(quantum-mechanics:angular-momentum:spatial)= # 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 $$\langle \mathbf{r} | \hat{\mathbf{L}} = \hat{\mathbf{e}}_a \underbrace{\left( - i \, \hbar \, \varepsilon_{abc}\, r_b \, \partial_c \langle \mathbf{r} | \right) }_{= \langle \mathbf{r} | \hat{L}_a }\ .$$ ````{dropdown} Properties :open: **Composition of 2 components of the angular momentum operator.** Using Cartesian coordinates $$\begin{aligned} \hat{L}_a \hat{L}_b & = - \hbar^2 r_b \partial_a + \hbar^2 \delta_{ab} r_c \partial_c - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ \end{aligned}$$ ```{dropdown} Details $$\begin{aligned} \hat{L}_a \hat{L}_b & = - \hbar^2 \varepsilon_{acd} r_c \partial_d \left( \varepsilon_{bef} r_e \partial_f \right) = \\ & = - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c \left( \delta_{de} \partial_f + r_e \partial_{df} \right) = \\ & = - \hbar^2 \varepsilon_{acd} \varepsilon_{bdf} r_c \partial_f - \hbar^2 \varepsilon_{acd} \varepsilon_{bdf} r_c r_e \partial_{df} = \\ & = - \hbar^2 \left( \delta_{af} \delta_{cb} - \delta_{ab} \delta_{cf} \right) r_c \partial_f - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ & = - \hbar^2 r_b \partial_a + \hbar^2 \delta_{ab} r_c \partial_c - \hbar^2 \varepsilon_{acd} \varepsilon_{bef} r_c r_e \partial_{df} = \\ \end{aligned}$$ 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, $$\begin{aligned} |\hat{L}|^2 & = 2 \hbar^2 r_a \partial_a - \hbar^2 r_b r_b \partial_{aa} + \hbar^2 r_a r_b \partial_{ab} \ . \end{aligned}$$ ```{dropdown} Details $$\begin{aligned} |\hat{L}|^2 & = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \\ & = \hat{L}_a \hat{L}_a = \\ & = -\hbar^2 r_a \partial_a + 3 \hbar^2 r_a \partial_a - \hbar^2 ( \delta_{ce} \delta_{df} - \delta_{cf} \delta_{de} ) r_c r_e \partial_{df} = \\ & = 2 \hbar^2 r_a \partial_a - \hbar^2 r_b r_b \partial_{aa} + \hbar^2 r_a r_b \partial_{ab} \ . \end{aligned}$$ ``` **Commutations with magnitude.** $$[ \hat{L}^2, \hat{L}_a ] = 0$$ ```{dropdown} Details :open: ``` The same hold for any integer power of $\hat{L}_a$, like $\hat{L}_a^2$, as it can be easily proved $$\begin{aligned} \left[ \hat{L}^2, \hat{L}^2_a \right] & = \hat{L}^2 \hat{L}_a^2 - \hat{L}_a^2 \hat{L}^2 = \\ & = \hat{L}^2 \hat{L}_a \hat{L}_a - \hat{L}_a^2 \hat{L}^2 = \\ & = \hat{L}_a \hat{L}^2 \hat{L}_a - \hat{L}_a \hat{L}_a \hat{L}^2 = \\ & = \hat{L}_a [ \hat{L}^2, \hat{L}_a ] = 0 \ . \end{aligned}$$ Analogous relations in **classical mechanics** $$ [ L_a, L_b ] = i \hbar \varepsilon_{abc} \hat{L}_c \qquad , \qquad \{ L_a, L_b \} = \varepsilon_{abc} L_c \\ $$ provides another example of the relation between Poisson brackets in classical mechanics and commutator in quantum mechanics (**Dirac**) $$\{ A, B \} \quad \leftrightarrow \quad \frac{1}{i \hbar} [ \hat{A}, \hat{B} ]$$ ```` ## 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 $\hat{L}_z$, * ```{dropdown} **$\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 $\ell \psi = - i \hbar \mathbf{r} \times \nabla \psi$. ... ``` ```{dropdown} **$\langle \mathbf{r} | \hat{\mathbf{L}}\ $ using spherical coordinates.** With the set of spherical coordinates $\psi(r,\theta,\phi)$ $$\begin{aligned} \mathbf{r} \times \nabla \psi & = \mathbf{r} \times \left[ \partial_r \psi \, \hat{\mathbf{r}} + \frac{1}{r} \partial_\theta \psi \, \hat{\boldsymbol\theta} + \frac{1}{r \sin \theta} \partial_{\phi} \psi \hat{\boldsymbol\phi} \right] = \\ & = \partial_\theta \psi \, \hat{\boldsymbol\phi} - \frac{1}{\sin \theta} \partial_\phi \psi \, \hat{\boldsymbol\theta} \ . \end{aligned}$$ 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 $$\hat{\mathbf{z}} \cdot \left( \mathbf{r} \times \nabla \psi \right) = \partial_\phi \psi$$ ``` ```{dropdown} Spectral decomposition of the $z$-component operator, $\hat{L}_z$ :open: $$\ell_z \psi_z(r,\theta,\phi) = -i \hbar \partial_\phi \psi_z(r,\theta,\phi)$$ 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. $\hat{L}_z | m_l \rangle = \hbar m_l | m_l \rangle$. ``` ```{dropdown} Spectral decomposition of commuting operators :open: 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 $\forall i$. ... ``` ```{dropdown} Relation between $\ \hat{L}_z\ $ and $\ \hat{L}^2 \$ :open: $$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$$ 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 $$\begin{aligned} \left[\hat{L}_y, \hat{L}_z \right] & = \hat{L}_y \hat{L}_z - \hat{L}_z \hat{L}_y = i \hbar \hat{L}_x \\ \left[\hat{L}_z, \hat{L}_x \right] & = \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z = i \hbar \hat{L}_y \end{aligned}$$ summing and subtracting $i (1)$ and $(2)$ them $$\begin{aligned} 0 & = - \hbar \hat{L}_x \mp i \hbar \hat{L}_y - i \hat{L}_y \hat{L}_z + i \hat{L}_z \hat{L}_y \pm \hat{L}_z \hat{L}_x \mp \hat{L}_x \hat{L}_z = \\ & = - ( \hat{L}_x \pm i \hat{L}_y ) ( \hbar \pm \hat{L}_z ) + \hat{L}_z ( i \hat{L}_y \pm \hat{L}_x ) \ , \end{aligned}$$ 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 $\psi$ $$\hat{L}_z ( \hat{L}_x \pm i \hat{L}_y ) \psi = ( \hat{L}_x \pm i \hat{L}_y ) ( \hat{L}_z \pm \hbar ) \psi = ( \ell_z \pm \hbar ) ( \hat{L}_x \pm i \hat{L}_y ) \psi$$ (eq:ang-mom:ladder:application) 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 $\ell$ $$\begin{aligned} 0 & = ( \ell_{z,min} + \hbar ) ( \ell^2 - \ell_{z,min}^2 + \hbar \ell_{z,min} ) \\ 0 & = ( \ell_{z,max} - \hbar ) ( \ell^2 - \ell_{z,max}^2 - \hbar \ell_{z,max} ) \\ \end{aligned}$$ Either $\ell_{z;min,max} = \mp \hbar$, or $$\begin{aligned} 0 & = \ell^2 - \ell_{z,min}^2 + \hbar \ell_{z,min} \\ 0 & = \ell^2 - \ell_{z,max}^2 - \hbar \ell_{z,max} \ , \end{aligned}$$ 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 $\ell_{z,min} = - \ell_{z,max}$. --- Following {eq}`eq:ang-mom:ladder:application`, 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 $\hat{L}_z$. --- $$\lambda\left(L_z\right) = \hbar m_l \qquad m_l \in $$ Now, for $\ell_{z,min} = -m_{l,max} \hbar$, and $\ell_{z,max} = m_{l,max} \hbar$, the corresponding eigenvalues $\ell^2$ are $$\begin{aligned} \ell^2 = \hbar^2 m_{l,max} \left( m_{l,max} + 1 \right) \ . \end{aligned}$$ ```