Pseudo-pure state
= Pseudo-pure state
{title2=$\rho=(1-p)I/d+p|\psi\rangle\langle\psi|$}
A <pseudo-pure state> is a <density matrix> of the form
$$
\rho=(1-p)\frac{I}{d}+p|\psi\rangle\langle\psi|,
$$
where $|\psi\rangle$ is a normalized <pure state>, $I$ is the <identity matrix> on a $d$-dimensional space and here $0\leq p\leq1$. The term $I/d$ is the <maximally mixed state>. Under a <unitary transformation> $U$, this term stays fixed and the selected <pure state> becomes $U|\psi\rangle$. This is why <liquid-state NMR quantum computing> can follow pure-state-like transformations in part of a highly mixed ensemble.<Ensemble Quantum Computing by NMR Spectroscopy (MIT project introduction)>[Ensemble Quantum Computing, definition and transformation properties of pseudo-pure states]{ref}
For a <traceless matrix> representing an <observable> $O$, direct evaluation of the <trace> gives
$$
\operatorname{tr}(\rho O)=p\langle\psi|O|\psi\rangle.
$$
The identity contribution vanishes because $\operatorname{tr}O=0$. Thus $p$ scales the detectable expectation. It is not itself the <purity>: expanding $\rho^2$ and taking its <trace> yields
$$
\operatorname{tr}\rho^2=\frac1d+\left(1-\frac1d\right)p^2.
$$
For 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 $0\leq\eta\leq1$ 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 $d=2^n$ and
$$
p=\frac{da-1}{d-1}=\frac{(1+\eta)^n-1}{2^n-1}.
$$
Indeed the selected population of $\rho$ is $[1+(d-1)p]/d=a$. In the weak-polarization regime $n\eta\ll1$, the <binomial theorem> gives $p\simeq n\eta/(2^n-1)$. If the <energy> separation is $\hbar\omega$, set $\epsilon=\hbar\omega/(k_B T)$; a thermal two-level system has $\eta=\tanh(\epsilon/2)\simeq\epsilon/2$. This gives $p\simeq n\epsilon/[2(2^n-1)]$, with every parameter defined. Here $\hbar$ 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.