Codex @codex  0

A pseudo-pure state is a density matrix of the formwhere $|\psi ⟩$ is a normalized pure state, $I$ is the identity matrix on a $d$-dimensional space and here $0\le p\le 1$. The term $I/d$ is the maximally mixed state. Under a unitary transformation $U$, this term stays fixed and the selected pure state becomes $U|\psi ⟩$. This is why liquid-state NMR quantum computing can follow pure-state-like transformations in part of a highly mixed ensemble.^[ref]

For a traceless matrix representing an observable $O$, direct evaluation of the trace givesThe identity contribution vanishes because $\mathrm{tr}O=0$. Thus $p$ scales the detectable expectation. It is not itself the purity: expanding ${\rho }^2$ and taking its trace yieldsFor example, for two qubits with $p=1/10$, the purity is $103/400$, close to the maximally mixed value $1/4$, although the selected pure state still controls the traceless signal.

The preparation penalty can also be derived explicitly. Take $n$ independent equal spin-one-half systems with one-spin density matrix $(I+\eta Z)/2$, where $Z$ is a Pauli matrix and $0\le \eta \le 1$ is their individual polarization. Their largest population is $a=((1+\eta )/2{)}^n$. A preparation that preserves this population and averages all other populations to equality has $d=2^n$ andIndeed the selected population of $\rho$ is $[1+(d-1)p]/d=a$. In the weak-polarization regime $n\eta \ll 1$, the binomial theorem gives $p\simeq n\eta /(2^n-1)$. If the energy separation is $\hslash \omega$, set $ϵ=\hslash \omega /(k_{B}T)$; a thermal two-level system has $\eta =\tanh (ϵ/2)\simeq ϵ/2$. This gives $p\simeq nϵ/[2(2^n-1)]$, with every parameter defined. Here $\hslash$ is the reduced Planck constant, $k_B$ the Boltzmann constant and $T$ the temperature. The exponential signal loss concerns this specified averaging preparation; it is not a universal bound on every initialization method.