21.5.3. Auxiliary Results on Operators for the Schrödinger Model of the Hydrogen Atom

1. All components of \(\hat{\mathbf{L}}\) commute with the kinetic term \(\frac{\hat{p}^2}{2m}\) and with \(\hat{L}^2\), proof

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

2. All components of \(\hat{\mathbf{L}}\) commute with the potential \(\hat{V}_{\text{Coulomb}}\), proof

   \[[\hat{V}_{\text{Coulomb}}, \hat{L}_b] = 0\]

3. Therefore, all components of \(\hat{\mathbf{L}}\) commute with the \(\text{H}\) atom Hamiltonian \(\hat{H}_{\text{H-atom}} = \frac{\hat{p}^2}{2m} + \hat{V}_{\text{Coulomb}}\):

   \[[\hat{H}_{\text{H-atom}}, \hat{L}_b] = 0\]

4. Simultaneous Commutative Set:

   \[\boxed{[\hat{H}_{\text{H-atom}}, \hat{L}^2] = 0, \quad [\hat{H}_{\text{H-atom}}, \hat{L}_a] = 0, \quad [\hat{L}^2, \hat{L}_a] = 0}\]

   i.e. the Hamiltonian \(\hat{H}_{\text{H-atom}}\), total angular momentum squared \(\hat{L}^2\), and any component \(\hat{L}_a\) all mutually commute, and thus share a set of common eigenvectors. For the commutation relation \([\hat{L}^2, \hat{L}_a] = 0\), see the section about spatial angular momentum.

21.5.3.1. Hamiltonian & Basic Definitions

The Hamiltonian of the electron in the Hydrogen atom with Coulomb potential is given by

\[\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}_{Coulomb} \ ,\]

that in space basis becomes

\[\langle \mathbf{r} | \hat{H} = - \frac{\hbar^2}{2 m} \nabla^2 \langle \mathbf{r} | - \frac{q^2}{4 \pi \varepsilon |\mathbf{r}|} \langle \mathbf{r} | \ .\]

Space Basis Representations

\[\langle \mathbf{r} | \Psi \rangle = \Psi(\mathbf{r}, t)\]

\[\langle \mathbf{r} | \hat{\mathbf{r}} = \mathbf{r} \langle \mathbf{r} | \quad \implies \quad \langle \mathbf{r} | \hat{\mathbf{r}} | \Psi \rangle = \mathbf{r} \, \Psi(\mathbf{r}, t)\]

\[\langle \mathbf{r} | \hat{\mathbf{p}} = -i\hbar \nabla_{\mathbf{r}} \langle \mathbf{r} | \quad \implies \quad \langle \mathbf{r} | \hat{\mathbf{p}} | \Psi \rangle = -i\hbar \nabla \Psi(\mathbf{r}, t)\]

Angular Momentum Operator

\[\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}\]

In Levi-Civita component notation, using Cartesian coordinates

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

In the space (position) basis (Cartesian coordinates):

\[\begin{split}\begin{aligned} \langle \mathbf{r} | \hat{\mathbf{L}} | \Psi \rangle & = \langle \mathbf{r} | \hat{\mathbf{e}}_a \hat{L}_a | \Psi \rangle = \\ & = \hat{\mathbf{e}}_a \varepsilon_{abc} \, r_b (-i\hbar \partial_c \langle \mathbf{r} | \Psi \rangle) = \\ & =-i\hbar \hat{\mathbf{e}}_a \varepsilon_{abc} \, r_b \, \partial_c \Psi(\mathbf{r},t) = \\ & = - i \hbar \mathbf{r} \times \nabla_{\mathbf{r}} \Psi(\mathbf{r},t) \ . \end{aligned}\end{split}\]

\(\langle \mathbf{r} | \hat{\mathbf{p}} | \Psi \rangle\)

\[\langle \mathbf{r} | \hat{\mathbf{p}} | \Psi \rangle = \int \langle \mathbf{r} | \hat{\mathbf{p}} | \mathbf{r}' \rangle \langle \mathbf{r}' | \Psi \rangle \, d\mathbf{r}' = \int \delta(\mathbf{r} - \mathbf{r}') \left(-i\hbar \nabla_{\mathbf{r}'} \Psi(\mathbf{r}', t)\right) d\mathbf{r}' = -i\hbar \nabla_{\mathbf{r}} \Psi(\mathbf{r}, t)\]

\(\langle \mathbf{r} | \hat{p}_b \hat{r}_a | \Psi \rangle\)

\[\langle \mathbf{r} | \hat{p}_b \hat{r}_a | \Psi \rangle = -i\hbar \partial_b \langle \mathbf{r} | \hat{r}_a | \Psi \rangle = -i\hbar \partial_b \left( r_a \langle \mathbf{r} | \Psi \rangle \right) = - i \hbar \partial_b \left( r_a \Psi(\mathbf{r},t) \right) = -i\hbar (\delta_{ab} \Psi + r_a \partial_b \Psi)\]

21.5.3.2. Canonical Commutation Relations (CCR)

Using postion basis and Cartesian coordinates, the CCR \([ \hat{\mathbf{r}}, \hat{\mathbf{P}} ] = i \hbar \mathbb{I}\) reads

\[[r_a, p_b] \Psi = r_a (-i\hbar \partial_b \Psi) - (-i\hbar \partial_b (r_a \Psi)) = i\hbar \, \delta_{ab} \Psi \ ,\]

i.e.

\[[r_a, p_b] = i\hbar \, \delta_{ab}\]

21.5.3.3. Commutators of Angular Momentum with Position and Momentum

21.5.3.3.1. Commutator \([r_a, L_b]\)

\[r_a L_b \Psi = r_a (-i\hbar \varepsilon_{bcd} \, r_c \partial_d \Psi) = -i\hbar \varepsilon_{bcd} \, r_a r_c \partial_d \Psi\]

\[L_b r_a \Psi = -i\hbar \varepsilon_{bcd} \, r_c \partial_d (r_a \Psi) = -i\hbar \varepsilon_{bcd} \, r_c \delta_{ad} \Psi - i\hbar \varepsilon_{bcd} \, r_c r_a \partial_d \Psi\]

Subtracting the two expressions gives:

\[[r_a, L_b] \Psi = i\hbar \varepsilon_{bcd} \, r_c \delta_{ad} \Psi = i\hbar \varepsilon_{bca} \, r_c \Psi \ ,\]

and thus

\[[r_a, L_b] =- i \hbar \, (\mathbf{r} \times)_{a b} \, \Psi \quad \left(\text{in general } \neq 0\right)\]

(Recall: \((\mathbf{a} \times \mathbf{b})_i = \varepsilon_{ijk} a_j b_k = \{ \mathbf{a}_{\times} \}_{ik} \cdot \{ \mathbf{b} \}_k\), i.e. \(\{ \mathbf{a}_{\times} \}_{ik} = \varepsilon_{ijk} a_j\))

21.5.3.3.2. Commutator \([p_a, L_b]\)

\[p_a L_b \Psi = -i\hbar \partial_a \left(-i\hbar \varepsilon_{bcd} \, r_c \partial_d \Psi\right) = -\hbar^2 \varepsilon_{bcd} \, \partial_a (r_c \partial_d \Psi) = -\hbar^2 \varepsilon_{bad} \, \partial_d \Psi - \hbar^2 \varepsilon_{bcd} \, r_c \partial_a \partial_d \Psi\]

\[L_b p_a \Psi = -i\hbar \varepsilon_{bcd} \, r_c \partial_d (-i\hbar \partial_a \Psi) = -\hbar^2 \varepsilon_{bcd} \, r_c \partial_d \partial_a \Psi\]

Subtracting the two expressions gives:

\[[p_a, L_b] \Psi = -\hbar^2 \varepsilon_{bad} \, \partial_d \Psi = \hbar^2 \varepsilon_{abd} \, \partial_d \Psi = i\hbar \varepsilon_{abc} \, p_c \Psi\]

Cartesian coordinates.

• \([p_x, L_x] \equiv 0\)

• \([p_x, L_y] = \hbar^2 \varepsilon_{xyz} \, \partial_z \Psi = i\hbar p_z\)

• \([p_x, L_z] = \hbar^2 \varepsilon_{xzy} \, \partial_y \Psi = -\hbar^2 \partial_y \Psi = -i\hbar p_y\)

---

21.5.3.3.3. Commutator \([\hat{p}^2, L_b]\)

\[[p^2, L_b] = \sum_a p_a p_a L_b - L_b \sum_a p_a p_a = \sum_a \Big( p_a [p_a, L_b] + [p_a, L_b] p_a \Big)\]

Using \([p_a, L_b] = i\hbar \varepsilon_{abc} p_c\):

\[\sum_a \left( p_a (i\hbar \varepsilon_{abc} p_c) + (i\hbar \varepsilon_{abc} p_c) p_a \right) = i\hbar \varepsilon_{abc} (p_a p_c + p_c p_a)\]

Since \(p_a p_c = p_c p_a\) (symmetric in \(a, c\)) and \(\varepsilon_{abc}\) is antisymmetric in \(a, c\), the sum vanishes identically:

\[[p^2, L_b] = 0 \quad \text{for all components } b\]

Example. For \(b = x\):

\[[p^2, L_x] = -i\hbar^3 \Big[ 1 \cdot \partial_z \partial_y \Psi - 1 \cdot \partial_y \partial_z \Psi \Big] = 0\]

21.5.3.4. Commutator of Coulomb Potential \(\hat{V}\) with Angular Momentum \(\hat{L}_b\)

In space basis, the Coulomb potential reads:

\[\langle \mathbf{r} | \hat{V} | \Psi \rangle = -\frac{q^2}{4\pi\varepsilon_0} \frac{1}{r} \Psi(\mathbf{r}, t) = -k \frac{1}{|\mathbf{r}|} \Psi(\mathbf{r}, t)\]

Let’s evaluate whether \(\hat{V}\) and \(\hat{L}_b\) commute:

1. Action of \(\hat{V} \hat{L}_b\):

   \[\langle \mathbf{r} | \hat{V} \hat{L}_b | \Psi \rangle = -k \frac{1}{\sqrt{x_a x_a}} \left( -i\hbar \varepsilon_{bcd} \, r_c \, \partial_d \Psi \right)\]

2. Action of \(\hat{L}_b \hat{V}\), with \(|\mathbf{r}| = \sqrt{ x_a x_a }\):

   \[\langle \mathbf{r} | \hat{L}_b \hat{V} | \Psi \rangle = -i\hbar \varepsilon_{bcd} \, r_c \, \partial_d \left( -k \frac{\Psi}{\sqrt{x_a x_a}} \right) = i\hbar k \varepsilon_{bcd} \, r_c \left[ -\frac{\partial_d |\mathbf{r}|}{|\mathbf{r}|^2} \Psi + \frac{1}{|\mathbf{r}|} \partial_d \Psi \right]\]

   Since \(\partial_d | \mathbf{r}| = \partial_d \sqrt{x_a x_a} = \frac{x_d}{|\mathbf{r}|}\):

   \[\langle \mathbf{r} | \hat{L}_b \hat{V} | \Psi \rangle = i\hbar k \varepsilon_{bcd} \, r_c \left[ -\frac{r_d}{|\mathbf{r}|^3} \Psi + \frac{1}{|\mathbf{r}|} \partial_d \Psi \right]\]

3. Commutator \([\hat{V}, \hat{L}_b]\):

   \[\langle \mathbf{r} | [\hat{V}, \hat{L}_b] | \Psi \rangle = -i\hbar k \, \frac{\varepsilon_{bcd} \, r_c r_d}{|\mathbf{r}|^3} \Psi = 0\]

   Why? The product \(r_c r_d\) is symmetric under swapping \(c \leftrightarrow d\), while \(\varepsilon_{bcd}\) is antisymmetric under swapping \(c \leftrightarrow d\). Summing over \(c, d\) gives zero!

   Example for \(b = x\). \(\varepsilon_{xcd} r_c r_d = \varepsilon_{xyz} y z + \varepsilon_{xzy} z y = yz - zy = 0\).