The short answer is that there is more to quantum information than "uncertainty". This is because there is more than one way to measure a state; and *that* is because there is more than one basis in which, in principle, you can store and retrieve information. Superpositions allow you to express information in a different basis than the computational basis — but **mixtures** describe the presence of a probabilistic element, no matter which basis you use to look at the state.

The longer answer is as follows —

Measurement as you have described it is specifically measurement in the computational basis. This is often described just as "measurement" for the sake of brevity, and large subsets of the community think in terms of this being the primary way to measure things. But in many physical systems, it is possible to *choose a measurement basis*.

A vector space over $\mathbb C$ has more than one basis (even more than one orthonormal basis), and on a mathematical level there isn't much that makes one basis more special than another, aside from what is convenient for the mathematician to think about. The same is true in quantum mechanics: unless you specify some specific dynamics, there is no basis which is more special than the others. That means that the computational basis
$$ \lvert 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \lvert 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
is not fundamentally different physically from another basis such as
$$ \lvert + \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \lvert - \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ -1 \end{bmatrix},$$
which is also an orthonormal basis. That means that there should be a way to "measure" a state $\lvert \psi \rangle \in \mathbb C^2$ in such a way that the probabilities of the outcomes depend on projections onto these states $\lvert + \rangle$ and $\lvert - \rangle$.

In some physical systems, the way one performs this measurement is to literally take the same apparatus and tilt it so that it is aligned with the X axis instead of the Z axis. Mathematically, the way we do this is to consider the projectors
$$ \Pi_+ = \lvert + \rangle\!\langle + \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \qquad \Pi_- = \lvert - \rangle\!\langle - \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$
and then to ask what the projections $\lvert \varphi_+ \rangle := \Pi_+ \lvert \psi \rangle$ and $\lvert \varphi_- \rangle := \Pi_- \lvert \psi \rangle$. The norm-squared of $\lvert \varphi_\pm \rangle$ determines the probability of "measuring $\lvert + \rangle$" and of "measuring $\lvert - \rangle$"; and normalising $\lvert \varphi_+ \rangle$ or $\lvert \varphi_- \rangle$ to have a norm of 1 yields the post-measurement state. (For a state on a single qubit, this will just be $\lvert + \rangle$ or $\lvert - \rangle$. More interesting post-measurement states may result if we consider multi-qubit states, and consider the projector $\Pi_+$ or $\Pi_-$ acting on one of many qubits.)

For density operators, one takes the state $\rho$ which you want to perform a measurement on, and consider $\rho_+ := \Pi_+ \rho \Pi_+$ and $\rho_- := \Pi_- \rho \Pi_-$. These operators may be sub-normalised in the same way that the states $\lvert \varphi_\pm \rangle$ might be, in the sense that they may have trace less than 1. The value of the trace of $\rho_\pm$ is the probability of obtaining the outcome $\lvert + \rangle$ or $\lvert - \rangle$ of the measurement; to renormalise, simply scale the projected operator to have trace 1.

Consider your state $\rho_2$ above. If you measure it with respect to the $\lvert \pm \rangle$ basis, what you will find is that $\rho_2 = \rho_{2,+} := \Pi_+ \rho_2 \Pi_+$. This means that projecting the operator with $\Pi_+$ does change the state, and that the probability of obtaining the outcome $\lvert + \rangle$ to the measurement is 1. If you do this instead with $\rho_1$, you will find a 50/50 chance of obtaining either $\lvert + \rangle$ or $\lvert - \rangle$. So the state $\rho_1$ is a mixed state, while $\rho_2 $ is not --- the difference being that $\rho_2$ has a definite outcome in a *different* measurement basis than the standard basis. You might say that $\rho_2$ stores a *definite* piece of information, albeit in a different basis than the computational basis.

More generally, a mixed state is one whose largest eigenvalue is less than 1, meaning that there is no basis in which you can measure it to get a definite outcome. Superpositions allow you to express information in a different basis than the computational basis; mixtures represent a degree of randomness about the state of the system you're considering, regardless of how you measure that system.

4

Possible duplicate of What's the difference between a pure and mixed quantum state?

– Mithrandir24601 – 2018-03-29T21:00:30.187This is probably also helpful: Physics SE: How is quantum superposition different from mixed state?

– v.tralala – 2019-02-16T15:35:20.233