5.2.3 Applications#

Prompts

What does first-order time-dependent perturbation theory compute in practice, and why can many applications be reduced to transition amplitudes and probabilities?
For a harmonic drive, why does the transition probability develop a resonance profile, and what controls the peak position and spectral width?
Why does the long-time limit lead to Fermi’s golden rule, and what makes the resonance condition \(E_f-E_i=\hbar\omega\) emerge in that limit?
For an adiabatic exponential turn-on, why is the transition profile Lorentzian in detuning, and how does the adiabatic limit connect to the stationary perturbative picture?
In linear response, why is \(\delta\hat{H}=-\hat{\boldsymbol{j}}\cdot\delta\boldsymbol{A}\) the right perturbation for weak electromagnetic driving, and how does this lead to the Kubo view of Hall conductivity?

Lecture Notes#

Overview#

Section 5.2.2 produced the Dyson series for the dressed Green’s function \(\hat{G}(t,t_0)\) (Eq. (194)). This subsection answers the experimentally meaningful question: if a system is prepared in \(\vert i\rangle\) at time \(t_0\), what is the probability to find it in \(\vert f\rangle\) at time \(t\)? The answer is the squared first-order matrix element of \(\hat{G}\), which evaluates differently for different shapes of \(\hat{V}(t)\). Three iconic results follow:

Fermi’s golden rule — sudden harmonic drive in the long-time limit gives a transition rate \(\propto\vert V_{fi}\vert^2\,\delta(E_f-E_i-\hbar\omega)\).
Adiabatic process — exponential ramp gives a Lorentzian in \(\Delta E\) and recovers the time-independent perturbation result (5.1.2) as \(\tau\to\infty\).
Kubo formula — replace the drive by a vector potential and the observable by a current; first-order linear response gives the conductivity, with quantized Hall response on filled Landau levels.

All three share the same one-line input, Eq. (196) below; the differences are entirely in the time profile of \(\hat{V}(t)\).

Transition Probability#

Prepare the system in an eigenstate \(\vert i\rangle\) of \(\hat{H}_0\) at time \(t_0\). After time \(t\), evolution under \(\hat{H}=\hat{H}_0+\hat{V}(t)\) takes it to \(\hat{G}(t,t_0)\vert i\rangle\). The probability to measure \(\vert f\rangle\) is

(195)#\[ \boxed{\; P_{i\to f}(t,t_0)=\bigl\vert\langle f\vert\hat{G}(t,t_0)\vert i\rangle\bigr\vert^2.\;} \]

For \(f\neq i\), the leading nonzero contribution comes from the first-order Dyson term in Eq. (194):

\[ \hat{G}(t,t_0)\simeq \hat{G}_0(t,t_0)-\frac{\mathrm{i}}{\hbar}\!\int_{t_0}^{t}\!\mathrm{d}t_1\,\hat{G}_0(t,t_1)\,\hat{V}(t_1)\,\hat{G}_0(t_1,t_0). \]

Sandwich between \(\langle f\vert\) and \(\vert i\rangle\). Since \(f\neq i\), \(\langle f\vert\hat{G}_0(t,t_0)\vert i\rangle=0\). Insert the spectral form Eq. (192) to extract Bohr phases:

Derivation: phase cancellation in the first-order amplitude

Eigenstate shortcuts give \(\hat{G}_0(t,t_1)\to\mathrm{e}^{-\mathrm{i}E_f(t-t_1)/\hbar}\) on \(\langle f\vert\) and \(\hat{G}_0(t_1,t_0)\vert i\rangle\to\mathrm{e}^{-\mathrm{i}E_i(t_1-t_0)/\hbar}\vert i\rangle\). Hence

\[ \langle f\vert\hat{G}(t,t_0)\vert i\rangle \simeq -\frac{\mathrm{i}}{\hbar}\!\int_{t_0}^{t}\!\mathrm{d}t_1\,\mathrm{e}^{-\mathrm{i}E_f(t-t_1)/\hbar}\,V_{fi}(t_1)\,\mathrm{e}^{-\mathrm{i}E_i(t_1-t_0)/\hbar}, \]

where \(V_{fi}(t_1)\equiv\langle f\vert\hat{V}(t_1)\vert i\rangle\). Collect the \(t_1\)-independent phases:

\[ \langle f\vert\hat{G}(t,t_0)\vert i\rangle \simeq -\frac{\mathrm{i}}{\hbar}\,\mathrm{e}^{-\mathrm{i}E_f t/\hbar}\,\mathrm{e}^{\mathrm{i}E_i t_0/\hbar}\!\int_{t_0}^{t}\!\mathrm{d}t_1\,V_{fi}(t_1)\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}, \]

with \(\omega_{fi}\equiv(E_f-E_i)/\hbar\). The leading \(t_1\)-independent phases drop out of \(\vert\cdot\vert^2\).

The transition probability to first order in \(\hat{V}\) is then

(196)#\[ \boxed{\; P_{i\to f}^{(1)}(t,t_0)=\frac{1}{\hbar^2}\left\vert\int_{t_0}^{t}\!\mathrm{d}t_1\,V_{fi}(t_1)\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\right\vert^{2}.\;} \]

This is the single time integral that the rest of the lecture evaluates for different time profiles of \(\hat{V}(t)\).

Fermi’s Golden Rule#

Take a sudden harmonic perturbation switched on at \(t_0=0\):

\[\begin{split} \hat{V}(t)=\begin{cases} \hat{V}\,\mathrm{e}^{-\mathrm{i}\omega t}, & t>0,\\[4pt] 0, & t\le 0, \end{cases} \end{split}\]

so \(V_{fi}(t_1)=V_{fi}\,\mathrm{e}^{-\mathrm{i}\omega t_1}\) with \(V_{fi}\equiv\langle f\vert\hat{V}\vert i\rangle\) time-independent. Substituting into Eq. (196):

\[ \int_0^t\!\mathrm{d}t_1\,\mathrm{e}^{\mathrm{i}(\omega_{fi}-\omega)t_1}=\frac{\mathrm{e}^{\mathrm{i}(\omega_{fi}-\omega)t}-1}{\mathrm{i}(\omega_{fi}-\omega)}, \]

and taking the squared modulus:

(197)#\[ \boxed{\; P_{i\to f}^{(1)}(t)=\frac{\vert V_{fi}\vert^{2}}{\hbar^{2}}\left[\frac{\sin\!\bigl((\omega_{fi}-\omega)t/2\bigr)}{(\omega_{fi}-\omega)/2}\right]^{2}.\;} \]

Derivation: exponential to sinc-squared

Let \(\alpha=(\omega_{fi}-\omega)/2\). Then

\[ \frac{\mathrm{e}^{\mathrm{i}2\alpha t}-1}{2\mathrm{i}\alpha} =\mathrm{e}^{\mathrm{i}\alpha t}\,\frac{\mathrm{e}^{\mathrm{i}\alpha t}-\mathrm{e}^{-\mathrm{i}\alpha t}}{2\mathrm{i}\alpha} =\mathrm{e}^{\mathrm{i}\alpha t}\,\frac{\sin(\alpha t)}{\alpha}, \]

so \(\vert(\mathrm{e}^{\mathrm{i}2\alpha t}-1)/(2\mathrm{i}\alpha)\vert^{2}=\sin^{2}(\alpha t)/\alpha^{2}\).

Properties of the sinc-squared kernel

Setting \(\alpha=(\omega_{fi}-\omega)/2\), the kernel \((\sin\alpha t/\alpha)^{2}\) has

a peak at \(\alpha=0\) (resonance) of height \(t^{2}\),
a width of order \(1/t\),
a total weight \(\int\mathrm{d}\alpha\,(\sin\alpha t/\alpha)^{2}=\pi t\) that grows linearly in \(t\).

The product (peak height)\(\times\)(width)\(\sim t\) is what makes a constant rate emerge after summing over a continuum of final states.

Long-time limit. Use the distribution identity

\[ \lim_{t\to\infty}\frac{1}{t}\left(\frac{\sin\alpha t}{\alpha}\right)^{\!2}=\pi\,\delta(\alpha)=2\pi\,\delta(2\alpha), \]

and define the transition rate \(W_{i\to f}\equiv\lim_{t\to\infty}P_{i\to f}^{(1)}(t)/t\). Then

Fermi’s golden rule

(198)#\[ \boxed{\; W_{i\to f}=\frac{2\pi}{\hbar}\,\vert\langle f\vert\hat{V}\vert i\rangle\vert^{2}\,\delta\!\bigl(E_f-E_i-\hbar\omega\bigr).\;} \]

The delta enforces resonance: the perturbation transfers probability between levels separated by \(\hbar\omega\). For static \(\hat{V}\) (\(\omega=0\)), the resonance is at \(E_f=E_i\).

Continuum of final states

Summing over final states with density of states \(\rho(E_f)\) (per unit energy),

\[ W_{i}=\int\mathrm{d}E_f\,\rho(E_f)\,W_{i\to f}=\frac{2\pi}{\hbar}\,\vert\langle f\vert\hat{V}\vert i\rangle\vert^{2}\,\rho(E_f)\Big\vert_{E_f=E_i+\hbar\omega}. \]

The matrix element is evaluated at the resonance shell selected by \(\delta(E_f-E_i-\hbar\omega)\).

“But \(P\to\infty\) as \(t\to\infty\)”

For a transition with \(\omega_{fi}\neq\omega\), the long-time condition means \(t\gg 1/\vert\omega_{fi}-\omega\vert\) — a microscopic time that can still be short compared with the depletion time \(\sim\hbar/\vert V_{fi}\vert\). The rate \(W_{i\to f}=P/t\) is meaningful in perturbation theory whenever \(P\ll 1\). Once \(P\) is no longer small, depletion of \(\vert i\rangle\) must be tracked separately (master equations, self-energy).

Discussion: time–energy resolution

The width \(\sim 2\pi/t\) of the sinc-squared peak is the finite-time energy resolution: a perturbation acting only for time \(t\) cannot distinguish energy differences smaller than \(2\pi\hbar/t\). Strict \(\delta\)-function energy conservation only emerges in the limit \(t\to\infty\). How does this connect to the time–energy uncertainty relation, and what changes if the perturbation is a finite pulse of duration \(T\)?

Adiabatic Process#

Now take an exponential ramp from the infinite past, switched off at \(t=0\):

\[\begin{split} \hat{V}(t)=\begin{cases} \hat{V}\,\mathrm{e}^{t/\tau}, & t<0,\\[4pt] 0, & t\ge 0, \end{cases}\qquad \tau>0. \end{split}\]

System prepared in \(\vert i\rangle\) in the infinite past; ask the transition probability to \(\vert f\rangle\) at \(t=0\). Substitute into Eq. (196) with \(t_0\to-\infty\) and \(t\to 0\):

(199)#\[ \boxed{\; P_{i\to f}=\frac{1}{\hbar^{2}}\left\vert\int_{-\infty}^{0}\!\mathrm{d}t_1\,V_{fi}\,\mathrm{e}^{t_1/\tau}\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\right\vert^{2} =\frac{\vert V_{fi}\vert^{2}}{(E_f-E_i)^{2}+(\hbar/\tau)^{2}}.\;} \]

Derivation: Lorentzian from the ramp integral

\[ \int_{-\infty}^{0}\!\mathrm{d}t_1\,\mathrm{e}^{(\mathrm{i}\omega_{fi}+1/\tau)t_1} =\Bigl[\frac{\mathrm{e}^{(\mathrm{i}\omega_{fi}+1/\tau)t_1}}{\mathrm{i}\omega_{fi}+1/\tau}\Bigr]_{-\infty}^{0} =\frac{1}{\mathrm{i}\omega_{fi}+1/\tau}, \]

(the lower limit vanishes because \(\tau>0\)). Then

\[ \left\vert\frac{V_{fi}}{\mathrm{i}\omega_{fi}+1/\tau}\right\vert^{2}\!\!/\hbar^{2} =\frac{\vert V_{fi}\vert^{2}}{\hbar^{2}\bigl(\omega_{fi}^{2}+1/\tau^{2}\bigr)} =\frac{\vert V_{fi}\vert^{2}}{(E_f-E_i)^{2}+(\hbar/\tau)^{2}}. \]

The result is a Lorentzian in \(\Delta E=E_f-E_i\), centered at zero with width \(\hbar/\tau\). States closer in energy hybridize more readily; the finite ramp time \(\tau\) sets an energy resolution and regularizes the singularity that would otherwise appear at \(\Delta E=0\).

Adiabatic limit \(\tau\to\infty\): matching to time-independent perturbation theory

As \(\tau\to\infty\) the perturbation is turned on infinitely slowly, and the eigenstate \(\vert i\rangle\) of \(\hat{H}_0\) evolves continuously into the corresponding eigenstate \(\vert i(V)\rangle\) of \(\hat{H}_0+\hat{V}\). To first order in \(\hat{V}\) (cf. 5.1.2),

\[ \vert i(V)\rangle = \vert i\rangle + \sum_{m\neq i}\vert m\rangle\,\frac{V_{mi}}{E_i-E_m}+\cdots, \]

\[ \vert\langle f\vert i(V)\rangle\vert^{2}=\frac{\vert V_{fi}\vert^{2}}{(E_i-E_f)^{2}}, \]

which matches the \(\tau\to\infty\) limit of Eq. (199). Time-dependent perturbation theory falls back to time-independent perturbation theory when the perturbation changes slowly enough.

For any realistic process, \(\tau\) is finite. The Lorentzian width \(\hbar/\tau\) is the uncertainty-principle resolution: the apparent singularity of the energy denominator in time-independent perturbation theory is smoothed out — a genuine physical fact, not an artefact of approximation.

Poll: when does the Lorentzian narrow to the static answer?

The transition probability \(P_{i\to f}\) from an exponential ramp is Eq. (199). As \(\tau\to\infty\) the Lorentzian narrows. Which statement best describes the limit?

(A) \(P_{i\to f}\to 0\) for every fixed \(\Delta E\neq 0\), but the integrated weight is finite and matches the static perturbation result \(\vert V_{fi}/(E_i-E_f)\vert^{2}\) at \(\Delta E\neq 0\).

(B) \(P_{i\to f}\to\infty\) at \(\Delta E=0\) regardless of \(V_{fi}\).

(D) The Lorentzian becomes a Gaussian.

Kubo Formula#

The same first-order machinery applies when the perturbation is a vector potential and the observable is a current. Couple the system to a uniform electric field switched on adiabatically (the same exponential ramp logic as the previous section, with \(\tau^{-1}\to 0^{+}\)):

\[\begin{split} \begin{split} \boldsymbol{E}(t)&=\boldsymbol{E}\,\mathrm{e}^{-\mathrm{i}(\omega+\mathrm{i}0^{+})t},\\ \delta\boldsymbol{A}(t)&=-\frac{\mathrm{i}\boldsymbol{E}}{\omega+\mathrm{i}0^{+}}\,\mathrm{e}^{-\mathrm{i}(\omega+\mathrm{i}0^{+})t},\\ \boldsymbol{E}(t)&=-\partial_t\delta\boldsymbol{A}(t). \end{split} \end{split}\]

Minimal coupling gives the perturbation

(200)#\[ \delta\hat{H}(t)=-\hat{\boldsymbol{j}}\cdot\delta\boldsymbol{A}(t). \]

To match the convention used in 4.3.3 Quantum Hall Effect, we keep the sample area \(A\) explicit (rather than setting \(A=1\)).

Derivation: Kubo formula

Each electron occupies a state \(\vert\alpha\rangle\) that evolves independently under \(\hat{H}_0 + \delta\hat{H}(t)\). We compute the response state-by-state, then sum over occupied states and divide by \(A\) to assemble the macroscopic current density \(\boldsymbol{j}\) via the current-density definition.

Step 1 — First-order response of one occupied state.

Each occupied single-particle state \(\vert\alpha\rangle\) evolves under the interaction-picture operator \(\hat{U}_{\mathcal{I}}(t)\) from §5.2.1. The expectation value of the single-particle current in that state is

\[ \langle\hat{j}_a(t)\rangle_\alpha \equiv \langle\alpha\vert\hat{U}_{\mathcal{I}}^\dagger(t)\,\hat{j}_a^{\mathcal{I}}(t)\,\hat{U}_{\mathcal{I}}(t)\vert\alpha\rangle. \]

Expanding to first order in \(\delta\hat{H}\) gives the standard first-order Kubo identity,

\[ \langle\hat{j}_a(t)\rangle_\alpha = \langle\alpha\vert\hat{j}_a^{\mathcal{I}}(t)\vert\alpha\rangle - \frac{\mathrm{i}}{\hbar}\int_{-\infty}^{t}\!\mathrm{d}t'\,\langle\alpha\vert[\hat{j}_a^{\mathcal{I}}(t),\delta\hat{H}_{\mathcal{I}}(t')]\vert\alpha\rangle, \]

with \(\hat{j}_a^{\mathcal{I}}(t) = \mathrm{e}^{\mathrm{i}\hat{H}_0 t/\hbar}\hat{j}_a\,\mathrm{e}^{-\mathrm{i}\hat{H}_0 t/\hbar}\).

Step 2 — Sum over occupied states; assemble the current density.

Substitute \(\delta\hat{H}_{\mathcal{I}}(t') = -\hat{j}_b^{\mathcal{I}}(t')\,\delta A_b(t')\), sum over occupied states, and divide by \(A\). The zeroth-order term \(\sum_{\alpha\in\mathrm{occ}}\langle\alpha\vert\hat{j}_a\vert\alpha\rangle = 0\) vanishes (no current without a driving field), leaving

\[ j_a(t) = \frac{\mathrm{i}}{\hbar A}\!\int_{-\infty}^{t}\!\mathrm{d}t'\,\delta A_b(t')\sum_{\alpha\in\mathrm{occ}}\langle\alpha\vert[\hat{j}_a^{\mathcal{I}}(t),\hat{j}_b^{\mathcal{I}}(t')]\vert\alpha\rangle. \]

Step 3 — Substitute the harmonic field.

Using \(\delta A_b(t') = (-\mathrm{i}E_b/(\omega+\mathrm{i}0^+))\,\mathrm{e}^{-\mathrm{i}(\omega+\mathrm{i}0^+)t'}\) and the change of variable \(\tau = t-t'\) (with time-translation invariance of the unperturbed system), the result \(j_a(t) = \sigma_{ab}(\omega)\,E_b\,\mathrm{e}^{-\mathrm{i}(\omega+\mathrm{i}0^+)t}\) defines

\[ \sigma_{ab}(\omega) = -\frac{1}{\hbar(\omega+\mathrm{i}0^+) A}\int_0^{\infty}\!\mathrm{d}\tau\,\mathrm{e}^{\mathrm{i}(\omega+\mathrm{i}0^+)\tau}\sum_{\alpha\in\mathrm{occ}}\langle\alpha\vert[\hat{j}_a^{\mathcal{I}}(\tau),\hat{j}_b^{\mathcal{I}}(0)]\vert\alpha\rangle. \]

Step 4 — Insert eigenbasis and evaluate the time integral.

Insert \(\sum_\gamma\vert\gamma\rangle\langle\gamma\vert\) between the two single-particle current operators and use the matrix-element identity \(\langle\alpha\vert\hat{j}_a^{\mathcal{I}}(\tau)\vert\gamma\rangle = \mathrm{e}^{-\mathrm{i}\omega_{\gamma\alpha}\tau}\,\langle\alpha\vert\hat{j}_a\vert\gamma\rangle\) on each \(\langle\alpha\vert\cdots\vert\alpha\rangle\) matrix element (with \(\omega_{\gamma\alpha} = (E_\gamma-E_\alpha)/\hbar\)),

\[ \sum_{\alpha\in\mathrm{occ}}\langle\alpha\vert[\hat{j}_a^{\mathcal{I}}(\tau),\hat{j}_b^{\mathcal{I}}(0)]\vert\alpha\rangle = \sum_{\alpha\in\mathrm{occ}}\sum_\gamma\Big(j^a_{\alpha\gamma}j^b_{\gamma\alpha}\,\mathrm{e}^{-\mathrm{i}\omega_{\gamma\alpha}\tau} - j^b_{\alpha\gamma}j^a_{\gamma\alpha}\,\mathrm{e}^{\mathrm{i}\omega_{\gamma\alpha}\tau}\Big), \]

with \(j^a_{\alpha\gamma} \equiv \langle\alpha\vert\hat{j}_a\vert\gamma\rangle\) a single-particle matrix element. The \(\tau\) integral gives \(\int_0^\infty\!\mathrm{d}\tau\,\mathrm{e}^{\mathrm{i}(\omega+\mathrm{i}0^+ \pm\omega_{\gamma\alpha})\tau} = \mathrm{i}/(\omega+\mathrm{i}0^+\pm\omega_{\gamma\alpha})\).

Step 5 — DC limit.

For \(\gamma = \alpha\) (intra-band), the integrand is finite but the prefactor \(1/\omega\) produces a pole — the Drude weight, which is absent for a gapped insulator and irrelevant to the antisymmetric Hall part. For \(\gamma\neq\alpha\) (inter-band), expand \(1/(\omega\pm\omega_{\gamma\alpha}) = \pm 1/\omega_{\gamma\alpha} - \omega/\omega_{\gamma\alpha}^2 + O(\omega^2)\). The \(\pm 1/\omega_{\gamma\alpha}\) piece, divided by \(\omega\), gives the symmetric (longitudinal) divergence that cancels in clean systems with current conservation. The \(-\omega/\omega_{\gamma\alpha}^2\) piece gives a finite antisymmetric contribution. Restricting \(\gamma\notin\mathrm{occ}\) — terms with \(\gamma\in\mathrm{occ}\) pair-cancel because both \(\alpha\) and \(\gamma\) run over the occupied set — yields

\[ \sigma_{xy} = \frac{\mathrm{i}\hbar}{A}\sum_{\alpha\in\mathrm{occ}}\sum_{\beta\notin\mathrm{occ}}\frac{\langle\alpha\vert\hat{j}_x\vert\beta\rangle\langle\beta\vert\hat{j}_y\vert\alpha\rangle - \langle\alpha\vert\hat{j}_y\vert\beta\rangle\langle\beta\vert\hat{j}_x\vert\alpha\rangle}{(E_\beta-E_\alpha)^2}, \]

the Kubo formula above. All matrix elements are single-particle (\(\hat{j}\)); the \(1/A\) traces back to using current density (current per unit area).

Substitute into Eq. (196) and compute the first-order response of the current \(\langle\hat{\boldsymbol{j}}\rangle\) on a filled-band ground state at \(T=0\). The result is the Hall conductivity:

Kubo formula (zero temperature)

(201)#\[ \sigma_{xy}=\frac{\mathrm{i}\hbar}{A} \sum_{\alpha\in\mathrm{occ}}\sum_{\beta\notin\mathrm{occ}} \frac{\langle\alpha\vert\hat{j}_x\vert\beta\rangle\langle\beta\vert\hat{j}_y\vert\alpha\rangle- \langle\alpha\vert\hat{j}_y\vert\beta\rangle\langle\beta\vert\hat{j}_x\vert\alpha\rangle} {(E_\beta-E_\alpha)^{2}}. \]

A transport coefficient written entirely in terms of virtual transitions between occupied \(\vert\alpha\rangle\) and empty \(\vert\beta\rangle\) states. The energy-denominator structure mirrors the time-independent perturbation theory of 5.1.2, now for a many-body filled ground state.

Evaluating Eq. (201) for \(\nu\) completely filled Landau levels gives the exact integer Hall quantization

\[ \sigma_{xy}=\nu\,\frac{e^2}{h}, \]

connecting first-order perturbation theory to a topologically protected observable (details in §4.3.3).

Summary#

Transition probability: \(P_{i\to f}=\vert\langle f\vert\hat{G}(t,t_0)\vert i\rangle\vert^{2}\); first order in \(\hat{V}\) collapses to the single time integral Eq. (196) involving the matrix element \(V_{fi}(t_1)\) and the Bohr phase \(\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\).
Fermi’s golden rule: sudden harmonic drive \(\hat{V}\mathrm{e}^{-\mathrm{i}\omega t}\) produces a sinc-squared resonance whose long-time limit is a rate \(\Gamma=\frac{2\pi}{\hbar}\vert V_{fi}\vert^{2}\,\delta(E_f-E_i-\hbar\omega)\).
Adiabatic process: exponential ramp \(\hat{V}\mathrm{e}^{t/\tau}\) produces a Lorentzian in \(\Delta E\) of width \(\hbar/\tau\); the \(\tau\to\infty\) limit recovers the energy-denominator language of 5.1.2.
Kubo formula: the same first-order machinery applied to \(\delta\hat{H}=-\hat{\boldsymbol{j}}\cdot\delta\boldsymbol{A}\) gives the Hall conductivity Eq. (201); integer-filled Landau levels yield the topological quantization \(\sigma_{xy}=\nu e^2/h\).

Homework#

1. Phase cancellation. Verify in detail the cancellation of the overall \(\mathrm{e}^{-\mathrm{i}E_f t/\hbar}\mathrm{e}^{\mathrm{i}E_i t_0/\hbar}\) phase in the first-order amplitude \(\langle f\vert\hat{G}(t,t_0)\vert i\rangle\), and conclude that \(P_{i\to f}^{(1)}\) depends only on \(V_{fi}(t_1)\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\). Why is this cancellation expected on general grounds?

2. Sinc-squared properties. From Eq. (197) with \(\alpha=(\omega_{fi}-\omega)/2\), derive each of the following:

(a) Peak height \(P^{(1)}\vert_{\alpha=0}=\vert V_{fi}\vert^{2}t^{2}/\hbar^{2}\).

(b) First zero at \(\alpha t=\pi\), i.e. width \(\Delta\alpha\sim\pi/t\).

Explain why (a)\(\times\)(b)\(\sim\)(c) is the algebraic origin of a constant rate in the long-time limit.

3. Sinc-to-delta. Prove the distributional identity

\[ \lim_{t\to\infty}\frac{1}{t}\left(\frac{\sin\alpha t}{\alpha}\right)^{\!2}=\pi\,\delta(\alpha). \]

Hint: act on a smooth test function \(g(\alpha)\) and use the change of variable \(u=\alpha t\) together with \(\int_{-\infty}^{\infty}(\sin u/u)^{2}\,\mathrm{d}u=\pi\).

4. Density of states. For free particles in three dimensions in a box of volume \(V\),

(a) Show that the density of states is \(\rho(E)=\dfrac{V m}{2\pi^{2}\hbar^{3}}\sqrt{2mE}\).

(b) Use Fermi’s golden rule with this \(\rho\) to express \(W_i\) in terms of \(\vert V_{fi}\vert^{2}\), the drive frequency \(\omega\), and the initial energy \(E_i\). How does \(W_i\) scale with \(E_i\) at fixed \(\vert V_{fi}\vert\)?

5. Adiabatic ramp Lorentzian. Derive Eq. (199) step by step, starting from \(\hat{V}(t)=\hat{V}\mathrm{e}^{t/\tau}\) for \(t<0\). State the FWHM in \(\omega_{fi}\) and in \(\Delta E=E_f-E_i\), and sketch the lineshape.

6. Adiabatic to static perturbation. Take the \(\tau\to\infty\) limit of Eq. (199) for fixed \(\Delta E\neq 0\), and compare with \(\vert\langle f\vert i(V)\rangle\vert^{2}\) from non-degenerate perturbation theory (5.1.2). Explain physically why the two answers must agree.

7. Three-level Raman (long-time limit). Continue the setup from HW 5.2.2.8: \(\hat{H}_0=\Delta\vert 3\rangle\langle 3\vert\) with \(E_1=E_2=0\), \(E_3=\Delta>0\), and \(\hat{V}(t)=\lambda(t)[(\vert 1\rangle+\vert 2\rangle)\langle 3\vert+\mathrm{h.c.}]\) with \(\lambda(t)=\lambda_0\cos(\omega t)\).

(a) Starting from the second-order amplitude \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle\) obtained in HW 5.2.2.8, evaluate the double time integral in the long-time limit \(\omega^{-1},\Delta^{-1}\ll t\ll\lambda_0^{-1}\). Identify which of the four oscillating terms contribute (those whose total exponent vanishes) and show

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle\approx\frac{\mathrm{i}\,\lambda_0^{2}\,t}{2}\,\frac{\Delta}{\Delta^{2}-\omega^{2}}. \]

(b) Compute \(P_{1\to 2}(t)=\vert\langle 2\vert\hat{G}(t,0)\vert 1\rangle\vert^{2}\) and identify the time scaling and the frequency dependence on \(\omega/\Delta\).

(c) Explain physically why the result is resonantly enhanced near \(\omega/\Delta=1\) and why the time scaling is \(t^{2}\) rather than \(t\) (compare to Fermi’s golden rule).

8. Minimal Kubo exercise. Take a two-level toy with \(\hat{H}_0=-\frac{1}{2}\hbar\omega_0\hat{Z}\), occupied \(\vert 0\rangle\), empty \(\vert 1\rangle\), and current operators \(\hat{j}_x=\hat{X}\), \(\hat{j}_y=\hat{Y}\) (without charge or geometric prefactors).

(a) Compute the four matrix elements entering the Kubo numerator.

(b) Evaluate

\[ \sigma_{xy}=\mathrm{i}\hbar\, \frac{\langle 0\vert\hat{j}_x\vert 1\rangle\langle 1\vert\hat{j}_y\vert 0\rangle-\langle 0\vert\hat{j}_y\vert 1\rangle\langle 1\vert\hat{j}_x\vert 0\rangle}{(E_1-E_0)^{2}}. \]

(c) Replace \(\hat{j}_y\to\hat{X}\) and show that \(\sigma_{xy}\) vanishes — i.e. why a Hall response requires non-commuting current operators.