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:
  1. 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 \(l\).

  2. 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 \( \hbar m \).

  3. 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 \( \hbar^2 l(l + 1) \).

  4. 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:

    \[ L_x = \frac{1}{2}(L_+ + L_-) \]
    \[ L_y = \frac{1}{2i}(L_+ - L_-) \]
  5. 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.

  6. 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:

  1. \( L_z \)
    \[ L_z = \hbar \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \end{pmatrix} \]
  2. \( L^2 \)
    \[ L^2 = \hbar^2 \frac{3}{4} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
  3. \( L_+ \)
    \[ L_+ = \hbar \begin{pmatrix} 0 & \sqrt{1} \\ 0 & 0 \end{pmatrix} \]
  4. \( L_- \)
    \[ L_- = \hbar \begin{pmatrix} 0 & 0 \\ \sqrt{1} & 0 \end{pmatrix} \]
  5. \( L_x \) 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:

  1. \( L_z \)
    \[ L_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \]
  2. \( L^2 \)
    \[ L^2 = \hbar^2 2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
  3. \( L_+ \)
    \[ L_+ = \hbar \begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix} \]
  4. \( L_- \)
    \[ L_- = \hbar \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix} \]
  5. \( L_x \) 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:

  1. \( L_z \)
    \[ L_z = \hbar \begin{pmatrix} \frac{3}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{3}{2} \end{pmatrix} \]
  2. \( L^2 \)
    \[ L^2 = \hbar^2 \frac{15}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
  3. \( L_+ \)
    \[ L_+ = \hbar \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
  4. \( L_- \)
    \[ L_- = \hbar \begin{pmatrix} 0 & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & 0 \\ 0 & \sqrt{2} & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]
  5. \( L_x \) 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.