16.1. Spatial Angular Momentum

Promoting the classical definition of angular momentum \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\), the angular momentum operator in quantum mechanics reads

\[\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}} = \hat{\mathbf{e}}_a \varepsilon_{abc} \hat{r} \hat{p}_c \ ,\]

and using space basis

\[\langle \mathbf{r} | \hat{\mathbf{L}} = \hat{\mathbf{e}}_a \underbrace{\left( - i \, \hbar \, \varepsilon_{abc}\, r_b \, \partial_c \langle \mathbf{r} | \right) }_{= \langle \mathbf{r} | \hat{L}_a }\ .\]

Properties

Composition of 2 components of the angular momentum operator. Using Cartesian coordinates

\[\begin{split}\begin{aligned} \hat{L}_a \hat{L}_b & = - \hbar^2 r_b \partial_a + \hbar^2 \delta_{ab} r_c \partial_c - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ \end{aligned}\end{split}\]

Details

\[\begin{split}\begin{aligned} \hat{L}_a \hat{L}_b & = - \hbar^2 \varepsilon_{acd} r_c \partial_d \left( \varepsilon_{bef} r_e \partial_f \right) = \\ & = - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c \left( \delta_{de} \partial_f + r_e \partial_{df} \right) = \\ & = - \hbar^2 \varepsilon_{acd} \varepsilon_{bdf} r_c \partial_f - \hbar^2 \varepsilon_{acd} \varepsilon_{bdf} r_c r_e \partial_{df} = \\ & = - \hbar^2 \left( \delta_{af} \delta_{cb} - \delta_{ab} \delta_{cf} \right) r_c \partial_f - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ & = - \hbar^2 r_b \partial_a + \hbar^2 \delta_{ab} r_c \partial_c - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ \end{aligned}\end{split}\]

The last term is symmetric w.r.t. \(a\), \(b\).

Commutators.

\[[ \hat{L}_a, \hat{L}_b ] = \hat{L}_a \hat{L}_b - \hat{L}_b \hat{L}_a = - \hbar^2 ( r_b\partial_a - r_a \partial_b) \]

If \(a = b\), the obvious result follows, as an operator commutes with itself. If \(a \ne b\),

\[[\hat{L}_a, \hat{L}_b ] = i \varepsilon_{abc} \hbar \hat{L}_c \ ,\]

or with a compact “vector notation”, \(\hat{\mathbf{L}} \times \hat{\mathbf{L}} = i \hbar \hat{\mathbf{L}}\).

Magnitude of the angular momentum, \(|\hat{L}|^2\). Using Cartesian coordinates,

\[\begin{aligned} |\hat{L}|^2 & = 2 \hbar^2 r_a \partial_a - \hbar^2 r_b r_b \partial_{aa} + \hbar^2 r_a r_b \partial_{ab} \ . \end{aligned}\]

Details

\[\begin{split}\begin{aligned} |\hat{L}|^2 & = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \\ & = \hat{L}_a \hat{L}_a = \\ & = -\hbar^2 r_a \partial_a + 3 \hbar^2 r_a \partial_a - \hbar^2 ( \delta_{ce} \delta_{df} - \delta_{cf} \delta_{de} ) r_c r_e \partial_{df} = \\ & = 2 \hbar^2 r_a \partial_a - \hbar^2 r_b r_b \partial_{aa} + \hbar^2 r_a r_b \partial_{ab} \ . \end{aligned}\end{split}\]

Commutations with magnitude.

\[[ \hat{L}^2, \hat{L}_a ] = 0\]

Details

The same hold for any integer power of \(\hat{L}_a\), like \(\hat{L}_a^2\), as it can be easily proved

\[\begin{split}\begin{aligned} \left[ \hat{L}^2, \hat{L}^2_a \right] & = \hat{L}^2 \hat{L}_a^2 - \hat{L}_a^2 \hat{L}^2 = \\ & = \hat{L}^2 \hat{L}_a \hat{L}_a - \hat{L}_a^2 \hat{L}^2 = \\ & = \hat{L}_a \hat{L}^2 \hat{L}_a - \hat{L}_a \hat{L}_a \hat{L}^2 = \\ & = \hat{L}_a [ \hat{L}^2, \hat{L}_a ] = 0 \ . \end{aligned}\end{split}\]

Analogous relations in classical mechanics

\[\begin{split} [ L_a, L_b ] = i \hbar \varepsilon_{abc} \hat{L}_c \qquad , \qquad \{ L_a, L_b \} = \varepsilon_{abc} L_c \\ \end{split}\]

provides another example of the relation between Poisson brackets in classical mechanics and commutator in quantum mechanics (Dirac)

\[\{ A, B \} \quad \leftrightarrow \quad \frac{1}{i \hbar} [ \hat{A}, \hat{B} ]\]

16.1.1. Eigenproblem of angular momentum operators

In this section, the eigenproblems of \(\hat{L}^2\) and \(\hat{L}_z\) are discussed. Here, all the operators and state functions are written in components using space basis, without explicitly writing the projection \(\langle \mathbf{r} |\).

• Using spherical coordinates, the \(2\pi\)-periodicity of the problem in \(\phi\) implies quantization in the spectral decomposition of \(\hat{L}_z\),

\(\langle \mathbf{r} | \hat{\mathbf{L}}\ \) using Cartesian coordinates.

\[\ell \psi = \hat{L}_a \psi = - i \hbar \varepsilon_{abc} r_b \partial_c \psi \ , \]

or \(\ell \psi = - i \hbar \mathbf{r} \times \nabla \psi\).

…

\(\langle \mathbf{r} | \hat{\mathbf{L}}\ \) using spherical coordinates.

With the set of spherical coordinates \(\psi(r,\theta,\phi)\)

\[\begin{split}\begin{aligned} \mathbf{r} \times \nabla \psi & = \mathbf{r} \times \left[ \partial_r \psi \, \hat{\mathbf{r}} + \frac{1}{r} \partial_\theta \psi \, \hat{\boldsymbol\theta} + \frac{1}{r \sin \theta} \partial_{\phi} \psi \hat{\boldsymbol\phi} \right] = \\ & = \partial_\theta \psi \, \hat{\boldsymbol\phi} - \frac{1}{\sin \theta} \partial_\phi \psi \, \hat{\boldsymbol\theta} \ . \end{aligned}\end{split}\]

Thus the \(z\)-component, with \(\hat{\mathbf{z}} \cdot \hat{\boldsymbol{\theta}} = - \sin \theta\), \(\hat{\mathbf{z}} \cdot \hat{\boldsymbol\phi} = 0\), can be written as

\[\hat{\mathbf{z}} \cdot \left( \mathbf{r} \times \nabla \psi \right) = \partial_\phi \psi\]

Spectral decomposition of the \(z\)-component operator, \(\hat{L}_z\)

\[\ell_z \psi_z(r,\theta,\phi) = -i \hbar \partial_\phi \psi_z(r,\theta,\phi)\]

The eigenfunctions read

\[\psi_{z,\ell}(r,\theta,\phi) = A(r,\theta) \, \exp\left( i \frac{\ell_z}{\hbar} \phi \right) \ .\]

Periodic condition \(\psi_z(0) = \psi_z(2 \pi)\) implies \(\frac{\ell_z}{\hbar} = m_l\), \(m_l \in \mathbb{Z}\).

Remark. The constraint \(m_l \in \mathbb{Z}\) is just one of the constraints on \(m_l\).

In order to keep the notation as intuitive as possible, the eigenvalue \(\ell_z\) and the corresponding eigenstate are denoted as \(\{ L_{z}, | L_z \rangle\} = \{ \hbar m_l, | m_l \rangle \}\), s.t. \(\hat{L}_z | m_l \rangle = \hbar m_l | m_l \rangle\).

Spectral decomposition of commuting operators

Let \(\hat{A} | a \rangle = a | a \rangle\), and \(\hat{B} | b \rangle = b | b \rangle\). If \(0 = [ \hat{A}, \hat{B} ] = \hat{A} \hat{B} - \hat{B} \hat{A}\). Pre-multiplying the first one by \(\hat{B}\),

\[a \hat{B} | a \rangle = \hat{B} \hat{A} | a \rangle = \hat{A} \hat{B} | a \rangle \ .\]

So that \(\hat{B} | a \rangle\) is an eigenstate of \(\hat{A}\) with eigenvalue \(a\).

• If \(a\) is an eigenvalue with algebraic multiplicity equal to \(1\), then \(\hat{B} | a \rangle\) must be proportional to the only eigenstate \(| a \rangle \) of the operator \(\hat{A}\) with eigenvalue \(a\), i.e.

  \[\hat{B} | a \rangle = \mu | a \rangle \ ,\]

  and thus \(| a \rangle\) is also an eigenstate of \(\hat{B}\).

• If \(a\) has algebraic multiplicity larger than \(1\), then \(\hat{B} | a_i \rangle\), can be a linear combination of all the eigenvectors \(| a_k \rangle\) with eigenvalues \(a_k = a\),

  \[\hat{B} | a_i \rangle = c_{ik} | a_k \rangle \ ,\]

  for \(\forall i\).

  …

Relation between \(\ \hat{L}_z\ \) and $\ \hat{L}^2 $

\[\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2\]

As \(\hat{L}^2\) commutes with the Cartesian components \(\hat{L}_a\), they share the same eigenvalues.The same holds for \(\hat{L}_a^2\)

\[\begin{aligned} \hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 \ , \end{aligned}\]

implies

\[\begin{aligned} \hat{L}^2 - \hat{L}_z^2 = \hat{L}_x^2 + \hat{L}_y^2 \ . \end{aligned}\]

Let \(\psi\) an eigenstate of \(\hat{L}^2\), \(\hat{L}_z\) with eigenvalues \(\ell^2\), \(\ell_z\), it follows

\[\left( \hat{L}_x^2 + \hat{L}_y^2 \right) \psi = \left( \hat{L}^2 - \hat{L}_z^2 \right) \psi = \left( \ell^2 - \ell_z^2 \right) \psi\]

As \(( \ell^2 - \ell_z^2 ) | \psi |^2 = \psi^* \hat{L}^2_x \psi = | \hat{L}_x \psi |^2 \ge 0\), it follows that \(\ell^2 - \ell^2_z \ge 0\), and thus \(-\ell \le \ell_z \le \ell\).

Combining

\[\begin{split}\begin{aligned} \left[\hat{L}_y, \hat{L}_z \right] & = \hat{L}_y \hat{L}_z - \hat{L}_z \hat{L}_y = i \hbar \hat{L}_x \\ \left[\hat{L}_z, \hat{L}_x \right] & = \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z = i \hbar \hat{L}_y \end{aligned}\end{split}\]

summing and subtracting \(i (1)\) and \((2)\) them

\[\begin{split}\begin{aligned} 0 & = - \hbar \hat{L}_x \mp i \hbar \hat{L}_y - i \hat{L}_y \hat{L}_z + i \hat{L}_z \hat{L}_y \pm \hat{L}_z \hat{L}_x \mp \hat{L}_x \hat{L}_z = \\ & = - ( \hat{L}_x \pm i \hat{L}_y ) ( \hbar \pm \hat{L}_z ) + \hat{L}_z ( i \hat{L}_y \pm \hat{L}_x ) \ , \end{aligned}\end{split}\]

and thus

\[\hat{L}_z ( i \hat{L}_y \pm \hat{L}_x ) = ( \hat{L}_x \pm i \hat{L}_y ) ( \hbar \pm \hat{L}_z )\]

or

\[\hat{L}_z ( \hat{L}_x \pm i \hat{L}_x ) = ( \hat{L}_x \pm i \hat{L}_y ) ( \hat{L}_z \pm \hbar )\]

Applying this operator to the eigenstate \(\psi\)

(16.1)\[\hat{L}_z ( \hat{L}_x \pm i \hat{L}_y ) \psi = ( \hat{L}_x \pm i \hat{L}_y ) ( \hat{L}_z \pm \hbar ) \psi = ( \ell_z \pm \hbar ) ( \hat{L}_x \pm i \hat{L}_y ) \psi\]

Thus, the state functions \(\psi^{\pm} = \left( \hat{L}_x \pm i \hat{L}_y \right) \psi\) are eigenstates of the operator \(\hat{L}_z\) with eigenvalues \(\ell_z \pm \hbar\). As \(\ell_z \in [ - \ell, \ell ]\), let’s call \(\ell_{z,min}\) and \(\ell_{z,max}\) the minimum and maximum values of \(\ell_z\). It follows that the operator \(\hat{L}^-\) has the smallest eigenvalue equal to \(\ell_{z,min} - \hbar\) and \(\hat{L}^+\) has the largest eigenvalue equal to \(\ell_{max,z} + \hbar\).

Thus, for \(\ell_{z,min}\), \(\ell_z - \hbar\) can’t be an eigenvalue of \(\hat{L}_z\), and thus the relation

\[\hat{L}_z \left( \hat{L}_x - i \hat{L}_y \right) \psi = ( \ell_{z,min} - \hbar ) ( \hat{L}_x - i \hat{L}_y ) \psi\]

implies that \(( \hat{L}_x - i \hat{L}_y ) \psi\) can’t be an eigenfunction of \(\hat{L}_z\), and thus it must be in the kernel or \(\hat{L}_z\), i.e.

\[0 = \hat{L}_z ( \hat{L}_x - i \hat{L}_y ) \psi_{\ell_z,min} = ( \hat{L}_x - i \hat{L}_y ) ( \hat{L}_z + \hbar ) \psi_{\ell_z,min} = ( \ell_{z,min} + \hbar ) ( \hat{L}_x - i \hat{L}_y ) \psi_{\ell,min}\]

A similar relation holds for the largest eigenvalue

\[0 = \hat{L}_z ( \hat{L}_x + i \hat{L}_y ) \psi_{\ell_z,max} = ( \hat{L}_x + i \hat{L}_y ) ( \hat{L}_z - \hbar ) \psi_{\ell_z,max} = ( \ell_{z,max} - \hbar ) ( \hat{L}_x + i \hat{L}_y ) \psi_{\ell,max}\]

Applying \(\hat{L}_x + i \hat{L}_y\) to the first equation, the operator becomes

\[( \hat{L}_x + i \hat{L}_y ) ( \hat{L}_x - i \hat{L}_y ) = \hat{L}_x^2 + \hat{L}_y^2 - i [ \hat{L}_x, \hat{L}_y ] = \hat{L}^2 - \hat{L}_z^2 + \hbar \hat{L}_z \ .\]

Applying \(\hat{L}_x - i \hat{L}_y\) to the second equation, the operator becomes

\[( \hat{L}_x - i \hat{L}_y ) ( \hat{L}_x + i \hat{L}_y ) = \hat{L}_x^2 + \hat{L}_y^2 + i [ \hat{L}_x, \hat{L}_y ] = \hat{L}^2 - \hat{L}_z^2 - \hbar \hat{L}_z \ .\]

Thus, the 2 equations are, for a given value of \(\ell\)

\[\begin{split}\begin{aligned} 0 & = ( \ell_{z,min} + \hbar ) ( \ell^2 - \ell_{z,min}^2 + \hbar \ell_{z,min} ) \\ 0 & = ( \ell_{z,max} - \hbar ) ( \ell^2 - \ell_{z,max}^2 - \hbar \ell_{z,max} ) \\ \end{aligned}\end{split}\]

Either \(\ell_{z;min,max} = \mp \hbar\), or

\[\begin{split}\begin{aligned} 0 & = \ell^2 - \ell_{z,min}^2 + \hbar \ell_{z,min} \\ 0 & = \ell^2 - \ell_{z,max}^2 - \hbar \ell_{z,max} \ , \end{aligned}\end{split}\]

and subtracting \(0 = (\ell_{z,max} + \ell_{z,min})(\ell_{z,max} - \ell_{z,min} + \hbar )\). As the content of the second bracket is always positive, then \(\ell_{z,min} = - \ell_{z,max}\).

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Following (16.1), if \(| \psi \rangle\) is the eigenvector of \(\hat{L}_z\) with eigenvalue \(\ell_z\), the wave functions \(\hat{L}^{\mp} | \psi \rangle\) are eigenvectors of \(\hat{L}_z\) with eigenvalues \(\ell_z \mp \hbar\). Thus, the set

\[\{ \ell_{z,min}, \ell_{z,min} + \hbar, \ell_{z,min} + 2 \hbar, \dots, \ell_{z,max}-\hbar, \ell_{z,max} \} = \{ - m_{l,max} \hbar, \dots , m_{l,max} \hbar \} \]

contains possible values of the eigenvalues of \(\hat{L}_z\).

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\[\lambda\left(L_z\right) = \hbar m_l \qquad m_l \in \]

Now, for \(\ell_{z,min} = -m_{l,max} \hbar\), and \(\ell_{z,max} = m_{l,max} \hbar\), the corresponding eigenvalues \(\ell^2\) are

\[\begin{aligned} \ell^2 = \hbar^2 m_{l,max} \left( m_{l,max} + 1 \right) \ . \end{aligned}\]