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Neutral Meson Oscillations (probabilities)

  1. Jun 26, 2015 #1

    ChrisVer

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    I am looking into the probability for : [itex] \mathcal{P}(B^0 \rightarrow B^0)[/itex].
    I said that if I start from a state [itex]|B^0> = \frac{1}{\sqrt{2}} (|B_L> +|B_H>)[/itex] with L(ight)/H(heavy) are the mass eigenstates, then after some time [itex]t[/itex] the state will evolve:
    [itex]|B^0(t) > = e^{-iHt} |B^0>= \frac{1}{\sqrt{2}} ( e^{-i(m_L - \frac{i}{2}\Gamma_L)t} |B_L> +e^{-i(m_H - \frac{i}{2}\Gamma_H)t} |B_H> ) [/itex]

    So far so good. Then the amplitude for [itex]B^0 \rightarrow B^0[/itex] will be given by:
    [itex]<B^0 | B^0(t)>=\frac{1}{2} (e^{-i(m_L - \frac{i}{2}\Gamma_L)t}+e^{-i(m_H - \frac{i}{2}\Gamma_H)t})[/itex]
    and the probability by:
    [itex]|<B^0|B^0(t)>|^2 = \frac{1}{4} \Big( e^{-\Gamma_H t} + e^{-\Gamma_L t} + 2e^{-(\Gamma_H+ \Gamma_L)t/2} \cos \Delta m t \Big) [/itex]
    If I'd neglect the [itex]\Delta \Gamma= \Gamma_L - \Gamma_H \Rightarrow \Gamma_L \approx \Gamma_H \equiv \Gamma[/itex] this would be written as:
    [itex]P= \frac{1}{2} e^{-\Gamma t} \Big( 1 + \cos \Delta m t \Big) [/itex]

    Unfortunately I'm said I have to prove that :
    [itex]P= \frac{1}{2 \tau} e^{-\Gamma t} \Big( 1 + \cos \Delta m t \Big) [/itex]
    I don't know where the [itex]\tau[/itex] comes from in the denominator. Any idea? In fact I don't even know why the probability should have "dimensions"...
     
    Last edited: Jun 26, 2015
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  3. Jun 26, 2015 #2

    mfb

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    I guess this is the probability that the particle decays as B^0 at that time. Which is the probability that the particle "is" a B^0 divided by the lifetime.
    This is not completely correct as some decays do not allow to specify the original type with certainty (basically every decay that is not semileptonic).
     
  4. Jun 26, 2015 #3

    ChrisVer

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    When you say original type you mean the [itex]B^0(t=0)[/itex]?
     
  5. Jun 26, 2015 #4

    ChrisVer

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  6. Jun 26, 2015 #5

    mfb

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    No, I mean the type at decay time. Which is not a well-defined thing in many analyses as you get interference between two different decays (from the mixed and unmixed state).
    See the "will decay as" description directly above equation 1.
     
  7. Jun 26, 2015 #6

    ChrisVer

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    So it's more like I start with a [itex]B^0[/itex] which can in general oscillate and exist in a [itex]B^0 /\bar{B}^0[/itex] state, and that probability gives the probability of that state components to decay... When there is a mixing it's impossible to figure out in non-semi leptonic decays what we had at the decay?
    To me the mixing depends on [itex]\Delta m[/itex]... for pretty small [itex]\Delta m[/itex] (compared to the lifetime and so the meaningful time interval [itex]t[/itex]), the mixing is done really slowly (as shown in the uploaded figures).
     

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  8. Jun 27, 2015 #7

    vanhees71

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    Isn't this simply the following? You have calculated the survival probability of your ##B_0## at time ##t## as a function of time, given that at ##t=0## you had a ##B_0## present. That's the usual decay law (it's worse to see, which approximations go into this, because in QT there cannot be strictly valid exponential decay laws; see, e.g., Sakurai):
    $$P(t)=\exp(-t/\tau).$$
    Then the decay rate is
    $$-\dot{P}=\frac{1}{\tau} P(t).$$
     
  9. Jun 27, 2015 #8

    mfb

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    Right.
    Also true, but your simplified model ignores that part.
    Sure, and you can see this in your formula. Larger Δm lead to faster oscillations.
     
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