# Time Dependent Perturbation Theory

Time-Dependent Potentials: $$H(t) = H_0 + V(t)$$.

$$|\psi(0)\rangle = |i\rangle$$. $$|\psi(t)\rangle = \sum_n c_n(t)|n\rangle$$. $$\mathcal{P}_{i\to f} = |\langle f|\psi(t)\rangle|^2$$. If $$|f\rangle = |m\rangle$$ then $$\mathcal{P}_{i\to f} = |c_m(t)|^2$$.

For two dimensional, with real off-diagonal $$V$$, let $$\alpha = c_1(t) + ic_2(t)$$. So, $$i\hbar\dot{\alpha} = v c_1 + iv c_2 = v\alpha$$. Thus, $$\dot{\alpha} = -i\omega \alpha \Rightarrow \alpha(t) = A\exp(-i\omega t) \Rightarrow c_1(t) + ic_2(t) = A(\cos(\omega t) - i\sin(\omega t))$$.

Recall: $$i\hbar \frac{d c_n(t)}{dt} = \sum_k V_{nk}(t)c_{k}(t)\exp(i\omega_{nk}t)$$.

$$c_n(t) = c_n^{(0)} + \lambda c_n^{(1)}(t) + \lambda^2\cdots$$. Feeding this in to our equation, $$i\hbar\left(\frac{dc_n^{(0)}}{dt} + \lambda\frac{dc_n^{(1)}}{dt}+\cdots\right) = \lambda\sum_k V_{nk}(c_k^{0}(t) + \lambda c_k^{(0)}(t)+\cdots)\exp(i\omega_{nk}t)$$

$$\lambda^0: \frac{d c_n^{(0)}}{dt} = 0 \Rightarrow c_n^{(0)} = const = \delta_{ni}$$.

$$\lambda^1: i\hbar\frac{dc_n^{(1)}}{dt} = \sum_k V_{nk}c_k^{(0)}(t)\exp(i\omega_{nk}t)$$

$$\lambda^r: i\hbar\frac{dc_n^{(r)}}{dt} = \sum_k V_{nk}c_k^{(r-1)}(t)\exp(i\omega_{nk}t)$$

$$\lambda^1: i\hbar\frac{dc_n^{(1)}}{dt} = \sum_k V_{nk}\delta_{ki}\exp(i\omega_{nk}t) = V_{ni}\exp(i\omega_{ni}t) \Rightarrow c_n^{(1)}(t) = \frac{1}{i\hbar}\int_0^t V_{ni}(t')\exp(i\omega_{ni}t')dt'$$

$$\mathcal{P}_{i\to f}(t) = |\langle f|\Psi(t)\rangle|^2 = |c_f(t)|^2 = |c_f^{(0)} + \lambda c_f^{(1)}(t) + \cdots|^2$$

Note for HW:

• If $$|i\rangle = |f\rangle \Rightarrow c_f^{(0)} = 1$$ and $$\mathcal{P}_{i\to f}(t) = |\langle f|\Psi(t)\rangle|^2 = |1 + \lambda \frac{1}{i\hbar}\int_0^t V_{ni}(t')\exp(i\omega_{ni}t')dt' + \cdots|^2$$
• If $$|i\rangle\neq |f\rangle$$ then $$c_f^{(0)}=0$$ and $$\mathcal{P}_{i\to f}(t) = |\langle f|\Psi(t)\rangle|^2 = |\lambda \frac{1}{i\hbar}\int_0^t V_{ni}(t')\exp(i\omega_{ni}t')dt' + \cdots|^2$$

### Alternative: Propogators

$$|\alpha,t_0;t\rangle = U_I(t,t_0)|\alpha,t_0;t_0\rangle_I$$.

$$c_n(t) = \langle n|U_I(t,t_0)|i\rangle$$.

$$i\hbar \frac{d}{dt}U_I(t,t_0) = V_IU_I(t,t_0)$$.

$$U(t_0,t_0) = \mathbb{I}$$

So, $$U(t,t_0) = \mathbb{I} + \frac{1}{i\hbar}\int_{t_0}^t V_I(t')U_I(t',t_0)dt'$$ The propogator on the inside can be written as, $$U_I(t',t_0)=\mathbb{I} + \frac{1}{i\hbar}\int_{t_0}^{t'} V_I(t'')U_I(t'',t_0)dt''$$. Using a Dyson series, an infinite recurrence, we can get an infinite recurrence to define the propogator. Using perturbation theory, we can then get $$U_I(t,t_0) = \mathbb{I} + \frac{1}{i\hbar}\int_{t_0}^tdt'V_I(t') + \left(\frac{1}{i\hbar}\right)^2\int_{t_0}^tdt' V_I(t')\int_{t_0}^{t'}dt'' V_I(t'') + \cdots$$ So, $$U_I(t,t_0) \approx \mathbb{I} + \frac{1}{i\hbar}\int_{t_0}^t V_I(t')dt'$$. Thus, $$\mathcal{P}_{i\to f} = |c_n(t)|^2 = |\langle n|U_I|i\rangle|^2 = |\langle n|1 + \frac{1}{i\hbar}\int_{t_0}^t dt' V_i(t')|i\rangle|^2 = |\delta_{ni} + \frac{1}{i\hbar}\int_{t_0}^tdt' V_{ni}\exp(i\omega_{ni}t)|^2$$. This gives us the same result as the perturbation theory collection and recurrence.

## Special Cases

### Step Function

$$V\neq f(t)$$. It is constant on either side of $$t=t_0$$. I.e. $$V = V_0\Theta(t-t_0) + V_i$$.

Assuming $$V = 0$$ before $$t=0$$. $$\mathcal{P}_{i\to f} = \frac{1}{\hbar^2}|V_{fi}|^2|\int_0^tdt'\exp(i\omega_{fi}t')|^2 = \frac{1}{\hbar^2}|V_{fi}|^2\frac{1}{\omega_{fi}^2}(\exp(i\omega_{fi}t)-1)^2 = \frac{4}{\omega_{fi}^2\hbar^2}|V_{fi}|^2\sin^2\left(\frac{\omega_{fi}t}{2}\right)$$.

## Transistions Between Continuum States

See: Ionization Discrete spectra transitions to continuum spectra from the energy being greater than the energy well.

Discrete spectra: $$H_0\Psi_n(\vec{r}) = E\Psi_n(\vec{r})$$

Continuous spectra: $$H_0\Psi_\alpha(\vec{r}) = E\Psi_\alpha(\vec{r})$$

$$\Psi(\vec{r},t) = \exp\left(-\frac{i}{\hbar}Ht\right)\Psi(\vec{r})$$

$$\int \Psi_{n'}^*\Psi_n d^3\vec{r} = \delta_{n'n}$$ $$\int \Psi_{\alpha'}^*\Psi_\alpha d^3\vec{r} = \delta(\alpha'-\alpha)$$ $$\int \Psi_n^*\Psi_\alpha d^3\vec{r} = 0$$

Closure: $$\sum_n |\Psi_n\rangle\langle\Psi_n| + \int d\alpha |\Psi_\alpha\rangle\langle\Psi_\alpha| = \mathbb{I}$$

At $$t=0$$ we turn on $$V(t)$$. SE: $$i\hbar \frac{\partial \Psi}{\partial t} = (H_0+V(t))\Psi$$

Before: $$\Psi(\vec{r},t) = \sum_n c_n(t)\exp\left(-\frac{i}{\hbar}E_n t\right)\Psi_n(\vec{r})$$.

Now: $$\Psi(\vec{r},t) = \sum_n c_n(t)\exp\left(-\frac{i}{\hbar}E_n t\right)\Psi_n(\vec{r}) + \int d\alpha c_\alpha(t)\exp\left(-\frac{i}{\hbar}E_\alpha t\right)\Psi_\alpha(\vec{r})$$. Normalization becomes: $$\sum_n |c_n|^2 + \int d\alpha |c_\alpha|^2 = 1$$. We now plug in this Ansatz into the SE.

So, $$i\hbar \left(\sum_n \frac{d c_n}{dt}\exp\left(-\frac{i}{\hbar}E_n t\right)\Psi_n(\vec{r}) + \int d\alpha \frac{d c_\alpha}{dt}\exp\left(-\frac{i}{\hbar}E_\alpha t\right)\Psi_\alpha(\vec{r}) \right) = \sum_n c_n(t)\exp\left(-\frac{i}{\hbar}E_n t\right)V(\vec{r},t)\Psi_n(\vec{r}) + \int d\alpha c_\alpha(t)\exp\left(-\frac{i}{\hbar}E_\alpha t\right)V(\vec{r},t)\Psi_\alpha(\vec{r})$$.

By orthogonality: $$i\hbar \frac{d c_n}{dt} = \sum_m c_m(t)\exp\left(-\frac{i}{\hbar}\omega_{mn} t\right)V_{mn} + \int d\alpha c_\alpha(t)\exp\left(-\frac{i}{\hbar}\omega_{m\alpha} t\right)V_{m\alpha}$$.

$$i\hbar \frac{d c_\beta}{dt} = \sum_n c_n(t)\exp\left(-\frac{i}{\hbar}\omega_{\beta n} t\right)V_{\beta n} + \int d\alpha c_\alpha(t)\exp\left(-\frac{i}{\hbar}\omega_{\beta \alpha} t\right)V_{\beta \alpha}$$.

$$V_{\kappa \ell} = \int d^3\vec{r} \Psi^*_{\kappa}V(\vec{r},t)\Psi_{\ell}$$.

From Perturbation Theory: $$c_n(t) = c_n^{(0)} + \lambda c_n^{(1)} + \cdots$$. $$c_\alpha(t) = c_\alpha^{(0)} + \lambda c_\alpha^{(1)} + \cdots$$.

For $$0^{th}$$ order: $$\frac{d c_{\kappa'}^{(0)}}{dt} = 0 \Rightarrow c_{\kappa'}^{(0)} = \mathbb{C}$$ are constant. For $$1^{st}$$ order: $$i\hbar \frac{d c_{\kappa'}^{(1)}}{dt} = \sum_n c_n^{(0)}\exp(i\omega_{\kappa' n}V_{\kappa' n}) + \int d\alpha c_\alpha^{(0)}\exp(i\omega_{\kappa' \alpha}V_{\kappa' \alpha})$$.

Assume we are at some state $$k$$ at $$t=0$$.

Then, $$c_n^{(0)} = c_\alpha^{(0)} = \delta_{n,k}+\delta(k-\alpha)$$. Assume $$k$$ is a discrete state. Then, $$\delta(k-\alpha) = 0$$.

So, $$i\hbar \frac{d c_n^{(1)}}{dt} = \exp(i\omega_{nk}t)V_{nk}(t)$$. Then, $$i\hbar \frac{d c_\alpha^{(1)}}{dt} = \exp(i\omega_{\alpha k}t)V_{\alpha k}(t)$$. Thus, $$c_\alpha^{(1)}(t) = \frac{1}{i\hbar}\int_0^t dt' V_{\alpha k}(t')\exp(i\omega_{\alpha k}t')$$

Remember: $$\mathcal{P}_{k\to\ell} = |c_\ell(t)|^2$$. For continuum, $$\mathcal{P}_{k\in[\alpha,\alpha+d\alpha]} = \int_{B(E)} |c_\alpha^{(1)}(t)|^2\mathcal{P}_\alpha(E) dE$$, the 1 superscript I am uneasy about. E.g. $$\mathcal{P}_{i\to f} = \frac{1}{\hbar^2}\int_{B(E)}\left|\int_0^t V_{\alpha k}(t')\exp(i\omega_{\alpha k}t')dt'\right|^2\mathcal{P}_\alpha(E)dE$$.

We determine (domain of energies $$B(E)$$ for which the particle ends up after transition) from the density of states, $$\mathcal{P}_\alpha(E)dE$$. For a discrete spectra, $$\mathcal{P}_n(E)dE = \delta(E-E_n)dE \Rightarrow \mathcal{P}_{n\to m} = |c_m(t)|^2$$.

If we had $$k=\beta$$, a continuous state. Then, $$c_n^{(0)} = c_\alpha^{(0)} = \delta_{n,\beta}+\delta(\beta-\alpha)$$. From our continuous state assumption, $$\delta_{n,\beta} = 0$$. So, $$i\hbar \frac{d c_n^{(1)}}{dt} = \exp(i\omega_{n\beta}t)V_{n\beta}(t)$$. Then, $$i\hbar \frac{d c_\alpha^{(1)}}{dt} = \exp(i\omega_{\alpha \beta}t)V_{\alpha \beta}(t)$$. Thus, $$c_\alpha^{(1)}(t) = \frac{1}{i\hbar}\int_0^t dt' V_{\alpha \beta}(t')\exp(i\omega_{\alpha \beta}t')$$

### Constant Perturbation

A pulse: Let $$V(\vec{r},t)$$ be turned on at a time $$t$$ and off a time $$T$$ later. $$V(\vec{r},t) = \begin{cases}V(\vec{r}) & 0\leq t\leq T\\0 & \text{otherwise}\end{cases}$$.

$$V_{\alpha\beta}(t') = \int_{\mathbb{R}^3}d^3\vec{r}\Psi_\alpha^*(\vec{r},t)V(\vec{r},t)\Psi_\beta(\vec{r},t)$$. So, $$\mathcal{P}_{\beta\to\alpha} = \frac{1}{\hbar^2}\int_{B(E)}|V_{\alpha\beta}|^2\mathcal{P}_{\alpha}(E)\left|\int_0^Tdt'\exp(i\omega_{\alpha\beta}t')\right|^2dE = \frac{1}{\hbar^2}\int_{B(E)}|V_{\alpha\beta}|^2\mathcal{P}_{\alpha}(E)\left|\text{sinc}^2\left(\frac{\omega_{\alpha\beta}}{2}T\right)\right|^2dE$$. Note that $$E_\alpha = E$$ in this case.

### Fermi’s Golden Rule - See Notes

$$\mathcal{P}_{\beta\to B(E)} = \frac{2\pi}{\hbar}|V_{\alpha\beta}|^2\rho_\alpha(E_F)$$ with $$B(E) = [E_f-\varepsilon/2,E_f+\varepsilon/2]$$ and $$V(\vec{r},t) = (\Theta(t)-\Theta(T-t))V(\vec{r})$$ which gives $$E_f\approx E_i$$.

For $$V=V_0\exp(i\omega t)$$ then $$E_f \approx E_i\pm\hbar\omega$$.

Energy non-conserving have zero transition probability rates. Energy conserving have constant transition probability rates.

Note, that spin increases the density of states!

### Some Applications of Fermi’s Golden Rule

#### Example - Scattering off Potential

Free particle (spinless) scattered by a ’localized enough’ $$V(\vec{r})$$. So, $$H=H_0 + V(\vec{r})$$. $$H_0 = \frac{p^2}{2m} \Rightarrow H_0\Psi_{\vec{k}}(\vec{r}) = K_{\vec{k}}\Psi_{\vec{k}}(\vec{r})$$. So, $$\Psi_{\vec{\kappa}}(\vec{r}) = \frac{1}{\sqrt{\tilde{V}}}\exp(i\vec{k}\cdot\vec{r}), E_{\vec{k}} = \frac{\hbar^2k^2}{2m}$$. For energy conserving, $$|k_i| = |k_f|$$.

$$\mathcal{P}_{\vec{k}_i\to \vec{k}_f} = \frac{2\pi}{\hbar}|V_{\vec{k}_f\vec{k}_i}|^2\rho(E_f)$$. $$V_{\vec{k}_f\vec{k}_i}$$ and $$\rho(E_f)$$ need to be computed. $$V_{\vec{k}_f\vec{k}_i} = \frac{1}{\tilde{V}}\int \exp(-i(\vec{k}_f-\vec{k}_i)\cdot\vec{r})V(\vec{r})d^3\vec{r} = V_{fourier}(\vec{k}_f-\vec{k}_i) = V_{fourier}(\vec{q})$$. Let $$\vec{q} = \vec{k}_f-\vec{k}_i$$. Ansatz: $$\rho(E_f) \propto 4\pi |k_i|^2$$. Note: implicitly from the setup we have volume chunks of $$\tilde{V}$$ where the wave function is the same so $$\exp(i\vec{k}\cdot(\vec{r}+\vec{L})) = \exp(i\vec{k}\cdot\vec{r})$$. So, $$\vec{k}\cdot\vec{L}=2\pi n$$. Hence, $$k_i = \frac{2\pi}{L_i}n_i$$ and so $$k=\frac{2\pi}{L}n$$. So, we have a grid of states of spacing $$2\pi/L$$ side length hence the volume is $$\left(\frac{2\pi}{L}\right)^3 = \frac{(2\pi)^3}{\tilde{V}}$$. Then, $$dN = \rho(\vec{k})d^3\vec{k} = \frac{1}{\frac{(2\pi)^3}{\tilde{V}}}d^3\vec{k} = \frac{\tilde{V}}{(2\pi)^3}k^2d\kappa d\Omega = \rho(E)dEd\Omega$$. Then, $$\rho(E) = \rho(\vec{k})k^2 \frac{dk}{dE} = \frac{\tilde{V}}{(2\pi)^3}\frac{mk}{\hbar^2}$$.

Therefore, $$\mathcal{P}_{\vec{k}_i\to\vec{k}_f}=\frac{2\pi}{\hbar}\frac{1}{\tilde{V}^2}|V_{fourier}(\vec{q})|^2 \frac{\tilde{V}}{(2\pi)^3}\frac{m}{\hbar^2}\frac{\sqrt{2mE}}{\hbar}=\frac{2^{3/2}\pi m^{3/2}}{\hbar^4}\frac{1}{\tilde{V}^2}|V_{fourier}(\vec{q})|^2 \frac{\tilde{V}}{(2\pi)^3}\sqrt{E} = \frac{2\pi}{\hbar\tilde{V}}|V_{fourier}(\vec{q})|^2\frac{m}{(2\pi)^3\hbar^2}k_f$$.

Differential Cross Section: $$\frac{d\sigma}{d\Omega} = \frac{\mathcal{P}_{\vec{k}_i\to\vec{k}_f}}{\frac{\text{Number,1 in our case}}{\tilde{V}}v_i} = \frac{\mathcal{P}_{\vec{k}_i\to\vec{k}_f}}{\frac{\text{Number,1 in our case}}{\tilde{V}}\frac{\hbar k_i}{m}} = \frac{m^2}{(2\pi)^2\hbar^4}|V_{fourier}(\vec{q})|^2$$. $$|\vec{q}|^2 = 4k_f\sin^2\frac{\theta}{2}$$.

Say we have a Coulomb potential: $$V(\vec{r}) = \frac{Z_1Z_2e^2}{r}$$, it is not ’localized enough’ for this method. Must go to zero faster than 1/r. So, if we have the Yukawa potential, $$\exp(-\alpha r)/r$$ we can localize it, solve it, then set $$\alpha\to 0$$ to get a solution at the very end. This gives $$V_{Fourier}(\vec{q}) = \frac{Z_1Z_2e^2}{(2\pi)^{3/2}}\frac{4\pi}{q^2}$$ and $$\frac{d\sigma}{d\Omega} = \frac{Z_1^2Z_2^2e^4}{16E^2\sin^4\frac{\theta}{2}}$$ - Rutherford cross section.

#### Example - Electromagnetic Fields

Direct the radiation along the y-axis with $$\omega = ck$$. So, direct the electric field along the z-axis and the magnetic field along the x-axis. $$\vec{A}(\vec{r},t) = \left(A_0\exp(i(ky-\omega t)) + A_0^*\exp(-i(ky-\omega t))\right)\hat{z}$$. So, $$\vec{E} = -\frac{\partial}{\partial t}\vec{A}(\vec{r}, t) = i\omega\left(A_0\exp(i(ky-\omega t)) + A_0^*\exp(-i(ky-\omega t))\right)\hat{z}$$ and $$\vec{B} = \nabla\times\vec{A} = ik(A_0\exp(i(ky-\omega t)) - A_0^*\exp(-i(ky-\omega t)))\hat{x}$$. Let $$i\omega A_0 = \frac{E_0}{2}$$ and $$ik A_0 = \frac{B_0}{2}$$. Then, $$\vec{E} = E_0\hat{z}\cos(ky-\omega t)$$ and $$\vec{B} = B_0\hat{x}\cos(ky-\omega t)$$. With $$\frac{E_0}{B_0} = \frac{\omega}{k} = c$$.

Constructing our Hamiltonian: $$H = H_{matter} + H_{EM} \Rightarrow \mathcal{E} = \mathcal{E}_{matter}\otimes\mathcal{E}_{EM}$$. $$H_{EM} = \sum_\lambda \hbar\omega_\lambda a_\lambda^\dagger a_\lambda = \sum_\lambda \hbar\omega_\lambda N_\lambda$$. Starting from classical Hamiltonian, $$\mathcal{H} = \vec{p}\cdot\vec{r}-\mathcal{L}$$. Thus, $$H_{matter} = \frac{1}{2m}\left(\vec{p}-q\vec{A}\right)^2 + qU(\vec{r},t)$$. In class: $$H_{matter} = \sum\limits_i\frac{1}{2m_i}\left(\vec{p_i}-\frac{q_i}{c}\vec{A}\right)^2 + \sum\limits_{i If we consider Hydrogen-like atoms, then we only have 2 charge carriers. Let \(q=-|e|$$ and $$\vec{\mu} = \frac{e}{mc}\vec{S}$$. Then, $$H_{matter} = \frac{1}{2m}\left(\vec{p}+\frac{e}{c}\vec{A}\right)^2-\frac{Ze^2}{r}+\frac{e}{mc}\vec{S}\cdot\vec{B} = \left(\frac{p^2}{2m} - \frac{Ze^2}{r}\right) + \frac{e}{mc}\vec{p}\cdot\vec{A} + \frac{e}{mc}\vec{S}\cdot\vec{B} + \frac{e^2}{2mc^2}\vec{A}^2 = H_0 + H_1 + H_2 + H_3$$. Terms $$H_1$$ and $$H_2$$ are typically dominant, unless you are dealing with high-power lasers.

Comparing $$H_1 = \frac{e}{mc}\vec{p}\cdot\vec{A}\sim \frac{e}{mc}pA_0$$ and $$H_2 = \frac{e}{mc}\vec{S}\cdot\vec{B}\sim\frac{e}{mc}\hbar kA_0$$. Then, $$\frac{H_2}{H_1} = \frac{\hbar k}{p} = \frac{\hbar}{p}\frac{2\pi}{\lambda}\sim \frac{\Delta x}{\lambda} = \frac{a_0}{Z\lambda}$$. For visible light, $$\lambda\sim 500$$ nm. Since $$a_0\sim 0.5$$ A, then the ratio is much smaller than 1. Hence, $$H_1\gg H_2$$. Remember that $$p$$ is the momentum of the electron and $$\hbar k$$ is the momentum of the radiation.

Then, $$H_1 = \frac{e}{mc}\vec{p}\cdot\vec{A} = \frac{e}{mc}p_z(A_0\exp(i(ky-\omega t)) + A_0^*\exp(-i(ky-\omega t))) = V_0\exp(-i\omega t) + V_0^\dagger\exp(i\omega t)$$. Thus, we get an absorption, $$V_0$$, and emission, $$V_0^\dagger$$, terms.

In the case of absorption, using Fermi’s golden rule, $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\left|\frac{e}{mc}A_0\langle f|\exp(iky)p_z|i\rangle\right|^2$$ with $$E_f = E_i+\hbar\omega$$.

Approximating, $$ky\sim \frac{2\pi}{\lambda}\frac{a_0}{Z}\ll 1$$. So, we get $$\exp(iky)\approx 1 + iky - \cdots$$. To the zeroth order (electric dipole approximation), $$\exp(iky)\approx 1$$. So, $$H_1 \approx H_{DE} = \frac{e}{mc}p_z(A_0\exp(-i\omega t) + A_0^*\exp(i\omega t)) = \frac{e}{mc}p_z\left(\frac{E_0}{2i\omega}\exp(-i\omega t) - \frac{E_0}{2i\omega}\exp(i\omega t)\right)$$. Hence, $$\langle f|H_{DE}|i\rangle \Rightarrow \mathbb{C}\langle f|p_z|i\rangle = \mathbb{C}im\omega_{fi}\langle f|z|i\rangle$$. Then, $$\mathcal{P}_{i\to f} \approx \frac{2\pi}{\hbar}\left|\frac{e}{mc}A_0\langle f|p_z|i\rangle\right|^2 = \frac{2\pi}{\hbar}\left|\frac{e}{mc}A_0m\omega_{fi}\langle f|z|i\rangle\right|^2$$. So, $$\Delta \ell=\pm 1$$ and $$\Delta m = 0$$.

To the first order, say the electric dipole term is zero, then $$\exp(iky)\approx 1+iky$$. Then, consider $$H_1-H_{DE}=\frac{e}{mc}p_zA_0(iky)$$. Hence, $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\left|\frac{e}{mc}A_0\langle f|p_ziky|i\rangle\right|^2 = \frac{2\pi}{\hbar}\left|\frac{e}{mc}\frac{B_0}{2}\langle f|p_zy|i\rangle\right|^2 = \frac{2\pi}{\hbar}\left|\frac{e}{mc}\frac{B_0}{2}\langle f\left|\frac{1}{2}(p_zy-zp_y)+\frac{1}{2}(p_zy+zp_y)\right|i\rangle\right|^2= \frac{2\pi}{\hbar}\left|\frac{e}{mc}\frac{B_0}{2}\langle f\left|\frac{1}{2}L_x+\frac{1}{2}(p_zy+zp_y)\right|i\rangle\right|^2$$. Combining this with $$H_2$$ we get $$H_1+H_2 = H_{DE} + \frac{e}{mc}\frac{B_0}{4}(L_x+2S_x) + \frac{e}{mc}\frac{B_0}{4}(p_zy+zp_y) = H_{DE} + H_{DM} + H_{QE}$$.

Thus the selection rules are for the magnetic dipole, $$\Delta\ell=0, \Delta m=\pm 1,0,\Delta s=0, \Delta m_s=\pm 1,0$$ note that $$\Delta m$$ and $$\Delta m_s$$ cannot both be zero. This is expected to be less probable than the selection rules for electric dipole selection terms.

For $$H_2=\frac{e}{mc}\vec{S}\cdot\vec{B} = \frac{e}{mc}S_xB_0\cos(\omega t)$$, with $$S_x = \frac{1}{2}\left(S_+ + S_-\right)$$.

For the electric quadrapole, $$\langle f|p_zy+zp_y|i\rangle = \langle f|\frac{im}{\hbar}[H_0,z]y + \frac{im}{\hbar}z[H_0,y]|i\rangle = \frac{im}{\hbar}\langle f|[H_0,zy]|i\rangle = \frac{im}{\hbar}\langle f|H_0zy-zyH_0|i\rangle = im\omega_{fi}\langle f|zy|i\rangle$$. $$z\sim Y_1^0$$ and $$Y\sim Y_1^{\pm 1}$$ so $$zy\sim Y_2^{\pm 1}$$. Then the selection rules are $$\Delta m = \pm 1,\Delta\ell = \pm 2, \pm 1, 0$$. By parity, $$\Delta\ell = \pm 2, 0$$.

Example: Hydrogen has a bright green line of 557.7 nm which is due to the quadrapole interaction.

#### Electric Dipole Approximation

Absorption cross-section $$\sigma_{abs} = \frac{\text{Energy per unit time absorbed by atom}}{\text{Energy flux of radiation field}} = \frac{\hbar\omega\mathcal{P}_{i\to f}}{\frac{\omega^2}{2\pi c}|A_0|^2} = \frac{\frac{2\pi}{\hbar}\frac{e^2}{c^2}\omega_{fi}^2|A_0|^2|\langle f|z|i\rangle|^2}{\frac{1}{2\pi}\omega^2/c|A_0|^2} = 4\pi^2\frac{e^2}{\hbar c}\omega_{fi}|\langle f|z|i\rangle|^2 = 4\pi^2\alpha\omega_{fi}|\langle f|z|i\rangle|^2$$. Note that $$\omega_{fi}=\omega$$. Oscillator Strength: $$f_{fi} = \frac{2m\omega_{fi}}{\hbar}|\langle f|z|i\rangle|^2$$ such that $$\sum_f f_{fi}=1$$. Then, $$\sigma_{abs} = 4\pi^2\alpha\frac{\hbar}{2m}f_{fi} = 2\pi^2\alpha\frac{\hbar}{m}f_{fi}$$. This oscillator strength sum is also called Thomas-Reiche-Kuhn sum rule.

### Emission

If absorption looks like $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\frac{e^2}{m^2c^2} |A_0|^2|\langle f|\exp(iky)p_z|i\rangle|^2\delta(E_f-E_i-\hbar\omega)$$. So, for emission, $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\frac{e^2}{m^2c^2} |A_0|^2|\langle f|\exp(-iky)p_z|i\rangle|^2\delta(E_f-E_i+\hbar\omega)$$. For any polarization, $$\vec{\epsilon}$$, $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\frac{e^2}{m^2c^2} |A_0|^2|\langle f|\exp(-iky)\vec{\epsilon}^*\cdot\vec{p}|i\rangle|^2\delta(E_f-E_i+\hbar\omega)$$. If $$A_0=0$$ then we get no absorption or emission. Thus, we get stimulated emission.

We also have spontaneous emission which is not described by this model, needs a purely quantum treatment.

#### Spontaneous Emission

From before: $$\mathcal{P}_{i\to f} = \frac{2\pi}{\hbar}\frac{e^2}{m^2c^2} |A_0|^2|\langle f|\exp(-iky)\vec{\epsilon}^*\cdot\vec{p}|i\rangle|^2\delta(E_f-E_i+\hbar\omega)$$.

We need a fully QM treatment. Second Quantization. We need to write $$\vec{A},\vec{E},\vec{B}$$ in terms of creation and annihilation operators.

So, $$\hat{A}_{0,\lambda,\vec{k}} = \sqrt{\frac{2\pi\hbar c^2}{\omega_k}}a_{\lambda,\vec{k}}$$. So, $$H = \sum_{\vec{k}}\sum_{\lambda=1}^2\hbar\omega_k\left(a_{\lambda,\vec{k}}^\dagger a_{\lambda,\vec{k}}+\frac{1}{2}\right) = \sum_{\vec{k}}\sum_{\lambda=1}^2\hbar\omega_k\left(N_{\lambda,\vec{k}}^\dagger + \frac{1}{2}\right)$$. $$\lambda$$ relates to the polarization and so every wave can have 2 possible polarization components.

$$a_{\lambda,\vec{k}}^\dagger |n_{\lambda,\vec{k}}\rangle = \sqrt{n_{\lambda,\vec{k}}+1}|n_{\lambda,\vec{k}}+1\rangle$$. $$N_{\lambda,\vec{k}}|n_{\lambda,\vec{k}}\rangle = n_{\lambda,\vec{k}}|n_{\lambda,\vec{k}}\rangle$$.

$$E_n = \sum_{\vec{k}}\sum_{\lambda=1}^2\hbar\omega_k\left(n_{\lambda,\vec{k}} + \frac{1}{2}\right)$$.

$$\vec{A}(\vec{r},t) = \sum_{\vec{k}} \sum_{\lambda} \sqrt{\frac{2\pi\hbar c^2}{\omega_k\tilde{V}}}\left[a_{\lambda,\vec{k}}\exp(i(\vec{k}\cdot\vec{r}-\omega_kt))\vec{\epsilon}_\lambda + a_{\lambda,\vec{k}}^\dagger\exp(-i(\vec{k}\cdot\vec{r}-\omega_kt))\vec{\epsilon}_\lambda^*\right]$$.

Recall, $$H_1\sim \vec{p}\cdot\vec{A} = \frac{e}{m}\sqrt{\frac{2\pi\hbar}{\tilde{v}}}\sum_k\sum_\lambda\frac{1}{\sqrt{\omega_k}}\left[a_{\lambda,\vec{k}}\exp(i(\vec{k}\cdot\vec{r}-\omega_kt))\vec{\epsilon}_\lambda + a_{\lambda,\vec{k}}^\dagger\exp(-i(\vec{k}\cdot\vec{r}-\omega_kt))\vec{\epsilon}_\lambda^*\right]$$, hence we still have a harmonic perturbation; $$V_0\exp(-i\omega t)+V_0^\dagger\exp(i\omega t)$$ with $$V_{0,k,\lambda} = \frac{e}{m}\sqrt{\frac{2\pi\hbar}{\tilde{V}}}\frac{1}{\sqrt{\omega_k}}a_{\lambda,\vec{k}}\exp(i\vec{k}\cdot\vec{r})\vec{\epsilon}_\lambda\cdot\vec{p}$$.

Then, $$|\Phi_i\rangle = |\Psi_i\rangle\otimes|n_\lambda,\vec{k}\rangle$$. For Absorption, $$|\Phi_f\rangle = |\Psi_f\rangle\otimes|n_{\lambda,\vec{k}}-1\rangle$$. For Emission, $$|\Phi_f\rangle = |\Psi_f\rangle\otimes|n_{\lambda,\vec{k}}+1\rangle$$.

Recall, for Absorption: $$\langle\Phi_f|V_{0,\lambda,\vec{k}}|\Phi_i\rangle = \frac{e}{m}\sqrt{\frac{2\pi\hbar}{\omega_k\tilde{V}}}\sqrt{n_{\lambda,\vec{k}}}\langle\Psi_f|\exp(i\vec{k}\cdot\vec{r})\vec{\epsilon}_\lambda\cdot\vec{p}|\Psi_i\rangle$$. $$\mathcal{P}_{i\to f}= \frac{4\pi^2e^2}{m^2\omega_k\tilde{V}}n_{\lambda,\vec{k}}|\langle\Psi_f|\exp(i\vec{k}\cdot\vec{r})\vec{\epsilon}_\lambda\cdot\vec{p}|\Psi_i\rangle|^2\delta(E_f-E_i-\hbar\omega_k)$$.

Recall, for Emission: $$\langle\Phi_f|V_{0,\lambda,\vec{k}}|\Phi_i\rangle = \frac{e}{m}\sqrt{\frac{2\pi\hbar}{\omega_k\tilde{V}}}\sqrt{n_{\lambda,\vec{k}}+1}\langle\Psi_f|\exp(i\vec{k}\cdot\vec{r})\vec{\epsilon}_\lambda\cdot\vec{p}|\Psi_i\rangle$$. $$\mathcal{P}_{i\to f}= \frac{4\pi^2e^2}{m^2\omega_k\tilde{V}}(n_{\lambda,\vec{k}}+1)|\langle\Psi_f|\exp(-i\vec{k}\cdot\vec{r})\vec{\epsilon}_\lambda^*\cdot\vec{p}|\Psi_i\rangle|^2\delta(E_f-E_i+\hbar\omega_k)$$. Now, we have the plus 1 which accounts for spontaneous emission. I.e. if the field has no photons emission is still possible if the state is excited.

The semiclassical approach does not have this constant term since $$n_{\lambda,\vec{k}}\gg 1$$ and so the constant term disappears.

Consider stimulated emission with the electric dipole approximation: $$n=0$$ and $$\exp(i\vec{k}\cdot\vec{r})\approx 1$$. $$\mathcal{P}_{i\to f}= \frac{4\pi^2 \omega_{fi}}{\omega_k\tilde{V}}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2\delta(E_f-E_i+\hbar\omega_k)$$. Where $$\vec{d}_{fi}=\langle \Psi_f|-e\vec{r}|\Psi_i\rangle$$ is the dipole moment matrix element.

We now need the density of states. $$dN = \rho(\vec{k})d^3\vec{k} = \frac{\tilde{V}}{(2\pi)^3}\left(k^2dkd\Omega\right) = \frac{\tilde{V}\omega^2}{(2\pi c)^3}d\omega d\Omega$$.

The Transition Rate for Emission of a Photon in the solid angle $$d\Omega$$: $$dW_{i\to f}^{em} = \frac{\tilde{V}}{(2\pi c)^3}d\Omega\int\omega^2\mathcal{P}_{i\to f}d\omega = \frac{\tilde{V}}{(2\pi c)^3}d\Omega\int\omega^2\frac{4\pi^2\omega_{fi}}{\omega_k\tilde{V}}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2\delta(E_f-E_i+\hbar\omega_k)d\omega = \frac{\omega^3}{2\pi\hbar c^3}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2d\Omega$$. Note, $$\omega=\omega_k=\omega_{fi}$$ from the delta function.

Consider for each polarization, $$\sum_{\lambda=1}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2 = |\vec{d}_{fi}|^2-|(d_{fi})_3|^3$$. $$\sum_{\lambda=1}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2 = |\vec{\epsilon}_1^*\cdot(d_{fi})_1|^2 + |\vec{\epsilon}_2^*\cdot(d_{fi})_2|^2$$ where the subscripts can be x, y, or z. Hence, $$\sum_{\lambda=1}|\vec{\epsilon}_\lambda^*\cdot\vec{d}_{fi}|^2 = \frac{2}{3}|\vec{d}_{fi}|^2$$. Then, $$dW_{i\to f}^{em}=\frac{\omega^3}{3\pi\hbar c^3}|\vec{d}_{fi}|^3d\Omega$$. Integrating about the entire sphere, $$W_{i\to f}^{em} = \frac{4\omega^3}{3\hbar c^3}|\vec{d}_{fi}|^2$$.

If we have a power-meter, using our probability per unit time, $$I_{i\to f}^{em} = \hbar\omega W_{i\to f}^{em} = \frac{4\omega^4}{3c^3}|\vec{d}_{fi}|^2$$, which is mostly classical except the dipole matrix element which in the classical case is just the oscillating antennae dipole. This $$\omega^4$$ harkens to Raleigh scattering.

Consider the lifetime of an excited state, $$\tau = \frac{1}{\sum_f W_{i\to f}}$$.

#### Hydrogen Excited State

Lifetime of Hydrogen atom excited state of 2p $$\to$$ 1s.

$$\tau_{2p} = \frac{1}{W_{2p\to 1s}}$$. Which comes down to computing the dipole matrix element, $$|\langle 100|\vec{r}|21m\rangle|^2$$. Should get the same value for any $$m$$. With $$W_{2p\to 1s}^{em} = \frac{1}{3}\sum_{m=-1}^1 W_{2pm\to 1s}$$. The third arises due to no bias between the three possible choices. $$W_{2p\to 1s} = \frac{4}{3}\frac{e^2\omega^3_{2p\to 1s}}{\hbar c^3}|\vec{r}_{fi}|^2 = 1.6$$ ns.

#### Linewidths, Shapes, Intensities

$$|\Psi(t)\rangle = \sum_n c_n(t)\exp\left(-\frac{i}{\hbar}E_n t\right)|n\rangle$$.

So, $$i\hbar \frac{d c_n(t)}{dt} = \sum_k V_{nk}(t)c_k(t)\exp(i\omega_{nk}t)$$ giving $$\mathcal{P}_{i\to f}=|c_f(t)|^2$$.

For a two level system with the population all in one state under a resonant harmonic perturbation the time evolution oscillates between the two levels. Say $$V\sim \gamma\sin\omega t$$ then $$|c_1|^2\sim\cos^2\frac{\gamma t}{\hbar},|c_2|^2 = 1-|c_1|^2$$.

Motivation: What happens if we were in an excited state to start and include sponteneous emission to the coefficient without any perturbative potential?

$$\mathcal{P}_{2}(t+dt) = \mathcal{P}_2(t)(1-W_{21}^{em}dt) = \mathcal{P}_2(t)(1-\frac{dt}{\tau})$$. So, $$1-W_{21}^{em}dt$$ is the probability that no transition from 2 to 1 happens. $$\mathcal{P}_2(t) = \mathbb{C}\exp\left(-\frac{t}{\tau}\right)$$. If at $$t=0$$ we are in the 2 state, $$\mathcal{P}_2(t) = \exp\left(-\frac{t}{\tau}\right)$$. Then, $$c_2(t) = \exp\left(-\frac{t}{2\tau}\right)$$ hence $$\Psi_2(\vec{r},t) = c_2(t)\Psi_2(\vec{r})\exp\left(-\frac{i}{\hbar}E_2t\right)$$. Then, $$\Psi_2(\vec{r},t)=\Psi_2(\vec{r})\exp\left(-\frac{i}{\hbar}\left(E_2-\frac{i\hbar}{2\tau}\right)t\right)$$. From, $$\exp\left(-\frac{i}{\hbar}E_2t\right) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}a(E')\exp\left(-\frac{i}{\hbar}E't\right)dE' = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\left(\sqrt{2\pi\hbar}\delta(E_2-E')\right)\exp\left(-\frac{i}{\hbar}E't\right)dE'$$. From our expression, $$\exp\left(-\frac{i}{\hbar}\left(E_2-\frac{i\hbar}{2\tau}\right)t\right) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty a(E')\exp\left(-\frac{i}{\hbar}E't\right)dE'$$. Using an inverse fourier transform, assuming $$\Psi_2(t<0)=0$$, $$a(E') = \frac{1}{\sqrt{2\pi\hbar}}\int_0^\infty dt\exp\left(-\frac{i}{\hbar}\left(E_2-\frac{i\hbar}{2\tau}\right)t\right)\exp\left(\frac{i}{\hbar}E't\right) = \frac{1}{\sqrt{2\pi\hbar}}\frac{-i\hbar}{E_2-\frac{i\hbar}{2\tau}-E'}$$. The probability of finding the system in state 2 with definite energy $$E$$: $$|a(E)|^2 = \frac{\hbar}{2\pi}\frac{1}{(E_2-E)^2+\frac{\hbar^2}{4\tau^2}}$$, a Lorentzian. If we say $$E=E_1+\hbar\omega$$, then, $$|a(E)|^2 = \frac{\hbar}{2\pi}\frac{1}{(E_2-E_1-\hbar\omega)^2+\frac{\hbar^2}{4\tau^2}} = \frac{1}{\hbar}\frac{1}{(\omega_{21}-\omega)^2+\frac{\Gamma^2}{4\hbar^2}}$$ where $$\Gamma = \frac{\hbar}{\tau}$$. Note that $$\Gamma$$ relates to the width and so for an infinite lifetime, we get a delta function.

Constructing a line width, the normalized distribution is: $$f(\omega) = \frac{\Gamma^2/4\hbar^2}{(\omega_{21}-\omega)^2+\frac{\Gamma^2}{4\hbar^2}}$$, which peaks at the resonant frequency $$\omega_{21}$$.

Relating to the uncertainty, $$\Delta E\sim \Gamma$$ and $$\Delta t\sim \tau$$ then $$\Delta E\Delta t \sim \hbar$$.

If we have multiple pathways, $$\tau$$ decreases and so $$\Gamma$$ increases.

For hydrogen atom 2s $$\to$$ 1p, $$\Gamma = \frac{\hbar}{\tau}\sim 4\cdot10^{-7}$$ eV.

Remains Lorentzian.

$$\tau = \frac{1}{W_{total}^{em}}$$. With a new term in the sum $$W_{collision}\equiv W_c = nv\sigma$$. The anatomy of this term is the concentration times the average velocity times the collision cross section.

The collisional term can be much larger than the other transition rates and dominate to lead to $$\Gamma = \hbar/\tau\sim W_c$$.

Becomes a Gaussian. Broader peak but falls to zero quicker.

Can think of a bunch of little Lorentzians from the individual molecules that as an ensemble gives the Gaussian.

$$\omega = \omega_0(1\pm\frac{v}{c})^{-1}\approx\omega_0(1\mp\frac{v}{c})$$.

For $$N$$ number of atoms with velocity in $$[v,v+dv]$$, $$dN = N_0\exp\left(-\frac{Mv^2}{2k_BT}\right)dv$$, a Maxwell distribution.

From the intensity of light, $$I(\omega) = I(\omega_0)\exp\left(-\frac{Mc^2}{2kT}\left(\frac{\omega-\omega_0}{\omega_0}\right)^2\right)$$. So the width is, $$\Delta \omega^D = \frac{2\omega_0}{c}\left(\frac{2k_BT}{M}\ln 2\right)$$.

### Voight Profile

Gaussian + Lorentzian.

MIDTERM MATERIAL ENDS HERE.

Created: 2023-06-25 Sun 02:27

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