- #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"...
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"...
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