The Klein Gordon equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}$$ $$ \vec{p} = -i\hbar\nabla$$
acting on $\phi(x,t)$ to give,
$$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$
where the d'alembertian operator is defined here as $$ \Box = \frac{1}{c^2}\partial_t^2 - \nabla^2$$
We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ \eta_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have,
$$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.
To make life simple, we adopt to the convention,
where we equate the planck's constant and the speed of light equal to 1 (dimensionless number). The consequences are, the dimension of Length and Time are the same and the dimension of Mass is just the inverse of Length or Time. In addition, Energy and Momentum is measured in the same unit as of the Mass.
And, contravariant vectors are $ A^{\mu}= (A_0, \vec{A})$ and corresponding covariant transformation is $ A_{\mu}= \eta_{\mu\nu}A^{\nu} = (A_0, -\vec{A})$ where the differential is defined in reverse way
$ \partial_\mu = \left(\partial_0,\vec{\nabla}\right) $
[$\partial_0 = \partial_t$] and
$ \partial^{\mu}= \eta^{\mu\nu}\partial_{\nu} = ({\partial}_0,-\vec{\nabla})$
With these substitution we get our KG equation as, $$\left(\Box + m^2\right)\phi = 0 $$ where $p^\mu{p}_\mu = m^2 \\ \,\, E^2 = \omega^2 = m^2 + {|\vec{p}|^2} $
Since $m^2$ and d'Alembertian operator is invariant under Lorentz transformation, if the function $\phi$ satisfies the condition $ \phi'(x') = \phi(x)$ then the whole KG equation is invariant under Lorentz transformation.
But, there are some problems with this equation in the basic definition of $\phi(x)$. First of all it cannot be the wave function of the particle as it is in the case of Non-relativistic Schr\"{o}dinger equation.
The problem arises because of the probability statements. To understand its significance, let us just assume that $\phi(x)$ is the usual wave function.
Then, it says that the probability of finding the particle at a position x is the same as of the finding the particle in $x'$ position in some other reference frame [Lorentz invariance condition]. It implies the wave function should behave like a scalar quantity and independent of direction, which is not true in general for spin half particles. The properties of spin half particles depends on from which direction it is measured and change with respect to different orientations of the reference frame.
And the second and the most important problem is that the probability density definition changes completely and it allows for the weird possibility of the negative probability.
The continuity equation with the time component becomes, $$\partial_0\rho+\vec{\nabla}\cdot\vec{J} = 0 = \partial_\mu{J}^\mu$$ where $ \rho = \frac{1}{2} \left[\phi^*(\partial_0\phi) - (\partial_0\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
and \newline
$\vec{J} = \frac{-1}{2}\left[\phi^*(\nabla\phi) - (\nabla\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
where $\rho$ is the equivalent probability density as defined in non-relativistic case can now have positive as well as negative values, that is not compatible with the definition of probability - you can check it for monochromatic wave of the for $\phi = Ae^{\pm{ikx}}$. Thus, either we will have to abandon KG equation or should find an alternative way of description for its definition.
acting on $\phi(x,t)$ to give,
$$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$
where the d'alembertian operator is defined here as $$ \Box = \frac{1}{c^2}\partial_t^2 - \nabla^2$$
We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ \eta_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have,
$$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.
To make life simple, we adopt to the convention,
where we equate the planck's constant and the speed of light equal to 1 (dimensionless number). The consequences are, the dimension of Length and Time are the same and the dimension of Mass is just the inverse of Length or Time. In addition, Energy and Momentum is measured in the same unit as of the Mass.
And, contravariant vectors are $ A^{\mu}= (A_0, \vec{A})$ and corresponding covariant transformation is $ A_{\mu}= \eta_{\mu\nu}A^{\nu} = (A_0, -\vec{A})$ where the differential is defined in reverse way
$ \partial_\mu = \left(\partial_0,\vec{\nabla}\right) $
[$\partial_0 = \partial_t$] and
$ \partial^{\mu}= \eta^{\mu\nu}\partial_{\nu} = ({\partial}_0,-\vec{\nabla})$
With these substitution we get our KG equation as, $$\left(\Box + m^2\right)\phi = 0 $$ where $p^\mu{p}_\mu = m^2 \\ \,\, E^2 = \omega^2 = m^2 + {|\vec{p}|^2} $
Since $m^2$ and d'Alembertian operator is invariant under Lorentz transformation, if the function $\phi$ satisfies the condition $ \phi'(x') = \phi(x)$ then the whole KG equation is invariant under Lorentz transformation.
But, there are some problems with this equation in the basic definition of $\phi(x)$. First of all it cannot be the wave function of the particle as it is in the case of Non-relativistic Schr\"{o}dinger equation.
The problem arises because of the probability statements. To understand its significance, let us just assume that $\phi(x)$ is the usual wave function.
Then, it says that the probability of finding the particle at a position x is the same as of the finding the particle in $x'$ position in some other reference frame [Lorentz invariance condition]. It implies the wave function should behave like a scalar quantity and independent of direction, which is not true in general for spin half particles. The properties of spin half particles depends on from which direction it is measured and change with respect to different orientations of the reference frame.
And the second and the most important problem is that the probability density definition changes completely and it allows for the weird possibility of the negative probability.
The continuity equation with the time component becomes, $$\partial_0\rho+\vec{\nabla}\cdot\vec{J} = 0 = \partial_\mu{J}^\mu$$ where $ \rho = \frac{1}{2} \left[\phi^*(\partial_0\phi) - (\partial_0\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
and \newline
$\vec{J} = \frac{-1}{2}\left[\phi^*(\nabla\phi) - (\nabla\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
where $\rho$ is the equivalent probability density as defined in non-relativistic case can now have positive as well as negative values, that is not compatible with the definition of probability - you can check it for monochromatic wave of the for $\phi = Ae^{\pm{ikx}}$. Thus, either we will have to abandon KG equation or should find an alternative way of description for its definition.
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