Examples where mixed states are eigenstates

[oquen]

I have actually read so much about density matrix and eigenstates today. I just want to know what particular situations when mixed states are eigenstates. Can this occur?

Mixed states and eigenstates have one thing in common.. they have a value.. but I know mixed states aren't eigenstates.. what are the exceptions?

Just one paragraph of accurate reply is enough to guide me to many reading. Thank you.

 

[blue_leaf77]

Mixed state is not described by a vector, in fact it's an operator.

they have a value.

What kind of value?

 

[oquen]

Mixed state is not described by a vector, in fact it's an operator.

What kind of value?

When entangled pair is sent to Alice and Bob.. Alice can measure values of position or momentum... and having values of position and momentum are said to be in eigenstates. hence my confusion all day today.

 

[PeterDonis]

Can this occur?

No.

having values of position and momentum are said to be in eigenstates.

That depends on the interpretation of QM you are using. In a collapse interpretation, measuring an observable (position or momentum or anything else) collapses the measured system into an eigenstate of that observable (the one corresponding to the value that is measured). But in a no collapse interpretation, like the MWI, this is not the case.

 

[oquen]

No.
That depends on the interpretation of QM you are using. In a collapse interpretation, measuring an observable (position or momentum or anything else) collapses the measured system into an eigenstate of that observable (the one corresponding to the value that is measured). But in a no collapse interpretation, like the MWI, this is not the case.

So what is the relationship of mixed state to eigenstates? Again going to alice and bob.. when entangled pair is sent to Alice and Bob.. Alice can measure values of position or momentum... isn't this the same as saying the mixed state Alice measures is in an eigenstate? But you disagree mixed states can't be eigenstates.

Or is the following the answer. Mixed state only produce statistics.. so when Alice measures the subsystem.. she gets a mixed state statistically and an Eigenstate of value? Is this it?

 

[A. Neumaier]

Mixed states and eigenstates have one thing in common.. they have a value.. but I know mixed states aren't eigenstates.. what are the exceptions?

In a mixed state with density operator $\rho$, the hermitian operator $A$ has a definite value $\alpha$ iff $A\rho=\alpha\rho$. This is the correct form of the eigenstate condition for density operators.

This happens regularly when a state is modeled in a more complete fashion than just with one degree of freedom.

For example, an electron in a Stern-Gerlach experiment is typically in a mixed state with respect to position, but may still have a definite spin up. In the simple textbook description, only the spin degree of freedom is considered, and the electron appears to be in a pure state.

Similarly, experiments with photons are often in a mixed (thermal) state with respect to polarization, but for photons on demand, the number operator has the definite value 1.

 

[oquen]

No.
That depends on the interpretation of QM you are using. In a collapse interpretation, measuring an observable (position or momentum or anything else) collapses the measured system into an eigenstate of that observable (the one corresponding to the value that is measured). But in a no collapse interpretation, like the MWI, this is not the case.

Are you saying that in MWI, mixed state where born rule applied has a world/branch whereas in collapse interpretation, mixed state where born rule applied doesn't necessarily produce any collapse at all? and the only way for collapse to occur in the latter is when it is in an eigenstate of that observable?

 

[PeterDonis]

Are you saying that in MWI, mixed state where born rule applied has a world/branch whereas in collapse interpretation, mixed state where born rule applied doesn't necessarily produce any collapse at all? and the only way for collapse to occur in the latter is when it is in an eigenstate of that observable?

I'm not sure what this means. I think you need to spend some time with a QM textbook; your background does not seem to be sufficient for an "I" level discussion.

Thread closed.