Klein–Gordon equation

The Klein–Gordon equation (Klein–Fock–Gordon equation or sometimes Klein–Gordon–Fock equation) is a, related to the. It is second-order in space and time and manifestly. It is a quantized version of the relativistic. Its solutions include a, a field whose quanta are spinless particles. Its theoretical relevance is similar to that of the. Electromagnetic interactions can be incorporated, forming the topic of, but because common spinless particles like the s are unstable and also experience the strong interaction (with unknown interaction term in the ,) the practical utility is limited.

The equation can be put into the form of a Schrödinger equation. In this form it is expressed as two coupled differential equations, each of first order in time. The solutions have two components, reflecting the charge degree of freedom in relativity. It admits a conserved quantity, but this is not positive definite. The wave function cannot therefore be interpreted as a. The conserved quantity is instead interpreted as, and the norm squared of the wave function is interpreted as a. The equation describes all spinless particles with positive, negative, and zero charge.

Any solution of the free Dirac equation is, component-wise, a solution of the free Klein–Gordon equation.

The equation does not form the basis of a consistent quantum relativistic one-particle theory. There is no known such theory for particles of any spin. For full reconciliation of quantum mechanics with special relativity, is needed, in which the Klein–Gordon equation reemerges as the equation obeyed by the components of all free quantum fields. In quantum field theory, the solutions of the free (noninteracting) versions of the original equations still play a role. They are needed to build the Hilbert space and to express quantum field by using complete sets (spanning sets of Hilbert space) of wave functions.

Statement
The Klein–Gordon equation with mass parameter $$m$$ is
 * $$\frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \nabla^2 \psi + \frac{m^2 c^2}{\hbar^2} \psi = 0.$$

Solutions of the equation are complex-valued functions $$\psi(t, \mathbf{x})$$ of the time variable $$t$$ and space variables $$\mathbf{x}$$; the $$\nabla^2$$ acts on the space variables only.

The equation is often abbreviated as
 * $$(\Box + \mu^2) \psi = 0,$$

where $μ = mc/ħ$, and $□$ is the, defined by
 * $$\Box = -\eta^{\mu\nu} \partial_\mu \, \partial_\nu = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2.$$

(We are using the (−, +, +, +) .)

The Klein–Gordon equation is often written in :
 * $$-\partial_t^2 \psi + \nabla^2 \psi = m^2 \psi$$.

The form of the Klein–Gordon equation is derived by requiring that solutions
 * $$\psi = e^{-i\omega t + i k\cdot x } = e^{i k_\mu x^\mu}$$

of the equation obey the energy–momentum relation of special relativity:
 * $$-p_\mu p^\mu = E^2 - P^2 = \omega^2 - k^2 = -k_\mu k^\mu = m^2.$$

Unlike the Schrödinger equation, the Klein–Gordon equation admits two values of $ω$ for each $k$: one positive and one negative. Only by separating out the positive and negative frequency parts does one obtain an equation describing a relativistic wavefunction. For the time-independent case, the Klein–Gordon equation becomes
 * $$\left[ \nabla^2 - \frac{m^2 c^2}{\hbar^2} \right] \psi(\mathbf{r}) = 0,$$

which is formally the same as the homogeneous.

History
The equation was named after the physicists and, who in 1926 proposed that it describes relativistic electrons. Other authors making similar claims in that same year were, Johann Kudar, and , and. Although it turned out that modeling the electron's spin required the, the Klein–Gordon equation correctly describes the spinless relativistic s, like the. On 4 July 2012, European Organization for Nuclear Research announced the discovery of the. Since the is a spin-zero particle, it is the first observed ostensibly   to be described by the Klein–Gordon equation. Further experimentation and analysis is required to discern whether the observed is that of the  or a more exotic, possibly composite, form.

The Klein–Gordon equation was first considered as a quantum wave equation by in his search for an equation describing s. The equation is found in his notebooks from late 1925, and he appears to have prepared a manuscript applying it to the hydrogen atom. Yet, because it fails to take into account the electron's spin, the equation predicts the hydrogen atom's fine structure incorrectly, including overestimating the overall magnitude of the splitting pattern by a factor of $4n⁄2n − 1$ for the $n$-th energy level. The Dirac equation relativistic spectrum is, however, easily recovered if the orbital-momentum quantum number $l$ is replaced by total angular-momentum quantum number $j$. In January 1926, Schrödinger submitted for publication instead his equation, a non-relativistic approximation that predicts the Bohr energy levels of hydrogen without.

In 1926, soon after the Schrödinger equation was introduced, wrote an article about its generalization for the case of s, where s were dependent on, and independently derived this equation. Both Klein and Fock used Kaluza and Klein's method. Fock also determined the for the. The Klein–Gordon equation for a has a simple  solution.

Derivation
The non-relativistic equation for the energy of a free particle is
 * $$\frac{\mathbf{p}^2}{2 m} = E.$$

By quantizing this, we get the non-relativistic Schrödinger equation for a free particle:
 * $$\frac{\mathbf{\hat{p}}^2}{2m} \psi = \hat{E} \psi,$$

where
 * $$\mathbf{\hat{p}} = -i \hbar \mathbf{\nabla}$$

is the ($∇$ being the ), and
 * $$\hat{E} = i \hbar \frac{\partial}{\partial t}$$

is the.

The Schrödinger equation suffers from not being, meaning that it is inconsistent with.

It is natural to try to use the identity from special relativity describing the energy:


 * $$\sqrt{\mathbf{p}^2 c^2 + m^2 c^4} = E.$$

Then, just inserting the quantum-mechanical operators for momentum and energy yields the equation


 * $$\sqrt{(-i\hbar\mathbf{\nabla})^2 c^2 + m^2 c^4} \, \psi = i \hbar \frac{\partial}{\partial t}\psi.$$

The square root of a differential operator can be defined with the help of ations, but due to the asymmetry of space and time derivatives, Dirac found it impossible to include external electromagnetic fields in a relativistically invariant way. So he looked for another equation that can be modified in order to describe the action of electromagnetic forces. In addition, this equation, as it stands, is (see also Introduction to nonlocal equations).

Klein and Gordon instead began with the square of the above identity, i.e.
 * $$\mathbf{p}^2 c^2 + m^2 c^4 = E^2,$$

which, when quantized, gives
 * $$\left( (-i\hbar\mathbf{\nabla})^2 c^2 + m^2 c^4 \right) \psi = \left( i \hbar \frac{\partial}{\partial t} \right)^2 \psi,$$

which simplifies to
 * $$-\hbar^2 c^2 \mathbf{\nabla}^2 \psi + m^2 c^4 \psi = -\hbar^2 \frac{\partial^2}{\partial t^2} \psi.$$

Rearranging terms yields
 * $$\frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi + \frac{m^2 c^2}{\hbar^2} \psi = 0.$$

Since all reference to imaginary numbers has been eliminated from this equation, it can be applied to fields that are, as well as those that have.

Rewriting the first two terms using the inverse of the $diag(−c^{2}, 1, 1, 1)$, and writing the Einstein summation convention explicitly we get
 * $$-\eta^{\mu \nu} \partial_\mu \, \partial_\nu \psi \equiv

\sum_{\mu=0}^{\mu=3} \sum_{\nu=0}^{\nu=3} -\eta^{\mu\nu} \partial_\mu \, \partial_\nu \psi = \frac{1}{c^2} \partial_0^2 \psi - \sum_{\nu=1}^{\nu=3} \partial_\nu \, \partial_\nu\psi = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \mathbf{\nabla}^2 \psi.$$

Thus the Klein–Gordon equation can be written in a covariant notation. This often means an abbreviation in the form of
 * $$(\Box + \mu^2) \psi = 0,$$

where
 * $$\mu = \frac{mc}{\hbar},$$

and
 * $$\Box = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2.$$

This operator is called the.

Today this form is interpreted as the relativistic for -0 particles. Furthermore, any component of any solution to the free (for a  particle) is automatically a solution to the free Klein–Gordon equation. This generalizes to particles of any spin due extension to the. Furthermore, in, every component of every quantum field must satisfy the free Klein–Gordon equation, making the equation a generic expression of quantum fields.

Klein–Gordon equation in a potential
The Klein–Gordon equation can be generalized to describe a field in some potential $V(ψ)$ as
 * $$\Box \psi + \frac{\partial V}{\partial \psi} = 0.$$

Conserved current
The conserved current associated to the U(1) symmetry of a complex field $$\varphi(x) \in \mathbb{C}$$ satisfying the Klein–Gordon equation reads


 * $$\partial_\mu J^\mu(x) = 0, \quad J^\mu(x) \equiv \frac{e}{2m} \left( \, \varphi^*(x) \partial^\mu\varphi(x) - \varphi(x)\partial^\mu \varphi^*(x) \, \right).$$

The form of the conserved current can be derived systematically by applying to the U(1) symmetry. We will not do so here, but simply give a proof that this conserved current is correct.

Proof using algebraic manipulations from the KG equation From the Klein–Gordon equation for a complex field $$\varphi(x)$$ of mass $$m$$, written in covariant notation


 * $$(\square + \mu^2) \varphi(x) = 0$$

and its complex conjugate


 * $$(\square + \mu^2) \varphi^*(x) = 0,$$

we have, multiplying by the left respectively by $$\varphi^*(x)$$ and $$\varphi(x)$$ (and omitting for brevity the explicit $$x$$ dependence),


 * $$\varphi^*(\square + \mu^2) \varphi = 0,$$
 * $$\varphi(\square + \mu^2) \varphi^* = 0.$$

Subtracting the former from the latter, we obtain


 * $$\varphi^* \square \varphi - \varphi \square \varphi^* = 0,$$


 * $$\varphi^* \partial_\mu \partial^\mu \varphi - \varphi \partial_\mu \partial^\mu \varphi^* = 0,$$

then we also know


 * $$\partial_\mu ( \varphi^* \partial^\mu \varphi ) = \partial_\mu \varphi^* \partial^\mu \varphi + \varphi^* \partial_\mu \partial^\mu \varphi $$

from which we obtain the conservation law for the Klein–Gordon field:


 * $$\partial_\mu J^\mu(x) = 0, \quad J^\mu(x) \equiv \varphi^*(x) \partial^\mu\varphi(x) - \varphi(x)\partial^\mu \varphi^*(x).$$

Relativistic free particle solution
The Klein–Gordon equation for a free particle can be written as


 * $$\mathbf{\nabla}^2\psi - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi = \frac{m^2c^2}{\hbar^2} \psi.$$

We look for plane-wave solutions of the form


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

for some constant $ω ∈ ℝ$ and  $k ∈ ℝ^{3}$. Substitution gives the dispersion relation


 * $$-|\mathbf{k}|^2 + \frac{\omega^2}{c^2} = \frac{m^2c^2}{\hbar^2}.$$

Energy and momentum are seen to be proportional to $ω$ and $k$:


 * $$\langle\mathbf{p}\rangle = \langle\psi| -i\hbar\mathbf{\nabla} |\psi\rangle = \hbar\mathbf{k},$$
 * $$\langle E\rangle = \left\langle\psi\left| i\hbar\frac{\partial}{\partial t} \right|\psi\right\rangle = \hbar\omega.$$

So the dispersion relation is just the classic relativistic equation:
 * $$\langle E \rangle^2 = m^2 c^4 + \langle\mathbf{p}\rangle^2 c^2.$$

For massless particles, we may set $m = 0$, recovering the relationship between energy and momentum for massless particles:


 * $$\langle E\rangle = \big|\langle\mathbf{p}\rangle\big| c.$$

Action
The Klein–Gordon equation can also be derived by a method, considering the action


 * $$\mathcal{S} = \int \left( -\frac{\hbar^2}{m} \eta^{\mu \nu} \partial_\mu\bar\psi \,\partial_\nu \psi - m c^2 \bar\psi \psi \right) \mathrm{d}^4 x,$$

where $ψ$ is the Klein–Gordon field, and $m$ is its mass. The of $ψ$ is written $\overbar{ψ}$. If the scalar field is taken to be real-valued, then $\overbar{ψ} = ψ$, and it is customary to introduce a factor of 1/2 for both terms.

Applying the formula for the to the Lagrangian density (the quantity inside the integral), we can derive the  of the scalar field. It is


 * $$T^{\mu\nu} = \frac{\hbar^2}{m} \left (\eta^{\mu \alpha} \eta^{\nu \beta} + \eta^{\mu \beta} \eta^{\nu \alpha} - \eta^{\mu\nu} \eta^{\alpha \beta} \right ) \partial_\alpha \bar\psi \, \partial_\beta \psi - \eta^{\mu\nu} m c^2 \bar\psi \psi .$$

By integration of the time–time component $T^{00}$ over all space, one may show that both the positive- and negative-frequency plane-wave solutions can be physically associated with particles with positive energy. This is not the case for the Dirac equation and its energy–momentum tensor.

Classical field
Taking the non-relativistic limit ($v << c$) of a classical Klein-Gordon field $ψ(x, t)$ begins with the ansatz factoring the oscillatory term,


 * $$\psi(\mathbb x, t) = \phi(\mathbb x, t)\,e^{-\frac i \hbar m c^2 t}.$$

Defining the kinetic energy $$E' = E - mc^2$$, $$E' \ll mc^2$$ in the non-relativistic limit $v << c$, and hence


 * $$\left|i\hbar \frac{\partial\phi}{\partial t}\right| \approx E' \phi \ll mc^2\phi.$$

Applying this yields the non-relativistic limit of the second time derivative of $$\psi$$,


 * $$\frac{\partial\psi}{\partial t} \approx -i\frac{mc^2}{\hbar}\phi\,e^{-\frac i \hbar mc^2 t}$$
 * $$\frac{\partial^2\psi}{\partial t^2} \approx -\left( i\frac{2mc^2}{\hbar} \frac{\partial\phi}{\partial t} + \left(\frac{mc^2}{\hbar}\right)^2 \phi \right) e^{-\frac i \hbar mc^2 t}$$

Substituting into the free Klein-Gordon equation, $$c^{-2}\partial_t^2 \psi = \nabla^2 \psi - m^2 \psi$$, yields


 * $$-\frac{1}{c^2}\left( i\frac{2mc^2}{\hbar} \frac{\partial\phi}{\partial t} + \left(\frac{mc^2}{\hbar}\right)^2 \phi \right) e^{-\frac i \hbar mc^2 t} \approx \left(\nabla^2 - \left(\frac{mc^2}{\hbar}\right)^2\right)\phi\,e^{-\frac i \hbar mc^2 t}$$

which (by dividing out the exponential and subtracting the mass term) simplifies to


 * $$i\hbar\frac{\partial\phi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\phi.$$

This is a classical.

Quantum field
The analogous limit of a quantum Klein-Gordon field is complicated by the non-commutativity of the field operator. In the limit $v << c$, the decouple and behave as independent quantum s.

Electromagnetic interaction
There is a simple way to make any field interact with electromagnetism in a way: replace the derivative operators with the gauge-covariant derivative operators. This is because to maintain symmetry of the physical equations for the wavefunction $$\varphi$$ under a local U(1) gauge transformation $$\varphi \to \varphi' = \exp(i\theta)\varphi$$, where $$\theta(t, \textbf{x})$$ is a locally variable phase angle, which transformation redirects the wavefunction in the complex phase space defined by $$\exp(i\theta) = \cos\theta + i \sin\theta$$, it is required that ordinary derivatives $$\partial_\mu$$ be replaced by gauge-covariant derivatives $$D_\mu = \partial_\mu - ieA_\mu$$, while the gauge fields transform as $$eA_\mu \to eA'_\mu = eA_\mu + \partial_\mu\theta$$. The Klein–Gordon equation therefore becomes


 * $$D_\mu D^\mu \varphi = -(\partial_t - ie A_0)^2 \varphi + (\partial_i - ie A_i)^2 \varphi = m^2 \varphi$$

in, where $A$ is the vector potential. While it is possible to add many higher-order terms, for example,
 * $$D_\mu D^\mu \varphi + A F^{\mu\nu} D_\mu \varphi D_\nu (D_\alpha D^\alpha \varphi) = 0,$$

these terms are not in 3 + 1 dimensions.

The field equation for a charged scalar field multiplies by $i$, which means that the field must be complex. In order for a field to be charged, it must have two components that can rotate into each other, the real and imaginary parts.

The action for a massless charged scalar is the covariant version of the uncharged action:


 * $$S = \int_x \eta^{\mu\nu} \left( \partial_\mu \varphi^* + ie A_\mu \varphi^* \right) \left( \partial_\nu \varphi - ie A_\nu\varphi \right) = \int_x |D \varphi|^2.$$

Gravitational interaction
In, we include the effect of gravity by replacing partial with s, and the Klein–Gordon equation becomes (in the )


 * $$\begin{align}

0 &= - g^{\mu \nu} \nabla_{\mu} \nabla_{\nu} \psi + \dfrac {m^2 c^2}{\hbar^2} \psi = - g^{\mu \nu} \nabla_{\mu} (\partial_\nu \psi) + \dfrac {m^2 c^2}{\hbar^2} \psi \\ &= -g^{\mu \nu} \partial_\mu \partial_\nu \psi + g^{\mu \nu} \Gamma^{\sigma}{}_{\mu \nu} \partial_\sigma \psi + \dfrac {m^2 c^2}{\hbar^2} \psi, \end{align}$$

or equivalently,


 * $$\frac{-1}{\sqrt{-g}} \partial_\mu \left( g^{\mu \nu} \sqrt{-g} \partial_\nu \psi \right) + \frac {m^2 c^2}{\hbar^2} \psi = 0,$$

where $g^{αβ}$ is the inverse of the that is the gravitational potential field, g is the  of the metric tensor, $∇_{μ}$ is the, and $Γ^{σ}_{μν}$ is the  that is the gravitational.