Klein-Gordon equation

How to construct

Starting from the relativistic connection between energy and momentum:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^2=\bold p^2c^2+m^2c^4}

Substituting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E \rightarrow i\hbar \frac{\partial}{\partial t}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold p \rightarrow -i\hbar \nabla} , we get Klein-Gordon equation for free particles as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\hbar^2 \frac{\partial ^2\psi(\bold r, t)}{\partial t^2}=(-\hbar^2c^2\nabla^2+m^2c^4)\psi(\bold r, t)\qquad \qquad \qquad \qquad \qquad (1)}

Klein-Gordon can also be written as the following:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\square-K^2)\psi(\bold r, t)=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;\;(2)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}} is d'Alembert operator and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K=\frac{mc}{\hbar}} .

Equation (2) looks like a classical wave equation with an extra term Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K^2} .

Potentials couple to the Klein-Gordon equation in a manner analogous to classical four-vectors. A good example of this is the potential four-vector:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi ^{ \mu } = (\phi ; \bold{A} )}

Coupling this to the momentum four-vector,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p ^{ \mu } = (\frac{E}{c} ; \bold{p} )}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E \rightarrow i\hbar \frac{\partial}{\partial t}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold{p} \rightarrow -i\hbar \bold{\nabla}} in the quantum limit, we find the conjugate momentum four vector:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P^{\mu} = p^{\mu}-\frac{e}{c}\Phi^{\mu} = ( \frac{E}{c}-\frac{e}{c}\phi ; \bold{p}-\frac{e}{c}\bold{A} )}

Squaring the conjugate momentum four vector and multiplying by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^2} , we obtain the Klein-Gordon equation in an electromagnetic field by moving to the quantum limit and acting on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi } :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c^2 P^{\mu}P_{\mu} = m^2 c^4 = (E-e\phi)^2-(c\bold{p}-e\bold{A})^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow\left[ i\hbar \frac {\partial}{\partial t}-e\phi(\bold r, t) \right] ^2\psi(\bold r, t)=\left( \left[ -i\hbar\nabla-\frac{e}{c}\bold A(\bold r, t)\right] ^2c^2+m^2c^4\right) \psi(\bold r, t)}

The Klein-Gordon equation is second order in time. Therefore, to see how the states of a system evolve in time we need to know both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\bold r, t)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial\psi(\bold r, t)}{\partial t}} at a certain time. While in nonrelativistic quantum mechanics, we only need Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\bold r, t)}

Also because the Klein-Gordon equation is second order in time, it has the solutions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\bold r, t)=e^{i(\bold p \bold r - Et)/\hbar}} with either sign of energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=\pm c\sqrt{\bold p^2+m^2c^2}} . The negative energy solution of Klein-Gordon equation has a strange property that the energy decreases as the magnitude of the momentum increases. We will see that the negative energy solutions of Klein-Gordon equation describe antiparticles, while the positive energy solutions describe particles.

Continuity equation

Multiplying (1) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \psi^{*}} from the left, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{c^2}\psi^{*} \frac{\partial ^2\psi(\bold r, t)}{\partial t^2}=\psi^{*}(-\hbar^2\nabla^2+m^2c^2)\psi(\bold r, t)\qquad \qquad \qquad (3)}

Multiplying the complex conjugate form of (1) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \psi} from the left, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{c^2}\psi \frac{\partial ^2\psi^{*}(\bold r, t)}{\partial t^2}=\psi(-\hbar^2\nabla^2+m^2c^2)\psi^{*}(\bold r, t)\qquad \qquad \qquad (4)}

Subtracting (4) from (3), we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{c^2}\left( \psi^{*} \frac{\partial ^2\psi}{\partial t^2}-\psi \frac{\partial ^2\psi^{*}}{\partial t^2}\right) =\hbar^2\left( \psi\nabla^2\psi^{*}-\psi^{*}\nabla^2\psi\right) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow -\frac{\hbar^2}{c^2}\frac{\partial}{\partial t}\left( \psi^{*}\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^{*}}{\partial t}\right) +\hbar^2\nabla\left( \psi^{*}\nabla\psi-\psi\nabla\psi^{*}\right) =0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac {\partial}{\partial t}\left[ \frac {i\hbar}{2mc^2}\left( \psi^{*}\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^{*}}{\partial t}\right) \right] +\nabla \left[ \frac {\hbar}{2mi}\left( \psi^{*}\nabla\psi-\psi\nabla\psi^{*}\right) \right] =0}

this give us the continuity equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\partial \rho}{\partial t}+\nabla \bold j = 0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad(5)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho = \frac {i\hbar}{2mc^2}\left( \psi^{*}\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^{*}}{\partial t}\right) \qquad \qquad \qquad \qquad \qquad \; \; \; \;(6)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold j = \frac {\hbar}{2mi}(\psi^{*}\nabla\psi-\psi\nabla\psi^{*})\qquad \qquad \qquad \qquad \qquad \qquad \qquad \; \;(7)}

From (5) we can see that the integral of the density Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \rho} over all space is conserved. However, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold\rho} is not positively definite. Therefore, we can neither interpret Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \rho} as the particle probability density nor can we interpret Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold j} as the particle current. The appropriate interpretation are charge density for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\rho(\bold r,t)} and electric current for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\bold j(\bold r, t)} since charge density and electric current can be either positive or negative.

Using the Klein-Gordon equation in an electromagnetic field and the same procedure as before, it can be shown that the continuity equation still holds in an electromagnetic field with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold{j} (\bold{r} , t) = \frac{1}{2m} [\psi^{*}(-i\hbar \bold{\nabla} - \frac{e}{c} \bold{A} )\psi - \psi(-i\hbar \bold{\nabla} + \frac{e}{c} \bold{A} )\psi^{*}]}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho (\bold{r} , t) = \frac{1}{2mc^2} [\psi^{*}(i\hbar \frac{\partial}{\partial t} - e\phi )\psi - \psi(i\hbar \frac{\partial}{\partial t} + e\phi )\psi^{*}]}

Nonrelativistic limit

In nonrelativistic limit when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v \ll c} or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p \ll mc} , we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=[(pc)^2+(mc^2)^2]^{1/2}=mc^2\left[ 1+(\frac{p}{mc})^2\right] ^{1/2}\approx mc^2\left( 1+\frac{1}{2}\left( \frac{p}{mc}\right) ^2\right) =mc^2+\frac{p^2}{2m}}

So, the relativistic energy is different from classical energy by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle mc^2} , therefore, we can expect that if we write the solution of Klein-Gordon equation as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi e^{-imc^2t/\hbar}} and substitute it into Klein-Gordon equation, we will get Schrodinger equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \psi} .

Indeed, doing so we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\hbar ^2\frac {\partial ^2 \psi}{\partial t^2}e^{-imc^2t/\hbar}+2\frac {\partial \psi}{\partial t}imc^2 \hbar e^{-imc^2t/\hbar}+\psi m^2c^4 e^{-imc^2t/\hbar}=-\hbar ^2 c^2\nabla ^2 \psi e^{-imc^2t/\hbar}+m^2c^4 \psi e^{-imc^2t/\hbar}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow -\frac {\hbar ^2}{2mc^2} \frac{\partial ^2 \psi}{\partial t^2}+i\hbar \frac {\partial \psi}{\partial t}=-\frac {\hbar ^2}{2m}\nabla ^2 \psi}

In the nonrelativistic limit the first term is considered negligibly small. As a result, for free particles in this limit we get back the Schrodinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar \frac {\partial \psi}{\partial t}=-\frac {\hbar ^2}{2m} \nabla ^2 \psi}

Negative energy states and antiparticles

The solutions to the Klein-Gordon equation allow for both positive and negative energies. The positive energies are no cause for concern, but the negative energies seem counter-intuitive classically, as they allow for negative probability densities, spontaneous transitions from the positive energy states to the negative energy states, and particles to propagate both directions in time. We can't simply "drop" the negative energy states, as they form part of the complete set of solutions. One way to interpret these negative energies was proposed by Stükelberg and Feynman: The negative energy solutions describe positive energy antiparticles, which are conjugates to particles and have been experimentally observed.

By observing the Klein-Gordon equation with negative energy states, we can observe some properties of antiparticles. Consider a free particle at rest (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold{p} = 0} ). The solution to the Klein-Gordon equation in a rest frame K is then

$\psi ({\mathbf {r}},t)=e^{-{\frac {i}{\hbar }}mc^{2}t}$

If we change to a frame K' moving with velocity $-{\mathbf {v}}$ relative to K, the particle has momentum ${\mathbf {p}}={\frac {m{\mathbf {v}}}{\sqrt {1-v^{2}/c^{2}}}}$ and energy $E_{p}={\frac {mc^{2}}{\sqrt {1-v^{2}/c^{2}}}}$, and the wavefunction becomes

$\psi \prime ({\mathbf {r\prime }},t\prime )=e^{{\frac {i}{\hbar }}({\mathbf {p\cdot r\prime }}-E_{p}t\prime )}$

If we now follow this same procedure for a negative energy state, the free particle wavefunction will be

$\psi ({\mathbf {r}},t)=e^{{\frac {i}{\hbar }}mc^{2}t}$

and in the K' frame,

$\psi \prime ({\mathbf {r\prime }},t\prime )=e^{{\frac {-i}{\hbar }}({\mathbf {p\cdot r\prime }}-E_{p}t\prime )}$

Notice that the wavefunctions for the negative energy state are complex conjugates of the wavefunctions for the positive energy states. If we take the complex conjugate of the Klein-Gordon equation in an electromagnetic field, we find that

${\frac {1}{c^{2}}}[i\hbar {\frac {\partial }{\partial t}}+e\phi ]^{2}\psi ^{*}({\mathbf {r}},t)=([{\frac {\hbar }{i}}{\mathbf {\nabla }}+{\frac {e}{c}}{\mathbf {A}}]^{2}+m^{2}c^{2})\psi ^{*}({\mathbf {r}},t)$

This tells us that $\psi ^{*}({\mathbf {r}},t)$ is a wavefunction satisfying the Klein-Gordon equation for a particle of mass m and charge -e, which is, from observation, the antiparticle of a particle of mass m and charge e with wavefunction $\psi ({\mathbf {r}},t)$.

Now, the density and current in the K' frame for the particles will be

$\rho ({\mathbf {r\prime }},t\prime )={\frac {E_{p}}{mc^{2}}}$

and

${\mathbf {j}}({\mathbf {r\prime }},t\prime )={\frac {\mathbf {p}}{m}}$

and for the antiparticles the density and current will be

$\rho ({\mathbf {r\prime }},t\prime )={\frac {-E_{p}}{mc^{2}}}$

and

${\mathbf {j}}({\mathbf {r\prime }},t\prime )={\frac {-{\mathbf {p}}}{m}}$

This tells us that an antiparticle moving in one direction has a current moving in the opposite direction. Thus, a particle current in one direction is equivalent to an antiparticle current in the opposite direction. For charged particles, this is easy to reconcile since the charge of the particle and its antiparticle are opposite.

For neutrally charged particles, however, the difference between particle and antiparticle are a bit more subtle. By observation, we see that there are a few possibilities. First, the particle and its antiparticle may have a "charge" that is not electromagnetic, e.g. the $\displaystyle {K^{0}}$ and ${\overline {K^{0}}}$, which have a strangeness of 1 and -1, respectively. It may also be the case that the neutral particle is in fact its own antiparticle, as with the $\pi ^{0}$, in which case the wavefunction is always real, thus making the current and density zero.