Formula Sheet

This page collects formulas that are repeatedly used in measurement theory and open quantum systems. It is a lookup page, not a derivation page. Follow the links for assumptions, domains of validity, and physical interpretation.

Unless stated otherwise, $\rho$ is a density operator on a finite-dimensional Hilbert space, $\rho\ge 0$, and $\operatorname{Tr}\rho=1$. Infinite-dimensional formulas require the usual domain and convergence qualifications.

Trace and Adjoint Conventions

The trace pairing between observables and states is

$$\langle A\rangle_\rho = \operatorname{Tr}(A\rho).$$

The Hilbert–Schmidt adjoint $\Phi^\dagger$ of a linear map $\Phi$ is defined by

$$\operatorname{Tr}\!\left[A\,\Phi(\rho)\right] = \operatorname{Tr}\!\left[\Phi^\dagger(A)\rho\right].$$

Trace preservation of $\Phi$ is equivalent to unitality of $\Phi^\dagger$:

$$\operatorname{Tr}\Phi(\rho)=\operatorname{Tr}\rho \quad\Longleftrightarrow\quad \Phi^\dagger(I)=I.$$

Unitality of $\Phi$ means $\Phi(I)=I$. A channel can be trace preserving without being unital; amplitude damping is the standard example.

Projective Measurements

A projective measurement with outcomes $a$ is a projection-valued measure $\{P_a\}$:

$$P_aP_b=\delta_{ab}P_a, \qquad P_a^\dagger=P_a, \qquad \sum_a P_a=I.$$

The Born probability is

$$p(a)=\operatorname{Tr}(P_a\rho).$$

For an ideal selective measurement whose recorded outcome is $a$, the Lüders update is

$$\rho \mapsto \rho_a = \frac{P_a\rho P_a}{\operatorname{Tr}(P_a\rho)}.$$

If the apparatus performs the measurement but the outcome is ignored, the nonselective state is

$$\rho \mapsto \sum_a P_a\rho P_a.$$

For a pure input $\lvert\psi\rangle$, the probability and normalized post-measurement vector are

$$p(a) = \langle\psi\rvert P_a\lvert\psi\rangle, \qquad \lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle}{\sqrt{p(a)}}.$$

For the operational distinctions among selective, nonselective, and conditionally updated states, see Selective and Nonselective Measurements and State-Update Rules.

POVMs

A POVM with outcomes $x$ is a set of effects $\{E_x\}$ satisfying

$$E_x\ge 0, \qquad \sum_x E_x=I.$$

Outcome probabilities are

$$p(x)=\operatorname{Tr}(E_x\rho).$$

Every projective measurement is a special POVM with $E_x=P_x$, but most POVMs do not specify a unique state update. That extra information belongs to a quantum instrument.

For a two-outcome unsharp spin-$z$ measurement on a qubit, a common effect pair is

$$E_\pm = \frac{1}{2} \left( I\pm \eta\sigma_z \right), \qquad 0\le \eta\le 1.$$

The parameter $\eta=1$ gives the sharp projective measurement of $\sigma_z$; $\eta=0$ gives no information about the state.

See POVMs for the canonical explanation of effects as outcome-statistics objects.

Kraus Operations and Instruments

A completely positive operation for outcome $x$ can be written

$$\mathcal I_x(\rho) = \sum_\alpha M_{x\alpha}\rho M_{x\alpha}^\dagger .$$

The probability of outcome $x$ is

$$p(x) = \operatorname{Tr}\mathcal I_x(\rho) = \operatorname{Tr}(E_x\rho),$$

where the associated POVM effect is

$$E_x = \sum_\alpha M_{x\alpha}^\dagger M_{x\alpha}.$$

The selective post-measurement state is

$$\rho_x = \frac{\mathcal I_x(\rho)}{\operatorname{Tr}\mathcal I_x(\rho)}.$$

The nonselective channel obtained by ignoring the record is

$$\Phi(\rho) = \sum_x \mathcal I_x(\rho) = \sum_{x,\alpha}M_{x\alpha}\rho M_{x\alpha}^\dagger .$$

Trace preservation of the overall instrument is

$$\sum_{x,\alpha}M_{x\alpha}^\dagger M_{x\alpha}=I.$$

Each individual outcome operation is trace-nonincreasing:

$$\sum_\alpha M_{x\alpha}^\dagger M_{x\alpha}\le I.$$

See Kraus Operators for outcome updates and Quantum Instruments for the full outcome-resolved structure.

Quantum Channels

A quantum channel is a completely positive trace-preserving map. In Kraus form,

$$\Phi(\rho) = \sum_r K_r\rho K_r^\dagger, \qquad \sum_r K_r^\dagger K_r=I.$$

The same channel has many Kraus representations. If two minimal sets $\{K_r\}$ and $\{L_s\}$ represent the same channel, they are related by a unitary mixing matrix:

$$L_s = \sum_r u_{sr}K_r.$$

The adjoint channel acts on observables as

$$\Phi^\dagger(A) = \sum_r K_r^\dagger A K_r.$$

A channel is unital exactly when

$$\sum_r K_rK_r^\dagger=I.$$

A unitary channel is the special case

$$\Phi(\rho)=U\rho U^\dagger.$$

A convex mixture of unitary channels has the form

$$\Phi(\rho) = \sum_r p_r U_r\rho U_r^\dagger, \qquad p_r\ge 0, \qquad \sum_r p_r=1.$$

For the complete-positivity requirement and its physical origin, see Completely Positive Maps. For Kraus nonuniqueness and environmental realizations, see Kraus Representation.

Choi Matrix

For a channel $\Phi:\mathcal B(\mathcal H_\text{in})\to\mathcal B(\mathcal H_\text{out})$, choose an orthonormal basis $\{\lvert i\rangle\}$ of $\mathcal H_\text{in}$ and define

$$J_\Phi = \sum_{i,j} \Phi(\lvert i\rangle\langle j\rvert) \otimes \lvert i\rangle\langle j\rvert .$$

With this output-input convention:

$$\Phi\ \text{is completely positive} \quad\Longleftrightarrow\quad J_\Phi\ge 0,$$

and

$$\Phi\ \text{is trace preserving} \quad\Longleftrightarrow\quad \operatorname{Tr}_{\text{out}}J_\Phi=I_{\text{in}}.$$

If

$$J_\Phi = \sum_r \lvert K_r\rangle\!\rangle \langle\!\langle K_r\rvert,$$

where $\lvert K_r\rangle\!\rangle$ is vectorized consistently with the convention above, then

$$\Phi(\rho) = \sum_r K_r\rho K_r^\dagger.$$

Different books place the input and output factors in the opposite order. Always check which partial trace encodes trace preservation. See Choi Matrix for the canonical convention used here.

Reduced Dynamics

For a system $S$ and environment $E$, the exact reduced state is

$$\rho_S(t) = \operatorname{Tr}_E \!\left[ U_{SE}(t)\rho_{SE}(0)U_{SE}^\dagger(t) \right].$$

If the initial state factorizes as

$$\rho_{SE}(0)=\rho_S(0)\otimes\rho_E,$$

then the reduced evolution is a quantum channel:

$$\rho_S(t)=\Phi_t(\rho_S(0)).$$

If

$$\rho_E=\sum_\beta q_\beta \lvert e_\beta\rangle\langle e_\beta\rvert,$$

then one Kraus representation is

$$K_{\alpha\beta}(t) = \sqrt{q_\beta}\, \langle e_\alpha\rvert U_{SE}(t)\lvert e_\beta\rangle_E,$$

so that

$$\Phi_t(\rho_S) = \sum_{\alpha,\beta} K_{\alpha\beta}(t)\rho_S K_{\alpha\beta}^\dagger(t).$$

Initial system-environment correlations can prevent a reduced map on arbitrary system states from being a completely positive channel. See Initial Correlations for the precise caveat and Reduced Dynamics for the general construction.

Decoherence

The simplest decoherence calculation starts from a correlated state

$$\lvert\Psi\rangle_{SE} = \sum_i c_i \lvert s_i\rangle \lvert e_i\rangle.$$

Tracing out the environment gives

$$\rho_S = \sum_{i,j} c_i c_j^* \langle e_j|e_i\rangle \lvert s_i\rangle\langle s_j\rvert .$$

The coherence between $\lvert s_i\rangle$ and $\lvert s_j\rangle$ is multiplied by the environment overlap $\langle e_j|e_i\rangle$. Orthogonal environment records suppress off-diagonal terms in that basis.

A phase-damping channel in a preferred qubit basis has the form

$$\rho = \begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11} \end{pmatrix} \mapsto \begin{pmatrix} \rho_{00} & \eta\rho_{01} \\ \eta^*\rho_{10} & \rho_{11} \end{pmatrix},$$

with $|\eta|\le 1$ for complete positivity. A Markovian pure-dephasing model often has $\eta(t)=e^{-t/T_\phi}$ after separating Hamiltonian phase rotation.

See What Is Decoherence? for interpretation and Dephasing vs Dissipation for the distinction between coherence loss and energy exchange.

Common Noise Channels

For a qubit, the phase-damping channel can be written

$$\Phi_\eta(\rho) = \frac{1+\eta}{2}\rho + \frac{1-\eta}{2}\sigma_z\rho\sigma_z, \qquad -1\le\eta\le 1.$$

The corresponding Bloch-vector action is

$$(r_x,r_y,r_z) \mapsto (\eta r_x,\eta r_y,r_z).$$

A qubit depolarizing channel is often parameterized by a Bloch-vector shrink factor $\lambda$:

$$\Phi_\lambda(\rho) = \lambda\rho + (1-\lambda)\frac{I}{2}, \qquad -\frac{1}{3}\le\lambda\le 1.$$

Its Bloch-vector action is

$$\mathbf r\mapsto \lambda\mathbf r.$$

Zero-temperature amplitude damping with decay probability $p$ is represented by

$$K_0 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-p} \end{pmatrix}, \qquad K_1 = \begin{pmatrix} 0 & \sqrt p \\ 0 & 0 \end{pmatrix}.$$

For the convention $\lvert 1\rangle\to\lvert 0\rangle$, the matrix elements transform as

$$\begin{aligned} \rho_{11}'&=(1-p)\rho_{11},\\ \rho_{00}'&=\rho_{00}+p\rho_{11},\\ \rho_{01}'&=\sqrt{1-p}\,\rho_{01}. \end{aligned}$$

An erasure channel into an enlarged Hilbert space is

$$\Phi(\rho) = (1-p)\rho + p\,\operatorname{Tr}(\rho) \lvert e\rangle\langle e\rvert,$$

where $\lvert e\rangle$ is orthogonal to the input code space.

See Common Noise Channels for model-selection guidance and parameter conventions. For full convention maps, see Depolarizing Channel, Amplitude-Damping Channel, Erasure and Loss Channels, and Gaussian Channels.

Lindblad–GKSL Dynamics

The standard Markovian master equation for a density operator is

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[H,\rho] + \sum_\mu \left( L_\mu\rho L_\mu^\dagger - \frac{1}{2}\{L_\mu^\dagger L_\mu,\rho\} \right).$$

The dissipator notation is

$$\mathcal D[L]\rho = L\rho L^\dagger - \frac{1}{2}\{L^\dagger L,\rho\}.$$

The Heisenberg-picture adjoint evolution is

$$\frac{dA}{dt} = \frac{i}{\hbar}[H,A] + \sum_\mu \left( L_\mu^\dagger A L_\mu - \frac{1}{2}\{L_\mu^\dagger L_\mu,A\} \right).$$

The short-time channel generated by a Lindblad equation is

$$\rho(t+\Delta t) = \rho(t)+\Delta t\,\mathcal L(\rho(t))+O(\Delta t^2),$$

with jump Kraus operators

$$K_\mu=\sqrt{\Delta t}\,L_\mu$$

and no-jump Kraus operator

$$K_0 = I - \Delta t \left( \frac{i}{\hbar}H + \frac{1}{2}\sum_\mu L_\mu^\dagger L_\mu \right) + O(\Delta t^2).$$

See Lindblad–GKSL Equation for hypotheses and examples, Lindblad Operators for operator meanings, Quantum Dynamical Semigroups for the semigroup property, and Detailed Balance for thermal rate constraints.

Pure Dephasing Master Equation

See Pure Dephasing Master Equation for assumptions, rate conventions, and physical interpretations.

For a qubit with Hamiltonian $H=(\hbar\omega_0/2)\sigma_z$ and Lindblad operator

$$L_\phi = \sqrt{\frac{1}{2T_\phi}}\, \sigma_z,$$

the master equation is

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[H,\rho] + \frac{1}{2T_\phi} \left( \sigma_z\rho\sigma_z-\rho \right).$$

The off-diagonal element evolves as

$$\rho_{01}(t) = e^{-i\omega_0 t} e^{-t/T_\phi} \rho_{01}(0),$$

while populations in the $\sigma_z$ basis are unchanged.

Amplitude Damping Master Equation

See Amplitude Damping Master Equation for assumptions, jump interpretation, and finite-temperature extensions.

For zero-temperature decay $\lvert e\rangle\to\lvert g\rangle$ with rate $\Gamma$,

$$L=\sqrt{\Gamma}\,\sigma_-, \qquad H=\frac{\hbar\omega_0}{2}\sigma_z.$$

The excited-state population satisfies

$$\rho_{ee}(t)=e^{-\Gamma t}\rho_{ee}(0).$$

The coherence satisfies

$$\rho_{eg}(t) = e^{-i\omega_0t} e^{-\Gamma t/2} \rho_{eg}(0)$$

in the absence of additional pure dephasing.

At finite temperature, with downward and upward rates $\Gamma_\downarrow$ and $\Gamma_\uparrow$ related by detailed balance for a single thermal bath, the corresponding thermal master equation gives

$$\dot\rho_{ee} = -\Gamma_\downarrow\rho_{ee} + \Gamma_\uparrow\rho_{gg}.$$

The steady excited-state population is

$$\rho_{ee}^{\text{ss}} = \frac{\Gamma_\uparrow} {\Gamma_\uparrow+\Gamma_\downarrow}.$$

Pauli Rate Equation

See Pauli Rate Equations for the population-only limit of quantum master equations.

For populations $p_n$ in a preferred basis,

$$\dot p_n = \sum_{m\ne n} \left( W_{n\leftarrow m}p_m - W_{m\leftarrow n}p_n \right), \qquad W_{n\leftarrow m}\ge0.$$

With the column-vector convention $\dot{\mathbf p}=K\mathbf p$,

$$K_{nm}=W_{n\leftarrow m} \quad (n\ne m), \qquad K_{nn} = - \sum_{m\ne n} W_{m\leftarrow n}.$$

Then $\sum_nK_{nm}=0$, so total probability is conserved.

Bloch Equations

For a weakly damped two-level system without drive, the Bloch vector $\mathbf r=\langle\boldsymbol\sigma\rangle$ often obeys

$$\begin{aligned} \dot r_x&=-\frac{r_x}{T_2}-\omega_0 r_y,\\ \dot r_y&=\omega_0 r_x-\frac{r_y}{T_2},\\ \dot r_z&=-\frac{r_z-r_z^{\text{eq}}}{T_1}. \end{aligned}$$

The common relation among relaxation, decoherence, and pure dephasing times is

$$\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi}.$$

Here $T_1$ is population relaxation, $T_2$ is transverse coherence decay, and $T_\phi$ is pure dephasing. The relation assumes the standard weak-coupling Markovian qubit model and the conventions above.

For the driven two-level version with detuning, saturation, and fluorescence, see Optical Bloch Equations.

Damped Oscillator

For a harmonic oscillator coupled to a thermal bath with mean occupation $\bar n$, a common Lindblad master equation is

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[\hbar\omega a^\dagger a,\rho] + \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho.$$

The mean occupation evolves as

$$\frac{d}{dt}\langle a^\dagger a\rangle = -\kappa \left( \langle a^\dagger a\rangle-\bar n \right).$$

For zero temperature, $\bar n=0$, and the coherent amplitude decays as

$$\langle a(t)\rangle = e^{-i\omega t}e^{-\kappa t/2} \langle a(0)\rangle.$$

Noise Spectra

For the canonical explanation of noise-spectrum conventions and rate formulas, see Noise Spectra. For the equilibrium relation between noise and dissipative response, see Fluctuation–Dissipation Relation.

For a stationary bath operator $B(t)$, a two-sided correlation spectrum is often defined as

$$S_{BB}(\omega) = \int_{-\infty}^{\infty} dt\, e^{i\omega t} \langle B(t)B(0)\rangle.$$

The symmetrized spectrum is

$$S_{BB}^{\text{sym}}(\omega) = \frac{1}{2} \int_{-\infty}^{\infty} dt\, e^{i\omega t} \langle \{B(t),B(0)\}\rangle.$$

The retarded susceptibility is

$$\chi_{BB}^R(t) = \frac{i}{\hbar} \theta(t) \langle [B(t),B(0)]\rangle.$$

With these conventions, an equilibrium fluctuation-dissipation relation is commonly written

$$S_{BB}^{\text{sym}}(\omega) = \hbar \coth\!\left(\frac{\beta\hbar\omega}{2}\right) \operatorname{Im}\chi_{BB}^R(\omega).$$

Factor-of-$2\pi$, Fourier-transform, and susceptibility-sign conventions vary. When comparing formulas from different references, check definitions before comparing rates.

Approximation Reminders

For a system–bath Hamiltonian

$$H = H_S+H_B+H_I, \qquad H_I = \sum_\alpha A_\alpha\otimes B_\alpha,$$

the bath correlation functions are

$$C_{\alpha\beta}(t) = \operatorname{Tr}_B \!\left[ B_\alpha(t)B_\beta(0)\rho_B \right].$$

The Born Approximation treats system–bath correlations as weak enough that the bath remains approximately fixed. The Markov Approximation assumes bath correlations decay quickly compared with the system evolution being resolved. The Secular Approximation drops rapidly rotating terms so the generator becomes block diagonal in Bohr-frequency sectors and is typically completely positive in the resulting Lindblad form.

Useful time-scale inequalities are schematic, not theorems:

$$\tau_B\ll \tau_S, \qquad \tau_B\ll \Gamma^{-1}, \qquad |\omega-\omega'|^{-1}\ll \Gamma^{-1}$$

for bath correlation time $\tau_B$, relaxation rate $\Gamma$, and distinct Bohr frequencies $\omega,\omega'$. Near degeneracies, strong driving, structured reservoirs, and long bath memory can invalidate the simplest Markovian formula even when it looks algebraically reasonable.

Quick Checks

Trace preservation from Kraus operators

Show that $\Phi(\rho)=\sum_r K_r\rho K_r^\dagger$ is trace preserving if $\sum_rK_r^\dagger K_r=I$.

Solution

Use cyclicity of the trace:

$$\operatorname{Tr}\Phi(\rho) = \sum_r \operatorname{Tr}(K_r\rho K_r^\dagger) = \operatorname{Tr} \!\left[ \left(\sum_rK_r^\dagger K_r\right)\rho \right] = \operatorname{Tr}\rho.$$

POVM associated with a two-Kraus outcome

An outcome $x$ has Kraus operators $M_{x1}$ and $M_{x2}$. What effect $E_x$ gives its probability?

Solution

The probability is

$$p(x) = \operatorname{Tr} \!\left[ (M_{x1}\rho M_{x1}^\dagger +M_{x2}\rho M_{x2}^\dagger) \right].$$

Cyclicity gives

$$p(x) = \operatorname{Tr} \!\left[ (M_{x1}^\dagger M_{x1} +M_{x2}^\dagger M_{x2})\rho \right],$$

so

$$E_x=M_{x1}^\dagger M_{x1}+M_{x2}^\dagger M_{x2}.$$

T-one and T-two relation

For a qubit with $T_1^{-1}=\Gamma_\uparrow+\Gamma_\downarrow$ and pure-dephasing time $T_\phi$, what is $T_2^{-1}$ in the standard weak-coupling model?

Solution

Population relaxation contributes half its rate to transverse coherence decay, while pure dephasing contributes directly:

$$\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi} = \frac{\Gamma_\uparrow+\Gamma_\downarrow}{2} + \frac{1}{T_\phi}.$$

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer (1983).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010).
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010).
• M. Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Springer (2007).
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012).