這是一個Mathematica程式碼備忘錄,量子力學基礎略過

l

=

5

Jz

=

DiagonalMatrix

Reverse

Table

x

{

x

-

l

l

}]]];

Data

=

Table

Sqrt

l

l

+

1

-

x

^

2

+

Abs

x

]],

{

x

-

l

l

}];

Mat

=

DiagonalMatrix

Delete

Data

FirstPosition

Data

Max

Data

]]]];

Jm

=

Reverse

Map

Append

#

0

&

Append

Reverse

Mat

],

Table

0

{

x

2

l

}]]]];

Jp

=

Transpose

Jm

];

Jx

=

Jp

+

Jm

/

2

Jy

=

Jp

-

Jm

/

2

I

);

Jp

//

MatrixForm

Jm

//

MatrixForm

Jz

//

MatrixForm

Jy

//

MatrixForm

Jx

//

MatrixForm

l

為角量子數,

\hbar

為約化普朗克常數

l=\frac{1}{2}

時,

J_x、J_y、J_z

即電子自旋角動量算符

J_{x(\frac{1}{2})}=\hbar \left( \begin{array}{cc}  0 & 0.5 \\  0.5 & 0 \\ \end{array} \right)

J_{y(\frac{1}{2})}=\hbar\left( \begin{array}{cc}  0 & -0.5 i \\ 0.5 i & 0 \\ \end{array} \right)

J_{z(\frac{1}{2})}=\hbar\left( \begin{array}{cc}  0.5 & 0\\  0 & -0.5 \\ \end{array} \right)

l=1

時,即各種教材上熟知的:

J_{x(1)}=\hbar\left( \begin{array}{ccc}  0 & \frac{1}{\sqrt{2}} & 0 \\  \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\  0 & \frac{1}{\sqrt{2}} & 0 \\ \end{array} \right)

J_{y(1)}=\hbar\left( \begin{array}{ccc}  0 & -\frac{i}{\sqrt{2}} & 0 \\  \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\  0 & \frac{i}{\sqrt{2}} & 0 \\ \end{array} \right)

J_{z(1)}=\hbar\left( \begin{array}{ccc}  1 & 0 & 0 \\  0 & 0 & 0 \\  0 & 0 & -1 \\ \end{array} \right)

以此類推,

l=2

時:

J_{x(2)}=\hbar\left( \begin{array}{ccccc}  0 & 1 & 0 & 0 & 0 \\  1 & 0 & \sqrt{\frac{3}{2}} & 0 & 0 \\  0 & \sqrt{\frac{3}{2}} & 0 & \sqrt{\frac{3}{2}} & 0 \\  0 & 0 & \sqrt{\frac{3}{2}} & 0 & 1 \\  0 & 0 & 0 & 1 & 0 \\ \end{array} \right)

J_{y(2)}=\hbar\left( \begin{array}{ccccc}  0 & -i & 0 & 0 & 0 \\  i & 0 & -i \sqrt{\frac{3}{2}} & 0 & 0 \\  0 & i \sqrt{\frac{3}{2}} & 0 & -i \sqrt{\frac{3}{2}} & 0 \\  0 & 0 & i \sqrt{\frac{3}{2}} & 0 & -i \\  0 & 0 & 0 & i & 0 \\ \end{array} \right)

J_{z(2)}=\hbar\left( \begin{array}{ccccc}  2 & 0 & 0 & 0 & 0 \\  0 & 1 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & -1 & 0 \\  0 & 0 & 0 & 0 & -2 \\ \end{array} \right)

l=3

時:

J_{x(3)}=\hbar\left( \begin{array}{ccccccc}  0 & \sqrt{\frac{3}{2}} & 0 & 0 & 0 & 0 & 0 \\  \sqrt{\frac{3}{2}} & 0 & \sqrt{\frac{5}{2}} & 0 & 0 & 0 & 0 \\  0 & \sqrt{\frac{5}{2}} & 0 & \sqrt{3} & 0 & 0 & 0 \\  0 & 0 & \sqrt{3} & 0 & \sqrt{3} & 0 & 0 \\  0 & 0 & 0 & \sqrt{3} & 0 & \sqrt{\frac{5}{2}} & 0 \\  0 & 0 & 0 & 0 & \sqrt{\frac{5}{2}} & 0 & \sqrt{\frac{3}{2}} \\  0 & 0 & 0 & 0 & 0 & \sqrt{\frac{3}{2}} & 0 \\ \end{array} \right)

J_{y(3)}=\hbar\left( \begin{array}{ccccccc}  0 & -i \sqrt{\frac{3}{2}} & 0 & 0 & 0 & 0 & 0 \\  i \sqrt{\frac{3}{2}} & 0 & -i \sqrt{\frac{5}{2}} & 0 & 0 & 0 & 0 \\  0 & i \sqrt{\frac{5}{2}} & 0 & -i \sqrt{3} & 0 & 0 & 0 \\  0 & 0 & i \sqrt{3} & 0 & -i \sqrt{3} & 0 & 0 \\  0 & 0 & 0 & i \sqrt{3} & 0 & -i \sqrt{\frac{5}{2}} & 0 \\  0 & 0 & 0 & 0 & i \sqrt{\frac{5}{2}} & 0 & -i \sqrt{\frac{3}{2}} \\  0 & 0 & 0 & 0 & 0 & i \sqrt{\frac{3}{2}} & 0 \\ \end{array} \right)

J_{z(3)}=\hbar\left( \begin{array}{ccccccc}  3 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 2 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 1 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & -1 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & -2 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & -3 \\ \end{array} \right)

l=5

時:(在知乎Markdown公式編輯器的效能邊緣試探)

J_{x(5)}=\hbar\left( \begin{array}{ccccccccccc}  0 & \sqrt{\frac{5}{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \sqrt{\frac{5}{2}} & 0 & \frac{3}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & \frac{3}{\sqrt{2}} & 0 & \sqrt{6} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & \sqrt{6} & 0 & \sqrt{7} & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & \sqrt{7} & 0 & \sqrt{\frac{15}{2}} & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & \sqrt{\frac{15}{2}} & 0 & \sqrt{\frac{15}{2}} & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & \sqrt{\frac{15}{2}} & 0 & \sqrt{7} & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & \sqrt{7} & 0 & \sqrt{6} & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 & \frac{3}{\sqrt{2}} & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{3}{\sqrt{2}} & 0 & \sqrt{\frac{5}{2}} \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \sqrt{\frac{5}{2}} & 0 \\ \end{array} \right)

J_{y(5)}=\hbar\left( \begin{array}{ccccccccccc}  0 & -i \sqrt{\frac{5}{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  i \sqrt{\frac{5}{2}} & 0 & -\frac{3 i}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & \frac{3 i}{\sqrt{2}} & 0 & -i \sqrt{6} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & i \sqrt{6} & 0 & -i \sqrt{7} & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & i \sqrt{7} & 0 & -i \sqrt{\frac{15}{2}} & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & i \sqrt{\frac{15}{2}} & 0 & -i \sqrt{\frac{15}{2}} & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & i \sqrt{\frac{15}{2}} & 0 & -i \sqrt{7} & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & i \sqrt{7} & 0 & -i \sqrt{6} & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & i \sqrt{6} & 0 & -\frac{3 i}{\sqrt{2}} & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{3 i}{\sqrt{2}} & 0 & -i \sqrt{\frac{5}{2}} \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & i \sqrt{\frac{5}{2}} & 0 \\ \end{array} \right)

J_{z(5)}=\hbar\left( \begin{array}{ccccccccccc}  5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 0 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -3 & 0 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -4 & 0 \\  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -5 \\ \end{array} \right)

不同的角量子數取值可以導致不同的算符分量之間的特殊關係

唯一兩條與角量子數取值無關的分量關係為:

\vec{J}\times\vec{J}=i\hbar\vec{J}

(對易關係)

\hat{J}^2=l(l+1)\hbar^2I

(本徵方程)

(我覺得自己早應該先從矩陣的角度理解角動量算符)

還有上面那個龐大的矩陣主對角線兩側的數值分佈規律

在角量子數取值很大的情況下,散點圖即為:

關於角動量算符們

l=100

(其實基本上就是個圓了)

挖坑存程式碼,方便你我他

2019年2月28日續更

如果像自旋中對於Pauli矩陣的定義,重新定義算符:

\hat{\sigma}=\frac{\hat{J}}{l}

由此定義得到

\hat{\sigma_x}=\frac{\hat{J_x}}{l}

\hat{\sigma_y}=\frac{\hat{J_y}}{l}

\hat{\sigma_z}=\frac{\hat{J_z}}{l}

對於任意一個三維空間向量

\vec{p}=(p_x,p_y,p_z)

記此向量模長

p=\sqrt{p_x^2+p_y^2+p_z^2}

定義一個新的矩陣

\hat{A}=p_x\hat{\sigma_x}+p_y\hat{\sigma_y}+p_z\hat{\sigma_z}

則該矩陣的本徵值數量為

2l+1

,且均勻地分佈在此向量模長兩倍的線段上

例如,

l=\frac{1}{2}

時,

\hat{A}

的本徵值只有

\pm p

關於角動量算符們

l=1

時,本徵值有

\pm p,0

關於角動量算符們

l=\frac{3}{2}

時,本徵值有

\pm p,\pm\frac{1}{3}p

關於角動量算符們

l=2

時,本徵值有

\pm p,\pm\frac{1}{2}p,0

關於角動量算符們

以此類推