Preamble

Co-positivity of Tensors and Bounded-from-Below Conditions for CP-Conserving Two-Higgs-Doublet Potentials
Yisheng Song*

Abstract

In this paper, we derive analytic necessary and sufficient conditions for the CP-conserving two-Higgs-doublet potential to be bounded from below by employing the theory of tensor co-positivity. This is achieved by treating the potential as a quartic homogeneous polynomial in the moduli of the two Higgs doublet fields, where the relative 'angle' between the doublets is described by a misalignment parameter, and then solving three distinct minimization problems with respect to this misalignment. The final analytic conditions are established using the corresponding theory and methods of higher-order tensors.

Keywords: Co-positivity, Fourth-order tensors, Homogeneous polynomial, 2HDM

Introduction

The stability of multi-Higgs potentials has attracted considerable attention in the particle physics community since Lee first proposed the two-Higgs-doublet model (2HDM) in 1973 \cite{1}. Weinberg subsequently presented a general framework for multi-Higgs potentials in 1976 \cite{2}, which has since been studied extensively in hundreds of papers. Among the simplest extensions of the Standard Model, the 2HDM has been the subject of numerous investigations into its bounded-from-below (BFB) conditions, encompassing both CP-conserving and CP-violating scenarios. These studies have yielded various analytic BFB conditions, including those for CP-conserving 2HDM in Refs. \cite{3-9}, the most general 2HDM in Refs. \cite{3,6}, and both CP-conserving and CP-violating cases in Refs. \cite{5,6,9-15}. Numerical approaches have also been developed \cite{16}, along with many other references not cited here. Tree-level metastability bounds for the most general 2HDM are discussed in Ref. \cite{17}. More recently, Bahl et al. \cite{18} presented an analytic sufficient condition for the BFB of CP-violating 2HDM. However, simple analytic necessary and sufficient conditions have remained elusive even for the CP-conserving 2HDM until now. In this work, we develop a systematic technique to address this problem and provide a simple analytic expression (Theorem 1) for the bounded-from-below conditions in CP-conserving 2HDM.

For 2HDM with explicit CP conservation, all couplings in the Higgs potential are real \cite{1,3,14,17}. The scalar potential for such a model with Higgs doublets $\Phi_1$ and $\Phi_2$ is given by

$$
\begin{aligned}
V_H(\Phi_1, \Phi_2) &= V_2(\Phi_1, \Phi_2) + V_4(\Phi_1, \Phi_2), \
V_2(\Phi_1, \Phi_2) &= m_{11}^2|\Phi_1|^2 + m_{22}^2|\Phi_2|^2 - m_{12}^2(\Phi_1^\dagger\Phi_2 + \Phi_2^\dagger\Phi_1), \
V_4(\Phi_1, \Phi_2) &= \Lambda_1|\Phi_1|^4 + \Lambda_2|\Phi_2|^4 + \Lambda_3|\Phi_1|^2|\Phi_2|^2 \
&\quad + \Lambda_4(\Phi_1^\dagger\Phi_2)(\Phi_2^\dagger\Phi_1) + \frac{\Lambda_5}{2}[(\Phi_1^\dagger\Phi_2)^2 + (\Phi_2^\dagger\Phi_1)^2] \
&\quad + \Lambda_6|\Phi_1|^2(\Phi_1^\dagger\Phi_2 + \Phi_2^\dagger\Phi_1) + \Lambda_7|\Phi_2|^2(\Phi_1^\dagger\Phi_2 + \Phi_2^\dagger\Phi_1),
\end{aligned}
$$

where $\Phi^\dagger$ denotes the Hermitian conjugate of $\Phi$. The bounded-from-below condition for the scalar potential in 2HDM requires only the non-negativity of the quartic part $V_4$ \cite{14}, i.e., $V_4(\Phi_1, \Phi_2) \geq 0$.

In this paper, using the theory and methods of higher-order tensors, we present necessary and sufficient conditions for the BFB of the 2HDM potential with explicit CP conservation. Our main result is the following:

Theorem 1. Let $\Lambda_1 > 0$, $\Lambda_2 > 0$. Then $V_4(\Phi_1, \Phi_2) \geq 0$ if and only if

1. $\Lambda_6 = \Lambda_7 = 0$, $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$, and $\Lambda_3 + \Lambda_4 - |\Lambda_5| + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$;

2. $\Delta \geq 0$, $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$, $\Lambda_3 + \Lambda_4 - \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$, $|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2$, and either

(i) $-2\sqrt{\Lambda_1\Lambda_2} \leq \Lambda_3 + \Lambda_4 + \Lambda_5 \leq 6\sqrt{\Lambda_1\Lambda_2}$, or

(ii) $\Lambda_3 + \Lambda_4 + \Lambda_5 > 6\sqrt{\Lambda_1\Lambda_2}$ and $|\sqrt{\Lambda_2}\Lambda_6 + \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} - 2\Lambda_1\Lambda_2$,

where

$$
\Delta = 4(12\Lambda_1\Lambda_2 - 12\Lambda_6\Lambda_7 + (\Lambda_3 + \Lambda_4 + \Lambda_5)^2)^3 - (72\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5) + 36\Lambda_6\Lambda_7(\Lambda_3 + \Lambda_4 + \Lambda_5) - 2(\Lambda_3 + \Lambda_4 + \Lambda_5)^3 - 108\Lambda_1\Lambda_2^2 - 108\Lambda_2\Lambda_6^2)^2.
$$

2HDM Potential and Real Tensors

To establish our main result, we transform the polynomial $V_4(\Phi_1, \Phi_2)$ in two complex variables into a fourth-order symmetric real tensor, and then apply tensor theory to prove our conclusions.

Let $\phi_k = |\Phi_k|$ denote the modulus of $\Phi_k$ for $k = 1, 2$. Then $\Phi_1^\dagger\Phi_2 = \phi_1\phi_2\rho e^{i\theta}$ and $\Phi_2^\dagger\Phi_1 = \phi_1\phi_2\rho e^{-i\theta}$, where $i^2 = -1$ and $\rho \in [0, 1]$ is the orbit space parameter \cite{7,15}. Substituting these expressions yields

$$
\begin{aligned}
V_4(\Phi_1, \Phi_2) &= \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + \Lambda_3\phi_1^2\phi_2^2 + \Lambda_4\rho^2\phi_1^2\phi_2^2 \
&\quad + \frac{\Lambda_5}{2}\rho^2\phi_1^2\phi_2^2(e^{2i\theta} + e^{-2i\theta}) + \Lambda_6\phi_1^2(\phi_1\phi_2\rho e^{i\theta} + \phi_1\phi_2\rho e^{-i\theta}) \
&\quad + \Lambda_7\phi_2^2(\phi_1\phi_2\rho e^{i\theta} + \phi_1\phi_2\rho e^{-i\theta}) \
&= \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + (\Lambda_3 + \Lambda_4\rho^2 - \Lambda_5\rho^2)\phi_1^2\phi_2^2 + 2\Lambda_5\rho^2\phi_1^2\phi_2^2\cos 2\theta \
&\quad + 2\Lambda_6\rho\phi_1^3\phi_2\cos\theta + 2\Lambda_7\rho\phi_1\phi_2^3\cos\theta.
\end{aligned}
$$

Let $x = \cos\theta$, so $x \in [-1, 1]$ and

$$
V_4(\Phi_1, \Phi_2) = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + (\Lambda_3 + \Lambda_4\rho^2 - \Lambda_5\rho^2)\phi_1^2\phi_2^2 + 2\Lambda_5\rho^2\phi_1^2\phi_2^2x^2 + 2\rho(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2)\phi_1\phi_2x.
$$

This expression defines a fourth-order two-dimensional symmetric tensor $A(\rho, x) = (a_{ijkl})$ with parameter-dependent entries:

$$
a_{1111} = \Lambda_1, \quad a_{2222} = \Lambda_2, \quad a_{1122} = a_{2211} = \frac{1}{2}[\Lambda_3 + \Lambda_4\rho^2 + \Lambda_5\rho^2(2x^2 - 1)],
$$

$$
a_{1112} = a_{1121} = a_{1211} = a_{2111} = \Lambda_6\rho x, \quad a_{1222} = a_{2122} = a_{2212} = a_{2221} = \Lambda_7\rho x.
$$

Thus, the BFB condition for $V_H(\Phi_1, \Phi_2)$ can be transformed into the co-positivity condition for the fourth-order tensor $A(\rho, x)$.

The application of positive definiteness and co-positivity of fourth-order symmetric tensors to BFB conditions in particle physics models was introduced in Ref. \cite{7}. More recently, Refs. \cite{19-23} provided distinct sufficient conditions for the co-positivity of fourth-order three-dimensional symmetric tensors, which may be applicable to BFB conditions for scalar potentials in particle physics models.

Co-positivity of Matrices and Tensors

Co-positivity of matrices has been used to verify BFB conditions in particle physics models \cite{7,8}. A symmetric matrix $M = (m_{ij})$ is co-positive if the quadratic form $x^\top M x \geq 0$ for all non-negative vectors $x \in \mathbb{R}^n$. For a $2 \times 2$ symmetric matrix $M = (m_{ij})$, the co-positivity conditions were established in Ref. \cite{24, Lemma 2.1} (see also Hadeler \cite{25, Theorem 2} and Nadler \cite{26, Lemma 1} for further details). Specifically, a $2 \times 2$ symmetric matrix $M = (m_{ij})$ is co-positive if and only if

$$
m_{11} \geq 0, \quad m_{22} \geq 0, \quad \text{and} \quad m_{12} + \sqrt{m_{11}m_{22}} \geq 0.
$$

Co-positivity of symmetric tensors has also been explored for testing BFB conditions in particle physics models \cite{7}. An $m$th-order $n$-dimensional symmetric tensor $T = (t_{i_1\cdots i_m})$ (with $i_j = 1, 2, \ldots, n$) is co-positive \cite{28-32} if the $m$-degree homogeneous polynomial $T x^m \geq 0$ for all non-negative vectors $x \in \mathbb{R}^n$, where $x = (x_1, x_2, \ldots, x_n)^\top$ and

$$
T x^m = x^\top(T x^{m-1}) = \sum_{i_1,\ldots,i_m=1}^n t_{i_1\cdots i_m}x_{i_1}\cdots x_{i_m},
$$

with $T x^{m-1} = (y_1, y_2, \ldots, y_n)^\top$ being a vector whose entries are $y_k = (T x^{m-1})k = \sum$.}^n t_{k i_2\cdots i_m}x_{i_2}\cdots x_{i_m

Consider a quartic homogeneous real polynomial in two variables $x, y$:

$$
f(x, y) = a_0x^4 + a_1x^3y + a_2x^2y^2 + a_3xy^3 + a_4y^4.
$$

This polynomial defines a fourth-order two-dimensional symmetric tensor $T = (t_{ijkl})$ with entries

$$
t_{1111} = a_0, \quad t_{2222} = a_4, \quad t_{1122} = t_{1212} = t_{1221} = t_{2112} = t_{2121} = t_{2211} = \frac{a_2}{6},
$$

$$
t_{1112} = t_{1121} = t_{1211} = t_{2111} = \frac{a_1}{4}, \quad t_{1222} = t_{2122} = t_{2212} = t_{2221} = \frac{a_3}{4}.
$$

Assuming $a_0 > 0$ and $a_4 > 0$, the co-positivity of this tensor $T$ was established in Ref. \cite{20} (see also Refs. \cite{33,34}). We restate this result as Lemma 1.

Lemma 1. Let $a_0 > 0$ and $a_4 > 0$. Then $f(x, y) \geq 0$ for all $x \geq 0, y \geq 0$ if and only if one of the following conditions holds:

1. $\Delta \leq 0$ and $\frac{a_1}{\sqrt{a_0}} + \frac{a_3}{\sqrt{a_4}} > 0$;

2. $a_1 \geq 0$, $a_3 \geq 0$, and $2\sqrt{a_0a_4} + a_2 \geq 0$;

3. $\Delta \geq 0$, $|a_1\sqrt{a_4} - a_3\sqrt{a_0}| \leq 4\sqrt{a_0a_2a_4} + 2a_0a_4$, and either

(i) $-2\sqrt{a_0a_4} \leq a_2 \leq 6\sqrt{a_0a_4}$, or

(ii) $a_2 > 6\sqrt{a_0a_4}$ and $a_1\sqrt{a_4} + a_3\sqrt{a_0} \geq -4\sqrt{a_0a_2a_4} - 2a_0a_4$,

where

$$
\Delta = 4(12a_0a_4 - 3a_1a_3 + a_2^2)^3 - (72a_0a_2a_4 + 9a_1a_2a_3 - 2a_3^3 - 27a_0a_3^2 - 27a_1^2a_4)^2.
$$

Bounded-from-Below Conditions

In this section, we derive the BFB conditions for the 2HDM with explicit CP conservation, establishing our main result, Theorem 1.

The quartic part of the CP-conserving two-Higgs-doublet potential can be rewritten as

$$
V_4(\Phi_1, \Phi_2) = 2\Lambda_5\rho^2\phi_1^2\phi_2^2x^2 + \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + [\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2]\phi_1^2\phi_2^2 + 2\rho(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2)\phi_1\phi_2x.
$$

Without loss of generality, we assume $\Lambda_1 > 0$ and $\Lambda_2 > 0$. If $\phi_1 > 0, \phi_2 = 0$ (or $\phi_1 = 0, \phi_2 > 0$), then $V_H(\Phi_1, \Phi_2) = \Lambda_1\phi_1^4$ (or $\Lambda_2\phi_2^4$). We therefore consider $\phi_1 > 0, \phi_2 > 0$ in what follows. Define

$$
A = 2\Lambda_5\rho^2\phi_1^2\phi_2^2, \quad B = 2\rho(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2)\phi_1\phi_2, \quad C = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + [\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2]\phi_1^2\phi_2^2.
$$

Then $V_4(\Phi_1, \Phi_2)$ can be viewed as a quadratic function $f(x)$ in the variable $x \in [-1, 1]$:

$$
f(x) = V_4(\Phi_1, \Phi_2) = Ax^2 + Bx + C.
$$

When $A > 0$, the function $f(x)$ attains its minimum either at the interior point $x = -\frac{B}{2A} \in [-1, 1]$ with value $\frac{4AC - B^2}{4A}$, or at the boundary points $x = \pm 1$ with values $A + C \pm B$ when $-\frac{B}{2A} \notin [-1, 1]$. The behavior of $f(x)$ is illustrated in Figure 1.

When $A \leq 0$, $f(x)$ attains its minimum at $x = -1$ (if $B \geq 0$) or $x = 1$ (if $B < 0$), with corresponding values $A - B + C$ or $A + B + C$. This behavior is shown in Figure 2.

From this analysis, we obtain the following proposition.

Proposition 2. $V_4(\Phi_1, \Phi_2) \geq 0$ if and only if $C \geq 0$, $f(1) \geq 0$, and $f(-1) \geq 0$.

Proof. For necessity, $V_4(\Phi_1, \Phi_2) = f(x) \geq 0$ for all $x \in [-1, 1]$ implies that $f(1) \geq 0$ and $f(-1) \geq 0$. For sufficiency, a quadratic function $f(x)$ is non-negative on $[-1, 1]$ if and only if its minimum value on this interval is non-negative. The minimum of $f(x)$ is the smallest among three candidate values: $\frac{4AC - B^2}{4A}$ (if $-\frac{B}{2A} \in [-1, 1]$), $f(1)$, and $f(-1)$. When $A > 0$, the unique extremum point $-\frac{B}{2A}$ is the minimum point only if $-\frac{B}{2A} \in [-1, 1]$, which means $-2A \leq B \leq 2A$. By Proposition 3(2), this implies $B = 0$. Therefore, $C \geq 0$ ensures $f(-\frac{B}{2A}) = C \geq 0$, and consequently $f(x) \geq 0$ if $C \geq 0$, $f(1) \geq 0$, and $f(-1) \geq 0$.

The following results are straightforward to obtain; detailed proofs are provided in the Appendix.

Proposition 3. (1) $C \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$ if and only if $\Lambda_3 + \Lambda_4 - \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$ and $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$.

(2) If $-2A \leq B \leq 2A$ for all $\phi_1 \geq 0, \phi_2 \geq 0$, then $\Lambda_6 = 0$, $\Lambda_7 = 0$, and $\Lambda_5 \geq 0$, i.e., $B = 0$ and $A \geq 0$.

From equations (5) and (6), we can express $A \pm B + C$ as quadratics in $\rho$:

$$
A \pm B + C = 2\Lambda_5\rho^2\phi_1^2\phi_2^2 + \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + [\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2]\phi_1^2\phi_2^2 \pm 2\rho(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2)\phi_1\phi_2.
$$

Define

$$
s(\rho) = a\rho^2 + b\rho + c, \quad t(\rho) = a\rho^2 - b\rho + c,
$$

where

$$
a = (\Lambda_4 + \Lambda_5)\phi_1^2\phi_2^2, \quad b = 2(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2)\phi_1\phi_2, \quad c = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + \Lambda_3\phi_1^2\phi_2^2.
$$

For $\rho \in [0, 1]$, the behavior of $s(\rho)$ and $t(\rho)$ follows similar patterns as $f(x)$. When $a > 0$, $s(\rho)$ reaches its minimum either at the interior point $\rho = -\frac{b}{2a} \in [0, 1]$ with value $\frac{4ac - b^2}{4a}$, or at the boundary points $\rho = 0$ or $1$ with values $c$ or $a + b + c$ when $-\frac{b}{2a} \notin [0, 1]$. The cases $a \leq 0$ are analogous. The same analysis applies to $t(\rho)$.

Proposition 4. (1) $A + B + C \geq 0$ if and only if $c \geq 0$ and $a + b + c \geq 0$.

(2) $A - B + C \geq 0$ if and only if $c \geq 0$ and $a - b + c \geq 0$.

Proof. For (1), necessity follows from $s(0) = c \geq 0$ and $s(1) = a + b + c \geq 0$. Sufficiency follows because $s(\rho)$ is non-negative on $[0, 1]$ if and only if its minimum on this interval is non-negative. The minimum is the smallest among $\frac{4ac - b^2}{4a}$ (if $-\frac{b}{2a} \in [0, 1]$), $s(0)$, and $s(1)$. When $a > 0$, the extremum point $-\frac{b}{2a}$ is the minimum point only if $-\frac{b}{2a} \in [0, 1]$, which implies $-2a \leq b \leq 0$. By Proposition 5(2), this forces $b = 0$. Thus $c \geq 0$ ensures $s(-\frac{b}{2a}) = c \geq 0$, and $s(\rho) \geq 0$ follows from $c \geq 0$ and $s(1) \geq 0$. The proof of (2) is analogous.

The following results are straightforward to obtain; detailed proofs are provided in the Appendix.

Proposition 5. (1) $c \geq 0$ if and only if $\Lambda_1 \geq 0$, $\Lambda_2 \geq 0$, and $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$.

(2) If $-2a \leq b \leq 0$, then $\Lambda_6 = 0$, $\Lambda_7 = 0$, and $\Lambda_4 + \Lambda_5 \geq 0$, i.e., $b = 0$ and $a \geq 0$.

(3) If $0 \leq b \leq 2a$, then $\Lambda_6 = 0$, $\Lambda_7 = 0$, and $\Lambda_4 + \Lambda_5 \geq 0$, i.e., $b = 0$ and $a \geq 0$.

Next, we establish conditions for $a \pm b + c \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$.

Proposition 6. $a - b + c \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$ if and only if one of the following holds:

1. $\Delta \leq 0$ and $\frac{\Lambda_6}{\sqrt{\Lambda_2}} + \frac{\Lambda_7}{\sqrt{\Lambda_1}} < 0$;

2. $\Lambda_6 \leq 0$, $\Lambda_7 \leq 0$, and $\Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$;

3. $\Delta \geq 0$, $|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2$, and either

(i) $-2\sqrt{\Lambda_1\Lambda_2} \leq \Lambda_3 + \Lambda_4 + \Lambda_5 \leq 6\sqrt{\Lambda_1\Lambda_2}$, or

(ii) $\Lambda_3 + \Lambda_4 + \Lambda_5 > 6\sqrt{\Lambda_1\Lambda_2}$ and $|\sqrt{\Lambda_2}\Lambda_6 + \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} - 2\Lambda_1\Lambda_2$,

where $\Delta$ is defined as in Theorem 1.

Proof. From equations (8) and (9), we have

$$
a - b + c = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 - 2\Lambda_6\phi_1^3\phi_2 - 2\Lambda_7\phi_1\phi_2^3 + (\Lambda_3 + \Lambda_4 + \Lambda_5)\phi_1^2\phi_2^2 = f(\phi_1, \phi_2) = a_0\phi_1^4 + a_1\phi_1^3\phi_2 + a_2\phi_1^2\phi_2^2 + a_3\phi_1\phi_2^3 + a_4\phi_2^4,
$$

with $a_0 = \Lambda_1$, $a_1 = -2\Lambda_6$, $a_2 = \Lambda_3 + \Lambda_4 + \Lambda_5$, $a_3 = -2\Lambda_7$, and $a_4 = \Lambda_2$. Applying Lemma 1 on the co-positivity of quartic forms yields the stated conditions.

Similarly, the following result can be established.

Proposition 7. $a + b + c \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$ if and only if one of the following holds:

1. $\Delta \leq 0$ and $\frac{\Lambda_6}{\sqrt{\Lambda_2}} + \frac{\Lambda_7}{\sqrt{\Lambda_1}} > 0$;

2. $\Lambda_6 \geq 0$, $\Lambda_7 \geq 0$, and $\Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$;

3. $\Delta \geq 0$, $|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2$, and either

(i) $-2\sqrt{\Lambda_1\Lambda_2} \leq \Lambda_3 + \Lambda_4 + \Lambda_5 \leq 6\sqrt{\Lambda_1\Lambda_2}$, or

(ii) $\Lambda_3 + \Lambda_4 + \Lambda_5 > 6\sqrt{\Lambda_1\Lambda_2}$ and $|\sqrt{\Lambda_2}\Lambda_6 + \sqrt{\Lambda_1}\Lambda_7| \geq -2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} - 2\Lambda_1\Lambda_2$.

Combining Proposition 4(1) with Propositions 5(1) and 7 yields:

Proposition 8. $A + B + C \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$ if and only if $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$ and one of the following holds:

1. $\Delta \leq 0$ and $\frac{\Lambda_6}{\sqrt{\Lambda_2}} + \frac{\Lambda_7}{\sqrt{\Lambda_1}} > 0$;

2. $\Lambda_6 \geq 0$, $\Lambda_7 \geq 0$, and $\Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$;

3. $\Delta \geq 0$, $|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2$, and either

(i) $-2\sqrt{\Lambda_1\Lambda_2} \leq \Lambda_3 + \Lambda_4 + \Lambda_5 \leq 6\sqrt{\Lambda_1\Lambda_2}$, or

(ii) $\Lambda_3 + \Lambda_4 + \Lambda_5 > 6\sqrt{\Lambda_1\Lambda_2}$ and $|\sqrt{\Lambda_2}\Lambda_6 + \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} - 2\Lambda_1\Lambda_2$.

Similarly, applying Proposition 5(1) and Proposition 6 to Proposition 4(2) gives:

Proposition 9. $A - B + C \geq 0$ for all $\phi_1 \geq 0, \phi_2 \geq 0$ if and only if $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$ and one of the following holds:

1. $\Delta \leq 0$ and $\frac{\Lambda_6}{\sqrt{\Lambda_2}} + \frac{\Lambda_7}{\sqrt{\Lambda_1}} < 0$;

2. $\Lambda_6 \leq 0$, $\Lambda_7 \leq 0$, and $\Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$;

3. $\Delta \geq 0$, $|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2$, and either

(i) $-2\sqrt{\Lambda_1\Lambda_2} \leq \Lambda_3 + \Lambda_4 + \Lambda_5 \leq 6\sqrt{\Lambda_1\Lambda_2}$, or

(ii) $\Lambda_3 + \Lambda_4 + \Lambda_5 > 6\sqrt{\Lambda_1\Lambda_2}$ and $|\sqrt{\Lambda_2}\Lambda_6 + \sqrt{\Lambda_1}\Lambda_7| \leq 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} - 2\Lambda_1\Lambda_2$.

Combining Propositions 2, 3, 8, and 9 yields our main result, Theorem 1.

Proof of Theorem 1. From Propositions 2 and 3, $V_4(\Phi_1, \Phi_2) \geq 0$ if and only if

$$
\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0, \quad \Lambda_3 + \Lambda_4 - \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0,
$$

and either $A + B + C \geq 0$ or $A - B + C \geq 0$. Propositions 8(2) and 9(2) imply that when both $A + B + C \geq 0$ and $A - B + C \geq 0$ hold, we must have $\Lambda_6 = \Lambda_7 = 0$, along with

$$
\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0, \quad \Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0.
$$

These inequalities together are equivalent to

$$
\Lambda_6 = \Lambda_7 = 0, \quad \Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0, \quad \Lambda_3 + \Lambda_4 - |\Lambda_5| + 2\sqrt{\Lambda_1\Lambda_2} \geq 0,
$$

which establishes condition (1) of Theorem 1.

The inequalities $A - B + C \geq 0$ and $A + B + C \geq 0$ cannot simultaneously hold under the conditions of Proposition 8(1) and Proposition 9(1), as this would lead to a contradiction. Therefore, these cases are excluded.

Finally, conditions (2) and (3) of Theorem 1 follow directly from Propositions 8(3) and 9(3), completing the proof.

Conclusions

In summary, we have proven analytic necessary and sufficient conditions for the bounded-from-below conditions of the 2HDM potential with explicit CP conservation. Additionally, we have established the co-positivity of the fourth-order two-dimensional symmetric tensor $A(\rho, x) = (a_{ijkl})$ defined in (3) with parameters $\rho \in [0, 1]$ and $x \in [-1, 1]$.

Some Remarks

1. When $\Lambda_6 = \Lambda_7 = 0$, Theorem 1 reduces to the well-known results in Refs. \cite{3,9-15}. Specifically, $V_4^{Z_2}(\Phi_1, \Phi_2) \geq 0$ if and only if $\Lambda_1 > 0$, $\Lambda_2 > 0$, $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$, and $\Lambda_3 + \Lambda_4 - |\Lambda_5| + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$.

2. Ivanov \cite{6} derived necessary and sufficient BFB conditions for CP-conserving 2HDM potentials in terms of eigenvalues of a $4 \times 4$ matrix, but these conditions are not fully analytic. It is unclear how to implement them analytically, and numerical implementation may be computationally slow. Our results provide fully analytic conditions for CP-conserving potentials that are easy to code and presumably faster to evaluate, which is advantageous for large-scale scans of the 2HDM parameter space common in the literature.

3. For the BFB of CP-conserving 2HDM potentials, Kannike \cite{7} presented a sufficient condition (Eq. (88)) using Lagrange multipliers, which is not completely analytic. More recently, Bahl et al. \cite{18} gave stronger sufficient conditions (Eqs. (5.20) and (5.21)) based on Ulrich and Watson's result (Eq. (30)) \cite{34}. In this paper, we use an optimized version of Ulrich and Watson's result (Lemma 1) to derive analytic BFB conditions (for more details on Lemma 1, see Refs. \cite{20,33}). For example, consider the parameter set: $\Lambda_1 = \Lambda_2 = 1$, $\Lambda_3 = -1$, $\Lambda_4 = 2$, $\Lambda_5 = -1$, $\Lambda_6 = 1$, $\Lambda_7 = -1$. We have

$$
\begin{aligned}
\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} &= -1 + 2 > 0, \
\Lambda_3 + \Lambda_4 - \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} &= -1 + 2 - (-1) + 2 > 0, \
\Lambda_3 + \Lambda_4 + \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} &= -1 + 2 - 1 + 2 > 0, \
|\sqrt{\Lambda_2}\Lambda_6 - \sqrt{\Lambda_1}\Lambda_7| &= 2 < 2\sqrt{\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5)} + 2\Lambda_1\Lambda_2 = 2\sqrt{0} + 2 = 2, \
-2\sqrt{\Lambda_1\Lambda_2} &< \Lambda_3 + \Lambda_4 + \Lambda_5 = 0 < 6\sqrt{\Lambda_1\Lambda_2} = 6, \
\Delta &= 4(12\Lambda_1\Lambda_2 - 12\Lambda_6\Lambda_7 + (\Lambda_3 + \Lambda_4 + \Lambda_5)^2)^3 \
&\quad - (72\Lambda_1\Lambda_2(\Lambda_3 + \Lambda_4 + \Lambda_5) + 36\Lambda_6\Lambda_7(\Lambda_3 + \Lambda_4 + \Lambda_5) \
&\quad - 2(\Lambda_3 + \Lambda_4 + \Lambda_5)^3 - 108\Lambda_1\Lambda_2^2 - 108\Lambda_2\Lambda_6^2)^2 \
&= 4(12 + 12 + 0)^3 - (0 - 108 - 108)^2 > 0.
\end{aligned}
$$

These parameters satisfy condition (2) of Theorem 1, implying $V_4(\Phi_1, \Phi_2) \geq 0$. However, they violate condition Eq. (5.20) of Bahl et al. \cite{18}:

$$
\sqrt{\Lambda_1\Lambda_2} + \Lambda_3 - (|\Lambda_4| + |\Lambda_5|) + 4\sqrt{\Lambda_1\Lambda_2 - (\Lambda_3 + |\Lambda_4| + |\Lambda_5|)} = 3 - 2 > 0,
$$

but

$$
\sqrt{\Lambda_1\Lambda_2} + \Lambda_3 - (|\Lambda_4| + |\Lambda_5| + 4\sqrt{\Lambda_1\Lambda_2 - (\Lambda_3 + |\Lambda_4| + |\Lambda_5|)}) = 1 - 1 - (2 + 1 + 4) < 0.
$$

1. The quartic part of the general 2HDM potential can be expressed as

$$
\begin{aligned}
V_4(\Phi_1, \Phi_2) &= \Lambda_1(\Phi_1^\dagger\Phi_1)^2 + \Lambda_2(\Phi_2^\dagger\Phi_2)^2 + \Lambda_3(\Phi_1^\dagger\Phi_1)(\Phi_2^\dagger\Phi_2) \
&\quad + \Lambda_4(\Phi_1^\dagger\Phi_2)(\Phi_2^\dagger\Phi_1) + \frac{1}{2}[(\Phi_1^\dagger\Phi_2)^2 + (\Phi_2^\dagger\Phi_1)^2] \
&\quad + \Lambda_6(\Phi_1^\dagger\Phi_1)(\Phi_1^\dagger\Phi_2 + \Phi_2^\dagger\Phi_1) + \Lambda_7(\Phi_2^\dagger\Phi_2)(\Phi_1^\dagger\Phi_2 + \Phi_2^\dagger\Phi_1) \
&= \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + (\Lambda_3 + \Lambda_4\rho^2 + \rho^2(\text{Re}\Lambda_5\cos 2\theta + \text{Im}\Lambda_5\sin 2\theta))\phi_1^2\phi_2^2 \
&\quad + 2\rho(\text{Re}\Lambda_6\cos\theta + \text{Im}\Lambda_6\sin\theta)\phi_1^3\phi_2 + 2\rho(\text{Re}\Lambda_7\cos\theta + \text{Im}\Lambda_7\sin\theta)\phi_1\phi_2^3.
\end{aligned}
$$

Let $x = \cos\theta$ and $y = \sin\theta$. The general 2HDM potential thus involves three angular parameters $(\rho, x, y)$ with $x^2 + y^2 = 1$. The approach used in Theorem 1 cannot be directly applied to the general 2HDM potential. How to extend this method remains an open question deserving further investigation.

Data Availability Statement

This manuscript has no associated data or the data will not be deposited. [Authors' comment: This is a theoretical study and there are no external data associated with the manuscript.]

Acknowledgements

The authors would like to express their sincere thanks to the editors and anonymous referees for their constructive comments and valuable suggestions, and to Professors Igor P. Ivanov, K. G. Klimenko, and Garv Chauhan for useful discussions and for reading the manuscript.

This work was supported by the National Natural Science Foundation of P.R. China (Grant No. 12171064), by the Team Project of Innovation Leading Talent in Chongqing (No. CQYC20210309536), and by the Foundation of Chongqing Normal University (20XLB009).

Appendix

Proof of Proposition 3. (1) The condition $C \geq 0$ is

$$
C = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + [\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2]\phi_1^2\phi_2^2 \geq 0,
$$

which is equivalent to the co-positivity of the $2 \times 2$ matrix $M = (m_{ij})$ with entries $m_{11} = \Lambda_1$, $m_{22} = \Lambda_2$, and $m_{12} = m_{21} = \frac{1}{2}[\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2]$. By the co-positivity criterion for $2 \times 2$ matrices, this is equivalent to

$$
\Lambda_3 + (\Lambda_4 - \Lambda_5)\rho^2 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0.
$$

Since $\rho^2 \in [0, 1]$, this holds for all $\rho$ if and only if $\Lambda_3 + \Lambda_4 - \Lambda_5 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$ and $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$.

(2) From equation (5), we have

$$
B - 2A = 2(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 - 2\Lambda_5\rho\phi_1\phi_2)\rho\phi_1\phi_2,
$$

$$
B + 2A = 2(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 + 2\Lambda_5\rho\phi_1\phi_2)\rho\phi_1\phi_2.
$$

Thus,

$$
B - 2A \leq 0 \iff -\Lambda_6\phi_1^2 - \Lambda_7\phi_2^2 + 2\Lambda_5\rho\phi_1\phi_2 \geq 0,
$$

$$
B + 2A \geq 0 \iff \Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 + 2\Lambda_5\rho\phi_1\phi_2 \geq 0.
$$

The condition $B - 2A \leq 0$ is equivalent to the co-positivity of the matrix $M = (m_{ij})$ with entries $m_{11} = -\Lambda_6$, $m_{22} = -\Lambda_7$, and $m_{12} = m_{21} = \Lambda_5\rho$. By the co-positivity criterion, this gives

$$
B - 2A \leq 0 \iff \Lambda_6 \leq 0, \Lambda_7 \leq 0, \text{ and } \Lambda_5\rho + \sqrt{\Lambda_6\Lambda_7} \geq 0.
$$

Similarly,

$$
B + 2A \geq 0 \iff \Lambda_6 \geq 0, \Lambda_7 \geq 0, \text{ and } \Lambda_5\rho + \sqrt{\Lambda_6\Lambda_7} \geq 0.
$$

Therefore, the inequalities $-2A \leq B \leq 2A$ imply $\Lambda_6 = 0$, $\Lambda_7 = 0$, and $\Lambda_5 \geq 0$, i.e., $B = 0$ and $A \geq 0$.

Proof of Proposition 5. (1) The condition $c = \Lambda_1\phi_1^4 + \Lambda_2\phi_2^4 + \Lambda_3\phi_1^2\phi_2^2 \geq 0$ is equivalent to $\Lambda_1 \geq 0$, $\Lambda_2 \geq 0$, and $\Lambda_3 + 2\sqrt{\Lambda_1\Lambda_2} \geq 0$.

(2) From equation (8), we have

$$
b + 2a = 2(\Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 + (\Lambda_4 + \Lambda_5)\phi_1\phi_2)\phi_1\phi_2,
$$

$$
2a - b = 2(-\Lambda_6\phi_1^2 - \Lambda_7\phi_2^2 + (\Lambda_4 + \Lambda_5)\phi_1\phi_2)\phi_1\phi_2.
$$

Thus,

$$
b + 2a \geq 0 \iff \Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 + (\Lambda_4 + \Lambda_5)\phi_1\phi_2 \geq 0,
$$

$$
b \leq 0 \iff \Lambda_6\phi_1^2 + \Lambda_7\phi_2^2 \leq 0.
$$

These imply

$$
b + 2a \geq 0 \iff \Lambda_6 \geq 0, \Lambda_7 \geq 0, (\Lambda_4 + \Lambda_5) + 2\sqrt{\Lambda_6\Lambda_7} \geq 0,
$$

$$
b \leq 0 \iff \Lambda_6 \leq 0, \Lambda_7 \leq 0.
$$

Consequently, we obtain $\Lambda_4 + \Lambda_5 \geq 0$, $\Lambda_6 = 0$, $\Lambda_7 = 0$, i.e., $b = 0$ and $a \geq 0$.

The proof of (3) follows similarly.

References

[1] T. Lee, A Theory of Spontaneous T Violation, Phys. Rev. D 8, 1226 (1973); CP nonconservation and spontaneous symmetry breaking, Physics Reports, 9(2) 143-177 (1974)

[2] S. Weinberg, Gauge Theory of CP Nonconservation, Phys. Rev. Lett. 37, 657 (1976)

[3] N. G. Deshpande, E. Ma, Pattern of symmetry breaking with two Higgs doublets, Phys. Rev. D 18, 2574-2576 (1978)

[4] A. Barroso, P.M. Ferreira, I.P. Ivanov, R. Santos, Metastability bounds on the two Higgs doublet model, J. High Energy Phys. 06, 045 (2013)

[5] G. Chauhan, Vacuum stability and symmetry breaking in left-right symmetric model. J. High Energ. Phys. 2019, 137 (2019)

[6] I. P. Ivanov, Minkowski space structure of the Higgs potential in the two-Higgs-doublet model Phys. Rev. D 75, 035001; Erratum Phys. Rev. D 76, 039902 (2007)

[7] K. Kannike, Vacuum stability of a general scalar potential of a few fields. Eur. Phys. J. C, 76, 324 (2016); Erratum. Eur. Phys. J. C, 78, 355 (2018)

[8] K. Kannike, Vacuum stability conditions from copositivity criteria. Eur. Phys. J. C, 72, 2093 (2012)

[9] K. G. Klimenko, On Necessary and Sufficient Conditions for Some Higgs Potentials to Be Bounded From Below, Theor. Math. Phys. 62, 58-65 (1985)

[10] M. Nebot, Bounded masses in two Higgs doublets models, spontaneous CP violation and Z2 symmetry, Phys. Rev. D 102, 115002 (2020)

[11] S. Nie, M. Sher, Vacuum stability bounds in the two-Higgs doublet model, Phys. Lett. B 449(1-2), 89-92 (1999)

[12] S. Kanemura, T. Kasai, Y. Okada, Mass bounds of the lightest CP-even Higgs boson in the two-Higgs-doublet model, Phys. Lett. B 471(2-3), 182-190 (1999)

[13] D. Eriksson, J. Rathsman, O. Stal, 2HDMC-two-Higgs-doublet model calculator, Comput. Phys. Commun. 181(1), 189-205 (2010); Erratum, Comput. Phys. Commun. 181(5), (2010)

[14] G.C. Branco, P.M. Ferreira, L. Lavoura, M.N. Rebelo, M. Sher, J.P. Silva, Theory and phenomenology of two-Higgs-doublet models, Phys. Rep. 516, 1 (2012)

[15] I.F. Ginzburg, M. Krawczyk, Symmetries of two Higgs doublet model and CP violation, Phys. Rev. D 72, 115013 (2005)

[16] M. Maniatisa, A. von Manteuffelb, O. Nachtmannc, F. Nagel, Stability and symmetry breaking in the general two-Higgs-doublet model, Eur. Phys. J. C 48, 805-823 (2006)

[17] I. P. Ivanov and João P. Silva, Tree-level metastability bounds for the most general two Higgs doublet model. Phys. Rev. D 92, 055017 (2015)

[18] H. Bahl, M. Carena, N. M. Coyle, A. Ireland, Carlos E.M. Wagner, New Tools for Dissecting the General 2HDM, J. High Energ. Phys. 2023, 165 (2023)

[19] J. Liu, Y. Song, Copositivity for 3rd order symmetric tensors and applications. Bull. Malays. Math. Sci. Soc. 45(1), 133-152 (2022)

[20] Y. Song, X. Li, Copositivity for a class of fourth order symmetric tensors given by scalar dark matter, J. Optim Theory Appl. 195, 334–346 (2022)

[21] Y. Song, L. Qi, Analytical expressions of copositivity for fourth-order symmetric tensors. Analy. Appl., 19(5), 779-800 (2021)

[22] Y. Song, L. Qi, A necessary and sufficient condition of positive definiteness for 4th order symmetric tensors defined in particle physics. arXiv: 2011.11262 (2020)

[23] Y. Song, Positive definiteness for 4th order symmetric tensors and applications. Anal. Math. Phys. 11, 10 (2021)

[24] L.E. Andersson, G. Chang, T. Elfving, Criteria for copositive matrices using simplices and barycentric coordinates. Linear Algebra Appl. 5, 9-30 (1995)

[25] K.P. Hadeler, On copositive matrices. Linear Algebra Appl. 49, 79-89 (1983)

[26] E. Nadler, Nonnegativity of bivariate quadratic functions on a triangle. Comput. Aided Geom. D. 9, 195-205 (1992)

[27] H. Ishimori, T. Kobayashi, H. Ohki, Y. Shimizu, H. Okada, M. Tanimoto, Non-Abelian Discrete Symmetries in Particle Physics, Prog. Theor. Phys. Suppl. 183, 1-163 (2010)

[28] L. Qi, Eigenvalues of a real supersymmetric tensor, J. Symbolic Comput., 40(6) 1302-1324 (2005)

[29] L. Qi, Symmetric Nonnegative Tensors and Copositive Tensors, Linear Algebra Appl., 439, 228-238 (2013)

[30] Y. Song, L. Qi, Necessary and sufficient conditions for copositive tensors, Linear Multilinear A 63(1), 120-131 (2015)

[31] L. Qi, H. Chen, Y. Chen, Tensor Eigenvalues and Their Applications, Springer Singapore, 2018

[32] L. Qi, Z. Luo, Tensor Analysis: Spectral Theory and Special Tensors, SIAM, Philadelphia 2017

[33] L. Qi, Y. Song, X. Zhang, Positivity Conditions for Cubic, Quartic and Quintic Polynomials, J. Nonlinear Convex Anal. 23(2), 191-213 (2022)

[34] G. Ulrich, L.T. Watson, Positivity conditions for quartic polynomials. SIAM J. Sci. Comput. 15, 528-544 (1994)