 \fi
\pgfmathparse{2*asin(\the\dimen0/\MainRadius/2)}
\global\expandafter\let\csname p\i-angle\endcsname\pgfmathresult
}
% Draw the arrow arcs
\foreach \i [evaluate=\i as \nexti using {int(mod(\i, \MainNum)+1}]
in {1, ..., \MainNum} {
\pgfmathsetmacro\StartAngle{
(\i-1)*360/\MainNum + \MainStartAngle
+ \csname p\i-angle\endcsname
}
\pgfmathsetmacro\EndAngle{
(\nexti-1)*360/\MainNum + \MainStartAngle
- \csname p\nexti-angle\endcsname
}
\ifdim\EndAngle pt < \StartAngle pt
\pgfmathsetmacro\EndAngle{\EndAngle + 360}
\fi
\draw
(M) ++(\StartAngle:\MainRadius)
arc[start angle=\StartAngle, end angle=\EndAngle, radius=\MainRadius]
;
}
\end{tikzpicture}
\end{document}
![Result](https://i.stack.imgur.com/0Yixr.png)
Equidistant nodes
Instead of putting the nodes on the corners of a regular polygon, the nodes can be placed in such a way on the circle, that the arrows have equal lengths. The advantage is, that the main radius can be made smaller for a more compact presentation.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{bending}
\begin{document}
\begin{tikzpicture}[
->,
thick,
main node/.style={circle, fill=blue!20, draw, align=center},
]
\newcommand*{\MainRadius}{1.2cm}
\newcommand*{\MainStartAngle}{90}
\coordinate (M) at (0, 0);
\newcommand*{\MainAngleSum}{0}
\begin{scope}
% First the nodes are only set, but clipped to get their size.
% The final location is not yet known.
\clip (M);
\foreach \t [count=\i] in {A, Hello\\World, 3, foobar, $\cdot$} {
\global\expandafter\let\csname p\i-text\endcsname\t
\node[main node] (p\i) at (M) {\t};
%
\global\let\MainNum\i % the last assignment is number of nodes
% Calculate the angle between the equal sides of the triangle
% with side length \MainRadius, \MainRadius and radius of circle node
% Result is stored in \p1-angle, \p2-angle, ...
\pgfextracty{\dimen0 }{\pgfpointanchor{p\i}{north}}
\pgfextracty{\dimen2 }{\pgfpointanchor{p\i}{center}}
\dimen0=\dimexpr\dimen2 - \dimen0\relax
\ifdim\dimen0<0pt \dimen0 = -\dimen0 \fi
\pgfmathparse{2*asin(\the\dimen0/\MainRadius/2)}
\global\expandafter\let\csname p\i-angle\endcsname\pgfmathresult
\pgfmathparse{\MainAngleSum + 2*\csname p\i-angle\endcsname}
\global\let\MainAngleSum\pgfmathresult
}
\end{scope}
\pgfmathsetmacro\MainAngleStep{(360 - \MainAngleSum)/\MainNum}
% Draw the nodes and arrow arcs
\global\let\CurrentAngle\MainStartAngle
\foreach \i in {1, ..., \MainNum} {
\pgfmathsetmacro\AngleA{\CurrentAngle + \csname p\i-angle\endcsname}
\pgfmathsetmacro\AngleB{\AngleA + \MainAngleStep}
\draw
(M) +(\CurrentAngle:\MainRadius)
node[main node] {\csname p\i-text\endcsname}
+(\AngleA:\MainRadius)
arc[
start angle=\AngleA,
end angle=\AngleB,
radius=\MainRadius,
]
;
\ifnum\i<\MainNum
\pgfmathparse{
\AngleB
+ \csname p\the\numexpr\i+1\relax-angle\endcsname
}
\global\let\CurrentAngle\pgfmathresult
\fi
}
\end{tikzpicture}
\end{document}
![Result equidistant](https://i.stack.imgur.com/vvRkT.png)
Best Answer
A possible TikZ free solution with automatic width calculation and allowing multi-line texts (a TikZ solution is below):
The command
\EightTexts
has eight mandatory arguments and LaTeX performs all necessary calculations using simple conditional tests on the length of the corresponding strings. The code in the image was produced usingThe complete code:
A solution using TikZ has been required in comments; here it is: