Transformation of the density operator with partial decoherence I'm trying to understand the transformation of a quantum state with partial decoherence.
An article says that if $\rho = \left| \Psi \right>\left< \Psi \right| $ is the density operator of the state $\left| \Psi \right>$, a partial decoherence transforms $\rho$ into $\rho^D =(1-p)\rho + p\cdot\text{diag}(\rho)$, where $0\leq p \leq 1$ could be understood as the probability of measure the state (so $p=0$ means no decoherence and $p=1$ means total decoherence).
My first question is how the state (and not the density operator) is transformed. I suppose that I can do the inner product so $\left| \Psi \right>$ transforms into $\left| \Psi \right>^D= (1-p)\left| \Psi \right>+p\cdot \text{diag}(\left| \Psi \right>\left< \Psi \right|)\left| \Psi \right>$. Is this correct?
My second question is what does total decoherence means. As far as I know, total decoherence appears when you measure a quantum state and then it becomes a classical state. But for $p=1$, the transformation is $\rho^D_{p=1} = \text{diag}(\rho)$. Why is this equivalent?
 A: The form written in your article says that, with probability $p$, the state loses all of the off-diagonal elements in some particular basis; i.e., the state loses all "coherences" in a particular basis with probability $p$. The remaining probability $1-p$ leaves the state unchanged.
You cannot answer the question in terms of pure states, because this process inherently makes a state more mixed! That is why there is no final pure state $|\Psi\rangle^D$ with which you can express your final result. Both pure and mixed states are states; I urge you to avoid referring only to pure states like $|\psi\rangle$ as a state, because density operators $\rho$ are states, too. So the first answer is no, this is not correct.
Total decoherence in this case means that the state transforms into a mixed state with no off-diagonal elements ("coherences"). This means that it behaves exactly the same as a classical (non-interfering, probabilistic) mixture of the basis states, with no possible interference effects arising if you measure the state in some superposition basis.

If this is confusing to you, you should read more about the difference between a superposition and a mixture. An initial pure state might be $$|\psi\rangle=\sqrt{q}|\uparrow\rangle+e^{i\phi}\sqrt{1-q}|\downarrow\rangle.$$ After undergoing complete decoherence in this $\uparrow/\downarrow$ basis, the state becomes mixed:
$$\rho^D_{p=1}=q|\uparrow\rangle\langle\uparrow|+(1-q)|\downarrow\rangle\langle\downarrow|.$$ All of the phase information $\phi$ is lost in the decohered state. If you measure either state in the $\uparrow/\downarrow$ basis, you will get the same probability distribution: $q$ and $1-q$. But if you measure in some other basis, the statistics will be different. For example, if you measure in the $$|\pm\rangle=\frac{|\uparrow\rangle\pm|\downarrow\rangle}{\sqrt{2}}$$ basis, the pure superposition state will have probabilities
$$P_\pm(|\psi\rangle)=\left|\sqrt{q}\pm\sqrt{1-q}e^{i\phi}\right|^2/2=(1\pm 2\sqrt{q}\sqrt{1-q}\cos\phi)/2$$ while the mixed (classical-like) state will show no interference
$$P_\pm(\rho^D_{p=1})=\frac{q}{2}+\frac{1-q}{2}=\frac{1}{2}.$$
Another way of looking at this is expressing everything as a matrix in the $\uparrow/\downarrow$ basis and seeing that the decoherence simply decreases the magnitudes of the off-diagonal elements:
$$|\psi\rangle\langle\psi|=\begin{pmatrix}q&\sqrt{q}\sqrt{1-q}e^{i\phi}\\\sqrt{q}\sqrt{1-q}e^{-i\phi} &1-q\end{pmatrix}$$ transforms into
$$\rho^D=\begin{pmatrix}q&(1-p)\sqrt{q}\sqrt{1-q}e^{i\phi}\\(1-p)\sqrt{q}\sqrt{1-q}e^{-i\phi} &1-q\end{pmatrix}.$$ The off-diagonal terms shrink with increasing $p$. Again, "decoherence" is always with respect to a particular basis.
