- #1
barnflakes
- 156
- 4
I read in a book:
For a qubit defined as: [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex]
Since [tex]|0\rangle[/tex] and [tex]|1\rangle[/tex] are the eigenstates of [tex]\sigma_z[/tex] then measuring sigma_z will yield either [tex]|0\rangle[/tex] or [tex]|1\rangle[/tex]. Measuring [tex]\sigma_x[/tex] on the same qubit will give one of the eigenstates of [tex]\sigma_x[/tex], which are [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex] and [tex]\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)[/tex].
Only problem is I don't see how you can obtain these general/algebraic states? When I make a measurement on the qubit ie. ([tex](\frac{1}{\sqrt{2}}(\langle 0| +\langle1|))\sigma_x(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle))[/tex] I just obtain a number? How can I keep this algebraic structure and prove the above?
For a qubit defined as: [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex]
Since [tex]|0\rangle[/tex] and [tex]|1\rangle[/tex] are the eigenstates of [tex]\sigma_z[/tex] then measuring sigma_z will yield either [tex]|0\rangle[/tex] or [tex]|1\rangle[/tex]. Measuring [tex]\sigma_x[/tex] on the same qubit will give one of the eigenstates of [tex]\sigma_x[/tex], which are [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex] and [tex]\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)[/tex].
Only problem is I don't see how you can obtain these general/algebraic states? When I make a measurement on the qubit ie. ([tex](\frac{1}{\sqrt{2}}(\langle 0| +\langle1|))\sigma_x(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle))[/tex] I just obtain a number? How can I keep this algebraic structure and prove the above?