# Mixed states, 4

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How is quantum superposition different from mixed state?

The state

$\displaystyle{|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right)}$

is a pure state. Meaning, there’s not a 50% chance the system is in the state $\displaystyle{|\psi_1 \rangle }$ and a 50% it is in the state $\displaystyle{|\psi_2 \rangle}$. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state $\displaystyle{|\Psi \rangle}$.

The point is that these statements are all made before I make any measurements.

— edited Jan 20 ’15 at 9:54

— answered Oct 12 ’13 at 1:42

— Andrew

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Given a state, mixed or pure, you can compute the probability distribution $\displaystyle{P(\lambda_n)}$ for measuring eigenvalues $\displaystyle{\lambda_n}$, for any observable you want. The difference is the way you combine probabilities, in a quantum superposition you have complex numbers that can interfere. In a classical probability distribution things only add positively.

— Andrew Oct 12 ’13 at 14:41

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— How is quantum superposition different from mixed state?

— Physics StackExchange

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2019.04.23 Tuesday ACHK