回复：这里牛爹妈多，问个数学物理题。

来源: 湘西山民于 2012-11-14 17:34:37 [档案] [博客] [旧帖] [给我悄悄话] 本文已被阅读：次 (5810 bytes)

回答: 这里牛爹妈多，问个数学物理题。由恰好于 2012-11-14 14:41:16

<A|B> 是点积，结果是一个复数。

For a finite-dimensional vector space, using a fixed orthonormal basis, the inner product can be written as a matrix multiplication of a row vector with a column vector:

$\langle A | B \rangle = A_1^* B_1 + A_2^* B_2 + \cdots + A_N^* B_N = \begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ \vdots \\ B_N \end{pmatrix}$

|A><B| 是叉积，结果是一个矩阵。

Outer products

A convenient way to define linear operators on H is given by the outer product: if $\langle\phi|$ is a bra and $|\psi\rangle$ is a ket, the outer product

$|\phi\rang \lang \psi|$

For a finite-dimensional vector space, the outer product can be understood as simple matrix multiplication:

$|\phi \rangle \, \langle \psi | = \begin{pmatrix} \phi_1 \\ \phi_2 \\ \vdots \\ \phi_N \end{pmatrix} \begin{pmatrix} \psi_1^* & \psi_2^* & \cdots & \psi_N^* \end{pmatrix} = \begin{pmatrix} \phi_1 \psi_1^* & \phi_1 \psi_2^* & \cdots & \phi_1 \psi_N^* \\ \phi_2 \psi_1^* & \phi_2 \psi_2^* & \cdots & \phi_2 \psi_N^* \\ \vdots & \vdots & \ddots & \vdots \\ \phi_N \psi_1^* & \phi_N \psi_2^* & \cdots & \phi_N \psi_N^* \end{pmatrix}$