From: hackbard Date: Fri, 17 Feb 2012 22:43:04 +0000 (+0100) Subject: more math, however, not satisfying! X-Git-Url: https://hackdaworld.org/gitweb/?a=commitdiff_plain;h=41de34392edb34632ebcf19bfa55bdb746f2c70a;p=lectures%2Flatex.git more math, however, not satisfying! --- diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex index f327750..e3417bd 100644 --- a/physics_compact/math_app.tex +++ b/physics_compact/math_app.tex @@ -69,7 +69,10 @@ $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies (positive definite) \end{itemize} for $\vec{u},\vec{v}\in V$ and $\lambda\in K$. -Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$. +Taking the complex conjugate $(\cdot)^*$ is the map from +\begin{equation} +z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.} +\end{equation} \end{definition} \begin{remark} @@ -82,7 +85,7 @@ Due to conjugate symmetry, linearity in the first argument results in conjugate This is called a sesquilinear form. If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form. -The inner product $(\cdot,\cdot)$ provides a mapping +Furtermore, the inner product $(\cdot,\cdot)$ provides a mapping \begin{equation} V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}} \quad @@ -100,18 +103,21 @@ If the inner product is nondegenerate, i.e.\ $\forall\vec{u}\, (\vec{v},\vec{u} Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective. Then, $V$ is isomorphic to $V^{\dagger}$. Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$. +The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$. -In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument. -This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with a dual vector or linear functional of dual space $V^{\dagger}$ +Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument. +In this case, the antilinearity property is assigned to element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space indicating an isomorphism of $V$ to the conjugate complex of its dual space. \begin{equation} -(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v} -\text{ CHECK ! ! !} +[(\lambda\vec{v})^{\dagger},\vec{u}]= +[\varphi_{\lambda\vec{v}},\vec{u}]= +\varphi_{\lambda\vec{v}}(\vec{u})= +\lambda^*\varphi_{\vec{v}}(\vec{u})= +\lambda^*(\vec{v},\vec{u}) \end{equation} -or the conjugate transpose in matrix formalism +According to this, in matrix formalism, the dual vector is associated with the conjugate transpose. \begin{equation} -(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .} +(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \end{equation} -In doing so, the conjugate transpose is associated with the dual vector. \end{remark} \begin{definition}[Outer product]