Lorentz Invariance

We now study the Lorentz invariance of the Klein-Gordon equation. The operator $\Box\equiv\partial^\mu\partial_\mu\equiv\frac{\partial}{\partial x_\mu}\frac{\partial}{\partial x^\mu}$ is invariant under a Lorentz transformation because it is a scalar product of 4-vectors, $\partial^\mu$. Also, the mass, $m$, is a scalar. Now consider a transformation from an unprimed system to a primed system. In the transformed primed system

$$\left(\frac{\partial}{\partial {x^\prime}^\mu}\frac{\partial}{\partial x^\prime_\mu} + m^2\right)\psi^\prime(x^\prime_\mu) = 0 ,$$

$$\left(\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x_\mu} + m^2\right)\psi^\prime(x^\prime_\mu) = 0 ,$$

$$(\Box + m^2)\psi^\prime(x^\prime_\mu) = 0 .$$ (4.12)

Since $\psi(x_\mu)$ and $\psi^\prime(x^\prime_\mu)$ refer to the same space-time point and $\psi$ is a scalar, the Klein-Gordon equation is Lorentz invariant. Notice that $\psi^\prime(x^\prime_\mu)$ and $\psi^\prime(x_\mu)$ are different and are related by $x^\prime_\mu=a_\mu^{\ \nu} x_\nu$. Also $\psi(x_\mu)$ and $\psi^\prime(x_\mu)$ refer to two different points with coordinates $x_\mu$ in the old and new system, respectively.

Since the Klein-Gordon operator does not change under continuous Lorentz transformations, we can reason that the wave function is multiplied by a factor with absolute value of unity in these transformations. In the case of the coordinate transformation $x\rightarrow x^\prime=\hat{a}x$ (where $\hat{a}$ is the Lorentz operator) the transformation law of the wave function is

$$\psi(x) \rightarrow \psi^\prime(x^\prime) = \lambda\psi(x),$$ (4.13)

with $\vert\lambda\vert=1$. Since $\hat{a}$ is real, $\lambda$ must be real and hence $\lambda = \pm 1$.

If the Lorentz transformation is continuous (ie. rotation in 4-space), $\hat{a}$ depends continuously on some variables, say $\alpha_i$. For all $\alpha_i=0$, we must have the identity transformation and $\lambda=1$ holds. A wave function which does not change under spatial rotations can describe a particle with spin-0.

For space inversion, which is a discrete Lorentz transformation, ${x^\prime}^k=-x^k$ and ${x^\prime}^0=x^0$. Applying the space inversion operator twice leads to the identity transformation. Therefore $\lambda^2=1$ or $\lambda=\pm 1$. We define two states:

for the case $\lambda=+1$,

$$\psi^\prime(\vec{x}^\prime,t^\prime) = \psi^\prime(-\vec{x},t) = \psi(\vec{x},t) \Rightarrow \ \textrm{scalar} ,$$ (4.14)

for the case $\lambda=-1$,

$$\psi^\prime(\vec{x}^\prime,t^\prime) = \psi^\prime(-\vec{x},t) = -\psi(\vec{x},t) \Rightarrow \ \textrm{pseudoscalar} .$$ (4.15)

Therefore solutions of the Klein-Gordon equation are scalar or pseudoscalar, ie. invariant under spatial rotations and proper Lorentz transformations, and are invariant (scalar) or change sign (pseudoscalar) under space inversion. The pi-meson (pion) is an example of a pseudoscalar meson that obeys the Klein-Gordon equation.