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I Wave equation solution using Fourier Transform

  1. Mar 13, 2017 #1
    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 [itex]\Box \ \phi = 0[/itex]. 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:

    [tex]\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}})[/tex]

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

    This equation now has [itex]\mathbf{p}[/itex] 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

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

    If we now use the Fourier inversion formula we have that

    [tex]\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}}.[/tex]

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

    [tex]\hat{\phi}(\mathbf{p},-t)=\hat{\phi}^\ast(\mathbf{p},t)[/tex]

    This provides us with

    [tex]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}[/tex]

    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?
     
  2. jcsd
  3. Mar 16, 2017 #2
    Did you handle eipx correctly when you formed φ*(p,t) ?
     
  4. Mar 17, 2017 #3
    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 [itex]t[/itex] rather than [itex]\mathbf{x}[/itex] variable. This was the whole problem, so I think I'll post the correct solution. The correct reality condition is:

    [tex]\hat{\phi}(-\mathbf{p},t)=\hat{\phi}^\ast(\mathbf{p},t)[/tex]

    which translates into

    [tex]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}[/tex]

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

    [tex]\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}})[/tex]

    Now this can be split into two integrals

    [tex]\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}})[/tex]

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

    [tex]\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}})[/tex]

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

    [tex]\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})[/tex]
     
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