# Klein-Gordon equation

• I
Suppose φ is solution to Klein-Gordon equation, Multiplying it by -iφ* we get

$iφ^*\frac{\partial^2φ}{\partial t^2}-iφ^*∇^2φ+iφ^*m^2=0$ ..............(5)

Taking the complex conjugate of the Klein-Gordon equation and multiplying by -iφ we get

$iφ\frac{\partial^2φ^*}{\partial t^2}-iφ∇^2φ^*+iφm^2=0$]..............(6)

If we subtract the second from the first we obtain

$\frac{\partial}{\partial t}[i(φ^*\frac{\partial φ}{\partial t}-φ\frac{\partial φ^*}{\partial t})]+ ∇. [-i(φ^*∇φ-φ∇φ^*)]=0$...............(7)

This has the form of an equation of continuity

$\frac{\partial p}{\partial t}+ ∇.j = 0$.........................(8)

Qns 1: What does it means to have the complex conjugate of the KL equation? I know we obtain eqn(5) by multiplying the KL eqn with -iφ*, so that we can get the probability but why do we need to come up with the complex conjugate of the KL eqn which lead us to eqn(6)?

Qns 2:Subsequently what is the rationale behind subtracting the second from the first to get eqn (7)?

Qns 3: Isn't eqn(8) conservation of charge? How does it relates to the KL eqn which is obtain from the energy-momentum conservation eqn?

Last edited:
DuckAmuck

Strictly speaking, the Klein Gordon equation should really be considered as the equation of motion for a classical or quantum field, not a wave equation, which is why doing all of this seems hand-wavy if we interpret it as the latter.

In an effort to not be too unclear, but still hopefully illuminating, consider a classical (i.e. non-quantized) complex scalar field $\phi(x)$. You can separate this field into two real ones, or just treat it and its complex conjugate as two independent fields. The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$. The equations of motion from the Euler-Lagrange equations are simply $(\Box+m^2)\phi=0$ and$(\Box+m^2)\phi^*=0$ (the KG equation for the two fields).

So far so good. Now, the above mentioned Lagrangian has a global symmetry - if we make a transformation $x\rightarrow x$ and $\phi(x)\rightarrow e^{i\varphi}\phi(x)$, where $\varphi$ is real, but otherwise arbitrary, then the Lagrangian will remain unchanged because the complex conjugate transforms as $\phi^*(x)\rightarrow e^{-i\varphi}\phi^*(x)$, so they cancel out. This is one of the most elementary examples of a global symmetry (called $U(1)$), and there's a theorem, called Noether's theorem, relating global symmetries of a Lagrangian (or more generally, global symmetries of the action, $S=\int\mathcal{L} d^4 x$ ) to conserved currents (conserved in the sense $\partial_\mu j^\mu=0$).

The current you've obtained is exactly the one which comes from this symmetry. Once you quantize the fields, you discover that the charge (which is $Q=\int j^0 d^3x$) corresponding to the conserved current counts the number of particles minus the number of antiparticles (or vice versa, since if $j^\mu$ is conserved, then so is $-j^\mu$), which due to charge quantization is the same thing as charge conservation.

How does it relates to the KL eqn which is obtain from the energy-momentum conservation eqn?

The KG equation isn't quite "obtained" from energy-momentum conservation, it's just that plane wave solutions to it obey the relativistic energy-momentum relation. There is however a different conserved Noether current called the energy-momentum tensor which is a consequence of translational symmetry of the above Lagrangian, from which we can obtain the standard energy-momentum conservation. This current doesn't really have any relation to the previous one, they are quite independent from one another. Note that the Lagrangian of a real scalar field will not have $U(1)$ symmetry, but it will have translational symmetry, and hence a corresponding energy-momentum tensor.

tl;dr version - this is a hand-wavy way to rediscover properties of classical fields

DuckAmuck and Mentz114
Strictly speaking, the Klein Gordon equation should really be considered as the equation of motion for a classical or quantum field, not a wave equation, which is why doing all of this seems hand-wavy if we interpret it as the latter.

In an effort to not be too unclear, but still hopefully illuminating, consider a classical (i.e. non-quantized) complex scalar field $\phi(x)$. You can separate this field into two real ones, or just treat it and its complex conjugate as two independent fields. The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$. The equations of motion from the Euler-Lagrange equations are simply $(\Box+m^2)\phi=0$ and$(\Box+m^2)\phi^*=0$ (the KG equation for the two fields).

So far so good. Now, the above mentioned Lagrangian has a global symmetry - if we make a transformation $x\rightarrow x$ and $\phi(x)\rightarrow e^{i\varphi}\phi(x)$, where $\varphi$ is real, but otherwise arbitrary, then the Lagrangian will remain unchanged because the complex conjugate transforms as $\phi^*(x)\rightarrow e^{-i\varphi}\phi^*(x)$, so they cancel out. This is one of the most elementary examples of a global symmetry (called $U(1)$), and there's a theorem, called Noether's theorem, relating global symmetries of a Lagrangian (or more generally, global symmetries of the action, $S=\int\mathcal{L} d^4 x$ ) to conserved currents (conserved in the sense $\partial_\mu j^\mu=0$).

The current you've obtained is exactly the one which comes from this symmetry. Once you quantize the fields, you discover that the charge (which is $Q=\int j^0 d^3x$) corresponding to the conserved current counts the number of particles minus the number of antiparticles (or vice versa, since if $j^\mu$ is conserved, then so is $-j^\mu$), which due to charge quantization is the same thing as charge conservation.

The KG equation isn't quite "obtained" from energy-momentum conservation, it's just that plane wave solutions to it obey the relativistic energy-momentum relation. There is however a different conserved Noether current called the energy-momentum tensor which is a consequence of translational symmetry of the above Lagrangian, from which we can obtain the standard energy-momentum conservation. This current doesn't really have any relation to the previous one, they are quite independent from one another. Note that the Lagrangian of a real scalar field will not have $U(1)$ symmetry, but it will have translational symmetry, and hence a corresponding energy-momentum tensor.

tl;dr version - this is a hand-wavy way to rediscover properties of classical fields

How do you obtain this equation: "The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$."?

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
How do u obtain this equation: "The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$."?
It is not something you obtain. It is the definition of the Klein--Gordon Lagrangian density from which you derive the KG equation.

haushofer
How do you obtain this equation: "The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$."?
You can obtain it by considering a non-interacting field which is a scalar under the Lorentz group and contains up to second order differential equations of motion. Much more than this action is hard to write down, but try that for yourself.

stevendaryl
Staff Emeritus
How do you obtain this equation: "The Lagrangian describing the free fields is given by $\mathcal{L}=\frac{1}{2}\partial^\mu\phi^*\partial_\mu\phi-\frac{1}{2}m^2\phi^*\phi$."?

Lagrangians are not typically derived. They are just guessed at. Then you use them to derive equations of motion, and look at whether those equations of motion are sensible.

DuckAmuck and vanhees71