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Quantum Physics: observables, eigenstates and probability

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    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] .

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Oct 14, 2012 #2

    TSny

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    Hi sgsurrey.

    Since the question wants you to express the outcome of a measurement of [itex]\widehat{C}[/itex] in terms of [itex]u_{+}[/itex] and [itex]u_{-}[/itex] it might be a good idea to work in the basis [itex]u_{+}[/itex], [itex]u_{-}[/itex].
    Can you find the matrix representations of [itex]\widehat{A}[/itex], [itex]\widehat{B}[/itex] , and [itex]\widehat{C}[/itex] with respect to the [itex]u_{+}[/itex], [itex]u_{-}[/itex] basis?

    Once you have the matrix representation of [itex]\widehat{C}[/itex], you can use it to find the possible results of a measurement of [itex]\widehat{C}[/itex] and the probability of each of those results if the initial state is [itex]u_{+}[/itex].
     
    Last edited: Oct 14, 2012
  4. Oct 14, 2012 #3
    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 realised that this method is outlined a few pages after what I had previously studied in my textbook so far; I will read this now. I'll hopefully then proceed with an attempt at the solution.
     
  5. Oct 14, 2012 #4

    TSny

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    It's possible to get the answer to this question without matrix representations of the operators. The matrix approach is just kind of a nice way to go about it in my opinion.
     
  6. Oct 14, 2012 #5
    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:

    [itex]u_{+} = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)[/itex]
    [itex]u_{-} = \left(\begin{array}{c} 0 \\ 1 \end{array}\right)[/itex]
    [itex]v_{+} = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)[/itex]
    [itex]v_{-} = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ -1 \end{array}\right)[/itex]

    Then it seems I can express operator A as:

    [itex]\widehat{A} = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)[/itex]

    Now stuck on operator B...

    Are my above assumptions correct?
     
  7. Oct 14, 2012 #6
    It would appear that B is:

    [itex]\widehat{B} = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)[/itex]

    and thus C:

    [itex]\widehat{C} = \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)[/itex]

    There is probably a much easier way to obtain those matrices than the way I just used (trial and error)...
     
  8. Oct 14, 2012 #7

    TSny

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    Yes, that looks correct. Formally, you can get the matrix elements for C by noting that the C11 element is <u+|C|u+> = <u+|A|u+> + <u+|B|u+>. To evaluate <u+|B|u+>, express u+ in terms of the v vectors. Likewise, the C12 matrix element is <u+|C|u->, etc.
     
  9. Oct 14, 2012 #8

    TSny

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    Sorry, it's easier to first find the matrix for B and then just add A and B to find C as you did! You can find the matrix elements for B using the method I described for C.
     
  10. Oct 14, 2012 #9
    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:

    [itex]\widehat{C}u_{+} = \left(\begin{array}{c} 1 \\ 1 \end{array} \right)[/itex]

    Which clearly has a magnitude [itex]\pm\sqrt{2}[/itex] (which I have in the solutions for this problem)

    I'm unsure of how I find the state that the 'atom' is now in... any hints? I feel I am out of my depth with this problem, perhaps I should return to it at a later stage.
     
  11. Oct 14, 2012 #10

    TSny

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    Operating on u+ with C does not correspond to measuring C when the systems is in the state u+, so that's not going to be of much help.

    Can you fill in the blank here:

    A fundamental postulate of QM is that the result of a measurement of an observable C is always one of the ___________ of C.
     
  12. Oct 14, 2012 #11
    Ah.. got it, sorry, matrix notation getting me confused.

    A measurement of an observable will be one of the eigenvalues:

    [itex](A-λI)\psi = 0[/itex]

    [itex]\left|\begin{array}{cc} 1-λ & 1 \\ 1 & -1-λ \end{array}\right| = 0[/itex]

    λ = ±√2
     
  13. Oct 14, 2012 #12

    TSny

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    Right. And what will the state of the system be after the measurement?
     
  14. Oct 15, 2012 #13
    Then get:

    [itex]ω_{+}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\begin{array}{c} 1-\sqrt{2} \\ 1 \end{array}\right)[/itex]
    [itex]ω_{-}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\begin{array}{c} 1+\sqrt{2} \\ 1 \end{array}\right)[/itex]

    and probabilities:

    [itex]\left(\left<ω_{+}|u_{+}\right>\right)^2 = 0.146[/itex]
    [itex]\left(\left<ω_{-}|u_{+}\right>\right)^2 = 0.854[/itex]

    In the solution I have:
    [itex]ω_{+}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\begin{array}{c} 1 \\ -(1-\sqrt{2}) \end{array}\right)[/itex]
    [itex]ω_{-}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\begin{array}{c} 1 \\ -(1+\sqrt{2}) \end{array}\right)[/itex]
    and the probabilities for ω+ and ω- are opposite...

    Do you know if this is equivalent; or have I made a mistake?
     
    Last edited: Oct 15, 2012
  15. Oct 15, 2012 #14
    I'm now aware that I have made a mistake and will hopefully figure this out later.
     
  16. Oct 15, 2012 #15

    TSny

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    Looks good! That's what I got anyway.
     
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