$\rho_0 = \lvert 0\rangle\langle 0\rvert$ is the density matrix of the zero state. As a pure state its density matrix carries no more information than the ket, but the density matrix representation is the right starting point for studying what happens to $\lvert 0\rangle$ when it couples to an environment — when the ket description breaks down.
$$\rho_0 = \lvert 0\rangle\langle 0\rvert = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$$
The diagonal entries are occupation probabilities: the qubit is in $\lvert 0\rangle$ with certainty. The off-diagonals are the coherences; both are zero because $\lvert 0\rangle$ is a computational basis state with no superposition to lose.
Every single-qubit density matrix has a Bloch vector $\vec{r}$ defined by $\rho = \tfrac{1}{2}(I + \vec{r}\cdot\vec{\sigma})$. For $\rho_0$ the vector points to the north pole:
$$\vec{r} = (0,\,0,\,1) \qquad \rho_0 = \tfrac{1}{2}(I + Z) = \begin{pmatrix}1&0\\0&0\end{pmatrix}$$
The state is pure: $\text{tr}(\rho_0^2) = 1$ and $\lvert\vec{r}\rvert = 1$. The Pauli expectation values follow from $\langle P\rangle = \text{tr}(\rho_0 P)$:
$$\langle X\rangle = 0 \qquad \langle Y\rangle = 0 \qquad \langle Z\rangle = 1$$
The dephasing channel $\mathcal{E}_\phi(\rho) = (1-p)\rho + p\,Z\rho Z$ kills off-diagonal coherences while leaving populations on the diagonal intact. Applied to $\rho_0$:
$$\mathcal{E}_\phi(\rho_0) = (1-p)\rho_0 + p\,Z\rho_0 Z = (1-p)\lvert 0\rangle\langle 0\rvert + p\lvert 0\rangle\langle 0\rvert = \rho_0$$
$\rho_0$ is invariant under dephasing. This is because $\lvert 0\rangle$ is a $Z$ eigenstate — it already has no transverse coherence to destroy. In Bloch vector terms, dephasing contracts the $x$ and $y$ components toward zero while leaving $r_z$ fixed; since $\rho_0$ has $r_x = r_y = 0$ already, nothing changes. The same holds for $\rho_1 = \lvert 1\rangle\langle 1\rvert$. Dephasing only harms superposition states like $\lvert +\rangle$ and $\lvert -\rangle$.
A bit-flip channel $\mathcal{E}_X(\rho) = (1-p)\rho + p\,X\rho X$ flips $\lvert 0\rangle \to \lvert 1\rangle$ with probability $p$, producing an incoherent mixture:
$$\mathcal{E}_X(\rho_0) = (1-p)\lvert 0\rangle\langle 0\rvert + p\lvert 1\rangle\langle 1\rvert = \begin{pmatrix}1-p & 0\\0 & p\end{pmatrix}$$
The Bloch vector contracts along the $z$-axis: $\vec{r} = (0, 0, 1-2p)$. At $p = 1/2$ the state reaches the maximally mixed state $I/2$; at $p = 1$ the qubit is flipped to $\rho_1$ with certainty.
The depolarizing channel $\mathcal{E}_{\text{dep}}(\rho) = (1-p)\rho + p\,I/2$ mixes the state uniformly with the maximally mixed state:
$$\mathcal{E}_{\text{dep}}(\rho_0) = (1-p)\lvert 0\rangle\langle 0\rvert + \frac{p}{2}I = \begin{pmatrix}1 - p/2 & 0\\0 & p/2\end{pmatrix}$$
The Bloch vector shrinks uniformly: $\vec{r} = (0, 0, 1-p)$. All three Pauli expectation values scale by $(1-p)$, so the state loses purity isotropically. At $p = 1$ the state is fully mixed: $\text{tr}(\rho^2) = 1/2$.
Amplitude damping models energy relaxation — the $T_1$ process in which the excited state $\lvert 1\rangle$ decays to the ground state $\lvert 0\rangle$. The Kraus operators are:
$$K_0 = \begin{pmatrix}1 & 0\\0 & \sqrt{1-\gamma}\end{pmatrix} \qquad K_1 = \begin{pmatrix}0 & \sqrt{\gamma}\\0 & 0\end{pmatrix}$$
Applied to $\rho_0$: $K_0\rho_0 K_0^\dagger = \rho_0$ and $K_1\rho_0 K_1^\dagger = 0$, so $\mathcal{E}_{AD}(\rho_0) = \rho_0$.
$\rho_0$ is the unique fixed point of amplitude damping. Any initial state converges to $\rho_0$ under repeated application — it is the thermal ground state at zero temperature. This is why $\lvert 0\rangle$ is the standard reset target in quantum hardware: active cooling or measurement followed by conditional flip drives every qubit to $\rho_0$.