PHOT 301: Quantum Photonics
LECTURE 23-24
Michaël Barbier, Fall semester (2024-2025)
Common basis of eigenstates of angular momentum:
\[ \boxed{\hat{\bf L}^2|l, m\rangle = l(l+1)\hbar^2|l, m\rangle} \qquad \qquad \boxed{\hat{L}_z |l, m\rangle = m \hbar|l, m\rangle} \]
\[ \begin{aligned} Y_{lm}(\theta, \phi) = (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta) e^{im\phi} \end{aligned} \]
with the associated Legendre polynomials
\[ P^{m}_l(z) = \frac{(1-z^2)^m/2}{2^l l!} \frac{d^{m+l}}{dz^{m+l}}(z^2 - 1)^l \]
\[ \begin{aligned} Y_{lm}(\theta, \phi) = (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta) e^{im\phi} \end{aligned} \]
\[ P^{m}_l(z) = \frac{(1-z^2)^m/2}{2^l l!} \frac{d^{m+l}}{dz^{m+l}}(z^2 - 1)^l \]
Lower order \(P^{m}_l(z)\) polynomials:
| \(l\) | \(m=-2\) | \(m=-1\) | \(m=0\) | \(m=1\) | \(m=2\) |
|---|---|---|---|---|---|
| 0 | \(1\) | ||||
| 1 | \(-\frac{1}{2}(1-x^2)^{1/2}\) | \(x\) | \((1-x^2)^{1/2}\) | ||
| 2 | \(\frac{1}{8}(1 - x^2)\) | \(-\frac{1}{2}x(1 - x^2)^{1/2}\) | \(\frac{1}{2}(3x^2 - 1)\) | \(3x(1 - x^2)^{1/2}\) | \(3(1 - x^2)\) |
\[ \begin{aligned} Y_{lm}(\theta, \phi) = (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta) e^{im\phi} \end{aligned} \]
Use the identities \(\,\,\cos(m\phi) = \frac{e^{im\phi} + e^{-im\phi}}{2}\,\,\) and \(\,\,\sin(m\phi) = \frac{e^{im\phi} - e^{-im\phi}}{2i}\,\):
\[ \left\{ \begin{aligned} p_x(\theta, \phi) &= \Re\{Y_{lm}\} = (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta)\, \cos(m\phi)\\ p_y(\theta, \phi) &= \Im\{Y_{lm}\} = (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta)\, \sin(m\phi) \end{aligned} \right. \]
Solutions are given by Laguerre polynomials
\[ \left[-\frac{\hbar^2}{2\mu} \frac{\partial^2}{\partial r^2} + V_{e}(r)\right] \chi(r) = E\chi(r) \quad\textrm{ with}\quad V_e(r) = \frac{l(l+1)\hbar^2}{2\mu r^2} + V(r) \]
\[ \chi(s) = s^{l+1} L_{n-l-1}^{2l+1}(s) e^{-s/2}, \qquad s = \frac{2r}{na_0}, \qquad E_n = -\frac{Ry}{n^2} \]
Normalization factor for \(R(r) = \chi(s)/r\):
\[ 1 = \int^\infty_0 R^2(r) \, r^2\, dr = \int^\infty_0 s^{2l} \left( L^{2l+1}_{n-l-1}(s) \right) \, e^{-s} \, s^2\, ds = \frac{2\,n\,(n+l)!}{(n - l - 1)!} \]
The normalized radial solutions:
\[ \boxed{\quad R_{nl}(s) = \left[ \frac{(n - l - 1)!}{2\,n\,(n+l)!} \left(\frac{2}{n\,a_0}\right)^3 \right]^{1/2} s^l \, L^{2l+1}_{n-l-1}(s) \, e^{-s/2}\quad} \]
\[ \boxed{\begin{aligned} \quad R_{nl}(s) &= \left[ \frac{(n - l - 1)!}{2\,n\,(n+l)!} \left(\frac{2}{n\,a_0}\right)^3 \right]^{1/2} s^l \, L^{2l+1}_{n-l-1}(s) \, e^{-s/2}\quad\\ &\\ \qquad s &= \frac{2r}{na_0}, \qquad\,\,\,\, a_0 = \frac{4\pi\varepsilon_0\hbar^2}{e^2 \mu} = 0.529\,\textrm{A}\\ \qquad E_n &= -\frac{Ry}{n^2},\qquad Ry = \frac{\hbar^2}{2\mu a_0^2} = \frac{\mu}{2} \left(\frac{e^2}{4\pi\varepsilon_0\hbar}\right)^2 = 13.6\,\textrm{eV}\\ \end{aligned}} \]
The wave function \(\,\,\psi(\vec{r}) = R_{nl}(r) Y_{lm}(\theta, \phi)\)
\[ \boxed{\begin{aligned} & \\ \quad Y_{lm}(\theta, \phi) &= (-1)^{m+|m|} \left[ \frac{2l + 1}{4\pi} \,\frac{(l - |m|)!}{(l + |m|)!} \right]^{1/2} P^{|m|}_l(\cos\theta) e^{im\phi} \quad\\ & \\ R_{nl}(s) &= \left[ \frac{(n - l - 1)!}{2\,n\,(n+l)!} \left(\frac{2}{n\,a_0}\right)^3 \right]^{1/2} s^l \, L^{2l+1}_{n-l-1}(s) \, e^{-s/2}\\ & \\ \end{aligned}} \]
with \(s = \frac{2\,r}{n\,a_0}\,\), associated Legendre and associated Laguerre polynomials
\[ P^{m}_l(z) = \frac{(1-z^2)^m/2}{2^l l!} \frac{d^{m+l}}{dz^{m+l}}(z^2 - 1)^l, \qquad L_k^{\alpha}(z) = \sum_{j = 0}^{k}(-1)^j \begin{pmatrix}k+\alpha\\ k - j\end{pmatrix} \frac{x^j}{j!} \]
From Wikipedia
\[ E_s = \vec{\mu_e}\cdot\vec{B} = g \mu_B \vec{\sigma}\cdot\vec{B} \]
Quantum Mechanics: Corresponding Hamiltonian (convert to operators)
\[ \hat{H}_s \,=\, \frac{g \mu_B}{2} \hat{\vec{\sigma}}\cdot\vec{B} \,=\, \frac{g\mu_B}{2} \left[ B_x \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} + B_y \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} + B_z \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} \right] \]
Factor 1/2 comes from \(\hat{\vec{\sigma}}\) definition
Classically: Lorentz force:
\[ \vec{F} = q\vec{E} + q (\vec{v} \times \vec{B}) \]
\[ \begin{aligned} \textrm{Electric field:} \qquad &\vec{E} = -e\nabla V - \frac{\partial \vec{A}}{\partial t}\\ \textrm{Magnetic field:} \qquad &\vec{B} = \nabla \times \vec{A} \end{aligned} \]
Quantum Mechanics: use corresponding momentum operator \(\,\,\hat{\vec{p}} \longrightarrow \hat{\vec{p}} - e\vec{A}\)
Add both spin \(\frac{g \mu_B}{2} \hat{\vec{\sigma}}\cdot\vec{B}\) and vector potential \(\hat{\vec{p}} - e\vec{A}\).
\[ \left[ \frac{1}{2m_0} (\vec{\hat{p}} - e \vec{A})^2 \mathbb{1} + V \mathbb{1} + \frac{g \mu_B}{2} \hat{\vec{\sigma}}\cdot\vec{B} \right] \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r}, t) \]
where \(\,\,\Psi(\vec{r}, t) = \begin{pmatrix}\Psi_\uparrow(\vec{r}, t) \\ \Psi_\downarrow(\vec{r}, t) \end{pmatrix}\,\,\) is a spinor
The part \((\vec{\hat{p}} - e \vec{A})^2\) can be written as (operator precedence):
\[ (\vec{\hat{p}} - e \vec{A})\cdot(\vec{\hat{p}} - e \vec{A}) = \hat{p}^2 -e\vec{A}\cdot \hat{\vec{p}} + i\hbar e \nabla\cdot\vec{A} + e^2 A^2 \]
\[ \left[ \frac{1}{2m_0} (\vec{\hat{p}} - e \vec{A})^2 \mathbb{1} + V \mathbb{1} + \frac{g \mu_B}{2} \hat{\vec{\sigma}}\cdot\vec{B} \right] \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r}, t) \]
where a we choose vector potential \(\vec{A} = \frac{1}{2}\,B\,(-y, x, 0)\)
\[ \begin{aligned} (\vec{\hat{p}} - e \vec{A})^2 &= \hat{p}^2 -e\vec{A}\cdot \hat{\vec{p}} + i\hbar e \nabla\cdot\vec{A}\\ \end{aligned} \]
where we approximated \({e^2 A^2} \approx 0\)
\[ \begin{aligned} (\vec{\hat{p}} - e \vec{A})^2 &= \hat{p}^2 + i\hbar e \vec{A}\cdot \nabla + i\hbar e \nabla\cdot\vec{A}\\ &= \hat{p}^2 + \frac{i\hbar e B}{2}(-y, x, 0) \cdot (\hat{p}_x, \hat{p}_y, \hat{p}_z) + \frac{i\hbar e B}{2} (\hat{p}_x, \hat{p}_y, \hat{p}_z) \cdot (-y, x, 0)\\ &= \hat{p}^2 + \frac{\hbar^2 e B}{2}(-y \partial_x + x\partial_y) + \frac{\hbar^2 e B}{2}\, 0 \\ &= \hat{p}^2 + \frac{\hbar e B}{2} \hat{L}_z \end{aligned} \]
where the z-component of angular momentum: \(\hat{L}_z = -i\hbar (x\partial_x - y\partial_x)\)
Filling this in the Pauli equation:
\[ \left[ \left(\frac{\vec{\hat{p}}^2}{2m_0} + V\right) + \frac{\hbar e B}{4 m_0} \hat{L}_z + \frac{g \mu_B}{2} \hat{\vec{\sigma}}\cdot\vec{B} \right] \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r}, t) \]
Approximating \(g/2 \approx 1\) and using the Bohr magneton \(\mu_B = \frac{e\hbar}{m_0}\)
\[ \left[ \left(\frac{\vec{\hat{p}}^2}{2m_0} + V\right) + \frac{\mu_B\,B}{2\hbar} \left(\hat{L}_z + B \hat{\sigma}_z \right) \right] \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r}, t) \]
The first term is just the Hydrogen Hamiltonian \(\hat{H}_0\)
\[ \left[ \hat{H}_0 + \frac{\mu_B\,B}{2\hbar} \left(\hat{L}_z + \hat{\sigma}_z \right) \right] \Psi(\vec{r}, t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r}, t) \]
For stationary solutions we assume \(\Psi(\vec{r}, t) = \psi(\vec{r}) \, e^{-iEt/\hbar}\)
\[ \left[ \hat{H}_0 + \frac{\mu_B\,B}{2\hbar} \left(\hat{L}_z + \hat{\sigma}_z \right) \right] \psi(\vec{r}) = E \psi(\vec{r}) \]
Remember that \(\psi(\vec{r}) = \begin{pmatrix} \psi_\uparrow \\ \psi_\downarrow \end{pmatrix}\) and we have a matrix equation:
\[ \left[ \hat{H}_0\, \mathbb{1} + \frac{\mu_B\,B}{2\hbar} \left(\hat{L}_z \, \mathbb{1} + \hbar\, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right) \right] \begin{pmatrix} \psi_\uparrow(\vec{r}) \\ \psi_\downarrow(\vec{r}) \end{pmatrix}(\vec{r}) = E \begin{pmatrix} \psi_\uparrow(\vec{r}) \\ \psi_\downarrow(\vec{r}) \end{pmatrix} \]
This corresponds to two uncoupled equations for spin-up and spin-down:
\[ \begin{aligned} \left[ \hat{H}_0 + \frac{\mu_B\,B}{\hbar} \left(\hat{L}_z + \hbar \right)\right] \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ \left[ \hat{H}_0 + \frac{\mu_B\,B}{\hbar} \left(\hat{L}_z - \hbar \right)\right] \psi_\downarrow(\vec{r}) &= E \, \psi_\downarrow(\vec{r}) \\ \end{aligned} \]
Fill in solutions of the Hydrogen atom (eigenstates of \(\hat{H}_0\)):
\[ |n,l,m,\uparrow \rangle = \psi_\uparrow, \qquad \qquad |n,l,m,\downarrow \rangle = \psi_\downarrow \]
\[ \begin{aligned} \hat{H}_0 \psi_\uparrow(\vec{r}) + \frac{\mu_B\,B}{\hbar} \left(\hat{L}_z + \hbar \right) \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ \hat{H}_0 \psi_\downarrow(\vec{r}) + \frac{\mu_B\,B}{\hbar} \left(\hat{L}_z - \hbar \right) \psi_\downarrow(\vec{r}) &= E \, \psi_\downarrow(\vec{r}) \\ \end{aligned} \]
The solutions are eigenstates of the Hydrogen atom, and \(\hat{L}_z\)
\[ \begin{aligned} \hat{H}_0 |n,l,m,\uparrow (\downarrow) \rangle &= E_{n} |n,l,m,\uparrow (\downarrow)\rangle\\ \hat{L}_z |n,l,m,\uparrow \rangle &= m\hbar |n,l,m,\uparrow \rangle\\ \hat{L}_z |n,l,m,\downarrow \rangle &= -m\hbar |n,l,m,\downarrow \rangle\\ \end{aligned} \]
Fill in solutions of the Hydrogen atom (eigenstates of \(\hat{H}_0\)):
\[ \begin{aligned} E_{nl} \psi_\uparrow(\vec{r}) + \mu_B\,B (m + 1) \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ E_{nl} \psi_\uparrow(\vec{r}) + \mu_B\,B (m - 1) \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ \end{aligned} \]
Fill in solutions of the Hydrogen atom (eigenstates of \(\hat{H}_0\)):
\[ \begin{aligned} E_{nl} \psi_\uparrow(\vec{r}) + \mu_B\,B (m + 1) \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ E_{nl} \psi_\uparrow(\vec{r}) + \mu_B\,B (m - 1) \psi_\uparrow(\vec{r}) &= E \, \psi_\uparrow(\vec{r}) \\ \end{aligned} \]
The eigenenergies \(E = E_{nlm\uparrow}\) depend now also on the magnetic quantum number \(m\) and the spin \(\uparrow (\downarrow)\):
\[\boxed{\begin{aligned} E_{nlm\uparrow} = E_{nl} + \mu_B\,B (m + 1) \qquad \textrm{with}\quad\psi_\uparrow\\ E_{nlm\downarrow} = E_{nl} + \mu_B\,B (m - 1) \qquad \textrm{with}\quad\psi_\downarrow \end{aligned}} \]
\[\boxed{\begin{aligned} E_{nlm\uparrow} = E_{nl} + \mu_B\,B (m + 1) \qquad \textrm{with}\quad\psi_\uparrow\\ E_{nlm\downarrow} = E_{nl} + \mu_B\,B (m - 1) \qquad \textrm{with}\quad\psi_\downarrow \end{aligned}} \]
\[ \frac{2\pi}{\Delta \lambda} = \Delta f = \frac{\Delta\omega}{2\pi} = \Delta E/h \]
Lecture 23-24: The Pauli-equation