Undergrad 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})