# I Klein-Gordon equation

#### TimeRip496

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?

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#### kontejnjer

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

#### TimeRip496

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

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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.

"Klein-Gordon equation"

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