# Mixed States

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.

## Answers and Replies

kith
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.

DrClaude
Demystifier
Gold Member
The best way to understand why this cannot be done is to try to do it on an explicit simple example. Have you tried?

kith
Demystifier
Gold Member
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.

Last edited:
Muthumanimaran and Nugatory
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.

bhobba
Mentor
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