User: How to get the matrix of angular momentum quantum operators.
Assistant: To find the matrix representation of angular momentum quantum operators, you typically focus on the angular momentum operators \( L_x, L_y, L_z \) and the total angular momentum operator \( L^2 \). Here is a step-by-step outline of how to construct these matrices in the context of quantum mechanics:
-
Identify the Quantum State Basis:
Use the eigenstates of the \( L^2 \) and \( L_z \) operators. For a system with total angular momentum quantum number \( l \), the states are labeled as \( |l, m\rangle \), where \( m \) ranges from \(-l\) to \(l\).
-
Matrix Elements of \( L_z \):
The operator \( L_z \) is diagonal in the \( |l, m\rangle \) basis:
\[
\langle l, m' | L_z | l, m \rangle = \hbar m \delta_{m', m}
\]
Thus, the matrix representation of \( L_z \) is diagonal, with entries \( \hbar m \).
-
Matrix Elements of \( L^2 \):
Similarly, \( L^2 \) is also diagonal:
\[
\langle l, m' | L^2 | l, m \rangle = \hbar^2 l(l + 1) \delta_{m', m}
\]
The matrix will have the same structure, with the eigenvalue \( \hbar^2 l(l + 1) \).
-
Matrix Elements of \( L_x \) and \( L_y \):
These operators can be expressed in terms of the raising and lowering operators \( L_+ \) and \( L_- \):
\[
L_+ = \hbar \sqrt{(l - m)(l + m + 1)} |l, m + 1\rangle \langle l, m|
\]
\[
L_- = \hbar \sqrt{(l + m)(l - m + 1)} |l, m - 1\rangle \langle l, m|
\]
The relations for \( L_x \) and \( L_y \) are:
\[
L_x = \frac{1}{2}(L_+ + L_-)
\]
\[
L_y = \frac{1}{2i}(L_+ - L_-)
\]
-
Constructing the Matrices:
For a given \( l \), create the \( (2l + 1) \times (2l + 1) \) matrices for \( L_x \) and \( L_y \) by calculating the appropriate matrix elements using the above definitions. Each matrix will be filled based on the transition probabilities defined by the raising and lowering operators.
-
Example for \( l = 1 \):
For \( l = 1 \), the basis states are \( |1, 1\rangle, |1, 0\rangle, |1, -1\rangle \).
Construct \( L_z \) and \( L^2 \) as diagonal matrices:
\[
L_z = \hbar \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{pmatrix}, \quad
L^2 = \hbar^2 2 \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]
Calculate \( L_+ \) and \( L_- \) using the raising and lowering operators.
Assistant: By following these steps, you can derive the matrix representations of the angular momentum operators for any integer or half-integer \( l \).
User: Please also give me \( l = \frac{1}{2}, 1, \frac{3}{2} \). \( L_x, L_y, L_z, L_+ \) and \( L_- \) matrix form.
Assistant: Sure! Let us calculate the matrix forms of the angular momentum operators \( L_x, L_y, L_z, L_+, \) and \( L_- \) for \( l = \frac{1}{2}, 1, \) and \( \frac{3}{2} \).
For \( l = \frac{1}{2} \)
Basis States: \( | \frac{1}{2}, \frac{1}{2} \rangle, | \frac{1}{2}, -\frac{1}{2} \rangle \)
Matrix Forms:
- \( L_z \)
\[
L_z = \hbar \begin{pmatrix}
\frac{1}{2} & 0 \\
0 & -\frac{1}{2}
\end{pmatrix}
\]
- \( L^2 \)
\[
L^2 = \hbar^2 \frac{3}{4} \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\]
- \( L_+ \)
\[
L_+ = \hbar \begin{pmatrix}
0 & \sqrt{1} \\
0 & 0
\end{pmatrix}
\]
- \( L_- \)
\[
L_- = \hbar \begin{pmatrix}
0 & 0 \\
\sqrt{1} & 0
\end{pmatrix}
\]
- \( L_x \) and \( L_y \)
Using \( L_x = \frac{1}{2}(L_+ + L_-) \) and \( L_y = \frac{1}{2i}(L_+ - L_-) \):
\[
L_x = \frac{\hbar}{2} \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\]
\[
L_y = \frac{\hbar}{2} \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\]
For \( l = 1 \)
Basis States: \(|1, 1\rangle, |1, 0\rangle, |1, -1\rangle \)
Matrix Forms:
- \( L_z \)
\[
L_z = \hbar \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{pmatrix}
\]
- \( L^2 \)
\[
L^2 = \hbar^2 2 \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]
- \( L_+ \)
\[
L_+ = \hbar \begin{pmatrix}
0 & \sqrt{2} & 0 \\
0 & 0 & \sqrt{2} \\
0 & 0 & 0
\end{pmatrix}
\]
- \( L_- \)
\[
L_- = \hbar \begin{pmatrix}
0 & 0 & 0 \\
\sqrt{2} & 0 & 0 \\
0 & \sqrt{2} & 0
\end{pmatrix}
\]
- \( L_x \) and \( L_y \)
\[
L_x = \frac{\hbar}{2} \begin{pmatrix}
0 & \sqrt{2} & 0 \\
\sqrt{2} & 0 & \sqrt{2} \\
0 & \sqrt{2} & 0
\end{pmatrix}
\]
\[
L_y = \frac{\hbar}{2} \begin{pmatrix}
0 & -i\sqrt{2} & 0 \\
i\sqrt{2} & 0 & -i\sqrt{2} \\
0 & i\sqrt{2} & 0
\end{pmatrix}
\]
For \( l = \frac{3}{2} \)
Basis States: \( | \frac{3}{2}, \frac{3}{2} \rangle, | \frac{3}{2}, \frac{1}{2} \rangle, | \frac{3}{2}, -\frac{1}{2} \rangle, | \frac{3}{2}, -\frac{3}{2} \rangle \)
Matrix Forms:
- \( L_z \)
\[
L_z = \hbar \begin{pmatrix}
\frac{3}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & -\frac{1}{2} & 0 \\
0 & 0 & 0 & -\frac{3}{2}
\end{pmatrix}
\]
- \( L^2 \)
\[
L^2 = \hbar^2 \frac{15}{4} \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\]
- \( L_+ \)
\[
L_+ = \hbar \begin{pmatrix}
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & \sqrt{2} & 0 \\
0 & 0 & 0 & \sqrt{3} \\
0 & 0 & 0 & 0
\end{pmatrix}
\]
- \( L_- \)
\[
L_- = \hbar \begin{pmatrix}
0 & 0 & 0 & 0 \\
\sqrt{3} & 0 & 0 & 0 \\
0 & \sqrt{2} & 0 & 0 \\
0 & 0 & \sqrt{3} & 0
\end{pmatrix}
\]
- \( L_x \) and \( L_y \)
\[
L_x = \frac{\hbar}{2} \begin{pmatrix}
0 & \sqrt{3} & 0 & 0 \\
\sqrt{3} & 0 & \sqrt{2} & 0 \\
0 & \sqrt{2} & 0 & \sqrt{3} \\
0 & 0 & \sqrt{3} & 0
\end{pmatrix}
\]
\[
L_y = \frac{\hbar}{2} \begin{pmatrix}
0 & -i\sqrt{3} & 0 & 0 \\
i\sqrt{3} & 0 & -i\sqrt{2} & 0 \\
0 & i\sqrt{2} & 0 & -i\sqrt{3} \\
0 & 0 & i\sqrt{3} & 0
\end{pmatrix}
\]
Assistant: These matrices represent the angular momentum operators for the specified values of \( l \). The structure of the matrices changes with the value of \( l \) due to the increasing number of basis states.