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Measurements of neutrinos

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Neutrinos are created in states of one of two possible flavors, [itex]f_1[/itex] or [itex]f_2[/itex]. Each flavor state can be expressed as a linear combination of mass eigenstates with masses [itex]m_1[/itex] and [itex]m_2[/itex]
    [tex]
    |f_1\rangle = |m_1\rangle a_{11}+|m_2\rangle a_{21}
    [/tex]
    [tex]
    |f_2\rangle = |m_1\rangle a_{12} + |m_2\rangle a_{22}
    [/tex]
    The unitary matrix [itex]a_{ij}[/itex] is called the mixing matrix. The different mass of neutrinos with the same momentum move at different speeds. Eventually the light neutrino ([itex]m_1[/itex]) will outrun the heavier neutrino ([itex]m_2[/itex]). When the lighter neutrino reaches a detector only the flavor can be detected. what is the probability of measuring the flavor [itex]f_1[/itex]? After the flavor [itex]f_1[/itex] is measured and selected, what is the probability that the neutrino continues moving with the lighter mass ([itex]m_1[/itex])?


    2. Relevant equations
    [tex]
    P=\frac{\langle a|b\rangle\langle b|a\rangle}{\langle a|a\rangle\langle b|b\rangle}
    [/tex]

    3. The attempt at a solution
    I feel like the two questions are the same... It sounds like to me that a neutrino prepared in a state [itex]|m_1\rangle[/itex] addresses the detector and I want to measure the probability of it being in a state [itex]f_1[/itex]. I interpret this as
    [tex]
    \begin{align}
    P(f_1) &= \langle m_1|f_1\rangle\langle f_1|m_1\rangle \\
    &= (a_{11}\langle m_1|m_1\rangle + a_{21}\langle m_1|m_2\rangle)(a_{11}^{\ast}\langle m_1|m_1\rangle + a_{21}^{\ast}\langle m_2|m_1\rangle
    \end{align}
    [/tex]
    Then I assumed that [itex]\langle m_1|m_2\rangle=0[/itex] and [itex]\langle m_1|m_1\rangle=1[/itex] by orthogonality. Then
    [tex]
    P(f_1)=|a_{11}|^2
    [/tex]
    The problem is that this seems to me to be the way to answer both questions... Where is my misunderstanding?

    Thanks,

    PS I know it says this already but please don't tell me the answer, I really want to figure this out on my own, thanks.
     
  2. jcsd
  3. Sep 3, 2011 #2
    Perhaps I see it differently...

    Is this saying that now the state is [itex]|f_1\rangle[/itex] and we want to know the probability of state [itex]|m_1\rangle[/itex], that is
    [tex]
    \langle f_1|m_1\rangle\langle m_1|f_1\rangle ?
    [/tex]

    Thanks,
     
  4. Sep 3, 2011 #3

    vela

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    Yes, that's how I read it as well.
     
  5. Sep 3, 2011 #4
    Okay, then I took [itex]f_i=a_{ij}m_j[/itex] and solved for [itex]m_j=a^{-1}_{ij}f_i[/itex]. I then carried out
    [tex]
    \langle f_1|m_1\rangle\langle m_1|f_1\rangle=\frac{|a_{22}|^2}{|a_{11}a_{22}-a_{12}a_{21}|^2}
    [/tex]
    but since [itex]a_{ij}[/itex] is unitary that is
    [tex]
    =|a_{22}|^2
    [/tex]

    Does that seem correct?
     
  6. Sep 4, 2011 #5
    ignore that dross above... I'm still stumped, they seem to be asking for the same probability. To me it sounds like: there is a neutrino of mass [itex]m_1[/itex], what is the probability of it having flavor [itex]f_1[/itex]? Then: There is a neutrino of flavor [itex]f_1[/itex], what is the probability of it having mass [itex]m_1[/itex]? Both of these seem to be [itex]|\langle f_1|m_1\rangle|^2[/itex]...

    May I have a hint?
     
  7. Sep 4, 2011 #6

    vela

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    I think that's right, actually.
     
  8. Sep 4, 2011 #7
    You think the expressions for those probabilities are both [itex]|\langle f_1 | m_1\rangle|^2[/itex]? Why would someone write a question like that... :confused:?
     
  9. Sep 4, 2011 #8

    vela

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    You'll have to ask your instructor. :wink:
     
  10. Sep 4, 2011 #9
    Thanks for all your help vela :smile:
     
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