# 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.

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]