With $|e\rangle=(1,0)^T,\ |g\rangle=(0,1)^T$ the outer products are read straight off:

Multiplying, $\sigma^+\sigma^-=\begin{pmatrix}1&0\\0&0\end{pmatrix}=|e\rangle\langle e|$ and $\sigma^-\sigma^+=\begin{pmatrix}0&0\\0&1\end{pmatrix}=|g\rangle\langle g|$ — the two level projectors. Subtracting gives $[\sigma^+,\sigma^-]=|e\rangle\langle e|-|g\rangle\langle g|=\sigma_z$; adding gives $\{\sigma^+,\sigma^-\}=|e\rangle\langle e|+|g\rangle\langle g|=\mathbb 1$. With $\sigma^\pm=\tfrac12(\sigma_x\pm i\sigma_y)$ these are exactly the spin-$\tfrac12$ ladder relations of ~QM-11. Hence sigma_plus @ sigma_minus$=\mathrm{diag}(1,0)$ and sigma_plus @ sigma_minus - sigma_minus @ sigma_plus$=$sigma_z.

On resonance $H_\text{RWA}=\tfrac12\Omega\sigma_x$, and since $\sigma_x^2=\mathbb 1$ the propagator is a rotation:

Acting on $|g\rangle$ with $\sigma_x|g\rangle=|e\rangle$,

so $P_e=|\langle e|\psi\rangle|^2=\sin^2(\Omega t/2)$. At $\Omega t=\pi$ this is $\sin^2(\pi/2)=1$ (complete inversion — the $\pi$-pulse) and at $\Omega t=2\pi$ it is $\sin^2\pi=0$ (the return). Matching rabi_excited_population(np.pi, 1.0, 0.0)$=1.0$ and rabi_excited_population(2*np.pi, 1.0, 0.0)$\approx0$.

With detuning $H_\text{RWA}=\tfrac12(\delta\sigma_z+\Omega\sigma_x) =\tfrac12\Omega_R\,\hat n\!\cdot\!\vec\sigma$ is a spin in the tilted field $\hat n=(\Omega,0,\delta)/\Omega_R$, $\Omega_R=\sqrt{\Omega^2+\delta^2}$. The propagator $e^{-i(\Omega_R t/2)\hat n\cdot\vec\sigma}=\cos\tfrac{\Omega_R t}{2}\mathbb 1 -i\sin\tfrac{\Omega_R t}{2}\,\hat n\!\cdot\!\vec\sigma$ flips $|g\rangle\to|e\rangle$ with amplitude $\langle e|\hat n\!\cdot\!\vec\sigma|g\rangle=\Omega/\Omega_R$, so

The flop now runs at $\Omega_R>\Omega$ but its amplitude is capped at $\Omega^2/\Omega_R^2=\Omega^2/(\Omega^2+\delta^2)<1$, reached when $\Omega_R t=\pi$ — the off-resonant atom never fully inverts. For $\Omega=1,\delta=1.5$: $\Omega_R=\sqrt{3.25}=1.803$ and the peak is $1/3.25=0.3077$, i.e. generalized_rabi(1.0, 1.5)$=1.803$ and rabi_excited_population(np.pi/1.803, 1.0, 1.5)$=0.3077$.

In the interaction picture w.r.t. $H_A=\tfrac12\hbar\omega_a\sigma_z$ the ladder operators rotate, $\sigma^\pm(t)=e^{\pm i\omega_a t}\sigma^\pm$, and $\cos\omega_L t=\tfrac12(e^{i\omega_L t}+e^{-i\omega_L t})$, so

The RWA drops the counter-rotating pair, which oscillate at $\omega_a+\omega_L\approx 2\omega_a$: over a Rabi period $\sim1/\Omega$ they average to nearly zero when $\Omega,|\delta|\ll\omega_a$ (the residue is the small Bloch–Siegert shift). What remains, in the frame rotating at $\omega_L$, is the static $H_\text{RWA}=\tfrac12(\delta\sigma_z+\Omega\sigma_x)$. Because nothing else is discarded, evolving $|g\rangle$ under two_level_hamiltonian and projecting on $|e\rangle$ reproduces the closed-form rabi_excited_population to $10^{-9}$.

The free terms commute with $\hat N=a^\dagger a+\sigma^+\sigma^-$ trivially. For the interaction use $[a^\dagger a,a]=-a$ and $[\sigma^+\sigma^-,\sigma^+]=\sigma^+$ (since $|e\rangle\langle e|\,\sigma^+=\sigma^+$ but $\sigma^+|e\rangle\langle e|=0$):

and likewise $[\hat N,a^\dagger\sigma^-]=0$, so $[\hat N,H]=0$. The conserved $\hat N$ cannot mix different excitation numbers, and $a\sigma^+,\,a^\dagger\sigma^-$ connect only $|e,n\rangle\leftrightarrow|g,n+1\rangle$ (both with $\hat N=n+1$), so $H$ is block diagonal in those doublets. The off-diagonal element is

This is why jcm_hamiltonian(5,5.3,0.9,10) comes out Hermitian and commutes with $\hat N$ to $10^{-10}$.

In the doublet $(|e,n\rangle,|g,n+1\rangle)$ the diagonal energies are $\omega_c n+\tfrac12\omega_a$ and $\omega_c(n+1)-\tfrac12\omega_a$, with off-diagonal $g\sqrt{n+1}$ from P5. Their half-sum is $\omega_c(n+\tfrac12)$ and their difference is $\delta=\omega_a-\omega_c$, so the block is

The traceless part $\tfrac12(\delta\sigma_z+2g\sqrt{n+1}\,\sigma_x)$ has eigenvalues $\pm\tfrac12\sqrt{\delta^2+4g^2(n+1)}$, so

The gap $E_{n,+}-E_{n,-}=\sqrt{\delta^2+4g^2(n+1)}$ becomes $2g\sqrt{n+1}$ on resonance; at $n=0$ that is the vacuum Rabi splitting $2g$. For $g=1$ this gives vacuum_rabi_splitting(1.0)$=2.0$ and the $n=1$ gap $2\sqrt2=2.828$, matching the eigenvalues of jcm_hamiltonian's block.

On resonance the $|e,n\rangle$ component sits in the doublet with gap $2g\sqrt{n+1}$, so the atom started excited has each Fock weight oscillate as $\cos(2g\sqrt{n+1}\,t)$; weighting by the Poissonian $P_n=e^{-\bar n}\bar n^{\,n}/n!$,

Expanding $\sqrt{n+1}\approx\sqrt{\bar n}+(n-\bar n)/2\sqrt{\bar n}$ across the Poisson width $\sigma_n=\sqrt{\bar n}$, the tones dephase and the envelope collapses as $e^{-g^2t^2/2}$, i.e. $t_c\sim\sqrt2/g$ — independent of $\bar n$. They rephase when neighbouring frequencies, spaced $\Delta\omega\approx g/\sqrt{\bar n}$, slip by $2\pi$:

A classical field carries a single Rabi frequency and would flop forever; the discrete spread of $\sqrt{n+1}$ tones — hence the revival — is a fingerprint of field quantization. For $\alpha=4$ ($\bar n=16$), $t_r=8\pi=25.13$ (revival_time(1.0, 16.0)); the inversion is $\approx0$ at $t_r/2$ and revives to $\approx+0.44$ at $t_r$, matching resonant_inversion_series to $10^{-9}$.

Detailed balance $N_1B_{12}\rho=N_2(A_{21}+B_{21}\rho)$, solved for the field, gives

Insert the inverse Boltzmann ratio $N_1/N_2=(g_1/g_2)\,e^{h\nu/k_BT}$:

This can equal Planck's $\rho=\dfrac{8\pi h\nu^3/c^3}{e^{h\nu/k_BT}-1}$ for every $T$ only if the coefficient of $e^{h\nu/k_BT}$ is $1$, i.e. $g_1B_{12}=g_2B_{21}$, and the prefactor matches, $A_{21}/B_{21}=8\pi h\nu^3/c^3$. With those two relations the rate-balance density is Planck identically — independent of $A_{21},g_1,g_2$ — so detailed_balance_energy_density/planck_spectral_energy_density$=1.000000$, and the prefactor is einstein_A_over_B(c/248e-9)$\approx1.09\times10^{-12}$ J·s/m³.

Divide the stimulated rate by the spontaneous rate and use the first Einstein relation $A_{21}/B_{21}=8\pi h\nu^3/c^3$ together with the Planck $\rho$:

The two are equal when $e^{h\nu/k_BT}-1=1$, i.e. $e^{h\nu/k_BT}=2$, giving the crossover $h\nu=k_BT\ln2$. For the 248 nm KrF line at 300 K, $h\nu/k_BT\approx193\gg1$, so

Spontaneous emission wins by $\sim10^{84}$: heat cannot build coherent gain, only a non-thermal population inversion can. This is stimulated_to_spontaneous(c/248e-9, 300.0)$\approx1.0\times10^{-84}$, while at $\nu=k_BT\ln2/h$ the ratio is exactly $1.0$.

Let $x=h\nu/k_BT$. For $x\ll1$ ($h\nu\ll k_BT$) expand $e^{x}-1=x+O(x^2)$:

The $h$ cancels, so the Rayleigh–Jeans law is precisely the classical $h\to0$ limit; its $\nu^2$ growth makes $\int\rho\,d\nu$ diverge — the ultraviolet catastrophe. For $x\gg1$ ($h\nu\gg k_BT$), $e^{x}-1\to e^{x}$ and

the Wien exponential cutoff that quantization supplies to tame the divergence. At $\nu=10^9$ Hz, $T=300$ K we sit deep in Rayleigh–Jeans: planck_spectral_energy_density(1e9, 300.0)$\approx3.86\times10^{-27}$ J·s/m³, matching $8\pi\nu^2k_BT/c^3$ to $<0.1\%$.

The photon equation factors,

On the dark branch $n=0$, the inversion balance $\dot N=R-\gamma N=0$ gives $N=R/\gamma$, valid while $N\le\kappa/G$, i.e. up to $R_{\text{th}}=\gamma\kappa/G$. Past threshold the photon term pins $N=\kappa/G$; substituting into $\dot N=0$,

The inversion clamps at $\kappa/G$: every additional pump quantum is converted to a photon, never to more inversion. With $G=\kappa=\gamma=1$, laser_threshold(1,1,1)$=1.0$, and laser_steady_state returns $(0.5,0)$ at $R=0.5$, $(1,1)$ at $R=2$, $(1,4)$ at $R=5$ — $N$ frozen at $1$, $n$ rising linearly.

Use $g^{(2)}(0)=\langle n(n-1)\rangle/\langle n\rangle^2$ with $\langle n(n-1)\rangle=\langle n^2\rangle-\langle n\rangle$. (i) Bose–Einstein $p_n=\bar n^{\,n}/(1+\bar n)^{n+1}$ has $\langle n\rangle=\bar n$ and variance $\bar n+\bar n^2$, so $\langle n(n-1)\rangle=(\bar n+2\bar n^2)-\bar n=2\bar n^2$ and

(ii) Poisson has variance $=\bar n$, hence $\langle n(n-1)\rangle=\bar n^2$ and $g^{(2)}=\bar n^2/\bar n^2=1$. (iii) A Fock state $|n_0\rangle$ is sharp, $\langle n(n-1)\rangle=n_0(n_0-1)$, so $g^{(2)}=n_0(n_0-1)/n_0^2=1-1/n_0$. This falls below $1$, yet for any classical (positive-$P$) field $g^{(2)}(0)=1+\langle(\Delta I)^2\rangle/\langle I\rangle^2\ge1$, so $g^{(2)}<1$ has no positive-probability description. Confirmed by g2_from_distribution $=2$ (thermal) and $1$ (Poisson), with g2_fock(5)$=0.8$ and g2_fock(1)$=0.0$ (perfect antibunching).

$g^{(1)}(\tau)$ is the normalized field (amplitude) correlation, and the fringe visibility is $\mathcal V=|g^{(1)}(\tau)|$. A narrowband-filtered thermal source has a long coherence time $\tau_c\sim1/\Delta\nu$, so for $\tau\ll\tau_c$ it reaches $|g^{(1)}|\to1$ — the same value an ideal laser gives. Identical visibility means

Hanbury-Brown–Twiss instead correlates intensities, $g^{(2)}(0)=\langle n(n-1)\rangle/\langle n\rangle^2$: thermal photons arrive in bunches ($g^{(2)}=2$), coherent light is Poissonian ($g^{(2)}=1$). The ordering from most-bunched to most-antibunched is

Hence g2_thermal() $>$ g2_coherent() $>$ g2_fock(1), i.e. $2 > 1 > 0$.

Trace the generator. The commutator vanishes by cyclicity, $\mathrm{Tr}[H,\rho]=0$. For each dissipator, cyclicity gives $\mathrm{Tr}(L_k\rho L_k^\dagger)=\mathrm{Tr}(L_k^\dagger L_k\rho)$, and the anticommutator term yields $\mathrm{Tr}\{L_k^\dagger L_k,\rho\}=2\,\mathrm{Tr}(L_k^\dagger L_k\rho)$, so

The $-\tfrac12\{L_k^\dagger L_k,\rho\}$ "no-jump" term is exactly what cancels the gain term $L_k\rho L_k^\dagger$ under the trace. Drop it and

so probability would leak in and $\rho$ would stop being normalized. With the full generator lindblad_rhs gives $|\mathrm{Tr}\,\dot\rho|<10^{-12}$ for any rho, H.

With $L=\sqrt\gamma\,\sigma^-$, $\sigma^-=|g\rangle\langle e|$, and $L^\dagger L=\gamma|e\rangle\langle e|$, take the $\langle e|\cdots|e\rangle$ element of the dissipator. The gain term vanishes ($\langle e|\sigma^-=0$) and the anticommutator gives $-\tfrac\gamma2\langle e|\{|e\rangle\langle e|,\rho\}|e\rangle=-\tfrac\gamma2(2\rho_{ee})$, so

For the ground state the gain term now survives, $\gamma\langle g|\sigma^-\rho\sigma^+|g\rangle=\gamma\rho_{ee}$, while its anticommutator vanishes, giving $\dot\rho_{gg}=+\gamma\rho_{ee}$. Thus $\dot\rho_{ee}+\dot\rho_{gg}=0$: the population leaving $|e\rangle$ lands in $|g\rangle$, conserving the trace. Starting in $|e\rangle$, excited_population follows $e^{-\gamma t}$ to $10^{-6}$ and _fit_rate returns $\gamma=1.0000$.

Take the $\langle e|\cdots|g\rangle$ element of the same dissipator. The gain term again vanishes ($\langle e|\sigma^-=0$), and of the two anticommutator pieces only the one acting on the left index survives:

The population decay (P2) touched $|e\rangle$ on both sides — two factors, rate $\gamma$ — whereas the coherence touches it on only one side, hence exactly half: $|\rho_{eg}(t)|=|\rho_{eg}(0)|e^{-\gamma t/2}$, i.e. $T_2=2T_1$. The off-diagonal collapsing to zero is the loss of definite relative phase — decoherence, $\rho$ becoming diagonal (a classical mixture). Starting in $|+\rangle$, abs(coherence(...)) decays at rate $\gamma/2=0.5000$, giving $T_2=2T_1$.

For $L_\phi=\sqrt{\gamma_\phi/2}\,\sigma_z$ we have $L_\phi^\dagger L_\phi=\tfrac{\gamma_\phi}{2}\sigma_z^2=\tfrac{\gamma_\phi}{2}\mathbb 1$, so the no-jump term is $-\tfrac12\{\tfrac{\gamma_\phi}{2}\mathbb 1,\rho\}=-\tfrac{\gamma_\phi}{2}\rho$ and the dissipator collapses to

Since $(\sigma_z\rho\,\sigma_z)_{ij}=(\sigma_z)_{ii}(\sigma_z)_{jj}\,\rho_{ij}$, the diagonal entries get $(\pm1)^2=1$ — populations untouched — while the off-diagonal gets $(+1)(-1)=-1$, so $\mathcal D[L_\phi]\rho|_{eg}=\tfrac{\gamma_\phi}{2}(-\rho_{eg}-\rho_{eg})=-\gamma_\phi\rho_{eg}$. Adding the emission part $-\tfrac\gamma2\rho_{eg}$ from P3,

With $\gamma=1$, $\gamma_\phi=0.8$ the simulated coherence-decay rate is $0.5+0.8=1.3000$ and the populations are unchanged.

With $H=\hbar\Delta|e\rangle\langle e|+\tfrac{\hbar\Omega}{2}\sigma_x$ and $L=\sqrt\gamma\,\sigma^-$, the master equation gives the optical Bloch equations

Set both to zero. The first gives $\rho_{eg}=\dfrac{i\Omega\,(\rho_{ee}-\rho_{gg})/2}{\gamma/2+i\Delta}$; feeding its imaginary part into $\gamma\rho_{ee}=-\Omega\,\mathrm{Im}\,\rho_{eg}$ and writing $w=\rho_{ee}-\rho_{gg}=2\rho_{ee}-1$,

On resonance ($\Delta=0$) this is $\Omega^2/(\gamma^2+2\Omega^2)$; for $\Omega=2,\gamma=1$ it equals $4/9=0.4444$, matching both steady_state_excited_population(2,1,0) and the long-time evolve_lindblad value.

On resonance $\rho_{ee}^{\,\mathrm{ss}}(\Omega)=\dfrac{\Omega^2}{\gamma^2+2\Omega^2}$. Differentiate:

so it increases monotonically with the drive. But the denominator always exceeds $2\Omega^2$ (the $\gamma^2>0$ term), hence

Because $\rho_{ee}<\tfrac12$ means $\rho_{ee}<\rho_{gg}=1-\rho_{ee}$, a classical drive can at best equalize the populations, never invert them — which is why lasing needs a $\ge3$-level scheme (~QO-03). Numerically steady_state_excited_population is increasing in $\Omega$ and tends to $0.4999\to\tfrac12$.

Unitary evolution $\rho(t)=U\rho(0)U^\dagger$ is a similarity transform, so it leaves the eigenvalues — and the purity $\mathrm{Tr}\,\rho^2$ — invariant; a pure state ($\mathrm{Tr}\,\rho^2=1$) can never become mixed (~QM-20). The Lindblad generator is not a similarity transform. Starting from $|+\rangle$ with $H=0$, the coherence decays as $\rho_{eg}=\tfrac12e^{-\gamma t/2}$ while the populations redistribute, so

at intermediate times: the pure state has genuinely mixed, irreversibly (entropy produced, no unitary undoes it). Yet every snapshot stays Hermitian, unit-trace, and positive semidefinite — eigenvalues $\ge0$, a legitimate state at each instant, the meaning of CPTP. Along a driven, damped trajectory is_positive_semidefinite(rho) holds at every evolve_lindblad snapshot, and the undriven run relaxes to density_matrix(ket_g) with $\rho_{ee}<10^{-6}$.

With $X_1=\tfrac12(a+a^\dagger)$ and $X_2=\tfrac1{2i}(a-a^\dagger)$,

Robertson then gives $\operatorname{Var}X_1\operatorname{Var}X_2\ge\tfrac14\big|\langle[X_1,X_2]\rangle\big|^2=\tfrac14\cdot\tfrac14=\tfrac1{16}$. On the vacuum $a|0\rangle=0$, so $\langle X_1\rangle=\langle X_2\rangle=0$ and only the $aa^\dagger$ term survives, $\langle0|X_1^2|0\rangle=\tfrac14\langle0|aa^\dagger|0\rangle=\tfrac14$, and likewise $\langle X_2^2\rangle=\tfrac14$ — the bound is saturated. A coherent state is $|\alpha\rangle=D(\alpha)|0\rangle$ with $D^\dagger aD=a+\alpha$, a c-number shift that leaves the fluctuation operator $a-\langle a\rangle$ untouched, so every coherent state inherits the vacuum's $\operatorname{Var}X_1=\operatorname{Var}X_2=\tfrac14$. Matches quadrature_variances(vacuum_state())→(0.25, 0.25) and coherent_state(1.5)→(0.25, 0.25), product $\tfrac1{16}=0.0625$, with the $[X_1,X_2]$ interior $\tfrac{i}{2}$.

With $M=\tfrac12(\xi^{*}a^2-\xi a^{\dagger2})$ the first nested commutator is

and feeding it back, $[\,[a,M],M\,]=-\xi\,[a^\dagger,M]=-\xi\big(-\xi^{*}a\big)=|\xi|^2a=r^2a$. The two commutators map $a\!\to\!a^\dagger\!\to\!a$, so the Hadamard series closes: even orders rebuild $a$, odd orders $a^\dagger$,

and $\xi/r=e^{i\phi}$ gives $S^\dagger aS=a\cosh r-a^\dagger e^{i\phi}\sinh r$. Finally $M^\dagger=\tfrac12(\xi a^{\dagger2}-\xi^{*}a^2)=-M$, so $M$ is anti-Hermitian and $S=e^{M}$ is unitary. This is what squeeze_operator(0.9, 40, phi=0.7) confirms, $S^\dagger S=\mathbb1$ to $10^{-9}$, with squeeze_operator(0.0, N)$=\mathbb1$.

Take $\phi=0$, so P2 gives $S^\dagger aS=a\cosh r-a^\dagger\sinh r$ and (adjoint) $S^\dagger a^\dagger S=a^\dagger\cosh r-a\sinh r$. Adding and subtracting,

Since $|\xi\rangle=S|0\rangle$, $\operatorname{Var}_{|\xi\rangle}X_1=\langle0|(S^\dagger X_1S)^2|0\rangle-\langle0|S^\dagger X_1S|0\rangle^2=e^{-2r}\operatorname{Var}_0X_1=\tfrac14e^{-2r}$, and likewise $\operatorname{Var}X_2=\tfrac14e^{+2r}$, product $\tfrac1{16}$. So $X_1$ (for $\phi=0$) is the squeezed quadrature, dropping below $\tfrac14$; a general $\phi$ rotates the squeezed direction by $\phi/2$ in phase space, and $\phi=\pi$ swaps $X_1\leftrightarrow X_2$. In decibels $10\log_{10}(\operatorname{Var}X_1/\tfrac14)=10\log_{10}e^{-2r}=-8.686\,r$. At $r=0.6$ this is quadrature_variances(squeezed_vacuum(0.6))$\approx(0.0753,0.830)$, product $0.0625$, i.e. $-5.2$ dB; squeezed_vacuum(0.7, phi=pi) swaps the two.

The generator $M=\tfrac12(\xi^{*}a^2-\xi a^{\dagger2})$ moves population only in steps of $\pm2$ (the $a^2$ and $a^{\dagger2}$ pieces), so acting on the even state $|0\rangle$ it never reaches an odd $|n\rangle$: $P(2m+1)=0$. The disentangled form $|\xi\rangle=\tfrac1{\sqrt{\cosh r}}\sum_m\tfrac{\sqrt{(2m)!}}{2^mm!}(-e^{i\phi}\tanh r)^m|2m\rangle$ squares to

For the mean, conjugate with $S$ ($\phi=0$): $\langle n\rangle=\langle0|(S^\dagger a^\dagger S)(S^\dagger aS)|0\rangle$, and with $S^\dagger aS=a\cosh r-a^\dagger\sinh r$ only the $\sinh^2 r$ term survives the vacuum sandwich,

Hence photon_distribution(squeezed_vacuum(0.8)) has odd-$n$ sum $\sim10^{-17}$ and $P(0)=1/\cosh0.8=0.748$, while mean_photon_number$=$mean_photon(r)$=\sinh^2 r$ (e.g. $0.789$ at $r=0.8$).

Use $g^{(2)}(0)=\langle a^\dagger a^\dagger aa\rangle/\langle a^\dagger a\rangle^2=\langle n(n-1)\rangle/\langle n\rangle^2$. For $|1\rangle$, $n=1$ so $n(n-1)=0$ and $g^{(2)}(0)=0$ (perfectly antibunched), with $Q=\langle n\rangle(g^{(2)}(0)-1)=-1$. For a coherent state $a|\alpha\rangle=\alpha|\alpha\rangle$ gives $\langle a^\dagger a^\dagger aa\rangle=|\alpha|^4=\langle n\rangle^2$, so $g^{(2)}(0)=1$ and $Q=0$. For the squeezed vacuum the even-pair statistics give $\operatorname{Var}n=2\sinh^2 r\cosh^2 r$ on top of $\langle n\rangle=\sinh^2 r$, so

with $Q=2\cosh^2 r-1=\cosh 2r>0$ — super-Poissonian. So a state can be quadrature-squeezed (sub-SQL) yet bunched: the two nonclassicalities are independent. Matches g2_zero→0.0, 1.0, 4.268$=3+1/\sinh^2 0.8$ and mandel_q→$-1,\,0,\,+2.58$.

$S_2=\exp(\xi^{}ab-\xi a^\dagger b^\dagger)$ has a generator that adds or removes one photon in each* mode at once ($ab$ and $a^\dagger b^\dagger$), so from $|0,0\rangle$ it stays on the diagonal $|n,n\rangle$; the disentangling theorem resums this to

The geometric marginal gives $\langle n_a\rangle=\langle n_b\rangle=\sinh^2 r$, with $n_a=n_b$ always (perfect number correlation). With $\phi=0$ the two-mode Bogoliubov rule $S_2^\dagger aS_2=a\cosh r-b^\dagger\sinh r$ (and $a\!\leftrightarrow\! b$) yields $S_2^\dagger(X_a+X_b)S_2=(X_a+X_b)e^{-r}$, so

and identically for $Y_a-Y_b$. As $r\to\infty$ both $\to0$: $X_a+X_b$ and $Y_a-Y_b$ become simultaneously sharp even though $[X,Y]\neq0$ per mode — the EPR "elements of reality." Confirmed by two_mode_photon_distribution(...) being diagonal with $P(n,n)=(\tanh^2 0.6)^n/\cosh^2 0.6$, and epr_variances(0.6)→(0.151, 0.151), both $<0.5=\tfrac12 e^{-2\cdot0.6}$.

Compare adjacent terms of $P=\varepsilon_0(\chi^{(1)}E+\chi^{(2)}E^2+\chi^{(3)}E^3+\cdots)$:

since this expansion of the bound-electron response only breaks down when $E$ nears the atomic field $E_{\rm at}\sim10^{11}$–$10^{12}$ V/m that binds the electron. Thus $\chi^{(n+1)}/\chi^{(n)}\sim1/E_{\rm at}$, and with $\chi^{(1)}\sim1$: $\chi^{(2)}\sim1/E_{\rm at}\sim$ pm/V, $\chi^{(3)}\sim1/E_{\rm at}^2\sim10^{-22}\ \text{m}^2/\text{V}^2$. In a centrosymmetric medium, inversion sends $E\to-E$ and $P\to-P$; but the $\chi^{(2)}$ term gives $\chi^{(2)}(-E)^2=+\chi^{(2)}E^2$, which can equal the required $-\chi^{(2)}E^2$ only if $\chi^{(2)}=0$ (every even order dies; odd $\chi^{(3)}$ survives). So chi_polarization(E, chi1) is the pure linear $\varepsilon_0\chi^{(1)}E$, while the $\chi^{(2)}$-only term $\varepsilon_0\chi^{(2)}E^2$ quadruples ($\times4$) when $E$ doubles.

In the undepleted-pump, slowly-varying-envelope limit the harmonic amplitude obeys $dA_2/dz=i\kappa A_1^2e^{-i\Delta k\,z}$ with $A_1$ fixed. Integrating over the crystal,

The efficiency is $\eta\propto|A_2(L)|^2\propto L^2\operatorname{sinc}^2(\Delta kL/2)$. It is largest at $\Delta k=0$, where $\operatorname{sinc}0=1$ and every slice adds in phase ($\eta\propto L^2$); its first zero is where the argument first hits $\pi$, i.e. $\Delta kL/2=\pi\Rightarrow\Delta kL=2\pi$. Hence shg_efficiency(0.0, L)$=1.0$ (normalized peak), shg_efficiency(2π/L, L)$\approx0$, while at the half-way point shg_efficiency(π/L, L)$=\operatorname{sinc}^2(\pi/2)=(2/\pi)^2\approx0.405>0$.

With $k=n\omega/c$ and the harmonic at $2\omega$,

using $\omega=2\pi c/\lambda$ so that $2\omega/c=4\pi/\lambda$. The coherence length is the run over which the relative phase $\Delta k\,z$ accumulates $\pi$, $L_c=\pi/|\Delta k|$. For an index split $\Delta n=0.030$ at $\lambda=1.064\,\mu$m, $\Delta k=4\pi(0.030)/(1.064\,\mu\text{m})\approx3.54\times10^5\ \text{m}^{-1}$ and $L_c=\pi/\Delta k\approx8.9\,\mu$m. Beyond $L_c$ the harmonic is $\pi$ out of phase and power flows back to the fundamental, so conversion oscillates with spatial period $2L_c$ and $L_c$ is its half-period — exactly the $L=2L_c$ null of the §2 $\operatorname{sinc}^2$ curve. This is shg_phase_mismatch(1.50, 1.53, 1.064e-6)$\approx3.54\times10^5\ \text{m}^{-1}$ with coherence_length$\approx8.9\,\mu$m.

The photon flux in beam $i$ is $\Phi_i=I_i/\hbar\omega_i$. A lossless $\chi^{(2)}$ step destroys one pump photon ($\omega_3$) to make one signal ($\omega_1$) and one idler ($\omega_2$), so $d\Phi_1=d\Phi_2=-d\Phi_3$, i.e.

Writing $\Delta I_i=\hbar\omega_i\,\Delta\Phi_i$ with $\Delta\Phi_1=\Delta\Phi_2=-\Delta\Phi_3\equiv\Delta\Phi$, energy balance follows: $\Delta I_1+\Delta I_2+\Delta I_3=\hbar(\omega_1+\omega_2-\omega_3)\Delta\Phi=0$ because $\omega_3=\omega_1+\omega_2$. For $\lambda_p=532$ nm and $\lambda_s=800$ nm the idler obeys $1/\lambda_i=1/\lambda_p-1/\lambda_s=(800-532)/(532\cdot800\ \text{nm})$, giving $\lambda_i\approx1588$ nm. Hence manley_rowe_check(...) returns dPhi_signal$=$dPhi_idler$=-$dPhi_pump$=10^{24}$, energy_residual$\approx0$, and an idler at $\approx1588$ nm.

A wave traversing length $L$ accumulates phase $\phi=(2\pi/\lambda)nL$; with the Kerr index $n=n_0+n_2I$ the intensity-dependent part is the self-phase modulation

the beam phase-modulating itself. For fused silica $n_2=2.6\times10^{-20}\ \text{m}^2/\text{W}$ at $I=10^{15}\ \text{W/m}^2$ over $L=10^{-2}$ m at $\lambda=1.064\,\mu$m,

The accumulated $B=\int(2\pi/\lambda)n_2I\,dz$ is the "B-integral"; once $B\gtrsim1$ the brightest filaments pick up $\sim1$ rad of extra phase, the wavefront self-focuses, and the beam breaks up — damaging a high-power chain (~PK-04), so designers hold $B$ below order unity. Linear in both $I$ and $L$, this is kerr_phase(2.6e-20, 1e15, 1e-2, 1.064e-6)$\approx1.54$ rad.

Backscattered Stokes light beats with the pump to drive a sound wave; phase matching for a reflected wave ($\mathbf k_S\approx-\mathbf k_p$) needs phonon wavevector $q\approx2k_p=4\pi n/\lambda_p$, and for a linear sound branch $\Omega_B=v_a q$, so

The Stokes gain is the Lorentzian $g(\Omega)=g_0(\Gamma_B/2)^2/[(\Omega-\Omega_B)^2+(\Gamma_B/2)^2]$, peaking at $g_0$ when $\Omega=\Omega_B$ and falling to $g_0/2$ where $(\Omega-\Omega_B)^2=(\Gamma_B/2)^2$, i.e. at $\Omega_B\pm\Gamma_B/2$ — the two half-maximum points a full $\Gamma_B$ apart. For water, $\nu_B=2(1.33)(1480)/(532\times10^{-9})\approx7.4$ GHz. So brillouin_shift(1.33, 1480, 532e-9)$\approx7.4$ GHz, and brillouin_gain returns $g_0$ at line center and $g_0/2$ at $\Omega_B\pm\Gamma_B/2$ — the lineshape SBS_Project fits.

Both gains are the identical Lorentzian $g(\Omega)=g_0(\Gamma/2)^2/[(\Omega-\Omega_0)^2+(\Gamma/2)^2]$; they differ only in which excitation the optical beat drives. Brillouin scatters off an acoustic phonon, $\Omega=v_a q$, gapless and linear in $q$; at the backscatter wavevector $q\approx2k$ this is

because $v_a/c\sim10^{-5}$. Raman scatters off an optical/vibrational phonon whose frequency is set by intramolecular bond stiffness, nearly $q$-independent, with $\hbar\Omega_R\sim0.1$ eV $\Rightarrow\Omega_R\sim10^{14}\ \text{rad/s}$ (tens of THz). The ratio $\Omega_R/\Omega_B\sim10^4$–$10^5$ is large precisely because acoustic-branch frequencies vanish as $q\to0$ while optical-branch frequencies sit at a finite bond-stiffness gap. So raman_gain(Ω_R, Ω_R, γ_R) peaks at $g_0$ exactly like SBS, but at $\Omega_R\sim10^{14}$ rad/s $\gg$ the GHz Brillouin shift.