Time-Independent Perturbation Theory

Applications

Fine Structure of Hydrogen

Hamiltonian: $H_{0} = \frac{p^{2}}{2 μ} - \frac{e^{2}}{r} \Rightarrow E_{n} = - \frac{E_{I}}{n^{2}}$ .

Relativistic Schr\:odinger Equation: $[{(p^{μ} - \frac{e}{c} A^{μ})}^{2} - m^{2} c^{2}] Ψ = 0$ with $p^{μ} = i ℏ (c \frac{\partial}{\partial t}, - \nabla), A^{μ} = (ϕ, \vec{A})$ .

Spin-1/2 charged particles gives the Dirac equation: $[γ_{μ} (p^{μ} - \frac{e}{c} A^{μ}) - m c] ψ = 0$ . With, $γ^{0} = (\begin{matrix} I & 0 \\ 0 & - I \end{matrix}), \vec{γ} = (\begin{matrix} 0 & \vec{σ} \\ - \vec{σ} & 0 \end{matrix})$ . This gives energies of: $E_{n, j} = m_{e} c^{2} [1 + α^{2} (n - j - 1 / 2 + \sqrt{(j + 1 / 2)^{2} - α^{2}})^{- 2}]^{- 1 / 2} \approx m_{e} c^{2} - \frac{E_{I}}{n^{2}} - \frac{m_{e} c^{2} α^{4}}{2 n^{4}} (\frac{n}{j + 1 / 2} - \frac{3}{4}) + \dots$ . $j$ is from $\vec{J} = \vec{L} + \vec{S}$ . $- \frac{m_{e} c^{2} α^{4}}{2 n^{4}} (\frac{n}{j + 1 / 2} - \frac{3}{4})$ is the fine structure term.

From the Semi-classical Bohr model, we will see that $e^{-}$ in hydrogen atom is weakly relativisitic.

$\frac{μ v^{2}}{r} = \frac{e^{2}}{r^{2}}, μ v r = n ℏ$ .

For $n = 1$ , $v = \frac{ℏ}{μ a_{0}} = \frac{e^{2}}{ℏ}$ , $a_{0} = \frac{ℏ^{2}}{μ e^{2}}$ . For $\frac{v}{c} = \frac{e^{2}}{ℏ c} = α \approx \frac{1}{137}$ .

$V_{f i n e} = - \frac{p^{4}}{8 m_{e}^{3} c^{2}} + \frac{1}{2 m_{e}^{2} c^{2}} \frac{1}{r} \frac{d V (r)}{d r} \vec{L} \cdot \vec{S} + \frac{ℏ^{2}}{8 m_{e}^{2} c^{2}} \nabla^{2} V (r)$ . Numbering each term sequentially $V_{f i n e} = V_{m v} + V_{s o} + V_{D}$ .

$V_{m v}$ is due to mass and velocity $m (v)$ .
$V_{s o}$ is due to spin orbit coupling.
$V_{D}$ is Darwin term - the non-locality term

$V_{m v}$ . $E = c \sqrt{p^{2} + m_{e}^{2} c^{2}}$ . If $\frac{v}{c} ≪ 1$ then $E = m_{e} c^{2} \sqrt{1 + \frac{p^{2}}{m_{e}^{2} c^{2}}} \approx m_{e} c^{2} (1 + \frac{p^{2}}{2 m_{e}^{3} c^{2}} - \frac{p^{4}}{8 m_{e}^{4} c^{4}} + \dots) = m_{e} c^{2} + \frac{p^{2}}{2 m_{e}^{3} c^{2}} - \frac{p^{4}}{8 m_{e}^{3} c^{2}} + \dots$ . $\frac{\frac{p^{4}}{8 m_{e}^{3} c^{2}}}{\frac{p^{2}}{2 m_{e}}} = \frac{p^{2}}{4 m_{e} c^{2}} = \frac{1}{4} \frac{v^{c}}{c^{2}} = \frac{1}{4} α^{2}$ .
$V_{s o}$ . Consider stationary system $K$ $(x, y, z)$ for our proton and a moving system $K^{'}$ $(x^{'}, y^{'}, z^{'})$ for the electron which has velocity $\vec{v}$ . Then, a magnetic field ${\vec{B}}^{'} = - \frac{\vec{v}}{c} \times \vec{E}$ , with the electric field being due to the proton and ${\vec{B}}^{'}$ is the magnetic field the electron experiences. Now with the magnetic field, we have a magnetic moment of the electron, ${\vec{M}}_{S} = \frac{e}{m_{e} c} \vec{S}$ . So, $V_{S O} = - {\vec{M}}_{S} \cdot {\vec{B}}^{'} = - \frac{e}{m_{e} c} \vec{S} \cdot (\vec{v} \times \vec{E})$ . Consider $\vec{F} = e \vec{E} = - \frac{d V}{d r} \frac{\vec{r}}{r}$ . For the coulomb interaction, $\vec{F} = - \frac{e^{2}}{r^{3}} \vec{r}$ . So, $- \frac{e^{2}}{m_{e}^{2} c^{2} r^{3}} \vec{S} \cdot (\vec{p} \times \vec{r}) = \frac{e^{2}}{m_{e}^{2} c^{2} r^{3}} \vec{L} \cdot \vec{S}$ . The factor of two in the original term comes from restricting a degree of freedom. $\frac{V_{S O}}{\frac{e^{2}}{r}} \approx \frac{e^{2}}{m_{e} c^{2} r} ℏ^{2} \frac{r}{e^{2}} = \frac{ℏ^{2}}{m_{e}^{2} c^{2} r^{2}} \approx \frac{ℏ^{2}}{m_{e}^{2} c^{2} a_{0}^{2}} = \frac{e^{4}}{ℏ^{2} c^{2}} = α^{2}$ .
For $ℓ = 0$ this term is strongest, i.e. it is relevant to s-states. Because the electron will be bouncing around at small radii, the first derivative averages out but the second derivative becomes ’significant’. $V_{D} = \frac{ℏ^{2}}{8 m_{e}^{2} c^{2}} \nabla^{2} (\frac{- e^{2}}{r}) = - \frac{ℏ^{2} e^{2}}{8 m_{e}^{2} c^{2}} \nabla^{2} (\frac{1}{r}) = \frac{π e^{2} ℏ^{2}}{2 m_{e} c^{2}} δ (\vec{r})$ . $⟨ Ψ | V_{D} | Ψ ⟩ \sim \frac{π e^{2} ℏ^{2}}{2 m_{e}^{2} c^{2}} \int | Ψ |^{2} δ (\vec{r}) d^{3} r = \frac{π e^{2} ℏ^{2}}{2 m_{e}^{2} c^{2}} | Ψ (0) |^{2}$ . For the Hydrogen atom, $Ψ \sim r^{ℓ} \exp (\frac{- r}{n a_{0}}) Y_{ℓ}^{m}$ . So, this only gives non-zero for $ℓ = 0$ . From $\int | Ψ |^{2} d^{3} x = 1, | Ψ (0) |^{2} \approx \frac{3}{4 π a_{0}^{3}}$ . $\frac{⟨ V_{D} ⟩}{e^{2} / r} = \frac{3 π ℏ^{2}}{8 π m_{e}^{2} c^{2} a_{0}^{2}} = \frac{3}{8} α^{2}$

Hyperfine Structure

$V_{h f} = \frac{e}{m_{e} r^{3}} \vec{L} \cdot {\vec{M}}_{I} + \frac{1}{r^{3}} [3 ({\vec{M}}_{S} \cdot \hat{n}) ({\vec{M}}_{I} \cdot \hat{n}) - {\vec{M}}_{S} \cdot {\vec{M}}_{I}] + \frac{8 π}{3} {\vec{M}}_{S} \cdot {\vec{M}}_{I} δ (\vec{r})$ .

First term is the interaction between the electron’s magnetic field and the proton’s magnetic moment. The second term is the electron-proton dipole-dipole interaction. The third term is Fermi’s contact term.

Our state is then: $| n ℓ m s m_{s} I m_{I} ⟩ \to | n ℓ m; m_{s} m_{I} ⟩$ .

So, for a 1s state,

$⟨ 100; m_{s}^{'} m_{i}^{'} | \vec{L} \cot {\vec{M}}_{I} | 00 m_{s} m_{I} ⟩ = 0$
$⟨ 100; m_{s}^{'} m_{i}^{'} | \frac{3 γ_{s} γ_{I}}{r^{3}} (\vec{S} \cdot \hat{n}) (\vec{I} \cdot \hat{n}) - \frac{γ_{S} γ_{I}}{r^{3}} \vec{S} \cdot \vec{I} | 100; m_{s} m_{I} ⟩ = ⟨ ψ^{'} | f (r) [3 \sum_{i} S_{i} n_{i} \sum_{j} I_{j} n_{j} - \sum_{i j} S_{i} I_{j} δ_{i j}] | ψ ⟩ = ⟨ ψ^{'} | f (r) \sum_{i j} S_{i} I_{j} (3 n_{i} n_{j} - δ_{i j}) | ψ ⟩ = ⟨ ψ^{'} | f (r) \sum_{i j} S_{i} I_{j} Y_{2}^{q} | ψ ⟩ = ⟨ 100; m_{s}^{'} m_{I}^{'} | f (r) {S_{i} I_{j}} Y_{2}^{q} | 100 m_{s} m_{I} ⟩ = 0$ , $ℓ_{1} = 0$ from $ψ$ and $ℓ_{2} = 2$ from the spherical harmonic. Thus, $ℓ_{3}$ from $ψ^{'}$ must be $ℓ_{1} + ℓ_{2} = 2$ .
$\frac{8 π}{3} ⟨ 100; m_{s}^{'} m_{I}^{'} | γ_{S} γ_{I} \vec{S} \cdot \vec{I} δ (\vec{r}) | 100; m_{s} m_{I} ⟩ = \frac{8 π}{3} γ_{S} γ_{I} ⟨ 100 | δ (\vec{r}) | 100 ⟩ ⟨ m_{s}^{'} m_{I}^{'} | \vec{S} \cdot \vec{I} | m_{s} m_{I} ⟩ = \frac{8 π}{3} γ_{S} γ_{I} | Ψ_{1 s} (0) |^{2} ⟨ m_{s}^{'} m_{I}^{'} | \vec{S} \cdot \vec{I} | m_{s} m_{I} ⟩$ . $| m_{s} m_{i} ⟩ \to | F m_{F} ⟩, \vec{F} = \vec{I} + \vec{S}, F = {0, 1}$ .

We know $ν_{h f} = 1420405751.768 \pm 0.001$ Hz.

Lamb Shift

Beginning of QED. 1.06 GHz. Got Nobel prize in 1955.

See PRL 94, 022001 (in 2005).

For $n = 2$ . $2 p_{3 / 2}, 2 s_{1 / 2}$ . $1 s_{1 / 2} \to 2 p_{3 / 2}$ .

Applying Magnetic Field

Weak Magnetic Field

$H = H_{0} + V_{f i n e} + V_{h f} + V_{z}$ . The $z$ is for Zeeman.

The strength of the magnetic field is relative to the hyperfine.

For the weak field, we can start with our hyperfine perturbed solution and add the magnetic field perturbation.

Let the magnetic field have strength $B_{0}$ and and be aligned on z-axis. $V_{z} = - ({\vec{M}}_{L} + {\vec{M}}_{S} + {\vec{M}}_{I}) \cdot {\vec{B}}_{0} = - (\frac{e}{2 m_{e} c} \vec{L} + \frac{e}{m_{e} c} \vec{S} + \frac{| e | g_{p}}{2 m_{p} c} \vec{I})$ . The proton spin will have a much smaller effect. So, lets just consider the electron spin and orbital angular momentum. Let $V_{z} = ω_{0} (L_{z} + 2 S_{z}), ω_{0} = \frac{e}{2 m_{e} c} B_{0}$ .

Considering the 1-s state. $F = 1$ $(g_{n} = 3)$ has energy $A ℏ^{2} / 4$ more than the 1s state, $F = 0$ $(g_{n} = 1)$ has energy $- 3 A ℏ^{2} / 4$ more than the 1s state.

For $F = 0, m_{F} = 0$ , $E^{(1)} = ⟨ 100; F m_{F} | V_{z} | 100; F m_{F} ⟩ = ⟨ 100; 00 | ω_{0} (L_{z} + 2 S_{z}) | 100; 00 ⟩ = ⟨ 100; 00 | ω_{0} (L_{z} + 2 S_{z}) | 100; 00 ⟩ = ℏ ω_{0} ⟨ 00 | 10 ⟩ = 0$ . Recall: $| 00 ⟩ = \frac{1}{\sqrt{2}} (| 1 / 2 - 1 / 2 ⟩ - | - 1 / 21 / 2 ⟩)$ .

For $F = 1$ , $⟨ 100; F m_{F} | V_{z} | 100; F m_{F} ⟩$ . Find the 3x3 matrix. $⟨ 1 m_{F} | 2 ω_{0} S_{z} | 1 m_{F} ⟩$ .

$2 ω_{0} ⟨ 11 | S_{z} | 11 ⟩ = ℏ ω_{0}$ . $2 ω_{0} ⟨ 1 - 1 | S_{z} | 1 - 1 ⟩ = - ℏ ω_{0}$ . $2 ω_{0} ⟨ 10 | S_{z} | 10 ⟩ = 0$ . $2 ω_{0} ⟨ 10 | S_{z} | 11 ⟩ = 0$ .

So, the energy corrections are $\pm ℏ ω_{0}, 0$ .

Strong Magnetic Field

The hyperfine will be the perturbation.

$V_{z} = 2 ω_{0} S_{z}$ .

$| m_{s} m_{I} ⟩$ basis.

$V_{z} = ℏ ω_{0} (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$ .

$E^{(1)} = ℏ ω_{0} \Rightarrow | 1 / 2, \pm 1 / 2 ⟩$ . $E^{(1)} = - ℏ ω_{0} \Rightarrow | - 1 / 2, \pm 1 / 2 ⟩$ .

Perturbation is $A \vec{I} \cdot \vec{S} = A (I_{z} S_{z} + \frac{1}{2} (I_{+} S_{-} + I_{-} S_{+}))$ .

For $E = E_{1 s} \pm ℏ ω_{0}$ , $V_{h f} = (\begin{matrix} A ℏ^{2} / 4 & 0 \\ 0 & - A ℏ^{2} / 4 \end{matrix})$ .

Then, $E = E_{1 s} \pm ℏ ω_{0} \pm A \frac{ℏ^{2}}{4}$ . The signs of the first term follow the sign of $| m_{s} ⟩$ and the signs of the second term follow the sign of the $m_{I} ⟩$ .

Intermediate Strength Magnetic Field

Both the hyperfine and the weak field will be the perturbative potentials.