# 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.

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