Given the Klein Gordon equation $$\left(\Box +m^{2}\right)\phi(t,\mathbf{x})=0$$ it is possible to find a solution $\phi(t,\mathbf{x})$ by carrying out a Fourier decomposition of the scalar field $\phi$ at a given instant in time $t$, such that $$\phi(t,\mathbf{x})=\int\frac{d^{3}x}{(2\pi)^{3}}\tilde{\phi}\left(t,\mathbf{k}\right)e^{i\mathbf{k}\cdot\mathbf{x}}$$ where $\tilde{\phi}\left(t,\mathbf{k}\right)$ are the Fourier modes of the corresponding field $\phi(t,\mathbf{x})$.

From this we can calculate the required evolution of the Fourier modes $\tilde{\phi}\left(t,\mathbf{k}\right)$ such that at each instant in time $t$, $\phi(t,\mathbf{x})$ is a solution to the Klein Gordon equation. This can be done, following on from the above, as follows: $$\left(\Box +m^{2}\right)\phi(t,\mathbf{x})=\left(\Box +m^{2}\right)\int\frac{d^{3}x}{(2\pi)^{3}}\tilde{\phi}\left(t,\mathbf{k}\right)e^{i\mathbf{k}\cdot\mathbf{x}}\qquad\qquad\qquad\qquad\qquad\qquad\;\;\,\\ =\int\frac{d^{3}x}{(2\pi)^{3}}\left[\left(\partial^{2}_{t}+\mathbf{k}^{2}+m^{2}\right)\tilde{\phi}\left(t,\mathbf{k}\right)\right]e^{i\mathbf{k}\cdot\mathbf{x}} =0\\ \Rightarrow \left(\partial^{2}_{t}+\mathbf{k}^{2}+m^{2}\right)\tilde{\phi}\left(t,\mathbf{k}\right)=0 \qquad\qquad\qquad$$

Question: This is all well and good, but why is it that in this case we only perform a Fourier decomposition of the spatial part only, whereas in other cases, such as for finding solutions for propagators (Green's functions), we perform a Fourier decomposition over all 4 spacetime coordinates? [e.g. $$G(x-y)=\int\frac{d^{4}x}{(2\pi)^{4}}\tilde{G}\left(t,\mathbf{k}\right)e^{ik\cdot x}$$ (where in this case $k\cdot x\equiv k_{\mu}x^{\mu}$).]

Is it simply because when we construct the appropriate QFT for a scalar field we do so in the Heisenberg picture, or is there something else to it?

Apologies if this is a really dumb question but it's really been bugging me for a while and I want to get the reasoning straight in my mind!


1 Answer 1


Notation: $x=(t,\boldsymbol x)$; $k=(k_0,\boldsymbol k)$; $kx=k_0t-\boldsymbol k\cdot\boldsymbol x$; $\mathrm dx=\mathrm dt\;\mathrm d^3\boldsymbol x$; etc.

You can in principle perform the Fourier decomposition on both space and time variables, but to do so you'll need several properties of the Dirac's delta funciton:

The first one is: let $\xi\in\mathbb R$; then $$ \delta(f(\xi))=\sum_{f(\xi_i)=0} \frac{\delta(\xi-\xi_i)}{|f'(\xi_i)|} \tag{1} $$ where the sum is over every $\xi_i$ such that $f(\xi_i)=0$, ie, over the roots of $f(\xi)$.

The second one is that, given $g(\xi)$ a known function, the distributional solution of $g(\xi)f(\xi)=0$ is $f(\xi)=h(\xi)\delta(g(\xi))$ for an arbitrary function $h(\xi)$. If you believe these, then the Fourier decomposition is as follows:

Let $\phi(x)$ be the solution of $$ (\partial^2+m^2)\phi(x)=0 $$

Take the Fourier transform of the equation to find $$ (k^2-m^2)\phi(k)=0 \tag{2} $$ where $$ \phi(k)=\int \mathrm dx\ \mathrm e^{ikx} \phi(x) $$

As $\phi(x)$ is a distribution, the solution of $(2)$ is $\phi(k)=h(k)\delta(k^2-m^2)$ for an arbitrary function $h(k)$. Inverting the Fourier Transform, we find $$ \phi(x)=\int\mathrm dk\ \mathrm e^{-ikx}h(k)\delta(k^2-m^2) $$

Next, use $(1)$ to expand the delta over the roots of $k^2-m^2$. These roots are easily found to be $k_0=\pm \omega(\boldsymbol k)$, where $\omega(\boldsymbol k)=+(\boldsymbol k^2+m^2)^{1/2}$. Therefore, it is immediate to get $$ \phi(x)=\int\mathrm dk\ \mathrm e^{-ikx}h(k)\frac{1}{2\omega}\left[\delta(k_0-\omega)+\delta(k_0+\omega)\right] $$ and, after integrating over $\mathrm dk_0$ using the deltas, we find $$ \phi(x)=\int\frac{\mathrm d \boldsymbol k}{2\omega}\ \left[\mathrm e^{-i\omega t} \mathrm e^{i\boldsymbol k\cdot\boldsymbol x}h(\omega,\boldsymbol k)+\mathrm e^{+i\omega t} \mathrm e^{i\boldsymbol k\cdot\boldsymbol x}h(-\omega,\boldsymbol k)\right] $$

Finally, make the change of variable $\boldsymbol k\to-\boldsymbol k$ in the second term, which yeilds the usual expansion $$ \phi(x)=\int\frac{\mathrm d \boldsymbol k}{2\omega}\ \left[\mathrm e^{-ikx}a(\boldsymbol a)+\mathrm e^{+ikx}b^\dagger(\boldsymbol k)\right] $$ where I have defined $a(\boldsymbol k)=h(\omega,\boldsymbol k)$ and $b^\dagger(\boldsymbol k)=h(-\omega,-\boldsymbol k)$.

As you can see, the solution is the same as yours (modulo some irrelevant prefactor that can be reabsorbed into the definition of $h(k)$), though the algebraic procedure to find it is a bit harder.

  • $\begingroup$ I didnt write every step of the calculation because I assume you can fill in the details. If you find this too coarse please tell me and Ill try to improve the anser. $\endgroup$ Nov 2, 2015 at 17:14
  • $\begingroup$ Thanks for your answer. So is it simply that we choose how we Fourier transform depending on what is most convenient for the problem at hand? Also, is the solution to $(2)$ $\phi (k)=h(k)\delta(k^{2}-m^{2})$ simply because $(k^{2}-m^{2})\phi(k)=(k^{2}-m^{2})\delta(k^{2}-m^{2})h(k)=0$ which follows from $(k^{2}-m^{2})\delta(k^{2}-m^{2})=(k^{2}-k^{2})=0$? $\endgroup$
    – Will
    Nov 2, 2015 at 17:54
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    $\begingroup$ I assume that in your last equation you mean $a(\boldsymbol k)$ and not $a(\boldsymbol a)$. $\endgroup$
    – Sito
    Jun 2, 2020 at 12:16
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    $\begingroup$ @Sito Yep, thanks! $\endgroup$ Jun 23, 2020 at 11:13
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    $\begingroup$ @SummonedEgar under $k\to-k$ the integral $\int_a^b \mathrm dk$ maps into $\int^{-a}_{-b}\mathrm dk$. So it is invariant as long as $-a=b$, but not if the limits are generic. In our case the integration is over all of $\mathrm R$ which is invariant under inversions, so indeed it is invariant. $\endgroup$ Feb 2, 2021 at 20:42

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