Wave equation solution using Fourier Transform

[leo.]

I'm studying Quantum Field Theory and the first example being given in the textbook is the massless Klein Gordon field whose equation is just the wave equation $\Box \ \phi = 0$. The only problem is that I'm not being able to get the same solution as the book. In the book the author states that the general solution is:

$$\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}(a_p e^{-i\omega_p t + i\mathbf{x}\cdot \mathbf{p}}+a_p^\ast e^{i\omega_p t - i\mathbf{x}\cdot \mathbf{p}})$$

Now to get this I expandand $\Box \ \phi = (\partial_t^2 - \nabla^2)\phi$ and took the Fourier transform so that we get the equation $(\partial_t^2 + \omega_p^2)\hat{\phi}(\mathbf{p},t)=0$ where $\omega_p = |p|$.

This equation now has $\mathbf{p}$ just as a parameter and we can easily solve it since it is just the simple harmonic oscilator equation of motion. We end up with

$$\hat{\phi}(\mathbf{p},t)= a_p e^{-i\omega_p t}+b_p e^{i\omega_p t}$$

If we now use the Fourier inversion formula we have that

$$\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}\hat{\phi}(\mathbf{p},t)e^{i\mathbf{p}\cdot \mathbf{x}}=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}(a_p e^{-i\omega_p t}+b_p e^{i\omega_p t})e^{i\mathbf{p}\cdot \mathbf{x}}.$$

This is almost the result, but we need, however, to ensure $\phi$ is a real field. For that we need to apply the reality condition to the Fourier transform:

$$\hat{\phi}(\mathbf{p},-t)=\hat{\phi}^\ast(\mathbf{p},t)$$

This provides us with

$$a_p e^{i\omega_p t} + b_p e^{-i\omega_p t} = a_p^\ast e^{i\omega_p t} + b_p^\ast e^{-i\omega_p t}$$

Now I'm stuck here. I'm not getting how from this condition I can arrive naturally at the textbook answer. How can I proceed this to arrive at the same solution that the textbook presents?

 

[John Park]

Did you handle e^ip⋅x correctly when you formed φ*(p,t) ?

 

[leo.]

Actually not. Sometime after posting this here, I realize that I made a huge mistake: I messed up the coordinates and applied the reality condition as if the Fourier transform was on the $t$ rather than $\mathbf{x}$ variable. This was the whole problem, so I think I'll post the correct solution. The correct reality condition is:

$$\hat{\phi}(-\mathbf{p},t)=\hat{\phi}^\ast(\mathbf{p},t)$$

which translates into

$$a_{-p}e^{-i\omega_p t}+b_{-p}e^{i\omega_p t}=a_{p}^\ast e^{i\omega_p t}+b_p^\ast e^{-i\omega_p t}$$

then using the linear indepence of the functions $e^{-i\omega_p t}$ and $e^{i\omega_p t}$ which can be derived from the Wronskian determinant, we derive that $a_{-p}=b_p^\ast$. Thus the expansion becomes:

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-i\omega_p t}e^{i \mathbf{x}\cdot \mathbf{p}} + a_{-p}^\ast e^{i\omega_p t}e^{i\mathbf{x}\cdot \mathbf{p}})$$

Now this can be split into two integrals

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-i\omega_p t}e^{i \mathbf{x}\cdot \mathbf{p}})+\int \dfrac{d^3 p}{(2\pi)^3}(a_{-p}^\ast e^{i\omega_p t}e^{i\mathbf{x}\cdot \mathbf{p}})$$

Finally on the second integral we perform the change of variables $\mathbf{p}\mapsto -\mathbf{p}$, and we get from the change of variables formula, recalling that $\omega_p = |\mathbf{p}|=\omega_{-p}$ we get

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-i\omega_p t}e^{i \mathbf{x}\cdot \mathbf{p}})+\int \dfrac{d^3 p}{(2\pi)^3}(a_{p}^\ast e^{i\omega_p t}e^{-i\mathbf{x}\cdot \mathbf{p}})$$

And finally we can combine the two integrals using the Minkowski inner product $x^\mu p_\mu = \eta_{\mu\nu}x^\mu p^\nu$ with again $x^\mu = (t,\mathbf{x})^\mu$ and $p^\mu = (\omega_p, \mathbf{p})^\mu$. This directly leads to the correct answer:

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 p}{(2\pi)^3}(a_p e^{-i p_\mu x^\mu}+a_{p}^\ast e^{i p_\mu x^\mu})$$