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Entanglement monogamy limits how strongly one subsystem can be entangled with several independent partners.

Quantum binary hypothesis testing chooses a two-outcome measurement to distinguish two candidate density operators, trading the probabilities of the two kinds of error.

For ensemble $\{p_j,{\rho }_j\}$ with average $\bar{\rho }={\sum }_j{p}_j{\rho }_j$, the pretty good measurement has $M_j={\bar{\rho }}^{-1/2}{p}_j{\rho }_j{\bar{\rho }}^{-1/2}$ on the support of $\bar{\rho }$.

A bipartite state is separable when it is a convex combination ${\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$ of product states; otherwise it is entangled.

A quantum channel is a linear completely positive trace-preserving map between operator algebras.

When $E-\mu \gg k_{B}T$, the additive $\mp 1$ is negligible and both quantum distributions reduce to the Maxwell--Boltzmann factor $e^{-(E-\mu )/(k_{B}T)}$.

The mean occupation of a fermionic one-particle state of energy $E$ isAt zero temperature this becomes the step function $1_{E<E_F}$.

The density of states $g(E)$ is defined so that $g(E)dE$ counts one-particle states with energies in an interval of width $dE$. A sum over closely spaced states then becomes an integral weighted by $g(E)$.

The mean occupation of a bosonic one-particle state of energy $E$ is

When $E-\mu \gg k_{B}T$, either quantum occupation law reduces to the classical factor

For independent frequencies $1$ and $2$,so the eigenspace at energy $k+3/2$ has one basis state $|n,m⟩$ for each nonnegative solution of $n+2m=k$.

The bosonic ladder operators satisfy

When two masses occupy spatial superpositions, their branch-dependent Newtonian gravitational potential energy produces relative phases. If those phases cannot be separated into one phase for each mass, the interaction turns an initial product state into an entangled state.

If $U|\psi ⟩=e^{2\pi ij/N}|\psi ⟩$, controlled powers of $U$ applied to a uniform $N$-state control register produceThese states are the Fourier basis. Applying the inverse quantum Fourier transform and measuring returns $j$ with probability one.

For the cyclic shift $S|j⟩=|j+1\mathrm{mod}N⟩$ and the conventionthe Fourier states obeyConsequently $S={\mathrm{QFT}}_{N}D{\mathrm{QFT}}_N^{-1}$, where $D|k⟩=e^{-2\pi ik/N}|k⟩$.

For a function of period $r$ dividing $N$, prepare a uniform superposition of inputs, evaluate the function into a second register, and measure that register. The first register becomes a periodic coset state. Applying the quantum Fourier transform and measuring gives

Let $r$ divide $N$ andThe geometric sum in the quantum Fourier transform vanishes unless $c$ is a multiple of $N/r$. Thus measurement after the transform returns the $r$ valueswith equal probabilities $1/r$.

The root-of-unity filter givesThus ${\mathrm{QFT}}_N^4=I$, while the square is an involution whose eigenvalues lie in $\{1,-1\}$.

Without a preferred timelike Killing flow there is no canonical positive-frequency splitting. Different choices are related by a Bogoliubov transformation and generally define different particle notions and vacuum states.

Relative to a timelike Killing field $K$, a positive-frequency solution obeys $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. Such solutions define the one-particle space and annihilation operators of a stationary vacuum.