More questions about 2 state system measurements

[ianmgull]

Homework Statement

Homework Equations

I know that there are two eigenstates of the operator C:

|B> = (1 0) as a column vector with eigenvalue 1
|R> = (0 1) also a column vector with eigenvalue -1

The Attempt at a Solution

I'm attempting to solve part c (second image).

My initial thought was as follows:

Measuring the state, and getting a result of red means: C|psi> = -|psi> because -1 is the eigenvalue corresponding to red. I would then perform another measurement on the new state (with the negative sign in front) and get a measurement value corresponding to blue.

However this doesn't seem to work. I'm a little unsure if I'm supposed to think about a measurement as listed above, or if I'm supposed to do <psi|C|psi> instead. I'm also not sure if I should be using the projection operator instead of the color operator C.

Any advice would be greatly appreciated.

Thanks.

 

[ianmgull]

My concern is that I don't know if psi is given as an eigenstate of the operator C, so I don't know if I can assume that a measurement will return one of the two eigenvalues on psi.

 

[vela]

Measuring the state, and getting a result of red means: C|psi> = -|psi> because -1 is the eigenvalue corresponding to red. I would then perform another measurement on the new state (with the negative sign in front) and get a measurement value corresponding to blue.

A common misconception is to equate making a measurement with applying the observable to the state of the system. That's what you're doing here.

A measurement always results in one of the eigenvalues of the observable. After the measurement, the system is in an eigenstate of the observable corresponding to the result of the measurement.

 

[ianmgull]

Thanks for the reply. I'm not sure I understand though.

Is it not true that -|psi> is representative of the state after a measurement of down (or Red in this case)?

 

[vela]

No, that's not correct. Review postulate 3 on this page: http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html

 

[ianmgull]

What I've gathered from postulate 3, is that a measurement produces an eigenvalue (-1 in this case). After the measurement is made, the wavefunction collapses onto one of the eigenstates of the operator.

This being the case, is it true that after the measurement is made, that the wavefunction is equal to the eigenstate corresponding to Red?

 

[vela]

The terms eigenvalue and eigenvector are not synonymous. One is a scalar; the other is a vector.

If what you meant was that the system's state immediately after the measurement is an eigenstate of the observable corresponding to the outcome of the measurement, then that's correct.

 

[ianmgull]

Yes, that's what I meant (I edited my post).

In the original question, psi is the state of two entangled particles. Does that mean that after the measurement of Red is obtained (for the first particle), that psi now looks like the following?

|psi>= (1/sqrt2)(|R>|R> - |R>|B>)

So the measurement was performed on the first particle, and a value corresponding to the eigenstate |R> was obtained for that particle, so the total state can be written where the first particle is in this eigenstate?

 

[vela]

No, you just keep the part of the original state that is consistent with the outcome of the measurement. In this case, you'd just throw out the first term (and then normalize what's left).

 

[ianmgull]

So |psi>= a(|R>|B>) where a is a normalization constant?

If so, what is the justification for dropping the first term?

 

[vela]

It's one of the postulates of quantum mechanics.

 

[ianmgull]

Ok. Thank you for your help.