You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We consider neutral pseudoscalar mesons in an entangled or Einstein-Podolsky-Rosen state with $C=-1$. Due to quantum entanglement and antisymmetry of this state, the rates of the joint decays of the meson pair display various interesting features, as is well known. As functions of CP and CPT violating parameters, here we obtain some exact results about the joint decay rates and their asymmetries for a given time difference defined for joint decays to flavor eigenstates, as well as those for joint decays to CP eigenstates. The entanglement allows a meaningful and useful definition of the transition amplitude from a CP eigenstate of a meson. These results yield useful information and criteria on CP and CPT violations.'
author:
- Yu Shi
title: 'Some exact results on CP and CPT violations in a $C=-1$ entangled pseudoscalar neutral meson pair'
---

European Physical Journal C [**73**]{}, 2506 (2013).

Introduction
============

Pseudoscalar neutral mesons are ideal systems in studying CP violation, and in search for CPT violation, which is implied by the standard model extension [@kos1]. Pseudoscalar neutral mesons in an entangled state have special properties due to quantum entanglement or Einstein-Podolsky-Rosen correlation [@epr; @old; @test; @cern; @go; @shi1]. Nowadays, such entangled meson pairs are routinely produced in $\phi$ or $B$ factories [@domenico; @kloe; @babar; @belle]. Hence it is interesting and important to explore the use of the entangled pairs in CP and CPT problems [@domenico; @kloe; @babar; @belle; @bernabeu1; @dunietz; @buchanan; @dambrosio; @kobayashi; @kos2; @bernabeu2; @baji; @bernabeu; @petrov; @bigi; @soni; @kittle; @huang; @shi2].

Previous investigations often used some approximations up to the first order of the CP or CPT violating parameters. Reasonable as it is, it is of special value to have some exact results without approximations, since being exactly zero is qualitatively different from being approximately zero with higher orders of CP or CPT violating parameters neglected. Moreover, exact results are important in making comparisons between different approximations.

In this paper, we present some exact results on the use of the $C=-1$ entangled state of pseudoscalar mesons to examine CP and CPT violations or conservations. A recent calculation ignored direct CP violation and the violation of $\Delta {\cal F} =\Delta Q$ rule, where ${\cal F}$ is the flavor quantum number [@shi2].

The rest of this paper is organized as the following. In Sec. \[review\], we review the single-particle bases, the time evolution of a single neutral meson as well as that of an entangled neutral meson pair with total $C=-1$. In Sec. \[flavor\], we consider the decays of both mesons into flavor eigenstates, for which some results concerning CP and CPT are given in Sec. \[violation\]. In Sec. \[cp\], we consider decays of both mesons into CP eigenstates, for which some results concerning CP and CPT are given in Sec \[cp2\]. A summary is made in Sec. \[summary\].

A Review of Single Neutral Meson and Entangled Neutral Mesons \[review\]
========================================================================

A neutral pseudoscalar meson $M^0$ and its antiparticle $\bar{M}^0$ are eigenstates of parity $P$ both with eigenvalue $-1$, and of a characteristic flavor ${\cal F}$ with eigenvalues $\pm 1$. ${\cal F}$ is strangeness for $K^0$ and $\bar{K}^0$, beauty for $B_d^0$ and $\bar{B}_d^0$, charm for $D^0$ and $\bar{D}^0$, and strangeness or beauty (with a minus sign) for $B_s^0$ and $\bar{B}_s^0$. With the phase convention $C|M^0\rangle = -|\bar{M}^0\rangle$ and $C|\bar{M}^0\rangle = -|M^0\rangle$, the eigenstates of $CP$ are $$|M_\pm \rangle =
\frac{1}{\sqrt{2}}(|M^0\rangle \pm  |\bar{M}^0\rangle), \label{cptof}$$ with eigenvalues $\pm 1$. It should be noted that physically a single particle cannot be in the state $|M_\pm \rangle$, as CP is violated.

In the flavor basis, the effective mass matrix $H$ can be written as $$H=\left(
\begin{array}{cc}
H_{00} & H_{0\bar{0}} \\
H_{\bar{0}0} & H_{\bar{0}\bar{0}}
\end{array}\right),$$ where $H_{00} \equiv \langle M^0|H|M^0\rangle$, $H_{0\bar{0}}\equiv\langle M^0|H|\bar{M}^0\rangle $, $H_{\bar{0}0 } \equiv \langle\bar{ M}^0|H|M^0\rangle$, $H_{\bar{0}\bar{0}} \equiv \langle\bar{ M}^0|H|\bar{M}^0\rangle$.

Indirect T violation and CP violation are characterized by a nonzero parameter $\epsilon_M$ defined through $$\frac{q}{p} \equiv \sqrt{ \frac{H_{\bar{0}0}}{ H_{0\bar{0}}} } \equiv \frac{1-\epsilon_M}{1+\epsilon_M},$$ because if CP or T is conserved indirectly, then $\epsilon_M =0$.

Indirect CPT violation and CP violation are characterized by a nonzero parameter $\delta_M$ defined as [@chou] $$\delta_M \equiv \frac{ H_{\bar{0}\bar{0}} -H_{00}}{ \sqrt{ H_{0\bar{0}}H_{\bar{0}0} } } \neq 0,$$ because if CPT or CP is conserved indirectly, then $\delta_M =0$.

The eigenvalues of $H$ are $$\begin{aligned}
\lambda_S  \equiv m_S -i\Gamma_S/2= H_{00} +  \sqrt{ H_{0\bar{0}}H_{\bar{0}0} } (\sqrt{1+\frac{\delta_M^2}{4}} + \frac{\delta_M}{2}), \\
\lambda_L \equiv  m_L -i\Gamma_L/2
= H_{\bar{0}\bar{0}}-\sqrt{ H_{0\bar{0}}H_{\bar{0}0} }
(\sqrt{1+\frac{\delta_M^2}{4}} + \frac{\delta_M}{2}),\end{aligned}$$ corresponding, respectively, to the eigenstates $$\begin{aligned}
|M_S\rangle =
\frac{1}{\sqrt{|p_S|^2+|q_S|^2}}(p_S|M^0\rangle
+q_S|\bar{M}^0\rangle) = \frac{1}{\sqrt{1+|\epsilon_S|^2}}(|M_+\rangle
+\epsilon_S|M_-\rangle), \label{ms1}\\
|M_L\rangle  =
\frac{1}{\sqrt{|p_L|^2+|q_L|^2}}(p_L|M^0\rangle - q_L|\bar{M}^0\rangle) =  \frac{1}{\sqrt{1+|\epsilon_L|^2}}(
\epsilon_L|M_+\rangle+ |M_-\rangle), \label{ml1}\end{aligned}$$ with $$\begin{aligned}
x_S \equiv \frac{q_S}{p_S} \equiv \frac{1-\epsilon_S}{1+\epsilon_S}=
\frac{q}{p}(\sqrt{1+\frac{\delta_M^2}{4}}+\frac{\delta_M}{2}),\\
x_L \equiv \frac{q_L}{p_L} \equiv \frac{1-\epsilon_L}{1+\epsilon_L}=
\frac{q}{p}(\sqrt{1+\frac{\delta_M^2}{4}}-\frac{\delta_M}{2}),\end{aligned}$$ If $\epsilon_S=0$, $|M_S\rangle=|M_+\rangle$, with $CP |M_+\rangle = |M_+\rangle$. If $\epsilon_L=0$, $|M_L\rangle = |M_-\rangle$, with $CP |M_-\rangle = -|M_-\rangle$. Conversely $$\begin{aligned}
|M_+\rangle =
\frac{1}{1-\epsilon_S\epsilon_L}(\sqrt{1+|\epsilon_S|^2}|M_S\rangle-
\epsilon_S\sqrt{1+|\epsilon_L|^2}|M_L\rangle), \\
|M_-\rangle =
\frac{1}{1-\epsilon_S\epsilon_L}(\sqrt{1+|\epsilon_L|^2}|M_L\rangle-
\epsilon_L\sqrt{1+|\epsilon_S|^2}|M_S\rangle). \label{lm2}\end{aligned}$$

We often have the definitions $$\epsilon \equiv \frac{1}{2}(\epsilon_S+\epsilon_L),$$ $$\delta \equiv \frac{1}{2}(\epsilon_S-\epsilon_L).$$ Hence $$\epsilon = \frac{\epsilon_M}{1+\epsilon_M^2+(1-\epsilon_M^2)\sqrt{1+\frac{\delta_M^2}{4}}}
\approx \frac{\epsilon_M}{2},$$ $$\delta   = -\frac{(1-\epsilon_M^2)\delta_M}{1+\epsilon_M^2+(1-\epsilon_M^2)
\sqrt{1+\frac{\delta_M^2}{4}}}\approx -\frac{\delta_M}{2}.$$

Under the Wigner-Weisskopf approximation, the evolution of an arbitrary state of a pseudoscalar meson $|M(t)\rangle$, as a superposition of $|M^0\rangle$ and $|\bar{M}^0\rangle$, can be described by a Schrödinger equation $$i\frac{\partial}{\partial t}|M(t)\rangle = H |M(t)\rangle,$$ using which we can find the following time-dependent states.

Starting as the mass eigenstate $|M_S\rangle$, the state of a single meson evolves as $$|M_S(t) \rangle = e^{-i\lambda_S t} |M_S\rangle.$$ Starting as the mass eigenstate $|M_L\rangle$, the state evolves as $$|M_L(t) \rangle  =  e^{-i\lambda_L t} |M_L\rangle.$$

Starting as the flavor eigenstate $|M^0\rangle$, the state evolves as $$|M^0(t) \rangle = G_{00}(t) |M^0\rangle
+G_{0\bar{0}}(t)|\bar{M}^0\rangle, \label{m0t}$$ where $$\begin{aligned}
G_{00}(t) &\equiv& \frac{e^{-i\lambda_S t}+\Omega e^{-i\lambda_L t}}{1+\Omega},\\ G_{0\bar{0}}(t)& \equiv &\frac{x_S(e^{-i\lambda_S t}- e^{-i\lambda_L t})}{1+\Omega},\end{aligned}$$ with $$\Omega \equiv \frac{x_S}{x_L} = \frac{q_Sp_L}{p_Sq_L}.$$ Starting as the flavor eigenstate $|\bar{M}_0\rangle$, the state evolves as $$|\bar{M}^0(t) \rangle=
G_{\bar{0}0}(t) |M^0\rangle
+G_{\bar{0}\bar{0}}(t)|\bar{M}^0\rangle, \label{m0bart}$$ where $$\begin{aligned}
G_{\bar{0}0}(t)& \equiv &\frac{x_S^{-1}(e^{-i\lambda_S t}- e^{-i\lambda_L t})}{1+\Omega^{-1}},\\
G_{\bar{0}\bar{0}}(t)& \equiv&
\frac{ e^{-i\lambda_S t}+ \Omega^{-1} e^{-i\lambda_L t}}{1+\Omega^{-1}}.\end{aligned}$$

Physically an isolated single meson state cannot start as a CP eigenstate $|M_{\pm}\rangle$, as CP is indeed violated. There is no way of tagging one meson to be in such an eigenstate by measuring the other if the two are prepared as an entangled pair. However, as explained below, the initial state of a pair of entangled mesons with $C=-1$ is exactly a superposition of two terms, in each of which the two mesons are in different CP eigenstates. Therefore it is useful to consider the evolution of the state of a meson starting as a CP eigenstate. Starting as the CP eigenstate $|M_{+}\rangle$, the state evolves as $$|M_{+}(t)\rangle = F_{+ +}(t) |M_+\rangle + F_{ +-}(t) |M_-\rangle. \label{mtime}$$ where $$\begin{aligned}
F_{++}( t) &=& \frac{1}{2(1+\Omega)}[(1 +x_L^{-1} + x_S + \Omega) e^{-i\lambda_S  t} +(1 -x_L^{-1} - x_S + \Omega) e^{-i \lambda_L  t}],\\
F_{+-}(t) & =& \frac{1 + x_L^{-1} - x_S - \Omega}{2(1+\Omega)}(e^{-i\lambda_S t}- e^{-i \lambda_L t}).\end{aligned}$$ Starting as the CP eigenstate $|M_{-}\rangle$, the state evolves as $$|M_{-}(t)\rangle = F_{- +}(t) |M_+\rangle + F_{- -}(t) |M_-\rangle. \label{mtime2}$$ where $$\begin{aligned}
F_{-+}( t)& =& \frac{1 - x_L^{-1} + x_S - \Omega}{2(1+\Omega)}(e^{-i\lambda_S t} - e^{-i \lambda_L  t}),\\
F_{--}( t) &=& \frac{1}{2(1+\Omega)}[(1 - x_L^{-1}- x_S + \Omega) e^{-i\lambda_S  t} +(1 + x_L^{-1} + x_S + \Omega) e^{-i \lambda_L  t}].\end{aligned}$$

Now consider the $C=-1$ entangled state of a pair of pseudoscalar mesons, which may be referred to as Alice (a) and Bob (b), $$\begin{aligned}
|\Psi_-\rangle & = & \frac{1}{\sqrt{2}}(|M^0\rangle_a|\bar{M}^0\rangle_b
-|\bar{M}^0\rangle_a|M^0\rangle_b),\end{aligned}$$ which can be produced from a source of $J^{PC}=1^{--}$.

A remarkable feature of $|\Psi_-\rangle$ is that it is also [*exactly*]{} $$\begin{aligned}
|\Psi_-\rangle &=& \frac{1}{\sqrt{2}}(|M_-\rangle_a|M_+\rangle_b-
|M_+\rangle_a|M_-\rangle_b). \label{pm}\end{aligned}$$ This means that although a single meson cannot be in the state $|M_{\pm}\rangle$, an entangled pair of mesons with $C=-1$ is [*exactly*]{} in the state $\frac{1}{\sqrt{2}}(|M_-\rangle_a|M_+\rangle_b-
|M_+\rangle_a|M_-\rangle_b)$. There is no tagging here for the initial mesons, as only the final decay products are measured. This allows the use of (\[mtime\]) and (\[mtime2\]) for the entangled pair, because of the linearity of quantum evolution and that the two entangled but separated mesons are non-interacting.

Starting as $|\Psi_-\rangle$, the state of the entangled meson pair evolves and decays to or produces certain products at $t_a$ and $t_b$, which may or may not be equal. To account for this situation, one has $$\begin{aligned}
|\Psi_-(t_a,t_b)\rangle &=&\frac{1}{\sqrt{2}}(|M^0(t_a)\rangle_a|\bar{M}^0(t_b)\rangle_b
-|\bar{M}^0(t_a)\rangle_a|M^0(t_b)\rangle_b)\\
&=&\frac{1}{\sqrt{2}}(|M_-(t_a)\rangle_a|M_+(t_b)\rangle_b
-|M_+(t_a)\rangle_a|M_-(t_b)\rangle_b).\end{aligned}$$

The joint rate that Alice decays to $|\psi_a\rangle$ at $t_a$ while Bob decays to $|\psi_b\rangle$ at $t_b$ is $$I(\psi_a,t_a;\psi_b,t_b) = |\langle \psi_a,\psi_b|{\cal H}_a {\cal H}_b |\Psi_-(t_a,t_b)\rangle|^2,$$ where ${\cal H}_a$ and ${\cal H}_b$ represent the Hamiltonians governing the decays of $a$ and $b$, respectively, $$\begin{aligned}
\langle\psi_a,\psi_b|{\cal H}_a {\cal H}_b|\Psi_-(t_a,t_b)\rangle
&=&\frac{1}{\sqrt{2}}(\langle\psi_a|{\cal H}_a|M^0(t_a)\rangle_a \langle \psi_b|{\cal H}_b |\bar{M}_0(t_b)\rangle_b
- \langle\psi_a|{\cal H}_a|\bar{M}_0(t_a)\rangle_a \langle \psi_b |{\cal H}_b |M^0(t_b)\rangle_b) \\
&=&\frac{1}{\sqrt{2}}(\langle\psi_a|{\cal H}_a|M_-(t_a)\rangle_a \langle \psi_b|{\cal H}_b |M_+(t_b)\rangle_b
- \langle\psi_a|{\cal H}_a|M_+(t_a)\rangle_a \langle \psi_b |{\cal H}_b |M_-(t_b)\rangle_b).\end{aligned}$$

Decays into flavor eigenstates \[flavor\] 
==========================================

Suppose from the entangled state $|\Psi_-\rangle$, Alice and Bob each decays or transits to a two-valued flavor eigenstate. Let us generically denote the flavor eigenstates as $|l^{\pm}\rangle$, with eigenvalue $\pm 1$, respectively. Examples for $|l^+\rangle$ include the semileptonic decay products $M^-\bar{l}\nu$, $D^-D_S^+$, $D^-K^+$, $\pi^-D_S^+$, $\pi^-K^+$ from $M^0=B^0$, and $D_S^-\pi^+$, $D_S^-D^+$, $K^-\pi^+$, $K^-D^+$ from $M^0=B_S^0$. Examples for $|l^-\rangle$ include the semileptonic decay products $M^+ l\bar{\nu}$, $D^+D_S^-$, $D^+K^-$, $\pi^+D_S^-$, $\pi^+K^-$ from $\bar{M}^0=B^0$, and $D_S^+\pi^-$, $D_S^+D^-$, $K^+\pi^-$, $K^+D^-$ from $M^0=B_S^0$. In the CPLEAR experiment on kaons [@cern], $|l^+\rangle$ and $|l^-\rangle$ are products produced via interaction with bound nucleons.

One can calculate the amplitude of such a joint decay in which Alice decays to $| l_a^x\rangle $ at $t_a$ while Bob decays to $|l_b^y\rangle $ at $t_b$, where $x$ and $y$ each represents $\pm 1$, $$\begin{aligned}
\langle l_a^x,l_b^y|{\cal H}_a{\cal H}_b|\Psi_-(t_a,t_b)\rangle & = &   \frac{1}{\sqrt{2}}(\langle l_a^x|{\cal H}_a|M^0(t_a)\rangle_a \langle l_b^y|{\cal H}_b|\bar{M}^0(t_b)\rangle_b
-\langle l_a^x|{\cal H}_a|\bar{M}^0(t_a)\rangle_a  \langle l_b^y|{\cal H}_b|M^0(t_b)\rangle_b) \\
&=&  C(l_a^x,l_b^y) e^{-i(\lambda_St_a+\lambda_Lt_b)} + D(l_a^x,l_b^y) e^{-i(\lambda_Lt_a+\lambda_St_b)},\end{aligned}$$ with $$\begin{aligned}
 C(l_a^x,l_b^y)  & \equiv & \frac{1}{\sqrt{2}(1+\Omega)}(-x_L^{-1}r_a^x r_b^y + r_a^x\bar{r}_b^y - \Omega \bar{r}_a^x r_b^y + x_S \bar{r}_a^x\bar{r}_b^y), \\
 D(l_a^x,l_b^y)  & \equiv & \frac{1}{\sqrt{2}(1+\Omega)}(x_L^{-1}r_a^x r_b^y +\Omega  r_a^x\bar{r}_b^y - \bar{r}_a^x r_b^y - x_S \bar{r}_a^x\bar{r}_b^y),\end{aligned}$$ where $$\begin{aligned}
r^{x}_\alpha & \equiv & \langle l^x_\alpha|{\cal H}_\alpha|M^0\rangle_\alpha,\\
\bar{r}^{x}_\alpha & \equiv &  \langle l^x_\alpha|{\cal H}_\alpha|\bar{M}_0\rangle_\alpha,\end{aligned}$$ $\alpha=a,b$.

Therefore we obtain the joint rate $$\begin{array}{rcl}
I(l^x_a, t_a; l^y_b, t_b)&=&   e^{-(\Gamma_S+\Gamma_L)t_a}
\{|C|^2 e^{-\Gamma_L\Delta t}+ |D|^2 e^{-\Gamma_S\Delta t} \\
&& + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (C^*D)\cos (\Delta m \Delta t) + \Im (C^*D) \sin  (\Delta m \Delta t) ]\}, \\
\end{array}$$ where $C\equiv C(l_a^x,l_b^y)$, $D\equiv D(l_a^x,l_b^y)$.

In experiments, it is more convenient to use the integrated rate, herein defined as $$I'(l_a^x, l_b^y, \Delta t) = \int_0^\infty I(l_a^x, t_a; l_b^y, t_a+\Delta t) dt_a,$$ which is simply given by $I(l^x_a, t_a; l^y_b, t_a+\Delta t)$ as in (\[il\]), with $e^{-(\Gamma_S+\Gamma_L)t_a}$ replaced as $1/(\Gamma_S+\Gamma_L)$.

We focus on the situation that $|l_a^+\rangle = |l_b^+\rangle=|l^+\rangle$ and $|l_a^-\rangle = |l_b^-\rangle=|l^-\rangle$, hence $$\begin{aligned}
r_a^+=r_b^+  & \equiv  & R^+  =  a+b, \\
\bar{r}_a^+=\bar{r}_b^+ &  \equiv& S^+ =  c^* - d^*, \\
r_a^-=r_b^- & \equiv & S^- = c + d, \\
\bar{r}_a^- = \bar{r}_b^-  &\equiv&   R^- = a^*-b^*,\end{aligned}$$ where the quantities $a$, $b$, $c$ and $d$ are the ones usually defined in literature [@maiani; @domenico].

If CP is conserved directly, then we have $R^+=R^-$ and $S^+=S^-$. If CPT is conserved directly, then we have $(R^+)^*=R^-$ and $(S^+)^*=S^-$. If $\Delta {\cal F} = \Delta Q$ rule is respected, then we have $S^\pm =0$.

One obtains $$\begin{aligned}
C(l^+,l^+)  & = & -D(l^+,l^+)= \frac{1}{\sqrt{2}(1+\Omega)}
[-x_L^{-1}(R^+)^2 +(1-\Omega) R^+S^+ +x_S (S^+)^2], \\
C(l^-,l^-)  & = & -D(l^+,l^+)= \frac{1}{\sqrt{2}(1+\Omega)}
[-x_L^{-1}(S^-)^2 +(1-\Omega) R^-S^- +x_S (R^-)^2], \\
C(l^+,l^-)  & = &  -D(l^-,l^+)  = U  \\
D(l^+,l^-)  & = &  -C(l^-,l^+) =  V,\end{aligned}$$ where $$\begin{aligned}
U \equiv   \frac{1}{\sqrt{2}(1+\Omega)}
[-x_L^{-1}R^+ S^- + x_S S^+ R^- + R^+R^- -\Omega S^+ S^-],  \label{u0} \\
V \equiv  \frac{1}{\sqrt{2}(1+\Omega)}
[x_L^{-1}R^+ S^- - x_S S^+ R^- + \Omega R^+R^- - S^+ S^-]. \label{v0}\end{aligned}$$

Therefore we obtain $$\begin{array}{rcl}
I(l^+, t_a; l^+, t_b)&=& \frac{|-x_L^{-1}(R^+)^2 +(1-\Omega) R^+S^+ +x_S (S^+)^2|^2}{2|1+\Omega|^2} e^{-(\Gamma_S+\Gamma_L)t_a}
[e^{-\Gamma_L\Delta t} + e^{-\Gamma_S\Delta t} -2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \cos (\Delta m \Delta t)], \\
I(l^-, t_a; l^-, t_b)&=& \frac{|-x_L^{-1}(S^-)^2 +(1-\Omega) R^-S^- +x_S (R^-)^2|^2}{2|1+\Omega|^2} e^{-(\Gamma_S+\Gamma_L)t_a}
[e^{-\Gamma_L\Delta t} + e^{-\Gamma_S\Delta t} -2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \cos (\Delta m \Delta t)], \\
I(l^+, t_a; l^-, t_b) &=&   e^{-(\Gamma_S+\Gamma_L)t_a}
\{|U|^2 e^{-\Gamma_L\Delta t}+ |V|^2 e^{-\Gamma_S\Delta t} + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (U^*V)\cos (\Delta m \Delta t) + \Im (U^*V) \sin  (\Delta m \Delta t) ]\},
\\
I(l^-, t_a; l^+, t_b) &=&   e^{-(\Gamma_S+\Gamma_L)t_a}
\{|V|^2 e^{-\Gamma_L\Delta t}+ |U|^2 e^{-\Gamma_S\Delta t} + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (U^*V)\cos (\Delta m \Delta t) - \Im (U^*V) \sin  (\Delta m \Delta t) ]\},
\end{array} \label{il}$$ where $\Delta m \equiv m_L-m_S$, $\Delta t \equiv t_b-t_a $.

 CP and CPT violations in joint decays into flavor eigenstates \[violation\]
============================================================================

The above four joint rates can form some asymmetries among them. As $
I'(l^x, l^y, \Delta t)e^{-(\Gamma_L+\Gamma_S)t_a} = \frac{1}{\Gamma_L+\Gamma_S} I(l^x, t_a; l^y, t_b),$ the asymmetries defined for the instantaneous joint rate $I(l^x, t_a; l^y, t_a+\Delta t )$ and its integration $I'(l^x, l^y, \Delta t)$ for a specific $\Delta t$ are equal, and depend only on $\Delta t $, $$A(l^{x} l^{y}, l^{z}l^{w}, \Delta t) \equiv \frac{I[l^x,t_a; l^y, t_a+\Delta t ]-I[l^z,t_a; l^w, t_a+\Delta t ]}{I[l^x,t_a; l^y, t_a+\Delta t ]+ I[l^z,t_a; l^w, t_a+\Delta t ]}=  \frac{I'[l^x, l^y, \Delta t]-I'[l^z, l^w, \Delta t]}
{I'[l^x, l^y, \Delta t]+I'[l^z, l^w, \Delta t]}.$$

Equal-flavor asymmetry 
-----------------------

Now we consider the equal-flavor asymmetry, which is $$A(++,--, \Delta t) = \frac{|x_S (S^+)^2-x_L^{-1}(R^+)^2 +(1-\Omega) R^+S^+ |^2 - |x_S (R^-)^2-x_L^{-1}(S^-)^2 +(1-\Omega) R^-S^- |^2 }{|x_S (S^+)^2-x_L^{-1}(R^+)^2 +(1-\Omega) R^+S^+ |^2 + |x_S (R^-)^2-x_L^{-1}(S^-)^2 +(1-\Omega) R^-S^- |^2 }, \label{a10}$$ which, because of the antisymmetry of $|\Psi_-\rangle$, is analogous to the famous Kabir asymmetry defined for the difference between the transition rates from $M^0$ to $\bar{M}^0$ and that from $\bar{M}^0$ to $M^0$ [@bigi]. Note that $A(++,--, \Delta t)$ is a constant independent of $\Delta t$.

If CP is conserved directly, then $R^+ = R^-$, $S^+ = S^-$, consequently $$A(++,--, \Delta t) = \frac{|x_S (\zeta_+)^2-x_L^{-1} +(1-\Omega) \zeta_+ |^2 - |x_S-x_L^{-1}(\zeta_+)^2 +(1-\Omega) \zeta_+ |^2 }{|x_S (\zeta_+)^2-x_L^{-1} +(1-\Omega) \zeta_+ |^2 - |x_S-x_L^{-1}(\zeta_+)^2 +(1-\Omega) \zeta_+ |^2  },  \label{cpd}$$ where $$\zeta_\pm \equiv \frac{S^\pm}{R^\pm},$$ which characterizes the violation of $\Delta {\cal F} =\Delta Q$ rule.

If CPT is conserved directly, then $(R^+)^*=R^-$ and $(S^+)^*=S^-$, consequently $$A(++,--, \Delta t) = \frac{|x_S (\zeta_+)^2-x_L^{-1} +(1-\Omega) \zeta_+ |^2 - |x_S^*-(x_L^*)^{-1}(\zeta_+)^2 +(1-\Omega^*) \zeta_+ |^2 }{|x_S (\zeta_+)^2-x_L^{-1} +(1-\Omega) \zeta_+ |^2 + |x_S^*-(x_L^*)^{-1}(\zeta_+)^2 +(1-\Omega^*) \zeta_+ |^2  }.$$

We can now obtain $A(++,--, \Delta t)$ for various combination cases.

If both CP and CPT are conserved directly, then $R^+ = R^-$ and $S^+ = S^-$ are both real numbers, consequently $A(++,--, \Delta t)$ is still given by (\[cpd\]), now with $\zeta_+$ being real number.

If CP is conserved indirectly, no matter whether CPT is conserved indirectly, then $\epsilon_M=\delta_M=0$ and thus $x_S=x_L=\Omega=1$, consequently $$A(++,--, \Delta t) = \frac{|(S^+)^2-(R^+)^2  |^2 - |(R^-)^2-(S^-)^2 |^2 }{|(S^+)^2-(R^+)^2  |^2 + |(R^-)^2-(S^-)^2 |^2 }.$$

If CPT is conserved indirectly, then $ \delta_M=0$, thus $x_S=x_L =q/p$, and thus $\Omega =1$, consequently $$A(++,--, \Delta t) = \frac{|\frac{q}{p}(S^+)^2-\frac{p}{q}(R^+)^2  |^2 - |\frac{q}{p}(R^-)^2-\frac{p}{q}(S^-)^2 |^2 }{|\frac{q}{p}(S^+)^2-\frac{p}{q}(R^+)^2  |^2 + |\frac{q}{p}(R^-)^2-\frac{p}{q}(S^-)^2 |^2 }.$$

If CP is conserved both directly and indirectly, then $$A(++,--, \Delta t) = 0. \label{zero1}$$

If CPT is conserved both directly and indirectly, then $$\begin{aligned}
A(++,--, \Delta t) &  = & \frac{|\frac{q}{p}(\zeta_+)^2- \frac{p}{q}   |^2 - |\frac{q^*}{p^*}- \frac{p^*}{q^*}  (\zeta_+)^2   |^2 }{  |\frac{q}{p}(\zeta_+)^2- \frac{p}{q}   |^2 + |\frac{q^*}{p^*}- \frac{p^*}{q^*}  (\zeta_+)^2   |^2 } \\ & = & \frac{(|\frac{q}{p}|^2-|\frac{p}{q}|^2)(|\zeta_+|^4-1)}{
(|\frac{q}{p}|^2+|\frac{p}{q}|^2)(|\zeta_+|^4+1)-4\Re (\frac{qp^*}{pq^*}(\zeta_+)^2)}. \label{a1}\end{aligned}$$

If CP is conserved directly while CPT is conserved indirectly, then $$\begin{aligned}
A(++,--, \Delta t) & = &  \frac{|\frac{q}{p}(\zeta_+)^2- \frac{p}{q}   |^2 - |\frac{q}{p}- \frac{p}{q}  (\zeta_+)^2   |^2 }{  |\frac{q}{p}(\zeta_+)^2- \frac{p}{q}   |^2 + |\frac{q}{p}- \frac{p}{q}  (\zeta_+)^2   |^2 }  \\  & = & \frac{(|\frac{q}{p}|^2-|\frac{p}{q}|^2)(|\zeta_+|^4-1)- 4\Im (\frac{qp^*}{pq^*})\Im((\zeta_+)^2) }{
(|\frac{q}{p}|^2+|\frac{p}{q}|^2)(|\zeta_+|^4+1)-4\Re (\frac{qp^*}{pq^*})\Re((\zeta_+)^2)}.\end{aligned}$$

If CP is conserved indirectly while CPT is conserved directly, then $$A(++,--, \Delta t) = 0. \label{zero2}$$

Combining the above two cases of $A(++,--, \Delta t) = 0$, as stated in (\[zero1\]) and (\[zero2\]), we obtain the following exact theorems.

[**Theorem 1**]{} [*If the equal-flavor asymmetry $A(++,--,\Delta t) \neq 0$, then we have one or two of the following violations: (1) CP is violated indirectly, (2) both CP and CPT are violated directly. Therefore assuming direct $CPT$ conservation, nonzero $A(++,--, \Delta t)$ implies that CP must be violated indirectly.* ]{}

Moreover, under the assumption that CPT is conserved both directly and indirectly, if $A(++,--,\Delta t) \neq 0$, then both factors in the numerator in (\[a1\]) must be nonzero. Hence we have the following exact theorem, which is clearly consistent with Theorem 1, and is especially useful in testing T violation.

[**Theorem 2**]{} [*If the equal-flavor asymmetry $A(++,--,\Delta t) \neq 0$ while CPT is assumed to be conserved both directly and indirectly, then in addition to indirect CP violation, we can draw the following conclusions: (1) $|q/p| \neq 1$, i.e. T must also be violated indirectly; (2) $|\zeta_+| \neq 1$, i.e. $|\langle l^+ |{\cal H}|\bar{M}_0\rangle| \neq |\langle l^+|{\cal H}  |M^0\rangle|$, $|\langle  l^-|{\cal H}  |M^0\rangle | \neq |\langle l^- |{\cal H}|\bar{M}_0\rangle| $, despite $ \langle l^+|{\cal H}  |M^0\rangle = \langle l^- |{\cal H}|\bar{M}_0\rangle^*$ and $\langle l^+ |{\cal H}|\bar{M}_0\rangle  = \langle  l^-|{\cal H}  |M^0\rangle^* $.* ]{}

Unequal-flavor asymmetry 
-------------------------

Now we consider the unequal-flavor asymmetry $A(+-,-+, \Delta t)$, which is given in general by $$A(+-,-+, \Delta t) = \frac{(|U|^2-|V|^2)(e^{-\Gamma_L\Delta t}-e^{-\Gamma_S\Delta t})+4\Im(U^*V)\sin (\Delta m\Delta t)}{(|U|^2+|V|^2)(e^{-\Gamma_L\Delta t}+e^{-\Gamma_S\Delta t})+4\Re(U^*V)\cos (\Delta m\Delta t) }, \label{genuneq}$$ with $U$ and $V$ given in (\[u0\]) and (\[v0\]). $A(+-,-+, \Delta t=0)$ always vanishes exactly.

If CP is conserved directly, then $R^+ = R^-$, $S^+ = S^-$, consequently $$\begin{aligned}
U & = & \frac{1}{\sqrt{2}(1+\Omega)}
[(x_S-x_L^{-1}) R^+ S^+  + (R^+)^2 -\Omega (S^+)^2], \label{ucp0} \\
V  & = & \frac{1}{\sqrt{2}(1+\Omega)}
[(x_L^{-1}- x_S) R^+ S^+   + \Omega (R^+)^2 - (S^+)^2], \label{vcp0}\end{aligned}$$ hence $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) with the following replacement, $$\begin{array}{rcl}
|U|^2-|V|^2& \rightarrow &  (1-|\Omega|^2)(1-|\zeta_+|^4)+2\Re [(1+\Omega^*)(x_S-x_L^{-1})\zeta_+(1-{\zeta_+^*}^2)] -4 \Im({\zeta_+^*}^2)\Im \Omega, \\
\Im(U^*V) &\rightarrow & \Im[-(x_S-x_L^{-1})\zeta_+(1+\Omega^*)(1-{\zeta_+^*}^2)   -\zeta_+^2 - |\Omega|^2 {\zeta_+^*}^2 +\Omega +  \Omega^*|\zeta_+|^4],\\
|U|^2+|V|^2 &\rightarrow & 2|x_S-x_L^{-1}|^2|\zeta_+|^2  + (1+|\Omega|^2)(1+|\zeta_+|^4)+2\Re [(1-\Omega^*)(x_S-x_L^{-1})\zeta_+(1+{\zeta_+^*}^2)   ] -4 \Re({\zeta_+}^2)\Re \Omega,\\
\Re(U^*V)&\rightarrow & -|x_S-x_L^{-1}|^2|\zeta_+|^2 +\Re[-(x_S-x_L^{-1})\zeta_+(1-\Omega^*)(1+{\zeta_+^*}^2)   -\zeta_+^2 - |\Omega|^2 {\zeta_+^*}^2 +\Omega +  \Omega^*|\zeta_+|^4].
\end{array} \label{sub0}$$

Furthermore, if CP is conserved directly while CPT is conserved indirectly, then $$\begin{aligned}
U & = & \frac{1}{\sqrt{2}(1+\Omega)}
[(\frac{q}{p}-\frac{p}{q}) R^+ S^+  + (R^+)^2 - (S^+)^2], \label{ucp} \\
V  & = & \frac{1}{\sqrt{2}(1+\Omega)}
[(\frac{p}{q}- \frac{q}{p}) R^+ S^+   + (R^+)^2 - (S^+)^2], \label{vcp}\end{aligned}$$ consequently, $$A(+-,-+, \Delta t) =
 \frac{2 \Re [(\frac{q}{p}- \frac{p}{q})\zeta_+ (1-{\zeta_+^*}^2)](e^{-\Gamma_L\Delta t}-e^{-\Gamma_S\Delta t})-4\Im[(\frac{q}{p}- \frac{p}{q})\zeta_+(1- {\zeta_+^*}^2)] \sin (\Delta m\Delta t)}{  [ |\frac{q}{p}- \frac{p}{q}|^2 |\zeta_+|^2 +  |1- {\zeta_+}^2|^2 ]  (e^{-\Gamma_L\Delta t}+e^{-\Gamma_S\Delta t})+ 2[ -|\frac{q}{p}- \frac{p}{q}|^2 |\zeta_+|^2 +  |1- {\zeta_+}^2|^2 ] \cos (\Delta m\Delta t) }.$$

If CPT is conserved directly, then $(R^+)^*=R^-$ and $(S^+)^*=S^-$, consequently $$\begin{aligned}
U & = & \frac{1}{\sqrt{2}(1+\Omega)}
[-x_L^{-1} R^+ (S^+)^* +x_S (R^+)^* S^+  + |R^+|^2 -\Omega |S^+|^2], \\
V  & = & \frac{1}{\sqrt{2}(1+\Omega)}
[x_L^{-1} R^+ (S^+)^* - x_S (R^+)^* S^+  + \Omega |R^+|^2 - |S^+|^2],\end{aligned}$$ hence $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) with the following replacement, $|U|^2-|V|^2 \rightarrow   (1-|\Omega|^2)(1-|\zeta+|^4)+2\Re\{ x_S  (1+\Omega^*)\zeta_+ - x_L^{-1} (1-\Omega^*)\zeta_+^*  \}(1-|\zeta+|^2)$, $|U|^2 + |V|^2 \rightarrow  2 (|x_L^{-1}|^2+ |x_S|^2-2 \Re \Omega) |\zeta_+|^2 +  (1+ |\Omega|^2) (1+|\zeta_+|^4) -4 \Re ({x_L^{-1}}^*x_S \zeta_+^2 ) +2 \Re\{ [ (1-\Omega^*)x_S -(1-\Omega) {x_L^{-1}}^*]\zeta_+\} (1 +  |\zeta_+|^2)$, $U^* V \rightarrow -(1+|\Omega|^2 + |x_S|^2+|x_L^{-1}|^2) |\zeta_+|^2 + \Omega  + \Omega^* |\zeta_+|^4 + {x_L^{-1}}^* x_S {\zeta_+}^2 -(\Omega{x_L^{-1}}^* + x_S) \zeta_+  + ({x_L^{-1}}^*  + \Omega^* x_S ) \zeta_+ |\zeta_+|^2 + (\Omega x_S^* +x_L^{-1} )  {\zeta_+}^*  -( x_S^* +\Omega^* x_L^{-1} )  {\zeta_+}^* |\zeta_+|^2 + x_S^*x_L^{-1}  {{\zeta_+}^*}^2$.

Furthermore, if CPT is conserved directly while CP is conserved indirectly, then $$\begin{aligned}
U & = & \frac{1}{2\sqrt{2}}
[-R^+ (S^+)^*  + S^+(R^+)^* + |R^+|^2 - |S^+|^2], \\
V  & = & \frac{1}{\sqrt{2}}
[R^+ (S^+)^* -  S^+(R^+)^*  + |R^+|^2 - |S^+|^2].\end{aligned}$$ Consequently $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) under the replacement specified in the last paragraph now with $x_L=x_S=\Omega =1$, therefore $$A(+-,-+, \Delta t) =
 \frac{ 2(1-|\zeta_+|^2)\Re(\zeta_+)(e^{-\Gamma_L\Delta t}-e^{-\Gamma_S\Delta t})-4(1-|\zeta_+|^2)\Im(\zeta_+) \sin (\Delta m\Delta t)}{   |1-\zeta_+^2|^2 (e^{-\Gamma_L\Delta t}+e^{-\Gamma_S\Delta t})+ 2[ 1+|\zeta_+|^4 -4|\zeta_+|^2 + 2\Re (\zeta_+^2) ] \cos (\Delta m\Delta t) }.$$

If both CP and CPT are conserved directly, $R^+ = R^-$ and $S^+ = S^-$ are both real numbers, consequently $U$ and $V$ are given as (\[ucp0\]) and (\[vcp0\]), while $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) with the replacement (\[sub0\]), but now with $\zeta_+=\zeta_+^*=\Re(\zeta_+^*)$ while $\Im ({\zeta^*}^2)=0$.

If CPT is conserved indirectly, then $x_S=x_L =q/p$, $\Omega =1$, hence $$\begin{aligned}
U & = & \frac{1}{2\sqrt{2}}
[-\frac{p}{q} R^+ S^-  + \frac{q}{p} S^+R^- + R^+ R^- - S^+ S^-], \label{upq} \\
V & = & \frac{1}{\sqrt{2}}
[\frac{p}{q}R^+ S^- - \frac{q}{p} S^+R^-  + R^+ R^- - S^+ S^-].  \label{vpq}\end{aligned}$$ Consequently $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) with the following replacement, $|U|^2-|V|^2 \rightarrow   4 \Re (-\frac{p^*}{q^*}  {\zeta_-}^*  + \frac{p^*}{q^*} |\zeta_-|^2 \zeta_+  + \frac{q^*}{p^*} {\zeta+}^*  -   \frac{q^*}{p^*} |\zeta_+|^2 \zeta_-) $, $|U|^2 + |V|^2 \rightarrow  2 (|\frac{p}{q} \zeta_-|^2 + |\frac{q}{p} \zeta_+ |^2 + 1 + |\zeta_+\zeta_-|^2 ) - 4 \Re ( \frac{p^*q}{q^*p} {\zeta_-}^* \zeta_+  +  \zeta_+ \zeta_-)$, $\Re (U^*V) \rightarrow 1 -|\frac{p}{q} \zeta_-|^2 - |\frac{q}{p} \zeta_+ |^2  + |\zeta_+ \zeta_-|^2 +  2\Re ( \frac{p^*q}{q^*p} \zeta_+{\zeta_-}^*  - \zeta_+\zeta_- ) $, $\Im (U^*V) \rightarrow    2\Im ( \frac{p}{q} \zeta_- +  \frac{p^*}{q^*} |\zeta_-|^2 \zeta_+ - \frac{q}{p} \zeta_+ - \frac{q^*}{p^*} |\zeta_+|^2\zeta_- ) $.

If CP is conserved indirectly, then no matter whether CPT is conserved indirectly, $x_S=x_L=\Omega=1$, consequently $U$ and $V$ are given by (\[upq\]) and (\[vpq\]) with $p/q=1$, hence $A(+-,-+, \Delta t)$ is given by (\[genuneq\]) with the following replacement, $|U|^2-|V|^2 \rightarrow   4 \Re (-  {\zeta_-}^*  +  |\zeta_-|^2 \zeta_+  +   {\zeta+}^*  -   |\zeta_+|^2 \zeta_-) $, $|U|^2 + |V|^2 \rightarrow  2 (|  \zeta_-|^2 + | \zeta_+ |^2 + 1 + |\zeta_+\zeta_-|^2) - 8\Re(\zeta_-) \Re(\zeta_+)$, $\Re (U^*V) \rightarrow 1 -| \zeta_-|^2 - | \zeta_+ |^2  + |\zeta_+ \zeta_-|^2 +  4 \Im (   \zeta_+) \Im (\zeta_-)$, $\Im (U^*V) \rightarrow    2\Im ( \zeta_- +  |\zeta_-|^2 \zeta_+ -   \zeta_+ - |\zeta_+|^2\zeta_- ) $.

If CP is conserved both directly and indirectly, $$U  = V  =  \frac{1}{2\sqrt{2}}
[(R^+)^2 - (S^+)^2],  \label{uvcp2}$$ which, according to (\[genuneq\]), implies $$A(+-,-+, \Delta t)=0. \label{cpuneq}$$ Hence we have the following exact theorem.

[**Theorem 3**]{} [*If the unequal-flavor asymmetry $A(+-,-+,  \Delta t) \neq 0 $, then CP must be violated, directly or indirectly or both.* ]{}

If CPT is conserved both directly and indirectly, then $$\begin{aligned}
U & = & \frac{1}{2\sqrt{2} }
[-\frac{p}{q} R^+ (S^+)^* +\frac{q}{p}  (R^+)^* S^+  + |R^+|^2 - |S^+|^2], \\
V  & = & \frac{1}{2\sqrt{2} }
[\frac{p}{q} R^+ (S^+)^* - \frac{q}{p}  (R^+)^* S^+  + |R^+|^2 - |S^+|^2].\end{aligned}$$ Consequently, in obtaining $ A(+-,-+,  \Delta t)$, the replacement for the four functions of $U$ and $V$ is given by that for the case of direct CPT conservation with $\zeta_-=\zeta_+^*$. Therefore $$A(+-,-+, \Delta t) = \frac{(1-|\zeta_+|^2)\{ -2\Re [(\frac{p}{q}- \frac{q^*}{p^*})\zeta_-](e^{-\Gamma_L\Delta t}-e^{-\Gamma_S\Delta t})+4\Im[(\frac{p}{q}+ \frac{q^*}{p^*})\zeta_-] \sin (\Delta m\Delta t)}{E_+   (e^{-\Gamma_L\Delta t}+e^{-\Gamma_S\Delta t})+2 E_- \cos (\Delta m\Delta t) },$$ where $E_{\pm}= (1-|\zeta_+|^2)^2 \pm  |\frac{p}{q}\zeta_- - \frac{q}{p}\zeta_+|^2$. Therefore we have the following exact theorem.

[**Theorem 4**]{} [*If the unequal-flavor asymmetry $A(+-,-+,  \Delta t)\neq 0$ for $\Delta t \neq 0$ while CPT is assumed to be conserved both directly and indirectly, then we can draw the following conclusions: (1) $|\zeta_+| = |\zeta_-| \neq 1$, i.e., $|\langle l^+ |{\cal H}|\bar{M}_0\rangle|= |\langle  l^-|{\cal H}  |M^0\rangle | \neq |\langle l^+|{\cal H}  |M^0\rangle|= |\langle l^- |{\cal H}|\bar{M}_0\rangle| $; (2) moreover, $\zeta_- =\zeta_+^* \neq 0$, i.e., $\langle l^{-}|{\cal H} |M^0\rangle = \langle l^{+}|{\cal H}|\bar{M}^0\rangle^* \neq 0$, which means $\Delta {\cal F} =\Delta Q$ rule must be violated.* ]{}

Decays into CP eigenstates \[cp\] 
==================================

Now we consider the situation that the decay products of Alice and Bob are both CP eigenstates, as in KLOE experiment, where the rate of both kaons decaying to $\pi^+\pi^-$ was obtained up to a proportional factor [@kloe; @domenico]. We generically denote the CP eigenstates as $|h^+\rangle$ with eigenvalue $+1$ and $|h^-\rangle$ with eigenvalue $-1$. Examples for $|h^+\rangle$ include $\pi^+\pi^-$, $\pi^0\pi^0$, etc. Examples for $|h^-\rangle$ include $\pi^0\pi^0\pi^0$, etc. For convenience of discussions, we introduce the parameters defined as $$w_{\pm\alpha}^x \equiv \langle h^x_{\alpha}|{\cal H}_{\alpha}|M_{\pm}\rangle_\alpha,$$ where $x=\pm 1$. These parameters were not introduced previously, because an isolated single meson cannot be in the state $|M_\pm\rangle$ physically. However, the entangled state $|\Psi_-\rangle$ renders the introduction of these parameters meaningful and convenient, because the entangled state can be exactly written in terms of the CP basis, as in (\[pm\]), hence an entangled pair of mesons decay from superposition of direct products of CP eigenstates. No that there is no tagging in CP basis, only the final decay products, e.g. pions, are measured. Therefore the use of parameter $w_{\pm\alpha}^{x}$ is legitimate. This is a remarkable point we would like to exploit. The parameter $w_{\pm}^x$ may be measured by using entangled mesons. They cannot be directly measured by using single mesons, as there is no way to prepare a single meson in a CP eigenstate. Nevertheless, by using (\[cptof\]), we have $$w_{\pm}^x \equiv \frac{1}{\sqrt{2}}(\langle h^x|{\cal H}|M^0\rangle \pm\langle h^x|{\cal H}|\bar{M}^0\rangle),$$ where each term on the RHS can be measured for isolated single mesons. Analogously, $w_{\pm}^x$ can also be related to parameters for mass basis, by using (\[lm2\]), $$\begin{aligned}
w_+^x =
\frac{1}{1-\epsilon_S\epsilon_L}(\sqrt{1+|\epsilon_S|^2}\langle h^x|{\cal H}|M_S\rangle-
\epsilon_S\sqrt{1+|\epsilon_L|^2}\langle h^x|{\cal H}|M_L\rangle), \\
w_-^x  =
\frac{1}{1-\epsilon_S\epsilon_L}(\sqrt{1+|\epsilon_L|^2}\langle h^x|{\cal H}|M_L\rangle-
\epsilon_L\sqrt{1+|\epsilon_S|^2}\langle h^x|{\cal H}|M_S\rangle),\end{aligned}$$ where each term on the RHS of each identity can be measured for isolated single mesons.

One can calculate the amplitude of such a joint decay in which Alice decays to $| h_a^x\rangle $ at $t_a$ while Bob decays to $|h_b^y\rangle $ at $t_b$, where $x$ and $y$ each represents $\pm 1$, $$\begin{aligned}
\langle h_a^x,h_b^y|{\cal H}_a {\cal H}_b |\Psi_-(t_a,t_b)\rangle & = &   \frac{1}{\sqrt{2}}(\langle h_a^x|{\cal H}_a|M^+(t_a)\rangle_a \langle h_b^y|{\cal H}_b |M_-(t_b)\rangle_b
-\langle h_a^x|{\cal H}_a|M^-(t_a)\rangle_a \langle h_b^y|{\cal H}_b |M_+(t_b)\rangle_b) \\
&=&  M(h_a^x,h_b^y) e^{-i(\lambda_St_a+\lambda_Lt_b)} + N(h_a^x,h_b^y) e^{-i(\lambda_Lt_a+\lambda_St_b)},\end{aligned}$$ where $ M(h_a^x,h_b^y)   \equiv  [- (1 -x_L^{-1} + x_S-\Omega) w_{+a}^x w_{+b}^y
+ (1 +x_L^{-1} + x_S+\Omega)w_{+a}^x w_{-b}^y
- (1 -x_L^{-1} - x_S+\Omega)w_{-a}^x w_{+b}^y
+ (1 +x_L^{-1} - x_S-\Omega)w_{-a}^x w_{-b}^y ]/[2\sqrt{2}(1+\Omega)]$, $
N(h_a^x,h_b^y) \equiv  [ (1 -x_L^{-1} + x_S-\Omega)w_{+a}^x w_{+b}^y + (1- x_L^{-1}- x_S+\Omega) w_{+a}^x w_{-b}^y -
(1 +x_L^{-1} + x_S+\Omega)w_{-a}^x w_{+b}^y
- (1 +x_L^{-1} - x_S-\Omega)w_{-a}^x w_{-b}^y ]/[2\sqrt{2}(1+\Omega)]$.

Therefore we obtain the joint rate $$\begin{array}{rcl}
I(h^x_a, t_a; h^y_b, t_b)&=&   e^{-(\Gamma_S+\Gamma_L)t_a}
\{|M |^2 e^{-\Gamma_L\Delta t}+ |N |^2 e^{-\Gamma_S\Delta t} \\
&& + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (M^* N )\cos (\Delta m \Delta t) + \Im (M^*N) \sin  (\Delta m \Delta t) ]\}, \\
\end{array}$$ where $M\equiv M(h_a^x,h_b^y)$, $N\equiv N(h_a^x,h_b^y)$.

In experiments, it is more convenient to use the integrated rate $$I'(h_a^x,h_b^y, \Delta t) = \int_0^\infty I(h_a^x,t_a; h_b^y, t_a+\Delta t) dt_a,$$ which is simply given by $I(l^x_a, t_a; l^y_b, t_a+\Delta t)$ as in (\[il\]) with $e^{-(\Gamma_S+\Gamma_L)t_a}$ replaced as $1/(\Gamma_S+\Gamma_L)$.

We focus on the situation that $|h_a^+\rangle = |h_b^+\rangle=|h^+\rangle$ and $|h_a^-\rangle = |h_b^-\rangle=|h^-\rangle$, hence $$\begin{aligned}
w_{+a}^+=w_{+b}^+  & \equiv  & Q^+ , \\
w_{-a}^{+}=w_{-b}^{+} &  \equiv& X^{+}, \\
w_{+a}^-=w_{+b}^-  & \equiv  & X^-, \\
w_{-a}^{-}=w_{-b}^{-} &  \equiv& Q^{-}.\end{aligned}$$ It is straightforward to find the following properties. If CP is conserved directly, then $X^{\pm}=0$, hence $X^{\pm}$ is a parameter characterizing the direct CP violation. If CPT is conserved directly, then $X^{\pm}$ is purely imaginary, i.e. $X^{\pm}=-{X^{\pm}}^*$.

We define $$\xi_{\pm} \equiv \frac{X^{\pm}}{Q^{\pm}}.$$

$Q^{\pm}$ and $X^{\pm}$ are not directly measurable quantities, as $M_S$ and $M_L$, rather than $M_+$ and $M_-$, are physical. However, from (\[ms1\]) and (\[ml1\]), we have $$\begin{aligned}
Q^++\epsilon_S X^+ &=&\sqrt{1+|\epsilon_S|^2} \langle h^+|{\cal H}|M_S\rangle, \\
\epsilon_L X^- + Q^-&=&\sqrt{1+|\epsilon_L|^2} \langle h^-|{\cal H}| M_L\rangle.\end{aligned}$$ Therefore, $$\begin{aligned}
\eta_{h^+} & \equiv &\frac{\langle h^+|{\cal H}|M_L\rangle}{\langle h^+|{\cal H}|M_S\rangle} = \frac{\xi^++\epsilon_L}{1+\epsilon_S \xi^+ },\\
\eta_{h^-} & \equiv & \frac{\langle h^-|{\cal H}|M_S\rangle}{\langle h^-|{\cal H}|M_L\rangle} = \frac{\xi^- +\epsilon_S}{1+\epsilon_L \xi^-}.\end{aligned}$$ In the case of $|h^+\rangle = |\pi^+\pi^-\rangle$, $|\eta_{h^+}\rangle$ is just the well-known $$\eta_{+-} = \frac{\langle \pi^+\pi^-|{\cal H}|M_L\rangle}{\langle \pi^+\pi^-|{\cal H}|M_S\rangle}.$$ In the case of $|h^+\rangle = |\pi^0\pi^0\rangle$, $|\eta_{h^+}\rangle$ is just the well-known $$\eta_{00} = \frac{\langle \pi^0\pi^0|{\cal H}|M_L\rangle}{\langle \pi^0\pi^0|{\cal H}|M_S\rangle}.$$

One obtains $$\begin{aligned}
M(h^+,h^+)  & = & -N(h^+,h^+)= \frac{1}{2\sqrt{2}(1+\Omega)}
[-(1 -x_L^{-1}+ x_S-\Omega)(Q^+)^2  + 2(x_L^{-1} + x_S) Q^+X^+
\nonumber \\&&  + (1 +x_L^{-1} - x_S-\Omega)(X^+)^2], \\
M(h^-,h^-)  & = & -N(h^-,h^-)= \frac{1}{2\sqrt{2}(1+\Omega)}
[-(1 -x_L^{-1}+ x_S-\Omega)(X^-)^2  + 2(x_L^{-1} + x_S) X^-Q^-
\nonumber \\&&  + (1 +x_L^{-1} - x_S-\Omega)(Q^-)^2], \\
M(h^+,h^-)  & = & -M(h^-,h^+)\nonumber \\
&=&  \frac{1}{2\sqrt{2}(1+\Omega)}[(1 + x_S) Q^+  + (1- x_S) X^+][(1 +x_L^{-1})Q^- - (1 - x_L^{-1})X^-] \equiv Z, \\
N(h^+,h^-)  & = & -N(h^-,h^+)\nonumber \\
 &= & \frac{1}{2\sqrt{2}(1+\Omega)}[(1 - x_L^{-1}) Q^+  - (1 +x_L^{-1})X^+][ (1- x_S)Q^- +(1 + x_S) X^-] \equiv Y.\end{aligned}$$

Therefore, we obtain $$\begin{array}{rcl}
I(h^+, t_a; h^+, t_b) & = &
\frac{| -(1 -x_L^{-1}+ x_S-\Omega)(Q^+)^2  + 2(x_L^{-1} + x_S) Q^+X^+
+ (1 +x_L^{-1} - x_S-\Omega)(X^+)^2   |^2  }{8|1+\Omega|^2}   \\ &&\times e^{-(\Gamma_S+\Gamma_L)t_a} [e^{-\Gamma_S \Delta t} + e^{-\Gamma_L\Delta t} - 2 e^{-\frac{1}{2}(\Gamma_S+\Gamma_L)\Delta t} \cos(\Delta m \Delta t)], \\
I(h^-, t_a; h^-, t_b) & = &
\frac{|-(1 -x_L^{-1}+ x_S-\Omega)(X^-)^2  + 2(x_L^{-1} + x_S) X^-Q^-
+ (1 +x_L^{-1} - x_S-\Omega)(Q^-)^2  |^2  }{8|1+\Omega|^2}   \\ &&\times e^{-(\Gamma_S+\Gamma_L)t_a} [e^{-\Gamma_S \Delta t} + e^{-\Gamma_L\Delta t} - 2 e^{-\frac{1}{2}(\Gamma_S+\Gamma_L)\Delta t} \cos(\Delta m \Delta t)], \\
I(h^+, t_a; h^-, t_b) & = &  e^{-(\Gamma_S+\Gamma_L)t_a}
\{|Z|^2 e^{-\Gamma_L\Delta t}+ |Y|^2 e^{-\Gamma_S\Delta t} + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (Z^*Y)\cos (\Delta m \Delta t) + \Im (Z^*Y) \sin  (\Delta m \Delta t) ]\},
\\
I(h^-, t_a; h^+, t_b) &=&   e^{-(\Gamma_S+\Gamma_L)t_a}
\{|Y|^2 e^{-\Gamma_L\Delta t}+ |Z|^2 e^{-\Gamma_S\Delta t} + 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re (Z^*Y)\cos (\Delta m \Delta t) - \Im (Z^*Y) \sin  (\Delta m \Delta t) ]\}.
\end{array} \label{i2}$$

One can also consider the integrated rate $$I'(h^x, h^y, \Delta t) = \int_0^\infty I(h^x, t_a; h^y, t_a+\Delta t) dt_a, \label{i22}$$ which is simply given by $I(h^x, t_a; h^y, t_a+\Delta t)$ as in (\[i2\]), with $e^{-(\Gamma_S+\Gamma_L)t_a}$ replaced as $1/(\Gamma_S+\Gamma_L)$.

CP and CPT violations in decays in CP basis \[cp2\]
===================================================

First, it can be seen that the equal-CP decay rates $I[h^+, t_a; h^+, t_b]$ and $I[h^-, t_a; h^-, t_b]$ are both proportional $  e^{-\Gamma_S \Delta t} + e^{-\Gamma_L \Delta t} - 2 e^{-\frac{1}{2}(\Gamma_S+\Gamma_L)\Delta t} \cos(\Delta m \Delta t),$ as demonstrated by KLOE experimental data [@domenico; @kloe].

For $\Delta t=0$, we always have $I[h^+, t_a; h^+, t_a]=[h^-, t_a; h^-, t_a]=I'[h^+, h^+, 0]=[h^-,  h^-,0]=0$ no matter whether CP or CPT is violated.

The four joint rates in (\[i2\]) or (\[i22\]) can form some asymmetries between different modes of decays into CP eigenstates. As $
I'(h^x, h^y, \Delta t)e^{-(\Gamma_L+\Gamma_S)t_a} = \frac{1}{\Gamma_L+\Gamma_S} I(h^x, t_a; h^y, t_b),$ the asymmetries defined for the instantaneous joint probability $I$ and its integration $I'$ for a specific $\Delta t$ are equal, and depend only on $\Delta t \equiv t_b-t_a$, $$B(h^xh^y, h^yh^x, \Delta t) \equiv \frac{I[h^x,t_a; h^y,t_a+\Delta t ]-I[h^y,t_a; h^x,t_a+\Delta t]}{I[h^x,t_a; h^y,t_a+\Delta t]+ I[h^y,t_a; h^x,t_a+\Delta t]}=  \frac{I'[h^x, h^y, \Delta t]-I'[h^y, h^x, \Delta t]}
{I'[h^x, h^y, \Delta t]+I'[h^y, h^x, \Delta t]},$$ which is always time-independent.

In general, the equal-CP asymmetry is found to be $$B(++,--, \Delta t) = \frac{ P(Q^+,X^+)  - P(X^-,Q^-)  }{P(Q^+,X^+) + P(X^-,Q^-)  }, \label{bb10}$$ where $$P(\beta,\gamma) \equiv | -(1 -x_L^{-1}+ x_S-\Omega)\beta^2  + 2(x_L^{-1} + x_S)\beta\gamma + (1 +x_L^{-1} - x_S-\Omega)\gamma^2|^2.$$

[**Theorem 5**]{} [*The equal-CP asymmetry $B(++,--, \Delta t)$ is always a constant independent of $\Delta t$.* ]{}

In general, the unequal-CP asymmetry is found to be $$\begin{aligned}
B(+-,-+, \Delta t)& =& \frac{(|Z|^2-|Y|^2) ( e^{-\Gamma_L\Delta t} -e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Im(Z^*Y) \sin (\Delta m \Delta t) }{(|Z|^2+|Y|^2) ( e^{-\Gamma_L\Delta t} +e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Re(Z^*Y) \cos (\Delta m \Delta t) } \\
& =& \frac{(|\tilde{Z}|^2-|\tilde{Y}|^2) ( e^{-\Gamma_L\Delta t} -e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Im(\tilde{Z}^*\tilde{Y}) \sin (\Delta m \Delta t) }{(|\tilde{Z}|^2+|\tilde{Y}|^2) ( e^{-\Gamma_L\Delta t} +e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Re(\tilde{Z}^*\tilde{Y}) \cos (\Delta m \Delta t) }, \label{une}\end{aligned}$$ where $$\begin{aligned}
\tilde{Z} & \equiv & [(1 + x_S)  + (1- x_S) \xi^+][(1 +x_L^{-1})- (1 - x_L^{-1})\xi^-], \\
\tilde{Y} & \equiv & [(1 - x_L^{-1})    - (1 +x_L^{-1}) \xi^+][ (1- x_S)+(1 + x_S) \xi^-]. \label{zy1}\end{aligned}$$ When $\Delta t=0$, the unequal-CP asymmetry is always $0$ no matter whether CP or CPT is violated. Therefore we have the following result.

[**Theorem 6**]{} [*For $\Delta t =0$, the unequal-CP asymmetry $B(+-,-+, 0)$ vanishes, no matter whether CP or CPT is violated.* ]{}

If CP is conserved directly, then $X^{\pm}=0$, consequently $B(++,--, \Delta t)$ reduces to $$\begin{aligned}
B(++,--,\Delta t)
=  \frac{{|Q^+|}^4|(1 - x_L^{-1})(1 + x_S)|^2 - {|Q^-|}^4 |(1 +x_L^{-1})(1 - x_S)|^2 }{ {|Q^+|}^4|(1 - x_L^{-1})(1 + x_S)|^2 + {|Q^-|}^4 |(1 +x_L^{-1})(1 - x_S)|^2   }. \label{eqdirect} \end{aligned}$$ while $B(+-,-+,  \Delta t)$ reduces to $$\begin{aligned}
B(+-,-+,  \Delta t) = \frac{(1-|W|^2) ( e^{-\Gamma_L\Delta t} -e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t}\Im(W) \sin (\Delta m \Delta t) }{(1+|W|^2) ( e^{-\Gamma_L\Delta t} +e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t}  \Re(W)\cos (\Delta m \Delta t) }, \label{nod}\end{aligned}$$ where $W\equiv  \frac{(1-x_L^{-1})(1-x_S)}{(1+x_L^{-1})(1+x_S)}$.

If CP is conserved indirectly, no matter whether CPT is conserved indirectly, we have $x_S=x_L^{-1} =1$. Consequently, $M(h^+,h^+)   =  -N(h^+,h^+)= \frac{1}{\sqrt{2}}  Q^+X^+,$ $M(h^-,h^-)   =  -N(h^-,h^-)= \frac{1}{\sqrt{2}} X^-Q^-,$ $Z= M(h^+,h^-)   =  -M(h^-,h^+)=  \frac{1}{\sqrt{2}} Q^+ Q^- $, $Y= N(h^+,h^-)   =  -N(h^-,h^+) = -\frac{1}{\sqrt{2}} X^+  X^-$. Therefore $$\begin{array}{rcl}
I(h^+, t_a; h^+, t_b) & = &  \frac{|Q^+X^+|^2}{2}e^{-(\Gamma_S+\Gamma_L)t_a} [e^{-\Gamma_S \Delta t} + e^{-\Gamma_L\Delta t} - 2 e^{-\frac{1}{2}(\Gamma_S+\Gamma_L)\Delta t} \cos(\Delta m \Delta t)], \\
I(h^-, t_a; h^-, t_b) & = &\frac{|Q^-X^-|^2}{2}e^{-(\Gamma_S+\Gamma_L)t_a} [e^{-\Gamma_S \Delta t} + e^{-\Gamma_L\Delta t} - 2 e^{-\frac{1}{2}(\Gamma_S+\Gamma_L)\Delta t} \cos(\Delta m \Delta t)], \\
I(h^+, t_a; h^-, t_b) & = & \frac{1}{2}  e^{-(\Gamma_S+\Gamma_L)t_a}
\{|Q^+Q^-|^2 e^{-\Gamma_L\Delta t}+ |X^+X^-|^2 e^{-\Gamma_S\Delta t} \\
&&+ 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re ((Q^+Q^-)^*X^+X^-)\cos (\Delta m \Delta t) + \Im ((Q^+Q^-)^*X^+X^-) \sin  (\Delta m \Delta t) ]\},
\\
I(h^-, t_a; h^+, t_b) &=& \frac{1}{2}  e^{-(\Gamma_S+\Gamma_L)t_a}
\{|X^+X^-|^2 e^{-\Gamma_L\Delta t}+ |Q^+Q^-|^2 e^{-\Gamma_S\Delta t}  \\&&+ 2e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} [\Re ((Q^+Q^-)^*X^+X^-)\cos (\Delta m \Delta t) - \Im ((Q^+Q^-)^*X^+X^-) \sin  (\Delta m \Delta t) ]\},
\end{array} \label{indirect2}$$

Therefore if CP is conserved indirectly, no matter whether CPT is conserved indirectly, the equal-CP asymmetry is simplified to $$B(++,--, \Delta t) = \frac{|Q^+X^+|^2- |Q^-X^-|^2  }{|Q^+X^+|^2+ |Q^-X^-|^2  },  \label{eqcpas}$$ while the unequal-CP asymmetry is simplified to $$\begin{aligned}
B(+-,-+, \Delta t)& =& \frac{(1-|\xi^+\xi^-|^2) ( e^{-\Gamma_L\Delta t} -e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Im(\xi^+\xi^-) \sin (\Delta m \Delta t) }{( 1+|\xi^+\xi^-|^2  ) ( e^{-\Gamma_L\Delta t} +e^{-\Gamma_S\Delta t})+ 4 e^{-\frac{\Gamma_S+\Gamma_L}{2}\Delta t} \Re(\xi^+\xi^- ) \cos (\Delta m \Delta t) }. \label{uneq2}\end{aligned}$$

If CP is conserved both directly and indirectly, no matter whether CPT is conserved or not, we have $X^{\pm}=0$, $x_S=x_L^{-1}=1$, then $$\begin{array}{rcl}
I(h^+_a, t_a; h^+_b, t_b) & = &0, \\
I(h^-_a, t_a; h^-_b, t_b) & = &0, \\
I(h^+_a, t_a; h^-_b, t_b) & = & \frac{|Q^+Q^-|^2}{2} e^{-(\Gamma_S+\Gamma_L)t_a} e^{-\Gamma_L\Delta t},
\\
I(h^-, t_a; h^+, t_b) &=& \frac{|Q^+Q^-|^2}{2}   e^{-(\Gamma_S+\Gamma_L)t_a}  e^{-\Gamma_S\Delta t},
\end{array} \label{i2cp}$$ As $I(h^+_a, t_a; h^+_b, t_b)=I(h^-_a, t_a; h^-_b, t_b)=I'(h^+_a, h^+_b, \Delta t)=I(h^-_a,  h^-_b,\Delta t)=0$, it is meaningless to define equal-CP asymmetry in this case. In this case, the unequal-CP asymmetry is $$\begin{aligned}
B(+-,-+,  \Delta t) = \frac{ e^{-\Gamma_L\Delta t} -e^{-\Gamma_S\Delta t} }{  e^{-\Gamma_L\Delta t} +e^{-\Gamma_S\Delta t}  }, \label{noid}\end{aligned}$$

[**Theorem 7**]{} [*If any equal-CP joint decay rate is nonzero, then CP must be violated, directly or indirectly or both.* ]{}

If CPT is conserved indirectly, then $ \delta_M=0$, thus $x_S=x_L =q/p$, and thus $\Omega =1$, consequently the equal-CP asymmetry is given by (\[bb10\]) with $P(\beta,\gamma)$ simplified to $$P(\beta,\gamma) \equiv |2(\frac{p}{q} + \frac{q}{p})\beta\gamma + (\frac{p}{q} -\frac{q}{p})(\gamma^2-\beta^2)|^2. \label{eqcpt}$$ The unequal-CP asymmetry $B(+-,-+,  \Delta t)$ is given by (\[une\]) with $\tilde{Z}$ and $\tilde{Y}$ reduced to $$\begin{aligned}
\tilde{Z} & \equiv &   [(1 + \frac{q}{p})  + (1- \frac{q}{p}) \xi^+][(1 +\frac{p}{q} )- (1 - \frac{p}{q} )\xi^-], \\
\tilde{Y} & \equiv & [(1 - \frac{p}{q} )    - (1 +\frac{p}{q} ) \xi^+][ (1- \frac{q}{p})+(1 + \frac{q}{p}) \xi^-]. \label{zy2}\end{aligned}$$

If CPT is conserved indirectly while CP is conserved directly, $x_S=x_L=q/p$ while $X^{\pm}=0$, then the equal-CP asymmetry reduces to $$\begin{aligned}
B(++,--, \Delta t)
 =  \frac{{|Q^+|}^4 - {|Q^-|}^4 }{ {|Q^+|}^4 + {|Q^-|}^4   },  \label{bab}
 \end{aligned}$$ while the unequal-CP asymmetry $B(+-,-+,  \Delta t)$ is given by (\[nod\]) now with $W =  \frac{2-(\frac{p}{q}+\frac{q}{p})}{2+(\frac{p}{q}+\frac{q}{p})}$.

If CPT is conserved directly, then $X^{\pm}$ is purely imaginary, all the above results under other additional conditions are still valid, respectively, under the constraint that $X^{\pm}$ is purely imaginary.

summary \[summary\] 
====================

In this paper, we have studied the joint decays of a pair of two pseudoscalar neutral mesons in an entangled state of $C=-1$, as produced in $\phi$ and $B$ factories. We exactly calculated the rates of the joint decays into flavor eigenstates, taking into account direct CP violation and violation of $\Delta {\cal F} =\Delta Q$ rule. We obtained some exact results on how to extract information on CP and CPT violations from various asymmetries of the joint decays to flavor eigenstates, or joint decays to CP eigenstates. Measurement of such joint rates and asymmetries can be used to determine various parameters, including those of CP and CPT violations.

Remarkably, the special property of the entanglement of the $C=-1$ pair allows us to propose the meaningful and useful definition of the transition amplitude between a CP eigenstate of the meson and the decay product which is a CP eigenstate.

The equal-flavor asymmetry $A(++,--,\Delta t)$ and the equal-CP asymmetry $B(++,--,\Delta t)$ are both always independent of $\Delta t$, while the unequal-flavor asymmetry $A(+-,-+,  \Delta t)$ and the unequal-CP asymmetry $B(+-,-+,\Delta t)$ generically depend on $\Delta t$.

We have considered various cases of possible direct or indirect CP or CPT violation, and obtain exact expressions of various asymmetries in various cases. Some of these exact results lead to simple yet powerful conclusions stated as several theorems.

If the equal-flavor asymmetry $A(++,--,\Delta t) \neq 0$, then we have one or two of the following violations: (1) CP is violated indirectly, (2) both CP and CPT are violated directly. Therefore if one assumes direct $CPT$ conservation, $A(++,--, \Delta t)\neq 0$ implies that CP must be violated indirectly.

Moreover, if $A(++,--,\Delta t) \neq 0$ while CPT is assumed to be conserved both directly and indirectly, then in addition to indirect CP violation, we can draw the following conclusions: (1) T must also be violated indirectly; (2) $|\langle l^+ |{\cal H}|\bar{M}^0\rangle| = |\langle  l^-|{\cal H}  |M^0\rangle \neq |\langle l^+|{\cal H}  |M^0\rangle| = |\langle l^- |{\cal H}|\bar{M}^0\rangle| $.

If the unequal-flavor asymmetry $A(+-,-+,  \Delta t) \neq 0 $, then CP must be violated directly or indirectly.

Moreover, if $A(+-,-+,  \Delta t)\neq 0$ for $\Delta t \neq 0$ while CPT is assumed to be conserved both directly and indirectly, then we can draw the following conclusions: (1) $|\langle l^+ |{\cal H}|\bar{M}^0\rangle| = |\langle  l^-|{\cal H}  |M^0\rangle \neq |\langle l^+|{\cal H}  |M^0\rangle| = |\langle l^- |{\cal H}|\bar{M}^0\rangle| $; (2) $\langle l^{-}|{\cal H} |M^0\rangle = \langle l^{+}|{\cal H}|\bar{M}^0\rangle^* \neq 0$, which means $\Delta {\cal F} =\Delta Q$ rule must be violated.

For joint decays to CP eigenstates, in addition to various detailed results of the asymmetries, we have the following conclusions. The equal-time equal-CP rates $I[h^+,t_a;h^+,t_a]$ and $I[h^-,t_a;h^-,t_a]$ vanish, no matter whether CP or CPT is violated. At equal times $\Delta t =0$, the unequal-CP asymmetry $B(+-,-+, 0)$ vanishes, no matter whether CP or CPT is violated. On the other hand, if any equal-CP decay rate is nonzero, then CP must be violated.

The detailed expressions of the rates and symmetries can be used to determine the CP and CPT violating parameters. Clearly, these results are consequences of the well-known antisymmetry and entanglement of the state $|\Psi_-\rangle$. We hope these exact results are useful in studies on CP and CPT violations.

I am grateful to A. Di Domenico and other members of KLOE2, as well as Z. Huang, for useful discussions. This work was supported by the National Science Foundation of China (Grant No. 10875028).

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