# What's the difference between a pure and mixed quantum state?

As per my limited understanding, 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 of the quantum system.

However, it is mentioned that different distributions of pure states can generate equivalent mixed states. So how a combination of exact information can result in the combination of probabilities?

"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 $1 − p$.

A pure state is what one would naturally call a state of a system. Now imagine you have a qubit in a certain state, say the equal superposition of both its computational basis states, which is $\frac{\sqrt{2}}{2} \left( \left|0\right> + \left|1\right> \right)$. Then measure it in the computational basis. What state do you get as a result?

If you read the measurement result, you know which state you have. But if you discard that result, then you don't know which state the system is in (either it is in $\left|0\right>$ or in $\left|1\right>$). This is different from the superposition you had before (which was a pure state): It is a mixed state.

Pure state: Systems whose state is unequivocally defined by A state vector in other words, single state vector. And this has the complete information about the system.

Mixed state: System whose state cannot be defined unequivocally by single state vector. It only has limited or no knowledge about the state of the system.

In reality, we often deal with ensemble of systems and we repeat the experiment. In such, cases it might be difficult to prepare the system exactly the same way to any particular initial state. Under such scenario, mixed state come in handy.

• The claim that the state vector of a system contains the "complete information" about the system is only true under certain interpretations of quantum mechanics (e.g. it is false in de Broglie-Bohm theory). In any case it's a bit misleading - you certainly can't predict the result of arbitrary measurements, which is how I would interpret "complete information". – tparker Mar 28 '18 at 15:15

In density matrix formalism, this is a pure state $$\lvert 0 \rangle \langle 0 \vert = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$$. There is nothing but $$\lvert 0 \rangle$$.

This is a mixed state $$.5 \times \lvert 0 \rangle \langle 0 \vert + .5 \times \lvert 1 \rangle \langle 1 \vert = \begin{bmatrix}.5 & 0 \\ 0 & .5\end{bmatrix}$$. There is 50% chance that the state 'was' in $$\lvert 0 \rangle$$ and 50% chance that the state 'was' in $$\lvert 1 \rangle$$.

• The answer might give an impression that a superposition of up and down states also is a mix states. – dushyanth Oct 2 at 18:31
• you're right.. i'm changing it – Hasan Iqbal Oct 2 at 21:18
• Thanks for the edit. But I'm afraid you didn't get my question right and there still is same amount of room for confusion that the probabilistic nature is what seperates a mix state from pure state. That's why I was saying that a superposition state might be mistaken for a mixed state. 1/√2*(|0> + |1>) has 50% probability of being in |0> and 50% to be in |1>but it's not a mix state, but from your answer one might assume it is. – dushyanth Oct 3 at 22:33