- #1

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I know Mixed states corresponds to statistical mixture of pure states, If we are able to represent Pure state as a ray in Hilbert space, why we can't represent mixed states as ray or superposition of rays in Hilbert space.

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- Thread starter Muthumanimaran
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- #1

- 81

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I know Mixed states corresponds to statistical mixture of pure states, If we are able to represent Pure state as a ray in Hilbert space, why we can't represent mixed states as ray or superposition of rays in Hilbert space.

- #2

kith

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If you have a spin-1/2 particle in a given pure state, the outcome of a spin measurement depends on the direction you measure. For example, there's always a direction where you get one of the outcomes with certainty. For mixed states, such a direction doesn't exist.

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- #4

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Algebraically, it is closely related to the following claim. Not every matrix ##C_{ij}## can be written as ##A_iB_j##.why we can't represent mixed states as ray or superposition of rays in Hilbert space.

How to prove it? By counterexample. Assume ##C_{ij}=A_iB_j##. As an example, consider the case ##C_{12}=0##. Then either ##A_1=0## or ##B_2=0##. But if ##A_1=0## then ##C_{11}=0##, and if ##B_2=0## then ##C_{22}=0##. Therefore the assumption cannot be satisfied when ##C_{12}=0##, ##C_{11}\neq 0##, and ##C_{22}\neq 0##. Q.E.D.

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- #5

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Thank you, your answer is simple and elegant.

- #6

bhobba

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By definition quantum states are positive operators of unit trace. Forget this Hilbert space stuff - that's the kiddy version. By definition pure states are operators of the form |u><u|. Mixed states, again by definition, are convex sums of pure states. It can be shown all states are mixed or pure (start a new thread if you want to discuss it). Obviously pure states can be mapped to rays in a Hilbert space - but keep in mind in general states are operators.

Thanks

Bill

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