# Solution to Klein-Gordon equation

I have a sound grounding on ODE's, not that much on PDE's, i've read many books on QFT and most if not all come to the conclusion that the solution to the Klein-Gordon equation $$(\partial_{\mu}\partial^{\mu} + m^2)\varphi=0$$ is $$\varphi(\vec x,t)=e^{-ip\cdot x}$$ without derivation where $$p \cdot x=p_{\mu}x^{\mu}=Et-\vec p\cdot \vec x.$$

Which to me means that the characteristic equation $\partial_{\mu}\partial^{\mu} + m^2$ if it should be named like that, has the root $ip_{\mu}x^{\mu}$.

Could someone please show me how these covariant PDE's are solved?

• Well, technically, a general solution would be $\int d^{3}p e^{-it\sqrt{m^{2} + p^{2}}}e^{-i\,p_{i}x^{i}}$, since you can freely linearly superpose plane wave solutions. – Jerry Schirmer Jul 12 '14 at 18:25
• Is there a way that $\varphi(x,t)$ can be decomposed into the function product $F(x)G(t)$ then solved using separation of variables?? – pkjag Jul 12 '14 at 18:28
• You'll get a product of four exponentials, with constants $p_{\mu}$ and the constraint $p_{\mu}p^{\mu} = m^{2}$, so it'll amount to the same thing. – Jerry Schirmer Jul 12 '14 at 18:41
• @pkjag What do you mean by "characteristic equation"? What you wrote does not have an equal sign (and assuming ... = 0 it is just the KG equation). – Lurco Jul 12 '14 at 19:08
• @pkjag: write down $\phi(x^{\mu}) = A(t)B(x)C(y)D(z)$. Apply the Laplacian, and do the same trick that you do in Griffith's Quantum mechanics to isolate the variables into four ODEs. – Jerry Schirmer Jul 12 '14 at 19:47

The Klein-Gordon equation (($\partial_{\mu}\partial^{\mu} + m^2)\varphi=0$) that you have mentioned is only for free field $\phi$.

Now the solution $$\varphi(\vec x,t)=e^{-ip\cdot x}$$ obeys well the free field condition $E^2=\vec{p}^2+m^2$. To verify this put the above solution in (($\partial_{\mu}\partial^{\mu} + m^2)\varphi=0$).

ADDENDUM: say you have $$(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+m^2)\phi=0 \ \ \ -------(1)$$.

now take $\phi=f(t)X(x)$

then equ.(1) becomes $$(\frac{1}{f}\frac{\partial^2 f}{\partial t^2}-\frac{1}{X}\frac{\partial^2X}{\partial x^2}+m^2)=0$$ $$-\frac{1}{X}\frac{\partial^2X}{\partial x^2}+m^2=-\frac{1}{f}\frac{\partial^2 f}{\partial t^2} =A^2=const$$

So you have f=$e^{-iAt}$ after solving $\frac{\partial^2 f}{\partial t^2} +A^2f=0$

similarly you have X=$e^{-iBx}$ after solving $\frac{\partial^2 X}{\partial x^2} +B^2X=0$ where $B^2=A^2-m^2$

• i know that it's a solution but i wanna know how you solve such pde's – pkjag Jul 12 '14 at 18:23
• Sorry, I misunderstood your question. – user22180 Jul 12 '14 at 18:32
• I wanna solve the klein-gordon equation as a pde but i'm unfamiliar with the term $\partial_{\mu}\partial^{\mu}$, which i'm trying to relate to the $D$ differential operator in ODE – pkjag Jul 12 '14 at 18:36
• @pkjag Are you familiar with Einstein's summation convention and the Lorentz metric? $\partial_\mu \partial^\mu := \sum_{\mu=0}^3 g^{\mu\nu} \partial_\mu \partial_\nu = \partial^2_0-\Delta$ and $\partial_0 := \frac{1}{c}\partial_t$, $\Delta$ is the laplasian. So the entire differential operator is just the wave equation operator + $m^2$. – Lurco Jul 12 '14 at 19:03
• So $A$ is taken as $E$ and $p$ is $B$?? – pkjag Jul 12 '14 at 19:26