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The density matrix $\rho = \frac{1}{2}(|0\rangle \langle0|+|1\rangle \langle1|)$ describes a system which is in state $|0\rangle$ and in $|1\rangle$ with equal probability.
Also the state $|\psi\rangle= \frac{1}{\sqrt2}(|0\rangle+|1\rangle)$ describes a system which is in state $|0\rangle$ and in $|1\rangle$ with equal probability. But the density matrix corresponding to $|\psi\rangle$ is not $\rho$, there are some cross terms $(|0\rangle \langle1|,|1\rangle \langle0|)$. So how should we explain this?

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The two states $\rho$ and $|\psi\rangle$ describe quite different things. $\rho$ is, in essence, something classical, while $|\psi\rangle$ is a truly quantum state.

The clearest way to see the different is to do an experiment on these two cases. First, apply a Hadamard gate, and measure in the standard basis. For the mixed state, $$ \rho\xrightarrow{H}\rho $$ This has 50:50 outcomes on measurement. For the pure state, $$ |\psi\rangle\xrightarrow{H}|0\rangle $$ always gives the answer $0\rangle$.

So, we have an experiment that clearly distinguishes these two cases. They are different things.

Moral of the story: where quantum is concerned, it is insufficient to only think about probabilities in the computational basis. It is the coherences between those basis states that are the essential element of quantum behaviour.

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  • $\begingroup$ That's a good example, can you share any resource where more discussion is given on this? Thank you. $\endgroup$ Mar 4 at 10:48
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Just as a matter of clarifying terminology, as well as complementing the answers given so far; both are normally called states. The difference is just that one is a pure state, which is the same as a vector state, while the other is a mixed state, exactly because it can be understood as the convex mixture of two states. Pure states cannot be a convex mixture of any other state by definition.

I wouldn't particularly say that $\vert \psi \rangle \equiv \vert + \rangle$ is a truly quantum state because we know that there are many classical models that reproduce the statistics and the phenomenology of states as such: Ref1, Ref2. Nevertheless, it is a coherent state with respect to the computational basis, so it does particularly have some nonclassical aspect attached to it.

The state $\rho$ can be understood as simply preparing either the states $\vert 0 \rangle \langle 0 \vert$ or $\vert 1 \rangle \langle 1 \vert$ with equal probability, while the same kind of procedure cannot be used to prepare the states $\vert + \rangle \langle + \vert$ because they are, by definition, not convex mixtures of anyone else.

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The off-diagonal terms in your case, tell us how much $|1\rangle$ state is there together with $|0\rangle$ to in the same time (superposition).

Since the first case (mixed state) doesnt contain states from different bases together (in superposition), the off-diagonal terms are zero.

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