- #1
sgsurrey
- 8
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Homework Statement
Observable [itex]\widehat{A}[/itex] has eigenvalues [itex]\pm[/itex]1 with corresponding eigenfunctions [itex]u_{+}[/itex] and [itex]u_{-}[/itex]. Observable [itex]\widehat{B}[/itex] has eigenvalues [itex]\pm[/itex]1 with corresponding eigenfunctions [itex]v_{+}[/itex] and [itex]v_{-}[/itex].
The eigenfunctions are related by:
[itex]v_{+} = (u_{+} + u_{-})/\sqrt{2}[/itex]
[itex]v_{-} = (u_{+} - u_{-})/\sqrt{2}[/itex]
Show that [itex]\widehat{C}[/itex] =[itex]\widehat{A}[/itex] + [itex]\widehat{B}[/itex] is an observable and find the possible results of a measurement of [itex]\widehat{C}[/itex].
Find the probability of obtaining each result when a measurement of [itex]\widehat{C}[/itex] is performed on an atom in the state [itex]u_{+}[/itex] and the corresponding state of the atom immediately after the measurement in terms of [itex]u_{+}[/itex] and [itex]u_{-}[/itex] .
Homework Equations
[itex]\widehat{A}u_{\pm}=\pm u_{\pm}[/itex]
[itex]\widehat{B}v_{\pm}=\pm v_{\pm}[/itex]
[itex]v_{+} = (u_{+} + u_{-})/\sqrt{2}[/itex]
[itex]v_{-} = (u_{+} - u_{-})/\sqrt{2}[/itex]
[itex]u_{+} = (v_{+} + v_{-})/\sqrt{2}[/itex]
[itex]u_{-} = (v_{+} - v_{-})/\sqrt{2}[/itex]
The Attempt at a Solution
Showing that C is an observable I have assumed is as simple as the fact that it is a linear combination of the A and B operators, which are observables matching the requirement to have real eigenvalues.
I'm slightly puzzled by the measurement part of the question; I have tried simply:
[itex]\widehat{C}v_{+} = \widehat{A}v_{+} + \widehat{B}v_{+}
= \widehat{A}(u_{+} + u_{-})/\sqrt{2} + v_{+} = v_{-} + v_{+} = \sqrt{2}u_{+}[/itex]
I'm not sure I fully understand this however; I expected a result of the format:
[itex]\widehat{C}v_{+} = c_{+}v_{+}[/itex]
...where c is a 'result of the measurement', the eigenvalue of C. I'm not sure if it is correct to say that [itex]\sqrt{2}[/itex] is the eigenvalue if the resulting eigenfunction is not the same as the starting eigenfunction.
The answer I have been given for this part of the question is [itex]\pm\sqrt{2}[/itex], but I feel I have now simply found the simplest way to get this in an answer, without understanding why (had I not been lost from the outset I would have avoided looking at the answer in the first place).
For the later part of the question my attempted answer is so far from the solution that I am clueless on how to approach it at this point. I am hoping that with some insight into this first part of the question I might understand how to approach the rest of the question. I would very much appreciate any guidance.