Wave equation solution using Fourier Transform

• I

Main Question or Discussion Point

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?

Related Differential Equations News on Phys.org
Did you handle eipx correctly when you formed φ*(p,t) ?

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:

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