# 全同粒子

## 量子力學的描述

### 對稱與反對稱量子態

$|n_1\rang |n_2\rang \,\!$

$|n_1, n_2; S\rang \equiv \frac{1}{\sqrt{2}} \times \bigg( |n_1\rang |n_2\rang + |n_2\rang |n_1\rang \bigg) \,\!$

$|n_1, n_2; A\rang \equiv \frac{1}{\sqrt{2}} \times \bigg( |n_1\rang |n_2\rang - |n_2\rang |n_1\rang \bigg) \,\!$

$|n_1\rang |n_2\rang \ne |n_2\rang |n_1\rang\,\!$

### 交換對稱性

$|n_1, n_2; ?\rang = \mbox{constant} \times \bigg( |n_1\rang |n_2\rang + i |n_2\rang |n_1\rang \bigg) \,\!$

$P (|\psi\rang |\phi\rang ) \equiv |\phi\rang |\psi\rang \,\!$

$P|n_1, n_2; S\rang = + |n_1, n_2; S\rang\,\!$
$P|n_1, n_2; A\rang = - |n_1, n_2; A\rang\,\!$

$H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + U(|x_1 - x_2|) + V(x_1) + V(x_2) \,\!$

$\left[P, H\right] = 0\,\!$

### N 個粒子

$|n_1,\,n_2,\,\cdots,\,n_N; S\rang =\sqrt{\frac{\prod_j N_j!}{N!}} \sum_{p\in \mathrm{permutation}(N)} |n_{p_1}\rang |n_{p_2}\rang \cdots |n_{p_N}\rang \,\!$ ;

$|n_1,\,n_2,\,\cdots,\,n_N; A\rang = \frac{1}{\sqrt{N!}} \sum_{p\in \mathrm{permutation}(N)} \mathrm{sign}(p)|n_{p_1}\rang |n_{p_2}\rang \cdots |n_{p_N}\rang\ \,\!$

$\lang n_1,\,n_2,\,\cdots,\,n_N;S|n_1,\,n_2,\,\cdots,\,n_N; S\rang=1\,\!$
$\lang n_1,\,n_2,\,\cdots,\,n_N;A|n_1,\,n_2,\,\cdots,\,n_N; A\rang=1\,\!$

### 斯萊特行列式

$\Psi^{(A)}_{n_1 \cdots n_N} (x_1, \cdots x_N) = \frac{1}{\sqrt{N!}} \left| \begin{matrix} \psi_{n_1}(x_1) & \psi_{n_1}(x_2) & \cdots & \psi_{n_1}(x_N) \\ \psi_{n_2}(x_1) & \psi_{n_2}(x_2) & \cdots & \psi_{n_2}(x_N) \\ \cdots & \cdots & \cdots & \cdots \\ \psi_{n_N}(x_1) & \psi_{n_N}(x_2) & \cdots & \psi_{n_N}(x_N) \\ \end{matrix} \right| \,\!$

### 範例

#### 二個全同玻色子

\begin{align}|1,\,1; S\rang & = \frac{1}{\sqrt{2!2!}} (|1\rang |1\rang+|1\rang |1\rang) \\ & = |1\rang |1\rang \\ \end{align}\,\!
\begin{align} |1,\,2; S\rang & = \frac{1}{\sqrt{2!1!1!}} (|1\rang |2\rang+|2\rang |1\rang) \\ & = \frac{1}{\sqrt{2}} (|1\rang |2\rang+|2\rang |1\rang) \\ \end{align}\,\!

#### 三個全同玻色子

\begin{align}|1,\,1,\,1; S\rang & = \frac{1}{\sqrt{3!3!}} (|1\rang|1\rang |1\rang+|1\rang|1\rang |1\rang+|1\rang|1\rang|1\rang \\ & \qquad\qquad+|1\rang|1\rang |1\rang+|1\rang|1\rang |1\rang+|1\rang|1\rang |1\rang) \\ & = |1\rang |1\rang|1\rang \\ \end{align}\,\!
\begin{align} |1,\,1,\,2; S\rang & = \frac{1}{\sqrt{3!2!1!}} (|1\rang|1\rang|2\rang+|1\rang|2\rang|1\rang+|1\rang|1\rang|2\rang \\ & \qquad\qquad+|1\rang|2\rang|1\rang+|2\rang|1\rang|1\rang+|2\rang|1\rang|1\rang) \\ & = \frac{1}{\sqrt{3}} (|1\rang|1\rang|2\rang+|1\rang|2\rang|1\rang+|2\rang|1\rang|1\rang) \\ \end{align}\,\!
\begin{align} |1,\,2,\,3; S\rang & =\frac{1}{ \sqrt{3!1!1!1!}} (|1\rang|2\rang|3\rang+|1\rang|3\rang|2\rang+|2\rang|1\rang|3\rang \\ & \qquad\qquad+|2\rang|3\rang|1\rang+|3\rang|1\rang|2\rang+|3\rang|2\rang|1\rang) \\ & = \frac{1}{\sqrt{6}} (|1\rang|2\rang|3\rang+|1\rang|3\rang|2\rang+|2\rang|1\rang|3\rang \\ & \qquad\qquad+|2\rang|3\rang|1\rang+|3\rang|1\rang|2\rang+|3\rang|2\rang|1\rang) \\ \end{align}\,\!

#### 二個全同費米子

\begin{align}|1,\,1; A\rang & = \frac{1}{\sqrt{2!}} (|1\rang |1\rang - |1\rang |1\rang) \\ & = 0 \\ \end{align}\,\!
\begin{align} |1,\,2; A\rang & = \frac{1}{\sqrt{2!}} (|1\rang |2\rang - |2\rang |1\rang) \\ & = \frac{1}{\sqrt{2}} (|1\rang |2\rang - |2\rang |1\rang) \\ \end{align}\,\!

#### 三個全同費米子

\begin{align}|1,\,1,\,1; A\rang & = \frac{1}{\sqrt{3!}} (|1\rang|1\rang |1\rang - |1\rang|1\rang |1\rang+|1\rang|1\rang |1\rang \\ & \qquad\qquad- |1\rang|1\rang |1\rang+|1\rang|1\rang |1\rang - |1\rang|1\rang |1\rang) \\ & =0 \\ \end{align}\,\!
\begin{align} |1,\,1,\,2; A\rang & = \frac{1}{\sqrt{3!}} (|1\rang|1\rang|2\rang - |1\rang|2\rang|1\rang - |1\rang|1\rang|2\rang \\ & \qquad\qquad+|1\rang|2\rang|1\rang+|2\rang|1\rang|1\rang - |2\rang|1\rang|1\rang) \\ & =0 \\ \end{align}\,\!
\begin{align} |1,\,2,\,3; A\rang & =\frac{1}{ \sqrt{3!}} (|1\rang|2\rang|3\rang - |1\rang|3\rang|2\rang - |2\rang|1\rang|3\rang \\ & \qquad\qquad+|2\rang|3\rang|1\rang+|3\rang|1\rang|2\rang - |3\rang|2\rang|1\rang) \\ & = \frac{1}{\sqrt{6}} (|1\rang|2\rang|3\rang - |1\rang|3\rang|2\rang - |2\rang|1\rang|3\rang \\ & \qquad\qquad+|2\rang|3\rang|1\rang+|3\rang|1\rang|2\rang - |3\rang|2\rang|1\rang) \\ \end{align}\,\!

## 統計性質

• $|0\rangle|0\rangle\,\!$
• $|1\rangle|1\rangle\,\!$
• $|0\rangle|1\rangle\,\!$
• $|1\rangle|0\rangle\,\!$

• $|0\rangle|0\rangle\,\!$
• $|1\rangle|1\rangle\,\!$
• $\frac{1}{\sqrt{2}}(|0\rangle|1\rangle + |1\rangle|0\rangle)\,\!$

$\frac{1}{\sqrt{2}}(|0\rangle|1\rangle - |1\rangle|0\rangle)\,\!$

## 註釋

1. ^ 反對稱性波函數為 $[\sin(x)sin(3y)-sin(3x)sin(y)]/\sqrt{2},\qquad 0\le x,y \le \pi$ 。注意到在 $x=y$ 附近，機率輻絕對值很微小，兩個費米子趨向於彼此互相遠離對方。
2. ^ 對稱性波函數為 $-[\sin(x)sin(3y)+sin(3x)sin(y)]/\sqrt{2},\qquad 0\le x,y \le \pi$ 。注意到在 $x=y$ 附近，機率輻絕對值較大，兩個費米子趨向於彼此互相接近對方。