Math & Science ⇒ newcommand problem

[juliette]

Hello,

I'm wondering why my first line doesn't work, but the second one does ?

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\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{mathrsfs}

\newcommand{\bra}[1]{\langle #1 |}
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\ppp}[1]{\bra{\psi_0}}
\newcommand{\pp}[1]{\ket{\psi_0}}

\begin{document}

$F(t)=\ppp\rho(t)\pp$
$F(t)=\bra{\psi_0}\rho(t)\ket{\psi_0}$

\end{document}

[localghost]

It's in the definition of the last two commands for the Dirac notation.

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\newcommand{\ppp}[1]{\bra{\psi_0}}
\newcommand{\pp}[1]{\ket{\psi_0}}

You say that there is to pass one argument, but there is no according parameter. You have to correct that.

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\newcommand{\ppp}{\bra{\psi_0}}
\newcommand{\pp}{\ket{\psi_0}}

The mathtools package offers a mechanism for an easier definition of such delimiter pairs.

Code: Select all

\documentclass[11pt]{article}
\usepackage{mathtools}  % loads »amsmath« subsequently

\DeclarePairedDelimiter{\bra}{\langle}{\rvert}
\DeclarePairedDelimiter{\ket}{\lvert}{\rangle}

\begin{document}
  $F(t)=\bra{\psi_0}\rho(t)\ket{\psi_0}$
\end{document}

The package manual has the details.


Best regards
Thorsten

[juliette]

Thank you so much Thorsten!
That was very helpful! I will use your method from now on =)

juliette