Mixed States

  • #1
Why cannot we represent mixed states with a ray in a Hilbert space like a Pure state.
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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  • #2
kith
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Because the statistics don't work out.

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.
 
  • #3
Demystifier
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The best way to understand why this cannot be done is to try to do it on an explicit simple example. Have you tried?
 
  • #4
Demystifier
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why we can't represent mixed states as ray or superposition of rays in Hilbert space.
Algebraically, it is closely related to the following claim. Not every matrix ##C_{ij}## can be written as ##A_iB_j##.

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
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.[/QUOTE]

Thank you, your answer is simple and elegant.
 
  • #6
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No need to prove anything. Its simple.

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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