Properties of the matrix formalism for l = 1
Task number: 4579
We introduce the basis notation \(|l, \, m\rangle\) for the Hilbert space with \(l = 1\)
\[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \equiv |1, \, 1\rangle \, , \qquad \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \equiv |1, \, 0\rangle \, , \qquad \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \equiv |1, \, -1\rangle \, . \]In this basis, the angular momentum component operators and the squared angular momentum operator are represented by the matrices
\[ \hat L_x = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \, , \qquad \hat L_y = \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \, , \] \[ \hat L_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, , \qquad \hat L^2 = 2 \hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \, . \]Why are \(\hat L_z\) and \(\hat L^2\) diagonal? What is the meaning of the numbers on the diagonal?
Verify the validity of the following relations in the matrix representation \[\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2\,,\] \[\left [\hat L_j, \hat L_k \right ] = i \hbar \varepsilon_{jkl} \hat L_l\,,\] \[\left [\hat L_j, \hat L^2 \right ] = 0\,,\] \[\hat L_z \, |l, \, m\rangle = m\hbar \, |l, \, m\rangle\,,\] \[\hat L^2 |l, \, m\rangle = \hbar^2 \, l(l + 1) \, |l, \, m\rangle\,.\]
Note: The derivation of the matrix representation of these operators can be found in Matrix formalism for l = 1, Solution b).
Hint
Recall the definition of the commutator of two operators.
Solution
Since the given operators are expressed in the \(|l, \, m\rangle\) basis of eigenstates of \(\hat L_z\) and \(\hat L^2\), their matrices are diagonal. For the same reason, the elements on the diagonal are the eigenvalues of the respective operators.
To verify the relation \(\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2\), we directly substitute the matrices on the right‑hand side and simplify
\[ \hat L_x^2 + \hat L_y^2 + \hat L_z^2 = \] \[ = \left [ \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \right ]^2 + \left [ \frac{\hbar}{\sqrt2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \right ]^2 + \left [ \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \right ]^2 = \] \[ = \frac{\hbar^2}{2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} + \frac{\hbar^2}{2} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} + \] \[ + \hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} = \] \[ = \frac{\hbar^2}{2} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix} + \frac{\hbar^2}{2} \begin{pmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{pmatrix} + \frac{\hbar^2}{2} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \] \[ = \frac{\hbar^2}{2} \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = 2\hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \hat L^2 \, . \]This confirms the validity of the given relation in the matrix representation.
Next, we verify the relation \(\left [\hat L_j, \hat L_k \right ] = i \hbar \varepsilon_{jkl} \hat L_l\). We perform the calculation for the components \(\hat L_x, \, \hat L_y\). Again, this is a direct calculation, where we substitute the matrices into the definition of the commutator (see Solution to hint) and simplify
\[ \left [\hat L_x, \hat L_y \right ] = \hat L_x \hat L_y - \hat L_y \hat L_x = \] \[ = \frac{\hbar^2}{2} \left [ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} - \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \right ] = \] \[ = \frac{\hbar^2}{2} \left [ \begin{pmatrix} i & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & -i \end{pmatrix} - \begin{pmatrix} -i & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & i \end{pmatrix} \right ] = \frac{\hbar^2}{2} \begin{pmatrix} 2i & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2i \end{pmatrix} = \] \[ = i\hbar \, \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} = i\hbar \hat L_z \, . \]For the remaining two commutators, the calculation proceeds analogously. Thus, the commutation relations of the angular momentum components in the matrix representation retain the same form as in the coordinate representation.
We illustrate the calculation of the commutator between an angular momentum component and the square of the angular momentum operator for the \(x\)‑component. This is again done by direct substitution and simplification
\[ \left [\hat L_x, \hat L^2 \right ] = \hat L_x \hat L^2 - \hat L^2 \hat L_x = \] \[ = \frac{2\hbar^3}{\sqrt2} \left [ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \right ] = \] \[ = \frac{2\hbar^3}{\sqrt2} \left [ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \right ] = 0 \, . \]For the remaining two commutators, the calculation proceeds analogously. Any component of the angular momentum operator thus commutes with the square of the angular momentum operator even in the matrix representation.
We verify the eigenvalue relation for the \(z\)‑component of the angular momentum operator for \(|1, \, 1\rangle\). We directly substitute into the left‑hand side and simplify
\[ \hat L_z \, |1, \, 1\rangle = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \hbar \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \hbar |1, \, 1\rangle \, . \]Since we know that \(m = 1\), this confirms that
\[ \hat L_z \, |1, \, 1\rangle = 1 \cdot \hbar \, |1, \, 1\rangle \, . \]The calculation for \(m = 0, \, m = -1\) is completely analogous. Altogether, we have \(\hat L_z \, |l, \, m\rangle = m\hbar \, |l, \, m\rangle\).
The relation for the eigenvalues of the square of the angular momentum operator is again verified for \(|1, \, 1\rangle\) by direct substitution into the left‑hand side and simplification
\[ \hat L^2 |1, \, 1\rangle = 2 \hbar^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 2 \hbar^2 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 2 \hbar^2 |1, \, 1\rangle \, . \]Since the calculation was done for \(l = 1\), we can write
\[ \hat L^2 |1, \, 1\rangle = \hbar^2 \, l(l+1) \, |1, \, 1\rangle \, . \]The calculation for \(m = 0, \, m = -1\) is completely analogous. Altogether, we have \(\hat L^2 \, |l, \, m\rangle = \hbar^2 \, l(l+1) \, |l, \, m\rangle\).
Answer
The operators \(\hat L_z, \, \hat L^2\) are diagonal because they are expressed in the basis of their eigenstates. The elements on the diagonal are their eigenvalues.
All the relations given in the task assignment, which we know to hold in the coordinate representation, are also valid in the matrix representation.
