1。左矢與右矢

(a)右矢ket

態向量:

\left | \psi(t)  \right \rangle ,\left |\varphi(t)  \right \rangle ,...

本徵態:

\left | a_i  \right \rangle ,\left |x  \right \rangle ,\left |\overrightarrow{p}  \right \rangle,\left |lm \right \rangle,...

(b)左矢bra

態向量:

\left \langle \psi(t) \right |,\left \langle \varphi(t)  \right |  ,...

本徵態:

\left \langle a_i \right |,\left \langle x \right |  ,\left \langle \overrightarrow{p}  \right |,\left \langle lm  \right |,...

bra與ket相互對偶

2。內積(inner product)

\begin{eqnarray} \left \langle \varphi(t)  \right |\cdot \left | \psi(t)  \right \rangle & = & \left \langle\varphi(t) | \psi(t)  \right \rangle \\ \left \langle a_i  \right |\cdot \left | a_j \right \rangle & = & \left \langle a_i |a_j \right \rangle  \\ \end{eqnarray}

基本性質:

\left \langle\varphi | \psi \right \rangle \equiv\left \langle\psi |\varphi  \right \rangle ^*=Complex \ function

兩個態正交:

\left \langle\varphi | \psi \right \rangle=0 (\left | \varphi  \right \rangle ,\left | \psi  \right \rangle \ne0)

態的歸一化:

\left \langle\psi  | \psi \right \rangle =1

Schwartz不等式:

\left | \left \langle\varphi | \psi \right \rangle  \right | ^2 \le \left \langle\varphi | \varphi \right \rangle \left \langle\psi | \psi \right \rangle

三角不等式:

\sqrt{(\left \langle \psi \right | +\left \langle \varphi  \right | ) \left ( \left | \psi  \right \rangle +\left | \varphi  \right \rangle  \right ) } \le \sqrt{\left \langle\varphi | \varphi \right \rangle } +\sqrt{\left \langle\psi | \psi \right \rangle }

3。外積(outer product)

\left | \psi  \right \rangle\left \langle \varphi  \right | or \left | \varphi   \right \rangle\left \langle \psi \right |

外積是算符

投影算符

\hat{p}_\varphi=\left |\varphi  \right \rangle \left \langle \varphi  \right |

\begin{eqnarray} \hat{p}_\varphi\left | a  \right \rangle & = & \left \langle \varphi | a \right \rangle \left |\varphi  \right \rangle \\ \left \langle b \right | \hat{p}_\varphi & = & \left \langle b |  \varphi \right \rangle \left \langle \varphi \right |  \end{eqnarray}

4。線性算符與反線性算符

(a)線性算符

\hat{A}(c_1\left |\varphi _1 \right \rangle +c_2\left |\varphi _2\right \rangle)=c_1\hat{A}\left |\varphi _1 \right \rangle +c_2\hat{A}\left |\varphi _2\right \rangle

(b)反線性算符

\hat{B}(c_1\left |\varphi _1 \right \rangle +c_2\left |\varphi _2\right \rangle)=c_1^*\hat{B}\left |\varphi _1 \right \rangle +c_2^*\hat{B}\left |\varphi _2\right \rangle

5。厄米共軛算符與厄米算符、反厄米算符

(a)厄米共軛算符

\left | \psi

,定義

\left \langle  \psi

\hat{A}^\dagger

為算符

\hat{A}

的厄米共軛算符

對於任意的

\left | \varphi \right \rangle ,\left | \psi \right \rangle

,若算符

\hat{B}

滿足

\left \langle \psi \right | \hat{B}\left | \varphi \right \rangle =\left \langle \varphi \right | \hat{A}\left | \psi \right \rangle^*

,則

\hat{B }

\hat{A}

的厄米共軛算符,即

\hat{B}=\hat{A}^ \dagger

\left\{\begin{array}{l} (\hat{A} \hat{B})^{\dagger}=\hat{B}^{\dagger} \hat{A}^{\dagger} \\ \left(\hat{A}^{\dagger}\right)^{\dagger}=A \\ (|\psi\rangle\langle\varphi|)^{\dagger}=|\varphi\rangle\langle\psi| \end{array}\right.

(b)厄米算符(Hermitian operator)

\hat{A}=\hat{A}^ \dagger

 \hat{A}\left | a_i  \right \rangle  =a_i\left | a_i  \right \rangle,i=1,2,...

本徵值

a_i

為實數

正交性:厄米算符屬於不同本徵值得本徵態正交

\left \langle a_i |a_j  \right \rangle =\delta_{ij}

完備性:

 \left | \psi  (t)\right \rangle =\sum_{i}c_i(t)\left | a_i  \right \rangle

(c)反厄米算符

\hat{A}^ \dagger =-\hat{A}

6。么正算符

\hat{U}

(Unitary operator)

定義式:

\hat{U}^\dagger= \hat{U}^{-1}

\hat{U} \hat{U}^{\dagger}=\hat{U}^{+} \hat{U}=\mathbb{1}(單位算符)

么正算符作用下

\hat{U}|\psi\rangle \equiv\left|\psi^{\prime}\right\rangle, \quad U|\varphi\rangle \equiv\left|\varphi^{\prime}\right\rangle

,內積保持不變

\left\langle\psi^{\prime} \mid \varphi^{\prime}\right\rangle=\left\langle\psi\left|\hat{U}^{\dagger} \hat{U}\right| \varphi\right\rangle =\langle\psi \mid \varphi\rangle

么正算符的本徵值為模為1的複數

7。正規算符

\hat{N}

(normal operator)

定義式:

[\hat{N},\hat{N}^\dagger] =0

厄米算符、么正算符是正規算符

正規算符屬於其不同本徵值的不同本徵態正交

\hat{N}|n\rangle=n|n\rangle, \hat{N}|m\rangle=m|m\rangle  \Rightarrow \langle n \mid m\rangle=0 \quad(n \neq m)

譜展開

\hat{N}=\sum_{n} n|n\rangle\langle n|

看完不妨雙擊螢幕點個贊~

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