# Measurements of spin

1. May 8, 2010

### barnflakes

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?

2. May 10, 2010

### Edgardo

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.

3. May 10, 2010

### Fredrik

Staff Emeritus
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.

4. May 10, 2010

### barnflakes

Yes exactly this!

5. May 10, 2010

### 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)