# Klein gordon equation

1. May 15, 2008

I am being really thick here

I have this wave equation, the massless klien gordon equation

$$\partial_{\mu}\partial^{\mu}\phi(x)=0$$

where the summation over $$\mu$$ is over 0,1,2,3

the general solution is a superposition of plane waves yes? i.e

$$\phi(x)=\int d^4 p \overline{\phi}(p)exp(i p_{\mu}x^{\mu})$$

where $$\overline{\phi}$$ is the weighting function.

When you susbsitute this back into the klein gordon equation you get down two factors of p, i.e

$$p_{\mu}p^{\mu}$$ which equals zero. (mass shell constraint), thus satisfying the equation of motion.

My question is, is $$\overline{\phi}(p)$$ arbitrary? I don't really understand why this is so, let alone believe it.

Hope peeps understand the question.

2. May 15, 2008

### neu

The superpostion equation you wrote is simply the fourier transform of $$\phi (x)$$.

$$\phi (p)$$ are not arbitrary as $$\phi (p) = FT^{-1}[\phi (x)]$$

3. May 15, 2008

yeah I understand that.

I'm being really thick here.

I want the general solution $$\phi(x)$$ to the equation, ie as a superposition of plane wave solutions, but in fourier space.

Ultimately I want to know what $$\overline{\phi}(p)$$ is.

4. May 15, 2008

What I mean is, if $$\overline{\phi}(p)$$ is one weighting function whose fourier transform solves the klein gordon equation and $$\overline{\psi}(p)$$ is another different weighting function is the fourier transform of $$\overline{\psi}$$ also a solution.

5. May 15, 2008

### lbrits

Well, any $$\phi(p)$$ will make $$\phi(x)$$ a solution. These are merely coefficients in your Fourier expansion. However, once you write the Hamiltonian in terms of $$\phi(p)$$ and $$\pi(p)$$, you will find that it simplifies greatly (decoupled harmonic oscillators, one for each p), and quantization is the next step.

6. May 15, 2008

great as long as you're sure about that. That's what I hoped.

I have guessed some expression between [tex]\phi(p)[\tex] and a whole load of other stuff and I want to test my conjecture on the computer. So, presumably I can just invent some suitable function for [tex]\phi(p)[\tex] stick it into a c-program and check it works.

7. May 15, 2008

### nrqed

As others said, phi(p) is arbitrary. The way to see this is the following: plane waves (with the condition on p^2) are solutions of the equation and the equation is linear, therefore arbitrary linear combinations of plane waves will satisfy the equation. Therefore phi(p) is arbitrary.

8. May 15, 2008

### strangerep

Adding my \$0.02 to what others have already said, the $\overline{\phi}(p)$ is not
entirely "arbitrary", because you have imposed the mass-shell constraint $p_{\mu}p^{\mu}=0$.
Think of that as a "constraint hypersurface" in 4D momentum space.

I.e., $\overline{\phi}(p)$ is undefined for values of p which are not on
the constraint hypersurface.

Last edited: May 15, 2008
9. May 15, 2008

### nrqed

True but this is not a condition on the functional form $\overline{\phi}(p)$, it's a restriction on the argument p. I mean, as long as p is a valid p, any function phi(p) is valid, right?

A different consideration arises if we consider localized wavepackets phi(x). Then there must be a condition on phi(p).

10. May 15, 2008

Yes and yes.