Neutral Meson Oscillations (probabilities)

[ChrisVer]

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

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

Unfortunately I'm said I have to prove that :
$P= \frac{1}{2 \tau} e^{-\Gamma t} \Big( 1 + \cos \Delta m t \Big)$
I don't know where the $\tau$ comes from in the denominator. Any idea? In fact I don't even know why the probability should have "dimensions"...

 

[mfb]

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).

 

[ChrisVer]

When you say original type you mean the $B^0(t=0)$?

 

[ChrisVer]

The probability is also given in Eq(1) here
http://arxiv.org/pdf/hep-ex/9901035v1.pdf

 

[mfb]

When you say original type you mean the $B^0(t=0)$?

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).

The probability is also given in Eq(1) here
http://arxiv.org/pdf/hep-ex/9901035v1.pdf

See the "will decay as" description directly above equation 1.

 

[ChrisVer]

So it's more like I start with a $B^0$ which can in general oscillate and exist in a $B^0 /\bar{B}^0$ 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 $\Delta m$... for pretty small $\Delta m$ (compared to the lifetime and so the meaningful time interval $t$), the mixing is done really slowly (as shown in the uploaded figures).

 

[vanhees71]

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
$$-\dot{P}=\frac{1}{\tau} P(t).$$

 

[mfb]

So it's more like I start with a $B^0$ which can in general oscillate and exist in a $B^0 /\bar{B}^0$ state, and that probability gives the probability of that state components to decay...

Right.

When there is a mixing it's impossible to figure out in non-semi leptonic decays what we had at the decay?

Also true, but your simplified model ignores that part.

To me the mixing depends on $\Delta m$... for pretty small $\Delta m$ (compared to the lifetime and so the meaningful time interval $t$), the mixing is done really slowly (as shown in the uploaded figures).

Sure, and you can see this in your formula. Larger Δm lead to faster oscillations.