Neutral Meson Oscillations (probabilities)

In summary, the conversation discusses the probability for a B^0 particle to decay into a B^0 particle as a function of time. The probability is given by an exponential decay law with a decay rate of 1/2τ, where τ is the lifetime of the particle. The conversation also mentions the effects of mixing and oscillations on the decay probability.
  • #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"...
 
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  • #2
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).
 
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  • #3
When you say original type you mean the [itex]B^0(t=0)[/itex]?
 
  • #5
ChrisVer said:
When you say original type you mean the [itex]B^0(t=0)[/itex]?
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).
ChrisVer said:
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.
 
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  • #6
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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  • #7
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
$$-\dot{P}=\frac{1}{\tau} P(t).$$
 
  • #8
ChrisVer said:
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...
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 [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).
Sure, and you can see this in your formula. Larger Δm lead to faster oscillations.
 
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1. What are neutral meson oscillations?

Neutral meson oscillations refer to the phenomenon where particles known as neutral mesons spontaneously change into their antiparticles and back again. This process is governed by the principles of quantum mechanics and occurs due to the mixing of two different types of neutral mesons, known as the "neutral kaon system" and the "neutral B meson system".

2. What is the significance of neutral meson oscillations?

Neutral meson oscillations are important because they provide insight into the fundamental laws of physics, such as the laws of conservation of energy and charge. They also help scientists understand the nature of matter and antimatter and can potentially lead to the discovery of new particles and interactions.

3. How are neutral meson oscillations measured?

Neutral meson oscillations are measured by observing the decay products of the neutral mesons. Since the oscillation process is probabilistic, scientists use statistical methods to determine the likelihood of a neutral meson changing into its antiparticle and back again. This is done by analyzing large amounts of data collected from high-energy particle collisions.

4. What factors affect the probabilities of neutral meson oscillations?

The probabilities of neutral meson oscillations are affected by several factors, including the mass difference between the two neutral meson states, the strength of the interactions between the neutral mesons and other particles, and the presence of external fields or forces that can influence the oscillation process.

5. What implications do neutral meson oscillations have for our understanding of the universe?

Neutral meson oscillations provide important insights into the fundamental laws of physics and can potentially lead to the discovery of new particles and interactions. They also help scientists understand the asymmetry between matter and antimatter in the universe, as well as the origins of our universe and its evolution over time.

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