Measurement query

1. May 31, 2005

James Jackson

Just working through some more quantum information stuff, and have come accross a stumbling block - I'm clearly missing something obvious.

Consider a source emits states $|\Phi\rangle = \cos\theta |0\rangle + e^{i\phi}\sin\theta |1\rangle$ with fixed $\theta$ and random phases $\phi$, with equal probability for each phase.

How can I show that a measurement of the operator Z ($Z|0\rangle = |0\rangle , Z|1\rangle = -|1\rangle$) doesn't yield any information about the state emitted by the source?

2. May 31, 2005

seratend

Just by computing the probability values: P(Z=0,|psi>)=<psi||0><O||psi> and P(Z=-1,|psi>)=<psi||1><1||psi> (<0|1>=0)

Seratend.

3. May 31, 2005

James Jackson

Thanks, I was pretty sure it was something simple I was overlooking - wood for the trees and all that!

This leads on to another measurement question: Suppose a source emits two states:

$$|\Phi_1\rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\phi}|1\rangle)$$
$$|\Phi_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle - e^{i\phi}|1\rangle)$$

Where $\phi$ is an arbitary fixed phase. What measurement can be used to distinguish between the two states? They form an orthonormal set, so clearly can be distinguished, I just can't see 'how' to measure them.

4. May 31, 2005

seratend

Just define another observable Z' with the good set of eigenvectors. I think now you are able to guess what vectors you have to choose.

Seratend.

5. May 31, 2005

James Jackson

Ah of course. I was trying to express the states as linear combinations of the eigenvectors of 'standard' operators rather than define my own.

Thanks for the pointers.

6. May 31, 2005

James Jackson

Just to confirm, would it be correct to define the measurement

$$M=|\Phi_1\rangle\langle\Phi_1 |-|\Phi_2\rangle\langle\Phi_2 |$$ to measure the states above?

(Edited to change + to -)

Last edited: May 31, 2005
7. May 31, 2005

seratend

Yes for the eigenvalues +1 and -1 (but you are free to select others).

Seratend.