Representing Mixed States in Hilbert Space

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SUMMARY

Mixed states in quantum mechanics cannot be represented as rays in Hilbert space, unlike pure states, due to their statistical nature. Pure states correspond to specific rays, while mixed states are defined as convex combinations of pure states, which cannot be expressed as a single ray or superposition of rays. The discussion highlights that quantum states are positive operators of unit trace, with pure states represented as |u> PREREQUISITES

  • Understanding of quantum mechanics concepts, specifically pure and mixed states.
  • Familiarity with Hilbert space and its representation of quantum states.
  • Knowledge of positive operators and their role in quantum state representation.
  • Basic algebraic manipulation of matrices and operators.
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  • Study the mathematical foundations of Hilbert space in quantum mechanics.
  • Explore the properties of positive operators and their significance in quantum theory.
  • Learn about the implications of convex combinations in quantum state representation.
  • Investigate counterexamples in linear algebra related to matrix representations.
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Quantum physicists, students of quantum mechanics, and researchers interested in the mathematical foundations of quantum state representation.

Muthumanimaran
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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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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.
 
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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?
 
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Muthumanimaran said:
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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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.
 
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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