⚽ Goal: combine special relativity and quantum mechanics

First attempt

💼 Case: lets consider a spinless particle

• 🧠 Remember: in non-relativistic quantum mechanics we have the Schrodinger equation

  $$ i \hbar \frac{\partial}{\partial t } \ket{\psi(t)}=\frac{\hat{\vec p}^2}{2m} \ket{\psi(t)} $$

  For a free spinless particles in position space representation we get

  $$ i\hbar \frac{\partial}{\partial t} \psi(\vec r,t)=-\frac{\hbar ^2 \vec \nabla ^2}{2m} \psi (\vec r,t) $$

  • which was achieved using the following relations (expand)

🗒️ Notes:

• ❌ Is invariant under Galilean transformation not Lorentz transformations
• ❌ It does not encode the relativistic energy-momentum relationship $E^2=c^2 \vec p^2+m^2 c^4$

Lets naively try to replace $H\to i\hbar \partial /\partial t$ and $\vec p \to \hat{\vec p}=-i\hbar \vec \nabla$

• thus $H=(c^2 \vec p^2+m^4 c^4)^{1/2}$ becomes

  $$ i\hbar \frac{\partial}{\partial t} \ket{\psi(t)}=(c^2 \hat{\vec p}^2 +m^2 c^4)^{1/2}\ket{\psi(t)} $$

• Or in position space we would get 👹💀

  $$ i\hbar \frac{\partial}{\partial t} \psi(\vec r,t)=(-\hbar ^2 c^2 \nabla ^2+m^2 c^4)^{1/2} \psi(\vec r,t)=mc^2 \left ( 1-\frac{\hbar ^2 \nabla ^2}{m^2 c^2}\right )^{1/2}\psi(\vec r,t) $$

💎 Conclusion: this equation is bad, it is difficult to work with and doesn't have the same powers for time and space

Klein-Gordon equation

Lets instead work directly with $H^2=c^2 \vec p^2+m^2 c^4$

• Using our Hamiltonian and momentum transformations

  $$ H^2 \to \left ( i \hbar \frac{\partial}{\partial t} \right )^2 =- \hbar ^2 \frac{\partial ^2}{\partial t^2} \qquad \vec p \to \hat{\vec p} =-i\hbar \vec \nabla $$

  we get

  $$ -\hbar ^2 \frac{\partial ^2 }{\partial t^2} \ket{\psi(t)} =(c^2 \hat{\vec p} ^2 + m^2 c^4) \ket{\psi(t)} $$

• We can turn this into position space to get the free particle Klein-Gordon equation

  $$ \boxed{- \hbar ^2 \frac{\partial^2}{\partial t^2}\psi(\vec r,t) = (-c^2 \hbar ^2 \nabla ^2 + m^2 c^4) \psi(\vec r,t)} $$

  🗒️ Notes:

  • It has the desired spatial and time derivative symmetry (2nd order)
  • $\psi(\vec r,t)$ is a scalar field so it cannot describe relativistic spin particles like electrons
  • It is good at describing particles like pions and kaons which have spin 0

• We can rewrite the equation in its compact form as

  $$ \left ( \frac{1}{c^2} \frac{\partial ^2}{\partial t^2} - \nabla ^2 +\frac{m^2 c^2}{\hbar^2} \right )\psi(\vec r,t)=0 \quad \Rightarrow \quad \boxed{(\square + \mu^2) \psi(\vec r,t)=0} $$

  where $\square \equiv \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla ^2$ is the d’Alembertian and $\mu=\frac{mc}{\hbar}$ is defined in terms of the rest mass thus is Lorentz invariant

Free particle solutions

• We start with a standard wave ansatz propagating with momentum $\vec p$

  $$ \psi(\vec r,t)=Ne^{i(\vec p\cdot \vec r- E_{\vec p} t)/\hbar} $$

  where $N$ is some suitable normalisation

• Substituting this into the Clein Gordon equation we get

  $$ \begin{aligned} \frac{1}{c^2} \left ( \frac{-i E_{\vec p}}{\hbar } \right )^2+ \frac{\vec p^2}{\hbar^2}+ \frac{m^2 c^2}{\hbar^2}&=0 \\ \Rightarrow \qquad - E^2 _{\vec p}+ c^2 \vec p^2 + m^2 c^4&=0 \end{aligned} $$

💎 Conclusion: Our wave ansatz is a solution to the Klein-Gordon equation if