Klein-Gordon Equation – What Would Change if a Linear Term is Added in the Lagrangian?

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The usual Klein-Gordon Lagrangian reads

\begin{equation} \mathscr{L}= \frac{1}{2}( \partial _{\mu} \Phi \partial ^{\mu} \Phi -m^2 \Phi^2) \, . \tag1\end{equation}

Without additional symmetry beyond Lorentz symmetry, nothing forbids an additional linear term:

\begin{equation} \mathscr{L}= \frac{1}{2}( \partial _{\mu} \Phi \partial ^{\mu} \Phi -m^2 \Phi^2) – C \Phi \, , \tag2\end{equation}
where $C$ is some constant.

This modified Lagrangian leads to a modified Klein-Gordon equation

$$( \partial _{\mu} \partial ^{\mu}+m^2)\Phi =C \, .\tag3$$

What would be the interpretation of this modified Klein-Gordon equation? Why do we usually neglect the linear term and hence the possible constant in the Klein-Gordon equation?

Best Answer

Hint 1. What happens if you take the vacuum expectation value of equation $(3)$? You should find $\langle\Phi\rangle\neq 0$. Why is this a bad thing?

Hint 2. (pretty much the same thing as Hint 1 actually) What happens under the field redefinition $\Phi\to\Phi+C/m^2$?

Related Solutions

[Physics] Finding the expression for probability density (the Klein Gordon equation)

The Schr$\ddot{\rm o}$dinger equation is non-relativistic and for a free particle is derived from the Hamiltonian \begin{equation} H\boldsymbol{=} \dfrac{p^2}{2m} \tag{K-01}\label{eqK-01} \end{equation} by the transcription \begin{equation} H\boldsymbol{\longrightarrow} i\hbar\dfrac{\partial}{\partial t}\quad \text{and}\quad \mathbf{p}\boldsymbol{\longrightarrow} \boldsymbol{-}i\hbar\boldsymbol{\nabla} \tag{K-02}\label{eqK-02} \end{equation} so that \begin{equation} i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{+}\dfrac{\hbar^2}{2m}\nabla^2\psi\boldsymbol{=} 0 \tag{K-03}\label{eqK-03} \end{equation} For a first try to derive a relativistic quantum mechanical equation we make use of the property that according to the theory of special relativity the total energy $\;E\;$ and momenta $\;(p_x,p_y,p_z)\;$ transform as components of a contravariant four-vector \begin{equation} p^\mu\boldsymbol{=}\left(p^0,p^1,p^2,p^3\right)\boldsymbol{=}\left(\dfrac{E}{c},p_x,p_y,p_z\right) \tag{K-04}\label{eqK-04} \end{equation} of invariant length \begin{equation} \sum\limits_{\mu\boldsymbol{=}0}^{3}p_{\mu} p^{\mu}\boldsymbol{\equiv}p_{\mu} p^{\mu}\boldsymbol{=}\dfrac{E^2}{c^2}\boldsymbol{-}\mathbf{p}\boldsymbol{\cdot}\mathbf{p}\boldsymbol{\equiv}m^2c^2\tag{K-05}\label{eqK-05} \end{equation} where $\;m\;$ is the rest mass of the particle and $\;c\;$ the velocity of light in vacuum.

Following this it is natural to take as the Hamiltonian of a relativistic free particle \begin{equation} H\boldsymbol{=}\sqrt{p^{2}c^2\boldsymbol{+}m^2c^4} \tag{K-06}\label{eqK-06} \end{equation} and to write for a relativistic quantum analogue of \eqref{eqK-03} \begin{equation} i\hbar \dfrac{\partial \psi}{\partial t}\boldsymbol{=}\sqrt{\boldsymbol{-}\hbar^2c^2 \nabla^{2}\boldsymbol{+}m^2c^4}\,\psi \tag{K-07}\label{eqK-07} \end{equation} Facing the problem of interpreting the square root operator on the right in eq. \eqref{eqK-07} we simplify mathematics by removing this square root operator, so that \begin{equation} \left[\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-08}\label{eqK-08} \end{equation} or recognized as the classical wave equation \begin{equation} \left[\square\boldsymbol{+}\left(\dfrac{mc}{\hbar}\right)^2\right]\psi\boldsymbol{=}0 \tag{K-09}\label{eqK-09} \end{equation} where⁽¹⁾ \begin{equation} \square\boldsymbol{\equiv}\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\boldsymbol{-}\nabla^{2}\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu} \tag{K-10}\label{eqK-10} \end{equation}

Equation \eqref{eqK-09} is the Klein-Gordon equation for a free particle. With its complex conjugate we have

\begin{align} & \dfrac{1}{c^2}\dfrac{\partial^2 \psi\hphantom{^{\boldsymbol{*}}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi\hphantom{^{\boldsymbol{*}}}\boldsymbol{=} 0 \tag{K-11.1}\label{eqK-11.1}\\ &\dfrac{1}{c^2}\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\boldsymbol{-}\nabla^{2}\psi^{\boldsymbol{*}}\boldsymbol{+}\left(\dfrac{mc}{\hbar}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)^2\psi^{\boldsymbol{*}}\boldsymbol{=} 0 \tag{K-11.2}\label{eqK-11.2} \end{align} Multiplying them by $\;\psi^{\boldsymbol{*}},\psi\;$ respectively and subtracting side by side we have⁽²⁾ \begin{align} \dfrac{1}{c^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial^2 \psi}{\partial t^2}\boldsymbol{-}\psi\dfrac{\partial^2 \psi^{\boldsymbol{*}}}{\partial t^2}\right)\boldsymbol{-}\left(\psi^{\boldsymbol{*}}\nabla^{2}\psi\boldsymbol{-}\psi\nabla^{2}\psi^{\boldsymbol{*}}\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0\quad \boldsymbol{\Longrightarrow} \nonumber\\ \dfrac{1}{c^2}\dfrac{\partial}{\partial t}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\boldsymbol{+}\boldsymbol{\nabla \cdot}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)&\boldsymbol{=} 0 \tag{K-12}\label{eqK-12} \end{align} We multiply above equation by $\;i\hbar/2m\;$ in order to have real quantities on one hand and on the other hand to have an identical expression for the probability current density vector as that one from the Schr$\ddot{\rm o}$dinger equation
\begin{equation} \dfrac{\partial}{\partial t}\left[\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\right]\boldsymbol{+}\boldsymbol{\nabla \cdot}\left[\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\right]\boldsymbol{=} 0 \tag{K-13}\label{eqK-13} \end{equation} so \begin{equation} \dfrac{\partial \varrho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla \cdot}\boldsymbol{S}\boldsymbol{=} 0 \tag{K-14}\label{eqK-14} \end{equation} where \begin{equation} \boxed{\:\:\varrho\boldsymbol{\equiv}\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)\:\:}\quad \text{and} \quad \boxed{\:\:\boldsymbol{S}\boldsymbol{\equiv}\dfrac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla }\psi^{\boldsymbol{*}}\boldsymbol{-}\psi^{\boldsymbol{*}}\boldsymbol{\nabla }\psi\vphantom{\dfrac{\partial^2 \psi}{\partial t^2}}\right)\:\:} \tag{K-15}\label{eqK-15} \end{equation} We would like to interpret $\dfrac{i\hbar}{2mc^2}\left(\psi^{\boldsymbol{*}}\dfrac{\partial \psi}{\partial t}\boldsymbol{-}\psi\dfrac{\partial \psi^{\boldsymbol{*}}}{\partial t}\right)$ as a probability density $\varrho$. However, this is impossible, since it is not a positive definite expression.

^{(1)
We define
\begin{align}
\blacktriangleright x^\mu\boldsymbol{=}\left(ct,\mathbf{x}\right)&\blacktriangleright \nabla^\mu\boldsymbol{=}\partial^\mu\boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\boldsymbol{-}\boldsymbol{\nabla}\right)
\nonumber\\
&\blacktriangleright \nabla_\mu\boldsymbol{=}\partial_\mu\boldsymbol{=}\dfrac{\partial}{\partial x^\mu}\boldsymbol{=}\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\boldsymbol{+}\boldsymbol{\nabla}\right)\blacktriangleright\square \boldsymbol{=}\nabla^\mu\nabla_\mu \boldsymbol{=}\partial^\mu\partial_\mu \boldsymbol{=}\dfrac{\partial}{\partial x_\mu}\dfrac{\partial}{\partial x^\mu}
\nonumber
\end{align}}

^{(2)
If $\;\psi\;$ and $\;\mathbf{a}\;$ are scalar and vector functions in $\;\mathbb{R}^{3}$ then
\begin{equation}
\boldsymbol{\nabla \cdot}\left(\psi\mathbf{a}\right)\boldsymbol{=}\mathbf{a}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi\boldsymbol{+}\psi\boldsymbol{\nabla \cdot}\mathbf{a}
\nonumber
\end{equation}}

[Physics] Lagrangian of Klein Gordon equation

My question is, is it correct that I have to introduce the free index $\nu$ , or are there easier ways to do this?

You didn't introduce a free index $\nu$ as it is contracted with another one. A free index means that the index is not summed over. What you did was introduce a $dummy$ index to sum over.

The answer to your question, however, is no there is not an easier was to do this. In quantum field theory, this is as easy as it gets. The way to prove the derivative of $(\partial \Phi)^2$ is to do precisely what you did by using the definition $\partial_\mu \Phi \partial_\nu \Phi g^{\mu\nu}$ and then use the product rule.

The resulting Klein Gordon equation should not depend on what convention you use for the metric, as you can just multiply by a minus sign to get the relative minus signs correct.

The above, however, may give you the impression that equations of motion will not "look" different in different metrics. But that would be wrong. Maxwells equations are a fine example of this.

$$ \partial_\mu F^{\mu\nu} = j^\nu \qquad (+---) $$

and

$$ \partial_\mu F^{\mu\nu} = -j^\nu \qquad (-+++). $$

The equations look different but they are not. Indeed the "extra" minus sign in the latter equation comes from the fact that $A_\mu = (-\Phi, \mathbf{A})$ in the $(-+++)$ convention. So the equations are identical. I think that this answer may be of some use.

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Related Solutions

[Physics] Finding the expression for probability density (the Klein Gordon equation)

[Physics] Lagrangian of Klein Gordon equation

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