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Neutrino Oscillation Formula

  1. Jul 5, 2014 #1
    Hi all!

    I am not sure how to prove mathematically that the expression for the probability that a neutrino originally of flavor α will later be observed as having flavor β
    [itex]P_{α \rightarrow β}=\left|<\nu_{\beta}|\nu_{\alpha}(t)>\right|^2=\left|\sum_{i}U_{β i}^*U_{α i}e^{-iE_it}\right|^2[/itex] (1)
    can be equivalently written as
    [itex]P_{α \rightarrow β}=δ_{αβ}-4\sum_{i>j}Re(U_{β i}^*U_{α i}U_{β j}U_{α j}^*)\sin^2(\frac{Δm_{ij}^2L}{4E})+2\sum_{i>j}Im(U_{β i}^*U_{α i}U_{β j}U_{α j}^*)\sin(\frac{Δm_{ij}^2L}{4E})[/itex] (2)
    Whoever is not familiar with the notation and would still like to contribute "mathematically", all the variables and constants are explained perfectly in the wiki article: http://en.wikipedia.org/wiki/Neutrino_oscillation

    The proof has to be trivial, but here I am trying for two hours and still not able to show this for a general case.
    Last edited: Jul 5, 2014
  2. jcsd
  3. Jul 5, 2014 #2


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    which part of the formula is confusing you?
    how to express [itex]E_{i} t[/itex] in terms of [itex]L, m_{i}[/itex]?
    This is done by some easy calculations, I think you can find them easy if you search... eg
    eq 16.3
    and also by saying [itex] t= \frac{L}{c} [/itex] the time they traveled from distance L up to reaching us.
    with c=1, and 16.3 you get your result.

    or how to write the sums?
    Those sums are in fact two sums, of i and j with the same expressions just change the i's to js and taking the conjugate...
    After that you can start breaking the sums in such a way that you'll get something as the final result... for example for i=j you will evenutally get delta because the exponentials will cancel each other out (remind i=j again), and the U matrices are unitary, so you will get the delta Kroenicker by summing them:
    [itex] (U^{\dagger} U)_{ab} = \delta_{ab} [/itex] for U unitary.
    The rest is more tedious work, but in general that's how you get the Re and Im part (depending of what i,j configurations you sum).
  4. Jul 5, 2014 #3
    I wanna start with (1) and reach (2) step by step.
    [itex]P_{α \rightarrow β}=\left|\sum_{i}U_{β i}^*U_{α i}e^{-iE_it}\right|^2=\left|\sum_{i}U_{β i}^*U_{α i}e^{-iE_it}\right|^2=\left|\sum_{i}(Re(U_{β i}^*U_{α i})+iIm(U_{β i}^*U_{α i}))(\cos(E_it)-i\sin(E_it))\right|^2=[/itex]
    [itex]\left[\sum_{i}Re(U_{β i}^*U_{α i})\cos(E_it)+Im(U_{β i}^*U_{α i})\sin(E_it)\right]^2+\left[\sum_{i}Im(U_{β i}^*U_{α i})\cos(E_it)-Re(U_{β i}^*U_{α i})\sin(E_it)\right]^2[/itex]
    Is this correct so far? How do we further simplify this expression and get the double sums? Also some trigonometry is involved for sure.
  5. Jul 6, 2014 #4


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    why don't you write the square explicitly?

    [itex] |\sum_i U_{β i}^* U_{α i} e^{-iE_{i}t}|^{2} = \sum_{i} \sum_{j} U_{β i}^* U_{α i} U_{ α j}^{*} U_{β j} e^{-iE_{i}t} e^{+iE_{j}t} [/itex]

    [itex]= \sum_{i} \sum_{j} U_{β i}^* U_{α i} U_{α j}^{*} U_{β j} e^{-i (ΔE)_{ij}t} [/itex]

    that's in general how you proceed...
    Last edited: Jul 6, 2014
  6. Jul 6, 2014 #5
    So you simply write the square of the absolute value as the complex number times its conjugate and you get the Δm differences.
    I see how for i=j, the Kronecker delta arises. Sorry but I am too bad at this and I still cannot work out the rest of the terms. Can you take me by hand or show the derivation in detail?
  7. Jul 7, 2014 #6


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    because in most of the cases this is an exercises, you should just keep track of your indices and what you get....
    I guess could help you if you showed me what you did, where you've reached. In most of cases it's just a to-be-done-carefully calculation, and at one point you have to use trigonometric identities (if I recall well)...
    One thing is for sure, after you found the case of delta, you need also to add to this the cases where i is not equal to j.... and from that see what you get for that term...
  8. Jul 7, 2014 #7


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    For the remaining terms (##i \neq j##), you have one term for ##i > j## and one term for ##j < i##. Consider how they are related by writing out both sums and then changing the summation indices in one of them. After that you can make use of the relations
    z + z^* = 2 Re(z), \quad z - z^* = 2 Im(z)
    (Also note that ##\Delta E_{ij} = - \Delta E_{ji}## ...)
  9. Jul 18, 2014 #8
    Thanks for your replies. I have finally worked out the derivation!
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