5.2.3 Applications#

Worked solutions for the homework problems in the 5.2.3 Applications lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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?

Solution.

Start from the first-order Dyson term sandwiched between the eigenstates \(\langle f\vert\) and \(\vert i\rangle\) of \(\hat{H}_0\) (with \(f\neq i\), so the zeroth-order term \(\langle f\vert\hat{G}_0(t,t_0)\vert i\rangle=0\)):

\[ \langle f\vert\hat{G}(t,t_0)\vert i\rangle =-\frac{\mathrm{i}}{\hbar}\int_{t_0}^{t}\mathrm{d}t_1\, \langle f\vert\hat{G}_0(t,t_1)\,\hat{V}(t_1)\,\hat{G}_0(t_1,t_0)\vert i\rangle. \]

The bare propagator is diagonal in the \(\hat{H}_0\) eigenbasis, \(\hat{G}_0(t,t')=\sum_n\vert n\rangle\,\mathrm{e}^{-\mathrm{i}E_n(t-t')/\hbar}\,\langle n\vert\). Acting to the left on \(\langle f\vert\) and to the right on \(\vert i\rangle\) picks out single terms:

\[ \langle f\vert\hat{G}_0(t,t_1)=\mathrm{e}^{-\mathrm{i}E_f(t-t_1)/\hbar}\,\langle f\vert, \qquad \hat{G}_0(t_1,t_0)\vert i\rangle=\mathrm{e}^{-\mathrm{i}E_i(t_1-t_0)/\hbar}\,\vert i\rangle. \]

Substituting, and writing \(V_{fi}(t_1)\equiv\langle f\vert\hat{V}(t_1)\vert i\rangle\),

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

Now isolate the parts of the exponents that do not depend on the integration variable \(t_1\). Expand both exponents:

\[ -\frac{\mathrm{i}E_f(t-t_1)}{\hbar}-\frac{\mathrm{i}E_i(t_1-t_0)}{\hbar} =\underbrace{-\frac{\mathrm{i}E_f t}{\hbar}+\frac{\mathrm{i}E_i t_0}{\hbar}}_{t_1\text{-independent}} +\underbrace{\frac{\mathrm{i}(E_f-E_i)t_1}{\hbar}}_{t_1\text{-dependent}}. \]

The first group is a constant prefactor; the second is exactly \(\mathrm{i}\omega_{fi}t_1\). Pulling the constant out of the integral,

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

Take the squared modulus. The prefactor \(\mathrm{e}^{-\mathrm{i}E_f t/\hbar}\,\mathrm{e}^{\mathrm{i}E_i t_0/\hbar}\) is a product of two pure phases, each of unit modulus, so it contributes a factor \(1\) to \(\vert\cdot\vert^2\). Likewise \(\vert-\mathrm{i}/\hbar\vert^2=1/\hbar^2\). Hence

\[ P_{i\to f}^{(1)}(t,t_0)=\bigl\vert\langle f\vert\hat{G}(t,t_0)\vert i\rangle\bigr\vert^2 =\frac{1}{\hbar^2}\left\vert\int_{t_0}^{t}\mathrm{d}t_1\,V_{fi}(t_1)\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\right\vert^{2}, \]

which depends only on the combination \(V_{fi}(t_1)\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\).

Why this is expected on general grounds. A transition probability is a physical observable, while the two phases that cancelled are not. The factor \(\mathrm{e}^{-\mathrm{i}E_f t/\hbar}\) is the free evolution accrued by the final state and \(\mathrm{e}^{\mathrm{i}E_i t_0/\hbar}\) that of the initial state — the phases of \(\hat{G}_0\vert f\rangle\) and \(\hat{G}_0\vert i\rangle\). Two facts make them unobservable. First, a global phase on a state has no physical content: \(\vert i\rangle\) and \(\mathrm{e}^{\mathrm{i}\theta}\vert i\rangle\) describe the same physics, so an amplitude can only enter a probability through its modulus. Second, the zero of energy is a convention: shifting \(\hat{H}_0\to\hat{H}_0+c\hat{I}\) sends \(E_n\to E_n+c\), multiplying the amplitude by the extra phase \(\mathrm{e}^{-\mathrm{i}c(t-t_0)/\hbar}\), which again drops out of \(\vert\cdot\vert^2\). Only the energy difference \(E_f-E_i\), i.e. the Bohr frequency \(\omega_{fi}\), is convention-independent — and it is precisely \(\omega_{fi}\) that survives in \(P^{(1)}\). The cancellation is the statement that the answer respects both invariances.

2. Sinc-squared properties. From \(P_{i\to f}^{(1)}(t)=\frac{\vert V_{fi}\vert^2}{\hbar^2}\bigl[\sin((\omega_{fi}-\omega)t/2)/((\omega_{fi}-\omega)/2)\bigr]^2\) 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.

Solution.

With \(\alpha=(\omega_{fi}-\omega)/2\), \(P_{i\to f}^{(1)}(t)=\frac{\vert V_{fi}\vert^2}{\hbar^2}\bigl[\sin((\omega_{fi}-\omega)t/2)/((\omega_{fi}-\omega)/2)\bigr]^2\) reads

\[ P^{(1)}(t)=\frac{\vert V_{fi}\vert^{2}}{\hbar^{2}}\left(\frac{\sin\alpha t}{\alpha}\right)^{2}. \]

(a) Peak height. The kernel \((\sin\alpha t/\alpha)^2\) is largest at the resonance \(\alpha=0\). There \(\sin\alpha t\approx\alpha t-\tfrac16(\alpha t)^3+\cdots\), so \(\sin\alpha t/\alpha\to t\) as \(\alpha\to0\), and

\[ \left(\frac{\sin\alpha t}{\alpha}\right)^{2}\Bigg\vert_{\alpha\to0}=t^{2}, \qquad\Longrightarrow\qquad P^{(1)}\big\vert_{\alpha=0}=\frac{\vert V_{fi}\vert^{2}t^{2}}{\hbar^{2}}. \]

The peak grows quadratically in \(t\).

(b) Width. The kernel vanishes when \(\sin\alpha t=0\) with \(\alpha\neq0\), i.e. \(\alpha t=n\pi\) for integer \(n\neq0\). The central resonance peak is bounded by its first zeros at \(\alpha t=\pm\pi\), so

\[ \alpha_{\text{first zero}}=\frac{\pi}{t}, \qquad \Delta\alpha\sim\frac{\pi}{t}. \]

The peak narrows as \(1/t\): the longer the perturbation acts, the sharper the resonance condition becomes.

(c) Integrated weight. Compute \(\displaystyle I=\int_{-\infty}^{\infty}\mathrm{d}\alpha\,\left(\frac{\sin\alpha t}{\alpha}\right)^{2}\) by the substitution \(u=\alpha t\) (so \(\alpha=u/t\), \(\mathrm{d}\alpha=\mathrm{d}u/t\)):

\[ I=\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{t}\left(\frac{\sin u}{u/t}\right)^{2} =\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{t}\,t^{2}\left(\frac{\sin u}{u}\right)^{2} =t\int_{-\infty}^{\infty}\mathrm{d}u\,\left(\frac{\sin u}{u}\right)^{2}. \]

The remaining integral is the standard result \(\int_{-\infty}^{\infty}(\sin u/u)^{2}\,\mathrm{d}u=\pi\) (for instance, by Parseval applied to the rectangle-function Fourier pair, or by contour integration of \(\int(1-\cos2u)/u^2\)). Therefore

\[ I=\int_{-\infty}^{\infty}\mathrm{d}\alpha\,\left(\frac{\sin\alpha t}{\alpha}\right)^{2}=\pi t. \]

Why (a)×(b) ∼ (c) gives a constant rate. The three results say the resonance peak is a spike of height \(\propto t^2\) and width \(\propto1/t\), so its area scales as

\[ \text{(peak height)}\times\text{(width)}\sim t^{2}\cdot\frac{1}{t}=t, \]

matching the exact integrated weight \(\pi t\) found in (c). The transition rate is \(W_i=P/t\) summed over a continuum of final states. Summing replaces \(\sum_f\) by \(\int\mathrm{d}E_f\,\rho(E_f)\); because the kernel is sharply peaked, only the area under the peak contributes, and that area grows linearly in \(t\). Dividing the area (\(\propto t\)) by \(t\) leaves a quantity independent of \(t\) — a constant rate. The quadratic single-state growth \(t^2\) and the linear-in-\(t\) area are not in conflict: a fixed resonant state has \(P\propto t^2\) (until depletion matters), but a band of states each spends progressively less time on resonance as the peak narrows, and the net effect is a steady, \(t\)-independent rate. This is precisely the content of Fermi’s golden rule.

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 \(\int_{-\infty}^{\infty}(\sin u/u)^{2}\,\mathrm{d}u=\pi\).

Solution.

A distributional identity is a statement about what both sides do when integrated against an arbitrary smooth, rapidly decaying test function \(g(\alpha)\). So define

\[ L_t\equiv\int_{-\infty}^{\infty}\mathrm{d}\alpha\;\frac{1}{t}\left(\frac{\sin\alpha t}{\alpha}\right)^{2}g(\alpha), \]

and show \(\lim_{t\to\infty}L_t=\pi\,g(0)\), which is by definition the action of \(\pi\,\delta(\alpha)\).

Change variables to \(u=\alpha t\), so \(\alpha=u/t\) and \(\mathrm{d}\alpha=\mathrm{d}u/t\). The kernel transforms as

\[ \frac{1}{t}\left(\frac{\sin\alpha t}{\alpha}\right)^{2} =\frac{1}{t}\left(\frac{\sin u}{u/t}\right)^{2} =\frac{1}{t}\,t^{2}\left(\frac{\sin u}{u}\right)^{2} =t\left(\frac{\sin u}{u}\right)^{2}. \]

Hence

\[ L_t=\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{t}\;t\left(\frac{\sin u}{u}\right)^{2}g\!\left(\frac{u}{t}\right) =\int_{-\infty}^{\infty}\mathrm{d}u\,\left(\frac{\sin u}{u}\right)^{2}g\!\left(\frac{u}{t}\right). \]

The \(t\)-dependence now sits only inside the slowly-varying argument \(u/t\) of the test function. As \(t\to\infty\), for every fixed \(u\) one has \(u/t\to0\) and, since \(g\) is continuous, \(g(u/t)\to g(0)\). The factor \((\sin u/u)^{2}\) is non-negative and integrable, with \(\int(\sin u/u)^2\,\mathrm{d}u=\pi<\infty\), and \(\vert g(u/t)\vert\) is bounded by \(\sup\vert g\vert\); the dominating function \((\sin u/u)^2\sup\vert g\vert\) is integrable. The dominated convergence theorem therefore lets the limit pass through the integral:

\[ \lim_{t\to\infty}L_t =\int_{-\infty}^{\infty}\mathrm{d}u\,\left(\frac{\sin u}{u}\right)^{2}\, \lim_{t\to\infty}g\!\left(\frac{u}{t}\right) =g(0)\int_{-\infty}^{\infty}\mathrm{d}u\,\left(\frac{\sin u}{u}\right)^{2} =\pi\,g(0). \]

Since this holds for every test function \(g\),

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

Two remarks. The prefactor \(1/t\) is exactly what is needed: without it the kernel would have weight \(\pi t\to\infty\); with it the weight is the fixed \(\pi\), the hallmark of a properly normalised nascent delta. And the rescaling \(\alpha\to2\alpha\) gives the equivalent form quoted in the lecture, \(\lim_{t\to\infty}t^{-1}(\sin\alpha t/\alpha)^2=2\pi\,\delta(2\alpha)\), used to turn the sinc-squared into the energy-conserving delta of Fermi’s golden rule.

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 \(\vert V_{fi}\vert\)?

Solution.

(a) Counting states. A free particle in a cubic box of volume \(V=L^3\) with periodic boundary conditions has plane-wave eigenstates labelled by wavevectors \(\boldsymbol{k}=(2\pi/L)(n_x,n_y,n_z)\), \(n_i\in\mathbb{Z}\), with energy \(E=\hbar^2 k^2/(2m)\). The allowed \(\boldsymbol{k}\)-points form a cubic lattice of spacing \(2\pi/L\), so each occupies a \(k\)-space volume \((2\pi/L)^3\) and the number of states per unit \(k\)-space volume is \(V/(2\pi)^3\).

The number of states with energy below \(E\) — equivalently with wavevector magnitude below \(k=\sqrt{2mE}/\hbar\) — is the volume of a ball of radius \(k\) times that density:

\[ N(E)=\frac{V}{(2\pi)^{3}}\cdot\frac{4}{3}\pi k^{3} =\frac{V}{6\pi^{2}}\,k^{3} =\frac{V}{6\pi^{2}}\left(\frac{2mE}{\hbar^{2}}\right)^{3/2} =\frac{V}{6\pi^{2}\hbar^{3}}\,(2mE)^{3/2}. \]

The density of states is the energy derivative:

\[ \rho(E)=\frac{\mathrm{d}N}{\mathrm{d}E} =\frac{V}{6\pi^{2}\hbar^{3}}\,(2m)^{3/2}\cdot\frac{3}{2}\,E^{1/2} =\frac{V\,(2m)^{3/2}}{4\pi^{2}\hbar^{3}}\sqrt{E}. \]

Rewrite \((2m)^{3/2}=2m\sqrt{2m}\) to bring the result to the requested form:

\[ \rho(E)=\frac{V\cdot 2m\sqrt{2m}}{4\pi^{2}\hbar^{3}}\sqrt{E} =\frac{Vm}{2\pi^{2}\hbar^{3}}\sqrt{2mE}. \]

(This counts orbital states only; a spin-\(\tfrac12\) particle would carry an extra factor of \(2\), which the problem statement omits.)

(b) Transition rate into the continuum. Fermi’s golden rule for a transition into a continuum of final states, summed with the density of states, is

\[ W_i=\frac{2\pi}{\hbar}\,\vert V_{fi}\vert^{2}\,\rho(E_f)\Big\vert_{E_f=E_i+\hbar\omega}, \]

where the energy-conserving delta of \(W_{i\to f}=\frac{2\pi}{\hbar}\,\vert V_{fi}\vert^2\,\delta(E_f-E_i-\hbar\omega)\) has pinned the final energy to the resonance shell \(E_f=E_i+\hbar\omega\). Insert the density of states evaluated at that energy:

\[ W_i=\frac{2\pi}{\hbar}\,\vert V_{fi}\vert^{2}\, \frac{Vm}{2\pi^{2}\hbar^{3}}\sqrt{2m(E_i+\hbar\omega)} =\frac{Vm\,\vert V_{fi}\vert^{2}}{\pi\hbar^{4}}\sqrt{2m\,(E_i+\hbar\omega)}. \]

Scaling with \(E_i\). At fixed matrix element \(\vert V_{fi}\vert\) the rate depends on the initial energy only through the square root,

\[ W_i\propto\sqrt{E_i+\hbar\omega}. \]

For an initial energy large compared with the photon energy, \(E_i\gg\hbar\omega\), this reduces to \(W_i\propto\sqrt{E_i}\) — the rate grows with energy because the available phase space of final states (the density of states) grows as \(\sqrt{E}\). Near threshold, \(E_i+\hbar\omega\to0^{+}\), the rate is suppressed: there are simply very few states for the particle to scatter into.

5. Adiabatic ramp Lorentzian. Derive \(P_{i\to f}=\frac{\vert V_{fi}\vert^2}{(E_f-E_i)^2+(\hbar/\tau)^2}\) 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.

Solution.

The perturbation is switched on exponentially from the infinite past and off at \(t=0\):

\[ \hat{V}(t)=\hat{V}\,\mathrm{e}^{t/\tau}\quad(t<0),\qquad \tau>0, \]

with the system prepared in \(\vert i\rangle\) at \(t_0\to-\infty\) and the transition probability to \(\vert f\rangle\) asked at \(t=0\). The matrix element is \(V_{fi}(t_1)=V_{fi}\,\mathrm{e}^{t_1/\tau}\) with \(V_{fi}=\langle f\vert\hat{V}\vert i\rangle\) time-independent. Substitute into the master formula \(P_{i\to f}^{(1)}(t,t_0)=\frac{1}{\hbar^2}\bigl|\int_{t_0}^{t}\mathrm{d}t_1\,V_{fi}(t_1)\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\bigr|^2\) with \(t_0\to-\infty\), \(t\to0\):

\[ P_{i\to f}=\frac{1}{\hbar^{2}}\left\vert\int_{-\infty}^{0}\mathrm{d}t_1\, V_{fi}\,\mathrm{e}^{t_1/\tau}\,\mathrm{e}^{\mathrm{i}\omega_{fi}t_1}\right\vert^{2}. \]

Step 1 — combine the exponents. The two exponentials merge into a single exponential with complex rate \(\mathrm{i}\omega_{fi}+1/\tau\):

\[ \int_{-\infty}^{0}\mathrm{d}t_1\,\mathrm{e}^{(\mathrm{i}\omega_{fi}+1/\tau)t_1}. \]

Step 2 — evaluate the integral. The antiderivative is \(\mathrm{e}^{(\mathrm{i}\omega_{fi}+1/\tau)t_1}/(\mathrm{i}\omega_{fi}+1/\tau)\):

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

The lower limit contributes nothing: as \(t_1\to-\infty\) the modulus \(\vert\mathrm{e}^{(\mathrm{i}\omega_{fi}+1/\tau)t_1}\vert=\mathrm{e}^{t_1/\tau}\to0\) because \(\tau>0\). This is exactly the role of the adiabatic switch-on — it regulates the integral at the infinite past.

Step 3 — take the squared modulus. With \(\vert\mathrm{i}\omega_{fi}+1/\tau\vert^{2}=\omega_{fi}^{2}+1/\tau^{2}\),

\[ P_{i\to f}=\frac{\vert V_{fi}\vert^{2}}{\hbar^{2}}\, \left\vert\frac{1}{\mathrm{i}\omega_{fi}+1/\tau}\right\vert^{2} =\frac{\vert V_{fi}\vert^{2}}{\hbar^{2}\bigl(\omega_{fi}^{2}+1/\tau^{2}\bigr)}. \]

Step 4 — convert to energy. Multiply numerator and denominator by \(\hbar^{2}\) and use \(\hbar\omega_{fi}=E_f-E_i\):

\[ \boxed{\; P_{i\to f}=\frac{\vert V_{fi}\vert^{2}}{(E_f-E_i)^{2}+(\hbar/\tau)^{2}}\;} \]

which is \(P_{i\to f}=\frac{\vert V_{fi}\vert^2}{(E_f-E_i)^2+(\hbar/\tau)^2}\).

FWHM. The result is a Lorentzian of the generic form \(\mathcal{L}(x)=C/(x^2+\gamma^2)\), peaked at \(x=0\) with maximum \(C/\gamma^2\). The half-maximum is reached where \(x^2+\gamma^2=2\gamma^2\), i.e. \(x=\pm\gamma\), so the full width at half maximum is \(2\gamma\).

In terms of \(\Delta E=E_f-E_i\): the half-width parameter is \(\gamma=\hbar/\tau\), so \(\mathrm{FWHM}_{\Delta E}=2\hbar/\tau\).
In terms of the Bohr frequency \(\omega_{fi}\): from \(P\propto1/(\omega_{fi}^2+1/\tau^2)\) the half-width parameter is \(\gamma=1/\tau\), so \(\mathrm{FWHM}_{\omega_{fi}}=2/\tau\).

The two are consistent: \(\mathrm{FWHM}_{\Delta E}=\hbar\,\mathrm{FWHM}_{\omega_{fi}}\).

Lineshape sketch. As a function of \(\Delta E\), the curve is a symmetric bell centred at \(\Delta E=0\), with peak value \(\vert V_{fi}\vert^{2}\tau^{2}/\hbar^{2}\) and characteristic width \(2\hbar/\tau\); it falls off as \(1/(\Delta E)^2\) in the tails.

 P
 |          .-.            peak height = |V_fi|^2 tau^2 / hbar^2
 |         /   \
 |        /     \          FWHM = 2 hbar / tau
 |   ____/       \____
 +----------+----------- Delta E
            0

A faster ramp (smaller \(\tau\)) gives a broader, lower resonance; a slower ramp (larger \(\tau\)) gives a narrower, taller one. The width \(\hbar/\tau\) is the finite-time energy resolution: the ramp lasting a time \(\sim\tau\) cannot resolve energy mismatches smaller than \(\hbar/\tau\).

6. Adiabatic to static perturbation. Take the \(\tau\to\infty\) limit of \(P_{i\to f}=\frac{\vert V_{fi}\vert^2}{(E_f-E_i)^2+(\hbar/\tau)^2}\) 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.

Solution.

The limit. Start from \(P_{i\to f}=\frac{\vert V_{fi}\vert^2}{(E_f-E_i)^2+(\hbar/\tau)^2}\),

\[ P_{i\to f}=\frac{\vert V_{fi}\vert^{2}}{(E_f-E_i)^{2}+(\hbar/\tau)^{2}}. \]

Hold \(\Delta E=E_f-E_i\neq0\) fixed and let \(\tau\to\infty\). Then \(\hbar/\tau\to0\), the regulator term drops out of the denominator, and

\[ \lim_{\tau\to\infty}P_{i\to f} =\frac{\vert V_{fi}\vert^{2}}{(E_f-E_i)^{2}} \qquad(\Delta E\neq0). \]

Comparison with time-independent perturbation theory. In non-degenerate perturbation theory (5.1.2), the exact eigenstate of \(\hat{H}_0+\hat{V}\) that grows continuously out of the unperturbed eigenstate \(\vert i\rangle\) is, to first order in \(\hat{V}\),

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

Overlap with a different unperturbed eigenstate \(\vert f\rangle\) (\(f\neq i\)) picks out the single \(m=f\) term:

\[ \langle f\vert i(V)\rangle=\frac{V_{fi}}{E_i-E_f}+\cdots \qquad\Longrightarrow\qquad \vert\langle f\vert i(V)\rangle\vert^{2}=\frac{\vert V_{fi}\vert^{2}}{(E_i-E_f)^{2}}. \]

Since \((E_i-E_f)^2=(E_f-E_i)^2\), this is identical to the \(\tau\to\infty\) limit above. Time-dependent perturbation theory, run with an infinitely slow ramp, reproduces time-independent perturbation theory exactly.

Why they must agree. This is the adiabatic theorem at work. Switching \(\hat{V}\) on infinitely slowly means the Hamiltonian changes much more slowly than any internal timescale \(\hbar/\vert\Delta E\vert\) of the system. Under such slow change, a system that starts in an instantaneous eigenstate stays in the corresponding instantaneous eigenstate: the unperturbed eigenstate \(\vert i\rangle\) deforms continuously, with no real (energy-non-conserving) transitions, into the exact perturbed eigenstate \(\vert i(V)\rangle\). The state at \(t=0\) is therefore just \(\vert i(V)\rangle\), and the “transition probability” to \(\vert f\rangle\) is nothing but the squared admixture coefficient of \(\vert f\rangle\) inside that perturbed eigenstate — which is exactly what first-order time-independent perturbation theory computes. The dynamical (golden-rule) language and the static (energy-denominator) language are two descriptions of the same physics; the slow-ramp limit is the bridge that forces them to coincide. The finite-\(\tau\) Lorentzian width \(\hbar/\tau\) also shows that the apparent divergence of the energy denominator at \(\Delta E=0\) is not physical: any real, finite ramp smooths the resonance into a peak of finite height.

★ 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).

(d) Near-resonance limit and the rotating wave approximation. Set \(\omega \approx \Delta\) (close to but not exactly at single-photon resonance) and re-evaluate the second-order amplitude under the rotating wave approximation (RWA): of the four oscillating terms identified in (a), keep only the one whose two exponent frequencies are both slow in the limit \(\omega \to \Delta\), and drop the other three as fast-oscillating. Show that the RWA amplitude is

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx \frac{\mathrm{i}\,\lambda_0^{2}\,t}{4(\Delta - \omega)}, \]

and verify that this agrees with the full off-resonant result of (a) in the near-resonant limit \(\vert\Delta - \omega\vert \ll \Delta + \omega\).

Solution.

Convention. The level-3 energy \(E_3=\Delta\) enters the Bohr phase as \(\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)/\hbar}\). The target amplitude quoted in the problem statement compares \(\Delta\) directly with the drive frequency \(\omega\), so — consistently with HW 5.2.2.8 and the reference treatment — we adopt units with \(\hbar=1\); equivalently \(\Delta\) denotes the level-3 angular frequency and the Bohr frequency of the virtual transition is \(\omega_{31}=\Delta\). With this convention the Dyson factor at order \(k\) is \((-\mathrm{i})^k\).

(a) Setting up the double integral. From HW 5.2.2.8 the first non-vanishing contribution to \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle\) is second order, because \(\langle 2\vert\hat{G}_0\vert 1\rangle=0\) and \(\langle 2\vert\hat{V}\vert 1\rangle=0\); the only path is \(\vert 1\rangle\to\vert 3\rangle\to\vert 2\rangle\) through the virtual intermediate state \(\vert 3\rangle\). With matrix elements \(\langle 3\vert\hat{V}(t_1)\vert 1\rangle=\lambda(t_1)\), \(\langle 2\vert\hat{V}(t_2)\vert 3\rangle=\lambda(t_2)\), and the level-3 propagator \(\langle 3\vert\hat{G}_0(t_2,t_1)\vert 3\rangle=\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)}\),

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx-\lambda_0^{2}\int_{0}^{t}\mathrm{d}t_2\int_{0}^{t_2}\mathrm{d}t_1\, \cos(\omega t_2)\cos(\omega t_1)\,\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)}. \]

Expanding the cosines. Write \(\cos(\omega t)=\tfrac12(\mathrm{e}^{\mathrm{i}\omega t}+\mathrm{e}^{-\mathrm{i}\omega t})\). The product of the two cosines with the Bohr phase \(\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)}=\mathrm{e}^{\mathrm{i}\Delta t_1}\mathrm{e}^{-\mathrm{i}\Delta t_2}\) splits into four terms, each of the form \(\mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{\mathrm{i}\Omega_2 t_2}\) with

\[ \Omega_1=\Delta\pm\omega,\qquad\Omega_2=-\Delta\pm\omega . \]

Thus

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx-\frac{\lambda_0^{2}}{4}\sum_{\text{4 terms}} \int_{0}^{t}\mathrm{d}t_2\int_{0}^{t_2}\mathrm{d}t_1\, \mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{\mathrm{i}\Omega_2 t_2}. \]

The generic nested integral. Doing the inner integral first,

\[ \int_{0}^{t}\mathrm{d}t_2\int_{0}^{t_2}\mathrm{d}t_1\,\mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{\mathrm{i}\Omega_2 t_2} =\int_{0}^{t}\mathrm{d}t_2\,\frac{\mathrm{e}^{\mathrm{i}\Omega_1 t_2}-1}{\mathrm{i}\Omega_1}\,\mathrm{e}^{\mathrm{i}\Omega_2 t_2} =\frac{1}{\mathrm{i}\Omega_1}\int_{0}^{t}\mathrm{d}t_2\Bigl(\mathrm{e}^{\mathrm{i}(\Omega_1+\Omega_2)t_2}-\mathrm{e}^{\mathrm{i}\Omega_2 t_2}\Bigr). \]

Identifying the long-time-dominant terms — the secular discriminant. Define the nested integral

\[ I(\Omega_1,\Omega_2)=\int_0^t\!\mathrm{d}t_2\int_0^{t_2}\!\mathrm{d}t_1\,\mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{\mathrm{i}\Omega_2 t_2}. \]

It grows linearly in \(t\) iff at least one of \(\Omega_1\), \(\Omega_2\), or \(\Omega_1+\Omega_2\) vanishes; otherwise every contribution is a bounded oscillation. (Quick check: at \(\Omega_1=0\) the inner integral becomes \(t_2\) and the outer one delivers a \(t/(\mathrm{i}\Omega_2)\) piece; at \(\Omega_2=0\) a \(-t/(\mathrm{i}\Omega_1)\) piece; at \(\Omega_1+\Omega_2=0\) a \(t/(\mathrm{i}\Omega_1)\) piece. Away from these three branches, every \(\mathrm{e}^{\mathrm{i}\Omega t}-1\) factor stays bounded.) In the window \(\omega^{-1},\Delta^{-1}\ll t\ll\lambda_0^{-1}\) only the linear-in-\(t\) pieces survive.

With \(\Omega_1=\Delta\pm\omega\) and \(\Omega_2=-\Delta\pm\omega\):

\(\Omega_1=0\) or \(\Omega_2=0\) both require \(\omega=\Delta\) — single-photon resonance with the intermediate state \(\vert 3\rangle\), where the energy denominator collapses and perturbation theory itself breaks down. We exclude this point and treat \(\omega\neq\Delta\).
\(\Omega_1+\Omega_2=\pm\omega\pm\omega\) vanishes for the two mixed-sign combinations (one \(+\omega\), one \(-\omega\)); the two same-sign combinations give \(\pm 2\omega\neq 0\) and contribute only bounded oscillations. The mixed-sign condition is the two-photon Raman resonance — absorption of one drive quantum followed by emission of one, with zero net frequency change matching the degeneracy \(E_2=E_1\). For a resonant term, \(\Omega_2=-\Omega_1\) and

\[ \int_{0}^{t}\mathrm{d}t_2\int_{0}^{t_2}\mathrm{d}t_1\, \mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{-\mathrm{i}\Omega_1 t_2} =\frac{1}{\mathrm{i}\Omega_1}\int_{0}^{t}\mathrm{d}t_2\Bigl(1-\mathrm{e}^{-\mathrm{i}\Omega_1 t_2}\Bigr) =\frac{t}{\mathrm{i}\Omega_1}-\frac{\mathrm{e}^{-\mathrm{i}\Omega_1 t}-1}{\Omega_1^{2}} \;\xrightarrow{\;t\to\infty\;}\;\frac{t}{\mathrm{i}\Omega_1}. \]

The two contributing terms. They are:

\(t_1\) carries \(+\omega\), \(t_2\) carries \(-\omega\): \(\Omega_1=\Delta+\omega\).
\(t_1\) carries \(-\omega\), \(t_2\) carries \(+\omega\): \(\Omega_1=\Delta-\omega\).

Each contributes \(t/(\mathrm{i}\Omega_1)\), so

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx-\frac{\lambda_0^{2}}{4}\left(\frac{t}{\mathrm{i}(\Delta+\omega)}+\frac{t}{\mathrm{i}(\Delta-\omega)}\right). \]

Combine the two fractions over the common denominator \(\Delta^2-\omega^2\):

\[ \frac{1}{\Delta+\omega}+\frac{1}{\Delta-\omega} =\frac{(\Delta-\omega)+(\Delta+\omega)}{\Delta^{2}-\omega^{2}} =\frac{2\Delta}{\Delta^{2}-\omega^{2}}. \]

Therefore

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx-\frac{\lambda_0^{2}}{4}\cdot\frac{t}{\mathrm{i}}\cdot\frac{2\Delta}{\Delta^{2}-\omega^{2}} =-\frac{\lambda_0^{2}t}{2\mathrm{i}}\cdot\frac{\Delta}{\Delta^{2}-\omega^{2}}. \]

Using \(1/\mathrm{i}=-\mathrm{i}\), so \(-1/(2\mathrm{i})=\mathrm{i}/2\):

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

(b) Transition probability. Take the squared modulus. The amplitude is \(\mathrm{i}\) times a real quantity, so

\[ P_{1\to 2}(t)=\bigl\vert\langle 2\vert\hat{G}(t,0)\vert 1\rangle\bigr\vert^{2} =\left(\frac{\lambda_0^{2}t}{2}\right)^{2}\left(\frac{\Delta}{\Delta^{2}-\omega^{2}}\right)^{2} =\frac{\lambda_0^{4}\,t^{2}}{4}\,\frac{\Delta^{2}}{(\Delta^{2}-\omega^{2})^{2}}. \]

It is convenient to factor out \(\Delta\) and display the dependence on the ratio \(\omega/\Delta\):

\[ P_{1\to 2}(t)=\frac{\lambda_0^{4}\,t^{2}}{4\,\Delta^{2}}\, \frac{1}{\bigl(1-(\omega/\Delta)^{2}\bigr)^{2}}. \]

Time scaling: \(P_{1\to 2}\propto t^{2}\) — coherent quadratic growth.
Frequency dependence: \(P_{1\to 2}\propto\bigl(1-(\omega/\Delta)^2\bigr)^{-2}\) — it diverges as \(\omega/\Delta\to1\) and is strongly suppressed for \(\omega/\Delta\gg1\) or \(\omega/\Delta\ll1\).

(c) Physical interpretation.

Resonant enhancement near \(\omega/\Delta=1\). The transition \(\vert 1\rangle\to\vert 2\rangle\) proceeds through the virtual intermediate state \(\vert 3\rangle\) at energy \(\Delta\). The drive supplies quanta of energy \(\omega\). When \(\omega\approx\Delta\) a single drive quantum nearly matches the energy gap to \(\vert 3\rangle\): the intermediate state is nearly on-shell, the energy denominator \(\Delta\pm\omega\) for the near-resonant pathway becomes small, and the second-order amplitude is strongly enhanced. This is exactly the small-energy-denominator amplification familiar from time-independent perturbation theory, now for a two-step (Raman) process — \(\vert 1\rangle\) absorbs a quantum to reach \(\vert 3\rangle\), then emits one to settle into \(\vert 2\rangle\). Away from resonance the virtual excursion through \(\vert 3\rangle\) is energetically costly and the process is suppressed.

Why \(t^{2}\), not \(t\). Fermi’s golden rule produces a linear-in-\(t\) probability (a constant rate) because it describes an irreversible decay into a continuum of final states: the sinc-squared kernel, summed over a continuum, yields a delta function and a steady rate. Here, by contrast, \(\vert 2\rangle\) is a single discrete level, degenerate with the initial state \(\vert 1\rangle\) (\(E_1=E_2=0\)). There is no continuum, hence no density of states to integrate over and no \(\delta\)-function to generate a rate. The amplitude itself grows linearly in \(t\) — it is the coherent accumulation of a fixed two-photon amplitude — so the probability, its square, grows as \(t^{2}\). This is the same quadratic growth seen for a single resonant final state in HW 2(a) before any continuum sum. The \(t^2\) law signals coherent Rabi-like oscillation (valid only for \(t\ll\lambda_0^{-1}\), before the linearised amplitude saturates), whereas the linear \(t\) law of the golden rule signals incoherent decay.

(d) Near-resonance limit and the RWA. Inspect the four \((\Omega_1, \Omega_2)\) combinations with \(\Omega_1 = \Delta \pm \omega\) and \(\Omega_2 = -\Delta \pm \omega\) near \(\omega \approx \Delta\):

\((\Delta + \omega, -\Delta + \omega)\): \(\Omega_1 \approx 2\Delta\) fast, \(\Omega_2 \approx 0\) slow, sum \(\approx 2\omega\) fast.
\((\Delta + \omega, -\Delta - \omega)\): both exponents fast (\(\sim\pm 2\Delta\)); a mixed-sign sum-frequency contribution irrelevant near resonance.
\((\Delta - \omega, -\Delta + \omega)\): both exponents slow (\(\approx \pm 0\)), sum exactly zero.
\((\Delta - \omega, -\Delta - \omega)\): \(\Omega_1 \approx 0\) slow, \(\Omega_2 \approx -2\Delta\) fast.

The RWA discards any term whose exponent oscillates on the fast scale \(1/(2\Delta)\) during the time window \(t \gg \Delta^{-1}\) of interest — these contributions average to zero and give only bounded \(\mathcal{O}(1/\Delta)\) pieces. Of the four, only the third combination — with both exponents \(\sim(\Delta - \omega)t\) — survives. Following the generic nested-integral identity of (a) with \(\Omega_2 = -\Omega_1\) and \(\Omega_1 = \Delta - \omega\),

\[ \int_{0}^{t}\!\mathrm{d}t_2\!\int_{0}^{t_2}\!\mathrm{d}t_1\, \mathrm{e}^{\mathrm{i}(\Delta - \omega)t_1}\,\mathrm{e}^{\mathrm{i}(\omega - \Delta)t_2} \;\approx\; \frac{t}{\mathrm{i}(\Delta - \omega)} \]

in the regime \(\vert\Delta - \omega\vert^{-1} \ll t\). Multiplying by the \(-\lambda_0^{2}/4\) prefactor of the Dyson expansion,

\[ \langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx -\frac{\lambda_0^{2}}{4}\cdot\frac{t}{\mathrm{i}(\Delta - \omega)} = \frac{\mathrm{i}\,\lambda_0^{2}\,t}{4(\Delta - \omega)}. \]

Consistency with (a). The full off-resonant amplitude derived in (a) is \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle \approx \mathrm{i}\lambda_0^{2}\,t\,\Delta/[2(\Delta^{2} - \omega^{2})]\). In the near-resonant limit \(\vert\Delta - \omega\vert \ll \Delta + \omega\),

\[ \frac{\Delta}{\Delta^{2} - \omega^{2}} = \frac{\Delta}{(\Delta - \omega)(\Delta + \omega)} \approx \frac{\Delta}{2\Delta\,(\Delta - \omega)} = \frac{1}{2(\Delta - \omega)}, \]

so the off-resonant formula reduces to \(\mathrm{i}\lambda_0^{2}\,t/[4(\Delta - \omega)]\), matching the RWA result exactly.

Physical content of the RWA. The surviving combination is the near-resonant Raman pathway — a drive quantum of frequency \(\omega \approx \Delta\) absorbed at the \(\vert 1\rangle \to \vert 3\rangle\) leg, then re-emitted at the \(\vert 3\rangle \to \vert 2\rangle\) leg also near resonance. The discarded counter-rotating partners — absorption-absorption and emission-emission, both carrying \(\mathrm{e}^{\pm\mathrm{i}(\Delta + \omega)t}\) — are non-zero, but their contributions to the amplitude oscillate too rapidly to accumulate over the relevant time scale. RWA is the natural and most commonly used approximation in coherent-control settings (cavity QED, two-level atoms driven by near-resonant lasers, NV centres), precisely because near-resonant drive is the regime of interest in such systems. Away from resonance (\(\omega/\Delta\) not close to \(1\)), RWA must be abandoned and the full off-resonant calculation of (a) used instead — they agree in the overlap regime but differ as \(\omega/\Delta\) departs from unity.

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.

Solution.

Energies. With \(\hat{H}_0=-\frac12\hbar\omega_0\hat{Z}\) and the standard convention \(\hat{Z}\vert 0\rangle=+\vert 0\rangle\), \(\hat{Z}\vert 1\rangle=-\vert 1\rangle\):

\[ E_0=-\tfrac12\hbar\omega_0(+1)=-\tfrac12\hbar\omega_0,\qquad E_1=-\tfrac12\hbar\omega_0(-1)=+\tfrac12\hbar\omega_0, \]

so the gap is \(E_1-E_0=\hbar\omega_0\) and \((E_1-E_0)^{2}=\hbar^{2}\omega_0^{2}\). The state \(\vert 0\rangle\) is the lower-energy (occupied) state and \(\vert 1\rangle\) the empty one, as stated.

(a) The four matrix elements. Use the action of the Pauli operators, \(\hat{X}\vert 0\rangle=\vert 1\rangle\), \(\hat{X}\vert 1\rangle=\vert 0\rangle\), \(\hat{Y}\vert 0\rangle=\mathrm{i}\vert 1\rangle\), \(\hat{Y}\vert 1\rangle=-\mathrm{i}\vert 0\rangle\), together with orthonormality \(\langle 0\vert 0\rangle=\langle 1\vert 1\rangle=1\), \(\langle 0\vert 1\rangle=0\):

\[\begin{split} \begin{split} \langle 0\vert\hat{j}_x\vert 1\rangle&=\langle 0\vert\hat{X}\vert 1\rangle=\langle 0\vert 0\rangle=1,\\ \langle 1\vert\hat{j}_x\vert 0\rangle&=\langle 1\vert\hat{X}\vert 0\rangle=\langle 1\vert 1\rangle=1,\\ \langle 0\vert\hat{j}_y\vert 1\rangle&=\langle 0\vert\hat{Y}\vert 1\rangle=\langle 0\vert(-\mathrm{i}\vert 0\rangle)=-\mathrm{i},\\ \langle 1\vert\hat{j}_y\vert 0\rangle&=\langle 1\vert\hat{Y}\vert 0\rangle=\langle 1\vert(\mathrm{i}\vert 1\rangle)=+\mathrm{i}. \end{split} \end{split}\]

(As a check, \(\langle 1\vert\hat{j}_y\vert 0\rangle=\langle 0\vert\hat{j}_y\vert 1\rangle^{*}\), consistent with \(\hat{Y}\) being Hermitian.)

(b) The Hall conductivity. Assemble the numerator of the Kubo formula:

\[ \langle 0\vert\hat{j}_x\vert 1\rangle\langle 1\vert\hat{j}_y\vert 0\rangle -\langle 0\vert\hat{j}_y\vert 1\rangle\langle 1\vert\hat{j}_x\vert 0\rangle =(1)(+\mathrm{i})-(-\mathrm{i})(1)=\mathrm{i}+\mathrm{i}=2\mathrm{i}. \]

Divide by \((E_1-E_0)^{2}=\hbar^{2}\omega_0^{2}\) and multiply by \(\mathrm{i}\hbar\):

\[ \sigma_{xy}=\mathrm{i}\hbar\,\frac{2\mathrm{i}}{\hbar^{2}\omega_0^{2}} =\frac{2\mathrm{i}^{2}\hbar}{\hbar^{2}\omega_0^{2}} =\boxed{\;-\frac{2}{\hbar\,\omega_0^{2}}\;} \]

The result is real (as a physical conductivity must be) and non-zero. The toy exhibits a finite Hall response: the antisymmetric current–current correlation \(\langle\hat{j}_x\hat{j}_y\rangle-\langle\hat{j}_y\hat{j}_x\rangle\) is non-zero, fed by the virtual transition \(\vert 0\rangle\to\vert 1\rangle\to\vert 0\rangle\).

(c) Why non-commuting currents are required. Now set \(\hat{j}_y\to\hat{X}\), so both current operators are the same operator, \(\hat{j}_x=\hat{j}_y=\hat{X}\). The numerator becomes

\[ \langle 0\vert\hat{X}\vert 1\rangle\langle 1\vert\hat{X}\vert 0\rangle -\langle 0\vert\hat{X}\vert 1\rangle\langle 1\vert\hat{X}\vert 0\rangle=0, \]

identically — the two products are term-by-term equal, so their difference vanishes. Hence \(\sigma_{xy}=0\).

The structural reason: the Kubo numerator is antisymmetric under exchange \(\hat{j}_x\leftrightarrow\hat{j}_y\). Any antisymmetric bilinear of two identical (or, more generally, commuting and simultaneously diagonalisable) operators vanishes. A non-zero Hall conductivity therefore requires the two current operators to be genuinely distinct and non-commuting. In the working example \([\hat{j}_x,\hat{j}_y]=[\hat{X},\hat{Y}]=2\mathrm{i}\hat{Z}\neq0\), and it is exactly this commutator — a measure of the failure of \(\hat{j}_x\) and \(\hat{j}_y\) to share eigenstates — that supplies the imaginary, antisymmetric correlation responsible for the transverse response. Physically, a Hall current flows perpendicular to the applied field; that transverse, chirality-carrying response cannot be built from a single current direction, and the non-commutativity of \(\hat{j}_x\) and \(\hat{j}_y\) is the operator-level expression of the broken \(x\leftrightarrow y\) symmetry (time-reversal / chirality) that the Hall effect needs.