"A pure state is the quantum state where we have exact information about the quantum system. And the mixed state is the combination of probabilities of the information about the quantum state ... different distributions of pure states can generate equivalent mixed states. I did not understand how a combination of exact information can result in the combination of probabilities.".

On a Bloch sphere, pure states are represented by a point on the surface of the sphere, whereas mixed states are represented by an interior point. The completely mixed state of a single qubit ${{\frac {1}{2}}I_{2}\,}$ is represented by the center of the sphere, by symmetry. The purity of a state can be visualized as the degree in which it is close to the surface of the sphere.

In quantum mechanics, the state of a quantum system is represented by a state vector (or ket) $| \psi \rangle$. A quantum system with a state vector $| \psi \rangle$ is called a pure state. However, it is also possible for a system to be in a statistical ensemble of different state vectors: For example, there may be a 50% probability that the state vector is $| \psi_1 \rangle$ and a 50% chance that the state vector is $| \psi_2 \rangle$.

This system would be in a mixed state. The density matrix is especially useful for mixed states, because any state, pure or mixed, can be characterized by a single density matrix.

**Mathematical description**

The state vector $|\psi \rangle$ of a pure state completely determines the statistical behavior of a measurement. For concreteness, take an observable quantity, and let A be the associated observable operator that has a representation on the Hilbert space ${\mathcal {H}}$ of the quantum system. For any real-valued, analytical function $F$ defined on the real numbers, suppose that $F(A)$ is the result of applying $F$ to the outcome of a measurement. The expectation value of $F(A)$ is

$$\langle \psi | F(A) | \psi \rangle\, .$$

Now consider a mixed state prepared by statistically combining two different pure states $| \psi \rangle$ and $| \phi\rangle$, with the associated probabilities $p$ and $1 − p$, respectively. The associated probabilities mean that the preparation process for the quantum system ends in the state $|\psi \rangle$ with probability $p$ and in the state $|\phi\rangle$ with probability $1 − p$.

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This has been discussed many times on physics.SE, e.g. How is quantum superposition different from mixed state?.

– glS – 2018-03-28T09:31:58.247