Quantum Field Theory
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Relativistic Quantum Mechanics Klein-Gordon Equation
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Related De Donder-Weyl Theory

The Klein-Gordon Equation is a relativistic generalisation of the Schrordinger Equation to Relativistic Quantum Mechanics. In Quantum Field Theory, it is a relativistic field equation for a spin-0 field (Klein-Gordon Field).

HistoryEdit this section

This equation was actually discovered by Erwin Schrodinger before he discovered the Schrodinger Equation, that bears his name. However, for reasons which we will be discussing later, he soon discarded the equation and decided to use the non-relativistic energy instead, which resulted in the intrinsically not Lorentz-Invariant Schrodinger Equation.  ;

This equation was named after Oskar Klein and Walter Gordon, who proposed (1926) that it describes relativistic electrons. Vladimir Fock, Johann Kudar, Theophile de Donder and Frans H van den Dungen, and Louis de Broglie also made these claims.

We now know that the Dirac Equation is the actual field equation for these spin-1/2 electrons. The Klein-Gordon Equation however, does describe spin-0 particles, such as the Higgs Field/ .

StatementEdit this section

Klein-Gordon Equation

$ \left(\Box+\mu^2\right)\Psi=0 $

Derivation of Klein-Gordon Equation from Schrodinger's Wave Equation

In the Schrodinger formulation of Quantum Mechanics, quantisation is as follows:

Start with the Energy:

$ E=\frac{m}{2}{{\left( \frac{\text{d}x}{\text{d}t} \right)}^{2}}+U=\frac{{{p}^{2}}}{2m}+U $

Then, quantise as follows:

$ \begin{align} & E\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ & \left( \frac{{{p}^{2}}}{2m}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ \end{align} $

To be Lorentz-Invariant (It needs to be relativistic) start with the Special Relativistic Energy formula instead:

$ E=\sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+U $

$ \begin{align} & E\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ & \left( \sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t} \\ \end{align} $

$ \left( \sqrt{m_{0}^{2}c_{0}^{4}-{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}}+U \right)\Psi =i\hbar \frac{\partial \Psi }{\partial t} $

The Laplacian cannot be calculated under the square root, so instead, we use

$ {{E}^{2}}={{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}+2U\sqrt{{{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4}}+{{U}^{2}} $

We realise that this cannot be properly calculated (the same problems as before, the square root), so, at least for now, we let $ U = 0 $, i.e. only free particles.

$ {{E}^{2}}={{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4} $

$ \begin{align} & {{E}^{2}}\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & \left( {{p}^{2}}c_{0}^{2}+m_{0}^{2}c_{0}^{4} \right)\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & -{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}\Psi +m_{0}^{2}c_{0}^{4}\Psi =-{{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}} \\ & {{\hbar }^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}}-{{\hbar }^{2}}c_{0}^{2}{{\nabla }^{2}}\Psi +m_{0}^{2}c_{0}^{4}\Psi =0 \\ & {{\hbar }^{2}}c_{0}^{2}\left( \frac{1}{c_{0}^{2}}\frac{{{\partial }^{2}}\Psi }{\partial {{t}^{2}}}-{{\nabla }^{2}}\Psi \right)+m_{0}^{2}c_{0}^{4}\Psi =0 \\ & -{{\hbar }^{2}}c_{0}^{2}\left( -\frac{{{\partial }^{2}}\Psi }{\partial {{\left( {{c}_{0}}t \right)}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right)+m_{0}^{2}c_{0}^{4}\Psi =0 \\ & -{{\hbar }^{2}}{{\eta }^{\mu \nu }}{{\partial }_{\mu }}{{\partial }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0 \\ & \left( \square +{{\mu }^{2}} \right)\Psi =0 \\ \end{align} $

In a PotentialEdit this section

In a potential, the Klein-Gordon Equation obviously gets modified as follows:

$ \left(\Box+\mu^2\right)\Psi+\frac{\partial V}{\partial\psi}=0 $

Free Particle SolutionEdit this section

For the free particle solution described earlier, the solution for the wavefunction remains the same as in the Non-Relativistic case:

$ \psi(\mathbf{r}, t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} $

This is because the additional energy $ m_0c_0^2 $ only induces a very fast phase rotation of the state.

However, an additional constraint, called the Dispersion Relation, arises:

$ -k^2+\frac{\omega^2}{c^2}=\mu^2 $ \

For massless particles, this becomes $ k=\pm\frac{\omega}{c_0} $.

Gravitational InteractionEdit this section

To see the effect of gravitational interaction, use the form of the Klein-Gordon Equation:

$ -{{\hbar }^{2}}{{\eta }^{\mu \nu }}{{\partial }_{\mu }}{{\partial }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0 $

It is obvious to anyone with any knowledge about General Relativity that when considering gravitational interaction, one does transformations such as $ \eta_{\mu\nu}\to g_{\mu\nu},\partial\to\nabla $. Therefore, in the presence of a gravitational field, the Klein-Gordon Equation becomes:

$ -{{\hbar }^{2}}{{g }^{\mu \nu }}{{\nabla }_{\mu }}{{\nabla }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi =0 $

Or, with a potential,

$ -{{\hbar }^{2}}{{g }^{\mu \nu }}{{\nabla }_{\mu }}{{\nabla }_{\nu }}\Psi +m_{0}^{2}c_{0}^{2}\Psi + \frac{\partial V}{\partial\psi}=0 $

IssuesEdit this section

However, the Klein-Gordon Equation has issues when interpreted as a standard wave equation.

  • Since it is second-order in time (like the classical wave equation from Classical Lorentz-Invariant EM), probability density is not explicitly always conserved.
  • The energy is not forced to be positive.

The Dirac Equation was an attempt to solve this issue/.