A relativistic version of the Schrödinger equation.

Correctly describes spin 0 particles.

The most memorable version of the equation can be written as shown at Section "Klein-Gordon equation in Einstein notation" with Einstein notation and Planck units:

$∂_{i}∂_{i}ψ−m_{2}ψ=0$

Has some issues which are solved by the Dirac equation:

- it has a second time derivative of the wave function. Therefore, to solve it we must specify not only the initial value of the wave equation, but also the derivative of the wave equation,As mentioned at Advanced quantum mechanics by Freeman Dyson (1951) and further clarified at: physics.stackexchange.com/questions/340023/cant-the-negative-probabilities-of-klein-gordon-equation-be-avoided, this would lead to negative probabilities.
- the modulus of the wave function is not constant and therefore not always one, and therefore cannot be interpreted as a probability density anymore
- since we are working with the square of the energy, we have both positive and negative value solutions. This is also a features of the Dirac equation however.

Bibliography:

- Video "Quantum Mechanics 12a - Dirac Equation I by ViaScience (2015)" at youtu.be/OCuaBmAzqek?t=600
- An Introduction to QED and QCD by Jeff Forshaw (1997) 1.2 "Relativistic Wave Equations" and 1.4 "The Klein Gordon Equation" gives some key ideas
- 2011 PHYS 485 lecture videos by Roger Moore from the University of Alberta at around 7:30
- www.youtube.com/watch?v=WqoIW85xwoU&list=PL54DF0652B30D99A4&index=65 "L2. The Klein-Gordon Equation" by doctorphys
- sites.ualberta.ca/~gingrich/courses/phys512/node21.html from Advanced quantum mechanics II by Douglas Gingrich (2004)

The Klein-Gordon equation directly uses a more naive relativistic energy guess of $p_{2}+m_{2}$ squared.

But since this is quantum mechanics, we feel like making $p$ into the "momentum operator", just like in the Schrödinger equation.

But we don't really know how to apply the momentum operator twice, because it is a gradient, so the first application goes from a scalar field to the vector field, and the second one...

So we just cheat and try to use the laplace operator instead because there's some squares on it:

$H=∇_{2}+m_{2}$

But then, we have to avoid taking the square root to reach a first derivative in time, because we don't know how to take the square root of that operator expression.

So the Klein-Gordon equation just takes the approach of using this squared Hamiltonian instead.

Since it is a Hamiltonian, and comparing it to the Schrödinger equation which looks like:
taking the Hamiltonian twice leads to:

$Hψ=i∂t∂ψ $

$H_{2}ψ=−∂_{2}t∂_{2}ψ $

We can contrast this with the Dirac equation, which instead attempts to explicitly construct an operator which squared coincides with the relativistic formula: derivation of the Dirac equation.

## Discussion (0)

Sign up or sign in create discussions.There are no discussions about this article yet.