# I Mixed States

1. Jun 2, 2016

### Muthumanimaran

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.

2. Jun 2, 2016

### 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.

3. Jun 2, 2016

### Demystifier

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. Jun 2, 2016

### Demystifier

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: Jun 2, 2016
5. Jun 2, 2016

### Muthumanimaran

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]

6. Jun 2, 2016

### Staff: 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