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量子力学基础-1

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1. 狄拉克符号

1.1 基矢

\(|0 \rang = \binom{1}{0}\) \(|1\rang = \binom{0}{1}\)

\(\lang 0| = (1~0)\) \(\lang 1| = (0~1)\)

1.2 态矢

\(| \psi \rangle = a|0\rangle + b|1\rangle\)

\(\lang \psi| = |\psi \rang ^{\dagger} = \Big( \left(|\psi \rang \right)^* \Big)^T = \Big( \left(|\psi \rang \right)^T \Big)^* = a^* \lang0| + b^*\lang1|\)

共轭,转置顺序可以互换.

2. 内积

\(\lang \varphi | \psi \rang = \lang \varphi || \psi \rang = | \varphi \rang^ \dagger |\psi\rang\) ,也可以表示为 \(( | \varphi \rang, |\psi \rang)\)

内积的性质:

(1) \((\cdot~, \cdot)\) 对第二个变量是线性的,即
\((|v\rang, \sum_{i}\lambda_i|w_i\rang) = \sum_i \lambda_i(|v\rang, |w_i\rang)\)
(2) \(\lang \varphi| \psi \rang = \Big(\lang \psi| \varphi \rang \Big)^*\)
推导如下:
\(\lang \varphi | \psi \rang = \lang \varphi|| \psi \rang = | \varphi\rang^ \dagger |\psi\rang\)
\(\lang \psi| \varphi \rang = \lang \psi|| \varphi \rang = | \psi\rang^ \dagger |\varphi\rang\)
\(\Big(\lang \psi| \varphi \rang \Big)^* = \Big(| \psi\rang^ \dagger |\varphi\rang \Big)^* = |\psi\rang^T |\varphi\rang^* = (|\varphi\rang^*)^T |\psi\rang = | \varphi\rang^ \dagger |\psi\rang\).

(3) 内积非负,当且仅当\(|v\rang=0\) 时取等号.
\(\lang i|j \rang = \delta_{ij} = \left\{\begin{matrix} 1~~~i=j \\ 0~~~~i\neq j \end{matrix}\right.\)

3. 算符

3.1 线性算符

\(A(|a\rang+|b\rang) = A|a\rang+A|b\rang\)

\(A(z|a\rang = zA|a\rang\)

3.2 特殊算符

恒等算符\(I\) \(I|a\rang = |a\rang\)

零算符0 \(0|a\rang = 0\)

3.3 本征值和本征矢

\(A|v\rang = \lambda|v\rang\) \(|v\rang 是 A 的本征矢量,\lambda 是相应的本征值.\)

如何求得本征值和本征矢量:

(1) \(A|v\rang =\lambda|v\rang = \lambda I |v\rang (I是等同算符)\)

(2) \((A-\lambda I)|v\rang = 0\)

(3) \(|A-\lambda I| = 0\),解得 \(\lambda\), 带入\((A-\lambda I)|v\rang = 0\),可求得 \(|v\rang\).

3.4 对易关系

\([A, B] = AB-BA\), 若\([A, B] = 0, 即 AB=BA,则称A与B是对易的.\)

\(\{A, B\} = AB+BA\), 若\({A, B} = 0, 即 AB=-BA,则称A与B是反对易的.\)

3.5 厄米算符

\(A = A^\dagger\), 则\(A\) 是厄米算符. (类似对称矩阵,\(A=A^T\))

\((AB)^{\dagger} = B^{\dagger} A^\dagger\)

推导:\((AB)^\dagger = ((AB)^*)^T) = ((AB)^T)^*) = (B^{T}A^T)^* = ((B^T)^*)((A^T)^*) = B^{\dagger}A^\dagger\)

厄米算符的本征值是实数.

3.6 外积算符

\(态矢 |u\rang自身的外积是 |u\rang \lang u|\).

\(\Big( |u\rang \lang u| \Big) |w\rang = |u\rang \Big(\lang u|w\rang \Big) =\lang u|w\rang |u\rang\), 得到的矢量与原来的态矢方向相同,但是系数是\(|u\rang 与|w\rang 的内积\), 所以算符\(|u\rang \lang u | 作用于态矢量|w\rang 可以看作 |w\rang 在 |u\rang 上的投影(分量).\)

3.7 投影算符

设空间 \(V\) 是\(n\)维, \(|i\rang (i=1,2,\cdots, n)\) 是空间\(V\) 的一组正交规一基,定义投影算符
\(P=\sum_{1}^{m}|i\rang\lang i|~~~~(i=1,2,\cdots,m),~~m\leq n\)
各个 \(|i\rang~(i=1,2,\cdots,m)\) 构成 \(V\) 的子空间 \(M\). 投影算符\(P\)作用于空间V中的任一态矢\(|w\rang\)相当于求取\(|w\rang\) 在子空间 \(M\) 中的分量. 显然,
(1). 当 \(m =n\)时,子空间\(M\)亦即原来的空间 \(V\). \(V\) 中任何态矢 \(|w\rang\) 在 \(V\)中的投影显然就是 \(|w\rang\) 本身,故
\(\sum_1^n |i\rang\lang i|w\rang = |w\rang\)
\(\sum_1^n |i\rang\lang i | = I\), \(I\) 是\(n\) 阶等同算符.

(2). 当 \(m<n\) 时,\(|w \rang\) 是空间 \(V\) 中的任意态矢,\(P|w \rang = \sum_1^m|i\rang \lang i|w\rang = |v\rang\) 是在 \(M\)中, \(|v\rang\) 在\(M\) 中的投影是 \(|v\rang\) 自身, 故
\(\left\{\begin{matrix} P(P|w\rang)= P|w\rang = |v\rang \\ P=P^2=P^3= \cdots \end{matrix}\right.\)

3.8 对角化及谱分解

\(A=\sum \lambda_i |i\rang \lang i | (i = 1,2, \cdots, n) = O = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}\)

其中,\(\lambda_i\) 是本征值,\(|i\rang\) 是对应的本征向量.

算符 \(A\) 能够对角化的充要条件是 \(A\) 是一个正规算符(normal operator), 即 \(AA^\dagger = A^{\dagger}A\).
\(A 能够对角化 \Longleftrightarrow AA^\dagger = A^{\dagger}A\)

3.9 幺正算符

幺正算符,也叫酉算符. \(U^\dagger = U^{-1}\),即 \(U^{\dagger}U = I\). (类似正交矩阵 \(AA^T = I\),\(A^T = A^{-1}\))

幺正算符的性质:

(1) 空间中任意两个矢量经幺正变换后内积保持不变.

推导如下:

​ 变换前的两个矢量,\(|s\rang\) 和\(|t\rang\),其内积为\(\lang s|t \rang\).

​ 以幺正算符作用于\(|s\rang\) 和\(|t\rang\) 后,\(U|s\rang\) 和\(U|t\rang\),其内积为 \(\Big(U|s\rang, U|t\rang \Big) = \lang s|U^{\dagger} U|t\rang = \lang s|I|t\rang = \lang s|(I|t\rang) = \lang s|t\rang\).

(2) 正交规一基经过幺正变换后变为另一组正交归一基.

​ 设\(|u_i\rang(i=1,2,\cdots,n)\) 是一组正交规一基,\(U\) 是幺正变换,\(U|u_i\rang = |w_i\rang\), 可得
\(\lang w_i|w_j \rang = \lang u_i| U^\dagger U | u_j \rang = \lang u_i | u_j \rang = \delta_{ij}\)
​ 反之,若 \(|u_i\rang\) 和 \(|w_i\rang\) 是两组归一正交基,则它们之间的变换是幺正变换. 即,\(A|u_i\rang = |w_i\rang\), A是幺正变换.

​ 推导如下:

​ 假设 \(A|u_i\rang = |w_i \rang\), 证明 \(A\) 是幺正算符.

​ $A|u_i\rang = |w_i \rang \Longrightarrow A|u_i\rang \lang u_i| = |w_i \rang \lang u_i| \Longrightarrow \sum_i A|u_i\rang \lang u_i| = \sum_i |w_i \rang \lang u_i| $

​ \(\Longrightarrow A\sum_i |u_i\rang \lang u_i| = \sum_i |w_i \rang \lang u_i| \Longrightarrow AI = \sum_i |w_i \rang \lang u_i| \Longrightarrow A = \sum_i |w_i \rang \lang u_i|\)

​ \(A^\dagger = \sum_i |u_i\rang \lang w_i|\)

​ \(A^{\dagger}A = \sum_i |u_i\rang \lang w_i| \sum_j |w_j \rang \lang u_j| = \sum_i \sum_j |u_i\rang \lang w_i|w_j \rang \lang u_j| = \sum_i |u_i\rang \lang u_i| = I\).

上式中 $I $ 是恒等算符,故 \(A\) 是幺正算符.

(3) 幺正算符的本征值都以1为模数,亦即可表示为\(e^{i\theta}\), \(\theta\) 是个实数.

3.10 泡利算符

\(\sigma_0 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) \(\sigma_ 1= \sigma_x = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)

\(\sigma_2= \sigma_y = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\) \(\sigma_3= \sigma_z = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)

泡利算符既是厄米算符,又是幺正算符,即\(\sigma = \sigma^\dagger = \sigma^{-1}\).

标签:lang,rang,psi,dagger,量子力学,sum,基础,varphi
来源: https://www.cnblogs.com/one2Four/p/15478632.html