# Using fourier analysis of the Klein Gordon equation

This question is more about a mathematical detail, and I am undoubtedly missing something very obvious. And note, I have sifted through the numerous questions on Fourier transform (FT) and the Klein-Gordon (KG) equation, and my questions isn't there. To be honest, it's not even KG specific, but just general FT methodology I think.

So, given the Klein-Gordon equation in natural units and 1D:

$$\dot \psi^2 - \psi' ^2 + m^2 \psi^2=0$$

Obtained by applying the Euler-Lagrange equations to a given Lagrangian, consistent with the system we want to study. And we have a corresponding Hamiltonian density

$$H = \frac{1}{2}\left(\pi ^2 + \psi ' ^2 + m^2 \psi ^2\right)$$

where $$\pi$$ is the conjugate momentum which turns out to just be $$\dot \psi$$ here.

Now if we want a real $$\psi$$, then a suitable Fourier expansion is

$$\psi (x,t) = \int \mathrm dk\, [a(k)e^{i(kx-\omega t)}+a^\star (k)e^{-i(kx-\omega t)}]$$

Now if one plugs this form into the Klein-Gordon equation, you get a dispersion relation $$\omega^2=k^2+m^2$$. But then this exact relation (with this $$\psi$$ expansion) makes the Hamiltonian density go to 0 everywhere!

In fact, now that I am comparing the Hamiltonian and the KG equation, the KG equation is exactly the statement that the Hamiltonian density vanishes everywhere, irrespective of the Fourier expansion of $$\psi$$, because of the form of the conjugate momentum.

How to interpret this?

• Your expression may give the lagrangian, not the hamiltonian. "may" because I did not check any signs. – my2cts Nov 5 '18 at 14:41
• Catch your mistake. Start from just one oscillator. The infinity of such in FT is a smokescreen. – Cosmas Zachos Nov 5 '18 at 14:54