Recent content by sgsurrey

I'm now aware that I have made a mistake and will hopefully figure this out later.

Then get: ω_{+}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\begin{array}{c} 1-\sqrt{2} \\ 1 \end{array}\right) ω_{-}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\begin{array}{c} 1+\sqrt{2} \\ 1 \end{array}\right) and probabilities: \left(\left<ω_{+}|u_{+}\right>\right)^2 = 0.146...

Ah.. got it, sorry, matrix notation getting me confused. A measurement of an observable will be one of the eigenvalues: (A-λI)\psi = 0 \left|\begin{array}{cc} 1-λ & 1 \\ 1 & -1-λ \end{array}\right| = 0 λ = ±√2

I'm now struggling to understand what this means. If I operate C on u+, or rather matrix multiply C with the vector, I get the vector: \widehat{C}u_{+} = \left(\begin{array}{c} 1 \\ 1 \end{array} \right) Which clearly has a magnitude \pm\sqrt{2} (which I have in the solutions for this...

It would appear that B is: \widehat{B} = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) and thus C: \widehat{C} = \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) There is probably a much easier way to obtain those matrices than the way I just used (trial and...

The matrix method sounds useful, so I'm reading up on it. As far as I can see so far I can represent each eigenfunction as a vector in terms of the 'u' basis: u_{+} = \left(\begin{array}{c} 1 \\ 0 \end{array}\right) u_{-} = \left(\begin{array}{c} 0 \\ 1 \end{array}\right) v_{+} =...

Thank you for your response. I'm a few weeks into this QM module, I've yet to cover the matrix representations and thus my understanding is lacking in this approach. Since you've suggested this I have realized that this method is outlined a few pages after what I had previously studied in my...

Homework Statement Observable \widehat{A} has eigenvalues \pm1 with corresponding eigenfunctions u_{+} and u_{-}. Observable \widehat{B} has eigenvalues \pm1 with corresponding eigenfunctions v_{+} and v_{-}. The eigenfunctions are related by: v_{+} = (u_{+} + u_{-})/\sqrt{2} v_{-} =...