Standard model and new physics contributions to K _L and K _S into four leptons

Abstract

We study the K _L and K _S decays into four leptons ($e\bar{e}e\bar{e}$, $\mu\bar{\mu}\mu\bar{\mu}$, $e\bar{e}\mu\bar{\mu}$) where we use a form factor motivated by vector meson dominance, and show the dependence of the branching ratios and spectra from the slopes. A precise determination of short-distance contribution to K _L →μμ is affected by our ignorance on the sign of the amplitude $\mathcal{A}(K_{L}\to\gamma\gamma)$ but we show a possibility to measure the sign of this amplitude by studying K _L and K _S decays in four leptons. We also investigate the effect of New Physics contributions for these decays.

Appendices

Appendix A: Detailed expressions for amplitudes

1.1 A.1 The K _S decays amplitudes

For the cases where ℓ ₁=ℓ ₂, there exist four diagrams that can be reduced to two different amplitudes $\mathcal{M}_{A}$ and $\mathcal{M}_{B}$ (Fig. 7),

$$\begin{aligned} \mathcal{M}_A & = e^2 \frac{\mathrm{F}_S ((p_1+p_2)^2,(p_3+p_4)^2 )}{(p_1+p_2)^2 (p_3+p_4)^2} \\ &\quad {}\times \bigl[ (p_1+p_2) \cdot(p_3+p_4) g^{\mu\nu} \\ &\quad {}- (p_1+p_2)^\nu (p_3+p_4)^\mu \bigr] \\ &\quad {}\times \bigl[ \bar{u}(p_1) \gamma_\mu v(p_2) \bigr] \bigl[ \bar {u}(p_3) \gamma_\nu v(p_4) \bigr] \end{aligned}$$

(A.1)

and

$$\begin{aligned} \mathcal{M}_B & = -e^2 \frac{\mathrm{F}_S ((p_3+p_2)^2,(p_1+p_4)^2 )}{(p_3+p_2)^2 (p_1+p_4)^2} \\ &\quad {}\times \bigl[ (p_2+p_3) \cdot(p_1+p_4) g^{\mu\nu} \\ &\quad {}- (p_2+p_3)^\nu (p_1+p_4)^\mu \bigr] \\ & \quad {}\times \bigl[ \bar{u}(p_3) \gamma_\mu v(p_2) \bigr] \bigl[ \bar {u}(p_1) \gamma_\nu v(p_4) \bigr]. \end{aligned}$$

(A.2)

Thus the total squared amplitude is given by (under symmetries considerations, $|\mathcal{M}_{A}|^{2} = |\mathcal{M}_{B}|^{2}$),

$$ | \mathcal{M} |^2 = \frac{1}{2} \bigl(|\mathcal{M}_A|^2 + \mathcal{M}_A \, \mathcal{M}_B^* \bigr). $$

(A.3)

Fig. 7

Amplitudes of K _S,L in four leptons

In the mixed case, ℓ ₁=μ and ℓ ₂=e, there are only two diagrams, and we have $|\mathcal{M}|^{2}=|\mathcal{M}_{A}|^{2}$.

1.2 A.2 The K _L decays amplitudes

The calculations of the amplitudes involving the K _L are identical in procedure that the ones for the K _S , we have to distinguish two kinds of amplitudes

$$\begin{aligned} \mathcal{M}_A & = e^2 \frac{\mathrm{F}_L ((p_1+p_2)^2,(p_3+p_4)^2 )}{(p_1+p_2)^2 (p_3+p_4)^2} \\ &\quad {}\times\varepsilon_{\mu\nu\rho\sigma} (p_1+p_2)^\rho (p_3+p_4)^\sigma \\ &\quad {}\times \bigl[ \bar{u}(p_1) \gamma^\mu v(p_2) \bigr] \bigl[ \bar {u}(p_3) \gamma^\nu v(p_4) \bigr] \end{aligned}$$

(A.4)

and

$$\begin{aligned} \mathcal{M}_B & =- e^2 \frac{\mathrm{F}_L ((p_3+p_2)^2,(p_1+p_4)^2 )}{(p_3+p_2)^2 (p_1+p_4)^2} \\ &\quad {}\times\varepsilon_{\mu\nu\rho\sigma} (p_2+p_3)^\rho (p_1+p_4)^\sigma \\ &\quad {}\times \bigl[ \bar{u}(p_3) \gamma^\mu v(p_2) \bigr] \bigl[ \bar {u}(p_1) \gamma^\nu v(p_4) \bigr]. \end{aligned}$$

(A.5)

The total amplitude is given by (under symmetries considerations, $|\mathcal{M}_{A}|^{2} = |\mathcal{M}_{B}|^{2}$),

$$ | \mathcal{M} |^2 = \frac{1}{2} \bigl(|\mathcal{M}_A|^2 + \mathcal{M}_A \, \mathcal{M}_B^* \bigr). $$

(A.6)

As before, in the mixed case, ℓ ₁=μ and ℓ ₂=e, there are only two diagrams, and we have $|\mathcal{M}|^{2}=|\mathcal{M}_{A}|^{2}$.

Appendix B: Bremsstrahlung CP-violating part

Using Low’s theorem [26], the amplitude $K_{S} (q)\to \mu (p_{-}) \bar{\mu} (p_{+}) \gamma^{*} (k)$ is the product of the $K_{S} (q)\to \mu (p_{-}) \bar{\mu} (p_{+})$ amplitude times the contribution of the soft photon radiated (Fig. 8):

$$ \begin{aligned}[b] &\mathcal{M} \bigl(K_S \to\mu\bar{\mu} \gamma^* \bigr) \\ &\quad =k^2\mathrm{F}_B^\mu(k,p_-,p_+)\; \epsilon_{\mu}(k) \mathcal{M}(K_S \to \mu\bar{\mu}), \end{aligned} $$

(B.1)

where $\mathcal{M}(K_{S} \to\mu\bar{\mu})$ is the decay amplitude of K _S into two muons

$$ \mathcal{M}(K_S \to\mu\bar{\mu}) = \mathcal{A}_{SD} \; \bar{u} \gamma _5 v $$

(B.2)

and

$$\begin{aligned} &\mathrm{F}_B^\mu(k,p_-,p_+) \\ &\quad = \frac{2e}{k^2} \biggl[ \frac{p_-^\mu}{k^2 +2 k \cdot p_-} + \frac {p_+^\mu}{k^2 -2 k \cdot p_+} \biggr]. \end{aligned}$$

(B.3)

Fig. 8

Bremsstrahlung amplitudes for K _S in four leptons

Now, we just have to contract with the muonic current $-i e \bar {u}\gamma^{\nu}v\;\epsilon^{*}_{\nu}$ to obtain the Bremsstrahlung contribution

$$\begin{aligned} &\mathcal{M}_{\mathrm{Brems.}} \\ &\quad = \mathcal{A}_{SD} \bigl\{ \mathrm{F}_B^\mu(q_1,p_3,p_4) \bigl[ \bar {u}(p_1) \gamma_\mu v(p_2) \bigr] \, \bigl[ \bar{u}(p_3) \gamma_5 v(p_4) \bigr] \\ & \qquad {}+ \mathrm{F}_B^\mu(q_2,p_1,p_2) \bigl[ \bar{u}(p_3) \gamma_\mu v(p_4) \bigr] \, \bigl[ \bar{u}(p_1) \gamma_5 v(p_2) \bigr] \\ & \qquad {}- \mathrm{F}_B^\mu(p_2+p_3,p_1,p_4) \bigl[ \bar{u}(p_3) \gamma_5 v(p_2) \bigr] \\ &\qquad {}\times \bigl[ \bar{u}(p_1) \gamma_\mu v(p_4) \bigr] \\ & \qquad {}- \mathrm{F}_B^\mu(p_1+p_4,p_3,p_2) \bigl[ \bar{u}(p_1) \gamma_\mu v(p_4) \bigr] \\ &\qquad {}\times \bigl[ \bar{u}(p_3) \gamma_5 v(p_2) \bigr] \bigr\} . \end{aligned}$$

(B.4)

In our case, $\operatorname{Re} \mathcal{A}_{SD}$ can be neglected, all the short-distance information is contained in $\operatorname{Im} \mathcal {A}_{SD}$ and we have [11]:

$$\begin{aligned} &\operatorname{Im} \mathcal{A}_{SD} \\ &\quad = - \frac{G_F \alpha_\mathrm{em}(M_Z)}{\pi\sin^2 \theta_W}\sqrt{2} m_\mu F_K \operatorname{Im} \bigl(V_{ts}^* V_{td} \bigr) Y(x_t), \end{aligned}$$

(B.5)

with $x_{t}=m_{t}^{2}/M_{W}^{2}$ and Y(x) is the Inami–Lin function:

$$ Y(x) = \frac{x}{8} \biggl[\frac{4-x}{1-x} + \frac{3x}{(1-x)^2} \ln x \biggr]. $$

(B.6)