Physlib.Particles.StandardModel.HiggsBoson.Potential

For a given Higgs potential $P$ characterized by the mass squared parameter $\mu^2$ and the quartic coupling coefficient $\lambda$, the function evaluates a Higgs field $\phi$ at a point $x$ in spacetime as: $$V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$$ where $\|\phi(x)\|^2$ is the pointwise squared norm of the Higgs field at the point $x$.

For a given Higgs potential $P$ and a Higgs field $\phi$, the function mapping each point $x$ in spacetime to the potential value $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$ is smooth ($C^\infty$). Here, $\|\phi(x)\|^2$ denotes the pointwise squared norm of the Higgs field at $x$.

Given a Higgs potential $P$ with mass squared coefficient $\mu^2$ and quartic coupling coefficient $\lambda$, its negation is a new potential defined by replacing these parameters with $-\mu^2$ and $-\lambda$ respectively.

For any Higgs potential $P$, Higgs field $\phi$, and point $x$ in spacetime, the value of the negated potential $P_{\text{neg}}$ (defined by replacing the mass squared coefficient $\mu^2$ with $-\mu^2$ and the quartic coupling $\lambda$ with $-\lambda$) evaluated at $x$ is equal to the negative of the value of the original potential $P$. That is, $$V_{P_{\text{neg}}}(\phi, x) = -V_P(\phi, x)$$ where $V_P(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$.

Given a Higgs potential $P$ with mass squared coefficient $\mu^2$, the mass squared coefficient of its negation $P_{\text{neg}}$ is equal to $-\mu^2$.

Let $P$ be a Higgs potential with quartic coupling coefficient $\lambda$. The quartic coupling coefficient of its negation $P_{\text{neg}}$ is equal to $-\lambda$.

For any point $x$ in spacetime, the Higgs potential $V(\phi, x)$ evaluated at the zero Higgs field $\phi = 0$ is zero, i.e., $V(0, x) = 0$.

Let $P$ be a Higgs potential characterized by the mass squared parameter $\mu^2$ and the quartic coupling coefficient $\lambda$. For any Higgs field $\phi$ and any point $x$ in spacetime, if $\lambda \neq 0$, the Higgs potential $V(\phi, x)$ satisfies the completed square identity: $$V(\phi, x) = \lambda \left( \|\phi(x)\|^2 - \frac{\mu^2}{2\lambda} \right)^2 - \frac{(\mu^2)^2}{4\lambda}$$ where $\|\phi(x)\|^2$ is the pointwise squared norm of the Higgs field at $x$.

For a Higgs field $\phi$ and a point $x$ in spacetime, let $\|\phi(x)\|^2$ denote the pointwise squared norm of the field, and let $V(\phi, x)$ be the Higgs potential characterized by the mass squared parameter $\mu^2$ and the quartic coupling coefficient $\lambda$. These quantities satisfy the following quadratic equation: $$\lambda (\|\phi(x)\|^2)^2 - \mu^2 \|\phi(x)\|^2 - V(\phi, x) = 0$$

Let $P$ be a Higgs potential characterized by the mass squared parameter $\mu^2$ and the quartic coupling coefficient $\lambda$. For any Higgs field $\phi$ and any point $x$ in spacetime, assuming $\lambda \neq 0$, the Higgs potential $V(\phi, x)$ is zero if and only if the value of the Higgs field at $x$ is zero ($\phi(x) = 0$) or the squared norm of the field at that point satisfies $\|\phi(x)\|^2 = \frac{\mu^2}{\lambda}$.

The discriminant of the quadratic equation associated with the Higgs potential for a Higgs field $\phi$ at a spacetime point $x$. Given the potential $V(\phi, x) = \lambda (\|\phi(x)\|^2)^2 - \mu^2 \|\phi(x)\|^2$, this value corresponds to the discriminant of the quadratic $a y^2 + b y + c = 0$ where $y = \|\phi(x)\|^2$, $a = \lambda$, $b = -\mu^2$, and $c = -V(\phi, x)$. It is given by the formula $\Delta = (-\mu^2)^2 - 4\lambda(-V(\phi, x)) = \mu^4 + 4\lambda V(\phi, x)$, where $\mu^2$ is the mass squared parameter and $\lambda$ is the quartic coupling coefficient.

For any Higgs field $\phi$ and any point $x$ in spacetime, let $V(\phi, x) = \lambda (\|\phi(x)\|^2)^2 - \mu^2 \|\phi(x)\|^2$ be the Higgs potential with quartic coupling coefficient $\lambda$ and mass squared parameter $\mu^2$. If $\lambda \neq 0$, then the discriminant of the associated quadratic equation, defined as $\Delta(\phi, x) = \mu^4 + 4\lambda V(\phi, x)$, is non-negative, i.e., $0 \leq \Delta(\phi, x)$.

Let $P$ be a Higgs potential characterized by a quartic coupling coefficient $\lambda \neq 0$ and a mass squared parameter $\mu^2$. For any Higgs field $\phi$ and any point $x$ in spacetime, the associated quadratic discriminant $\Delta(\phi, x) = \mu^4 + 4\lambda V(\phi, x)$ is equal to the product of its square root with itself: $$\Delta(\phi, x) = \sqrt{\Delta(\phi, x)} \cdot \sqrt{\Delta(\phi, x)}.$$

Let $P$ be a Higgs potential characterized by a quartic coupling coefficient $\lambda \neq 0$ and a mass squared parameter $\mu^2$. For any Higgs field $\phi$ and any point $x$ in spacetime, the quadratic discriminant $\Delta(\phi, x)$ associated with the potential is equal to zero if and only if the potential function $V(\phi, x)$ satisfies $$V(\phi, x) = -\frac{(\mu^2)^2}{4\lambda}.$$

Let $P$ be a Higgs potential characterized by a quartic coupling coefficient $\lambda \neq 0$ and a mass squared parameter $\mu^2$. For any Higgs field $\phi$ and any point $x$ in spacetime, the quadratic discriminant $\Delta(\phi, x)$ associated with the Higgs potential is zero if and only if the pointwise squared norm of the Higgs field $\|\phi(x)\|^2$ satisfies $$\|\phi(x)\|^2 = \frac{\mu^2}{2\lambda}.$$

Let $P$ be a Higgs potential characterized by a mass squared parameter $\mu^2$ and a quartic coupling coefficient $\lambda$. If $\lambda < 0$, then for any Higgs field $\phi$ and any point $x$ in spacetime, the potential function $V(\phi, x)$ is bounded above by $$V(\phi, x) \leq -\frac{(\mu^2)^2}{4\lambda}.$$

Let $P$ be a Higgs potential characterized by a mass squared parameter $\mu^2$ and a quartic coupling coefficient $\lambda$. If $\lambda > 0$, then for any Higgs field $\phi$ and any point $x$ in spacetime, the potential function $V(\phi, x)$ is bounded below by $$-\frac{(\mu^2)^2}{4\lambda} \leq V(\phi, x).$$

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda$, such that the potential of a Higgs field $\phi$ at a point $x$ in spacetime is given by $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$. If $\lambda < 0$, then for every Higgs field $\phi$ and point $x$ in spacetime, if the mass squared parameter is positive ($\mu^2 > 0$), then the potential is non-positive, i.e., $V(\phi, x) \leq 0$.

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda$, such that the potential of a Higgs field $\phi$ at a point $x$ in spacetime is given by $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$. If $\lambda > 0$, then for every Higgs field $\phi$ and point $x$ in spacetime, if the mass squared parameter is negative ($\mu^2 < 0$), the potential is non-negative, i.e., $V(\phi, x) \geq 0$.

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda < 0$. For any real number $c$, there exists a Higgs field $\phi$ and a spacetime point $x$ such that the potential $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$ is equal to $c$ if and only if one of the following two conditions holds: - $\mu^2 > 0$ and $c \leq 0$. This implies that if $\lambda$ is negative and $\mu^2$ is positive, the potential takes every non-positive value. - $\mu^2 \leq 0$ and $c \leq -\frac{(\mu^2)^2}{4\lambda}$. This implies that if $\lambda$ is negative and $\mu^2$ is non-positive, the potential takes every value less than or equal to the bound $-\frac{(\mu^2)^2}{4\lambda}$.

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda > 0$. For any real number $c$, there exists a Higgs field $\phi$ and a spacetime point $x$ such that the potential $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$ is equal to $c$ if and only if one of the following two conditions holds: - $\mu^2 < 0$ and $0 \le c$. This implies that if $\lambda$ is positive and $\mu^2$ is negative, the potential takes every non-negative value. - $0 \le \mu^2$ and $-\frac{(\mu^2)^2}{4\lambda} \le c$. This implies that if $\lambda$ is positive and $\mu^2$ is non-negative, the potential takes every value greater than or equal to its minimum bound $-\frac{(\mu^2)^2}{4\lambda}$.

For a given Higgs potential $P$, the property $P.\text{IsBounded}$ is true if the potential function $V(\phi, x)$ is bounded from below. Specifically, there exists a real constant $c$ such that for all Higgs fields $\phi$ and all points $x$ in spacetime, the potential satisfies $V(\phi, x) \geq c$.

Let $P$ be a Higgs potential with quartic coupling coefficient $\lambda$ and mass squared parameter $\mu^2$, defined by the function $V(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$ for a Higgs field $\phi$ and spacetime point $x$. If the potential $P$ is bounded from below (i.e., there exists a real constant $c$ such that $V(\phi, x) \geq c$ for all $\phi$ and $x$), then the quartic coefficient $\lambda$ is non-negative ($\lambda \geq 0$).

Let $P$ be a Higgs potential with quartic coupling coefficient $\lambda$. If $\lambda > 0$, then the potential $P$ is bounded from below. This means there exists a real constant $c$ such that for any Higgs field $\phi$ and any point $x$ in spacetime, the potential function satisfies $V(\phi, x) \geq c$.

Let $P$ be a Higgs potential with quartic coupling $\lambda$ and mass squared parameter $\mu^2$. If the quartic coupling $\lambda$ is zero, then the potential $P$ is bounded from below if and only if the mass squared parameter $\mu^2$ is non-positive ($\mu^2 \le 0$). This reflects the condition under which the term $-\mu^2 \|\phi\|^2$ remains bounded from below.

Let $V$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda$. If $\lambda > 0$ and $\mu^2 \le 0$, then for any Higgs field $\phi$ and spacetime point $x$, the pair $(\phi, x)$ is a global minimum of the potential function $(\phi, x) \mapsto V(\phi, x)$ if and only if $V(\phi, x) = 0$.

Let $V$ be a Higgs potential characterized by the mass squared parameter $\mu^2$ and the quartic coupling $\lambda$. Suppose $\lambda > 0$ and $\mu^2 \le 0$. For any Higgs field $\phi$ and spacetime point $x$, the pair $(\phi, x)$ is a global minimum of the potential function $(\phi, x) \mapsto V(\phi, x)$ if and only if the Higgs field vanishes at that point, i.e., $\phi(x) = 0$.

Let $P$ be a Higgs potential with mass squared parameter $\mu^2 \ge 0$ and quartic coupling $\lambda > 0$. For any Higgs field $\phi$ and spacetime point $x$, the pair $(\phi, x)$ is a global minimum of the potential function $V(\phi, x)$ if and only if $V(\phi, x) = -\frac{(\mu^2)^2}{4\lambda}$.

Let $P$ be a Higgs potential with quartic coupling $\lambda > 0$ and mass squared parameter $\mu^2 \ge 0$. For any Higgs field $\phi$ and point $x$ in spacetime, the pair $(\phi, x)$ is a global minimum of the potential function $V(\phi, x)$ if and only if the pointwise squared norm of the Higgs field satisfies $$\|\phi(x)\|^2 = \frac{\mu^2}{2\lambda}.$$

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda$. Suppose the quartic coupling is positive, $\lambda > 0$. For any Higgs field $\phi$ and spacetime point $x$, the configuration $(\phi, x)$ is a global minimum of the potential function $V(\phi, x)$ if and only if one of the following conditions holds: 1. $\mu^2 \ge 0$ and the squared norm of the Higgs field satisfies $\|\phi(x)\|^2 = \frac{\mu^2}{2\lambda}$. 2. $\mu^2 < 0$ and the Higgs field vanishes at that point, $\phi(x) = 0$.

For a given Higgs potential $P$, let $V_P(\phi, x)$ denote the potential function evaluating a Higgs field $\phi$ at a point $x$ in spacetime, defined by $V_P(\phi, x) = -\mu^2 \|\phi(x)\|^2 + \lambda (\|\phi(x)\|^2)^2$. Let $P_{\text{neg}}$ be the negated potential such that its value is $V_{P_{\text{neg}}}(\phi, x) = -V_P(\phi, x)$. A configuration $(\phi, x)$ is a global maximum of the Higgs potential $P$ if and only if it is a global minimum of the negated potential $P_{\text{neg}}$.

Let $P$ be a Higgs potential with mass squared parameter $\mu^2$ and quartic coupling $\lambda$. Suppose the quartic coupling is negative, $\lambda < 0$. For any Higgs field $\phi$ and spacetime point $x$, the configuration $(\phi, x)$ is a global maximum of the potential function $V(\phi, x)$ if and only if one of the following conditions holds: 1. $\mu^2 \le 0$ and the squared norm of the Higgs field satisfies $\|\phi(x)\|_H^2 = \frac{\mu^2}{2\lambda}$. 2. $\mu^2 > 0$ and the Higgs field vanishes at that point, $\phi(x) = 0$.