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The fixed-particle-number sector is spanned by $|1⟩=c_1^{†}|0⟩$ and $|2⟩=c_2^{†}|0⟩$. In this basis,The relevant quantum Gibbs state is normalized within this two-dimensional sector, not over all particle numbers. With $\beta =(k_{B}T{)}^{-1}$ and $x=\beta t$, the exponential of a Pauli vector givesThereforeIts eigenstates are $|\pm ⟩=(|1⟩\pm |2⟩)/\sqrt{2}$, with energies $\mp t$ and probabilities $p_{\pm }=e^{\pm x}/(2\cosh x)$. These are nonnegative and sum to one, so the matrix is Hermitian and has trace one. Its purity of a density operator isFor $t\ne 0$, the zero-temperature limit is the pure ground-state projector,with purity one. As $T\to \infty$, $x\to 0$ and $\rho \to I/2$, with purity $1/2$. At every strictly positive finite temperature and finite $t$, both eigenvalues are nonzero, so the state is mixed. If $t=0$, the two levels are degenerate and the Gibbs state is $I/2$ at all temperatures, including its zero-temperature limit.

Using dimensionless Von Neumann entropy, $S=-\mathrm{Tr}(\rho \ln \rho )$, givesFor nonzero $t$, $S\to 0$ as $T\to 0$, while $S\to \ln 2$ as $T\to \infty$. More precisely, the high-temperature expansion is $S=\ln 2-x^2/2+O(x^4)$, and the low-temperature form is $S\sim (1+2|x|)e^{-2|x|}$. The thermodynamic entropy includes an overall factor $k_B$.

Here the dimensionless spin one-half operator is $S_x={\sigma }_x/2$. Since the Pauli matrix obeys ${\sigma }_x^2=I$, the even and odd terms of the matrix exponential giveThis is the exponential of a Pauli vector for an $x$-axis rotation. With a physical spin angular momentum $S_x^{\mathrm{phys}}=\hslash {\sigma }_x/2$, the exponent is $-i\theta S_x^{\mathrm{phys}}/\hslash$.

Use the ordered orthonormal basis $(|1s⟩,|2p_x⟩)$. The projected Hamiltonian between pulses is $H_0=-\Delta {\sigma }_z$, so its free time-evolution operator through a period isAcross one instantaneous pulse, integrating the Schrödinger equation givesusing the exponential of a Pauli vector. The finite free Hamiltonian contributes vanishing evolution during the zero-duration kick. A period measured immediately before successive pulses therefore consists of the kick followed by free evolution. Composition givesThis is the Floquet operator at the specified phase of the pulse train. Actual hydrogenic energy levels are not opposite numbers: the displayed convention subtracts their midpoint, with $\Delta =(E_{2p}-E_{1s})/2$. Restoring that midpoint multiplies the result by an unobservable global phase. Likewise $E$ in the given Hamiltonian denotes the electric coupling per unit length; if it is instead a physical electric-field strength, its charge factor must be included in $E{x}_0$.

Multiplying the two matrix exponentials with the Pauli matrix multiplication law givesOne has $C^2+|\mathbf{v}{|}^2=1$. If $|\mathbf{v}|\ne 0$, put $\theta =2\arccos C$ and $\mathbf{n}=\mathbf{v}/|\mathbf{v}|$. The exponential of a Pauli vector gives $U_F^N=I\cos (N\theta /2)-i\mathbf{n}\cdot \mathbf{\sigma }\sin (N\theta /2)$. Its action on a Bloch vector is a rotation through $N\theta$ about $\mathbf{n}$, and its quasienergies are $\pm \hslash \theta /(2T)$ modulo $2\pi \hslash /T$. When $U_F=\pm I$, its axis is arbitrary and the Bloch vector is unchanged.

A quantum Gibbs state is the density operator of a canonical ensemble with inverse temperature $\beta =1/(k_{B}T)$. It exists when the canonical partition function $Z=\mathrm{Tr}(e^{-\beta H})$ is finite. The spectral theorem assigns probabilities $e^{-\beta E_j}/Z$ to an orthonormal eigenbasis of the Hamiltonian. An additive constant in the Hamiltonian cancels between numerator and denominator. For $H=-\mu B{\sigma }_z$, the exponential of a Pauli vector, continued to an imaginary argument, gives ${\rho }_{\beta }=[I+\tanh (\beta \mu B){\sigma }_z]/2$.