Measurements of spin

[barnflakes]

I read in a book:

For a qubit defined as: \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)

Since |0\rangle and |1\rangle are the eigenstates of \sigma_z then measuring sigma_z will yield either |0\rangle or |1\rangle. Measuring \sigma_x on the same qubit will give one of the eigenstates of \sigma_x, which are \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle).

Only problem is I don't see how you can obtain these general/algebraic states? When I make a measurement on the qubit ie. ((\frac{1}{\sqrt{2}}(\langle 0| +\langle1|))\sigma_x(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)) I just obtain a number? How can I keep this algebraic structure and prove the above?

[Edgardo]

What exactly is your question?

In the last example you have calculated \langle \psi | \sigma_x | \psi \rangle which is the expectation value of \sigma_x for the state |\psi \rangle. The expectation value is a number.

[Fredrik]

The question appears to be "How do you express the eigenvectors of \sigma_x as linear combinations of eigenvectors of \sigma_z?". I don't have time to answer that right now, so I'll leave it for someone else.

[barnflakes]

The question appears to be "How do you express the eigenvectors of \sigma_x as linear combinations of eigenvectors of \sigma_z?". I don't have time to answer that right now, so I'll leave it for someone else.

Yes exactly this!

[kanato]

The representation of \sigma_x in the basis of the eigenvectors of \sigma_z is
\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]
So if you diagonalize that you will get the eigenvectors, which will be (1,1) and (1,-1). The eigenvectors for \sigma_z in this basis are of course (1,0) and (0,1)