Solving the Massless Klein Gordon Equation

[bartadam]

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.

 

[neu]

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

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

Have I misunderstood your question?

 

[bartadam]

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.

 

[bartadam]

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.

 

[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.

 

[bartadam]

Well, any [tex]\phi(p)[\tex] will make $$\phi(x)[\tex]a solution$$[/tex]

[tex][tex] <br /> great as long as you're sure about that. That's what I hoped.<br /> <br /> 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 $$\phi(p)[\tex] stick it into a c-program and check it works.$$[/tex][/tex][/tex]

 

[nrqed]

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.

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.

 

[strangerep]

[...]
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. [...] is $\overline{\phi}(p)$ arbitrary? [...]

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.

 

[nrqed]

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.

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).

 

[strangerep]

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).

Yes and yes.