The Weak Eigenstates of the Neutral Kaon

The eigenstates of the neutral kaon in respect of strong and electromagnetic interaction are given by

$$\vert K_0\rangle = \vert d\overline{s}\rangle \quad \vert\overline{K}_0\rangle = \vert\overline{d} s\rangle$$ (1.1)

In absence of weak interaction strangeness is conserved, and since there are no lighter particles with $S\not= 0$, the above particles would be stable and distinctively observable. Weak interaction however can carry the s-quark over to a u-quark governed by the element $V_{us}$ of the CKM-matrix (see fig. 1.1).

Figure 1.1: weak transition of a s-quark to a u-quark

Similar transitions exist for $c$ and $t$ quarks. Since $K_0$ and $\overline{K}_0$ are neutral, the combination of quark-transitions like fig. 1.1 leads to a second order weak transition between $K_0$ and $\overline{K}_0$ states through Feynman graphs like that in fig. 1.2. Note that inside the loop may also appear $c$ and $t$ quarks - a fact that is crucial for the occurrence of CP violation discussed later in this chapter.

Figure 1.2: example Feynman graph for the transition $K_0 \leftrightarrow \overline {K}_0$ via weak interaction ($\Delta S = 2$); the vertices carry the corresponding elements of the CKM-matrix

For studying such transitions we consider an effective Hamiltonian $H$ which includes the weak interaction and acts on the strong eigenstates $\vert K_0\rangle$ and $\vert\overline{K}_0\rangle$. These states form an orthogonal basis of a 2-dimensional Hilbert space; since we are only interested in transitions between these eigenstates, we restrict ourself to this subspace of the whole Hilbert space, where $H$ can be written as a $2\times 2$-Matrix^1.1:

$$H = \left( \begin{array}{ccc} H_{11} & H_{12} \\ H_{21} & H... ...gle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right) \quad$$ (1.2)

Of course it is generally not guaranteed that states in this subspace stay there for $t \rightarrow \infty$ - physically speaking, the kaons might decay to other particles like pions, which are not part of this subspace. As a consequence, the matrix $H$ generally is not hermitian. However, it can always be split in a hermitian and anti-hermitian part:

$$H = M - \frac{i}{2}\Gamma$$ (1.3)

The matrices $M$ and $\Gamma$ are hermitian by definition:

$$M_{11}=M^*_{11} \quad M_{22}=M^*_{22} \quad M_{12}=M^*_{21}$$ (1.4)

$$\Gamma_{11}=\Gamma^*_{11} \quad \Gamma_{22}=\Gamma^*_{22} \quad \Gamma_{12}=\Gamma^*_{21}$$

Solving the Schrödinger equation $i\frac{d}{dt}\vert\Psi\rangle = H\vert\Psi\rangle$ one finds the solution

$$\vert\Psi_a(t)\rangle = e^{-i(M_a-\frac{i}{2}\Gamma_a)t}\vert\Psi_a\rangle$$ (1.5)

with the (real) eigenvalues $M_a,\Gamma_a$ and the eigenstates $\vert\Psi_a\rangle$. If one now calculates the probability $P_a$ to find the initial ($t=0$) state $\vert\Psi_a\rangle$ at some time $t$, one finds

$$P_a = \vert\langle\Psi_a(0)\vert\Psi_a(t)\rangle\vert^2 = e^{-\Gamma_a t}$$ (1.6)

which gives rise to the interpretation of $\Gamma$ as decay width, while $M$ is the mass of the eigenstate ($H$ shall be given in the rest frame of the kaon).

To discuss the properties of the eigenvalues and -states of $H$, it is not necessary to compute the matrix elements $H_{ij}$ (which would not be straightforward, because nonperturbative strong interactions had to be taken into account): fortunately, some fundamental symmetries alone are sufficient to get some interesting properties of the solutions.

First, CPT-Symmetry^1.2 gives [4]

$$M_{11}=M_{22} \quad \Gamma_{11}=\Gamma_{22}$$ (1.7)

Since the CPT-theorem guarantees CPT-Symmetry under very general assumptions on the theoretical side, and also holds very well experimentally, (1.7) can be regarded as exact. CP-symmetry, though, can be broken - in the standard model, for instance, CP-violation enters via a phase in the CKM-matrix^1.3. However, CP-violation turns out to be small experimentally, so it is worthwhile to study the limit of exact CP-conservation; in this case, one finds

$$M_{12}=M_{21} \quad \Gamma_{12}=\Gamma_{21}$$ (1.8)

if one defines the relative phases of $\vert K_0\rangle$ and $\vert\overline{K}_0\rangle$ accordingly^1.4. Together with (1.4) this means that $M_{12}$ and $\Gamma_{12}$ are real.

For the following, exact CP-conservation is assumed; the consequences of a (small) CP-violation are studied in the next section. Exact CP-Symmetry now leads to the matrix

$$H = \left( \begin{array}{ccc} H_{11} & H_{12} \\ H_{12} & H... ...d{array} \right) \quad H_{ij} = M_{ij}-\frac{i}{2}\Gamma_{ij}$$ (1.9)

with the eigenvalues

$$H_{1,2} = H_{11} \pm H_{12}$$ (1.10)

and the normalized eigenstates

$$\vert K_{1,2}\rangle = \frac{1}{\sqrt{2}}(\vert K_0\rangle \pm \vert\overline{K}_0\rangle )$$ (1.11)

These states are not only eigenstates to H, but also of CP:

$$CP\vert K_{1,2}\rangle = \pm \vert K_{1,2}\rangle )$$ (1.12)

The masses of the two kaon states are given by^1.5

$$M_{1,2} = M \mp \Delta \quad M=M_{11}, \Delta=-M_{12}$$ (1.13)

As the element $M_{12}$ is related to the energy of the weak transition $\vert K_0\rangle \leftrightarrow \vert\overline{K}_0\rangle$ while $M_{11}$ contains the binding energy of the strong interaction between the constituent quarks, it can be expected that $\Delta \ll M$. Indeed, experimentally one finds [34]

$$M = 497.672 \pm 0.031 \mbox{MeV} \quad \Delta=(1.745 \pm 0.004) \times 10^{-12} \mbox{MeV}$$ (1.14)

The situation is different for $\Gamma_{1,2}$: any decay involves a transition $s \rightarrow u$, so $\vert\Gamma_{12}\vert$ need not be much smaller than $\vert\Gamma_{11}\vert$. Actually, in $2^{nd}$ order perturbation theory it follows [5]^1.6 that

$$\Gamma_{11} \approx \Gamma_{12}$$ (1.15)

This leads to

$$\Gamma_{1} \approx 2\Gamma_{11} \quad \Gamma_{2} \approx 0$$ (1.16)

which means that the CP-odd state $\vert K_{2}\rangle$ is almost stable in comparison with the CP-even state $\vert K_{1}\rangle$, at least has a significantly longer mean lifetime. The physical explanation for this is that for the primary decay mode $K \rightarrow \pi\pi$, the $\pi \pi$ is CP-even, and therefore forbidden for the CP-odd kaon state $\vert K_{2}\rangle$ under our current assumption of exact CP-conservation.

Experimentally, one finds [34]

$$\tau_{1} = \Gamma_{1}^{-1} = (0.8935 \pm 0.0008) \times 10^... ... \tau_{2} = \Gamma_{2}^{-1} = (5.17 \pm 0.04) \times 10^{-8} s$$ (1.17)

To conclude: under exact CP-symmetry one has two weak kaon eigenstates mixed from the strong eigenstates like

$$\vert K_{1,2}\rangle = \frac{1}{\sqrt{2}}(\vert K_0\rangle \pm \vert\overline{K}_0\rangle )$$ (1.18)

with almost the same masses, but quite different lifetimes.