Neutrino Oscillation Formula

  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:

    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. ChrisVer

    ChrisVer 2,403
    Gold Member

    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. 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. ChrisVer

    ChrisVer 2,403
    Gold Member

    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. 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. ChrisVer

    ChrisVer 2,403
    Gold Member

    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. 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. Thanks for your replies. I have finally worked out the derivation!
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