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So I am trying to solve the Klein-Gordon equation using a Fourier transform of the spatial components only. The Klein-Gordon equation reads:

$$ (\partial ^2 + m^2)\phi(x) = 0. $$

If I let

$$ \phi(x) = \phi(t, \mathbf{x}) = \frac{1}{(2\pi)^{3/2}}\int \widetilde{\phi}(t,\mathbf{k})e^{-i \mathbf{k} \cdot \mathbf{x} } \mathrm{d}^3k,$$

I substitute this into the K-G equation to get

$$\frac{\partial^2 \widetilde{\phi}}{\partial t^2} +(\mathbf{k}^2+m^2)\widetilde{\phi}=0. $$

This is the differential equation for a simple harmonic oscillator so I can immediately write

$$ \widetilde{\phi}(t,\mathbf{k})=A(\mathbf{k})e^{i \omega_{\mathbf{k}}t}+B(\mathbf{k})e^{-i \omega_{\mathbf{k}} t}, $$

and therefore I find

$$ \phi(x)= \frac{1}{(2\pi)^{3/2}}\int \bigg(A(\mathbf{k})e^{i \omega_{\mathbf{k}}t-\mathbf{k} \cdot \mathbf{x}}+B(\mathbf{k})e^{-i \omega_{\mathbf{k}} t -\mathbf{k} \cdot \mathbf{x}}\bigg) \mathrm{d}^3k. $$

Now at this stage I get slightly confused. I have seen the general solution to the Klein-Gordon equation and it has a factor of $1/2 \omega_{\mathbf{k}}$ inside the integrand, any hints as to how I can proceed to this?

I have also read that in order to tidy notation up I would set $\mathbf{k} \rightarrow -\mathbf{k} $ in the second term of the integrand, but wouldn't this change $ \mathrm{d}^3k \rightarrow -\mathrm{d}^3k$ as the Jacobian determinant is $-1$, so surely it should be a difference of two terms?


The solution I am trying to get to:

$$ \phi(x)= \frac{1}{(2\pi)^{3/2}}\int \frac{1}{2\omega_{\mathbf{k}}} \bigg(A(\mathbf{k})e^{i \omega_{\mathbf{k}}t-\mathbf{k} \cdot \mathbf{x}}+B(\mathbf{k})e^{-i \omega_{\mathbf{k}} t +\mathbf{k} \cdot \mathbf{x}}\bigg) \mathrm{d}^3k. $$

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    $\begingroup$ Redefine your $A({\bf k})$ and $B({\bf k})$ to $\frac{1}{2\omega_{\bf k}} A({\bf k})$ and $\frac{1}{2\omega_{\bf k}} B({\bf k})$ respectively. $\endgroup$ – Prahar Dec 19 '17 at 6:30
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The general linear combination of solutions is $$ \phi(x)= \int\!\frac{\mathrm{d}^4k}{(2\pi)^4}\,\delta(k^2-m^2)\,\Theta(k^0)\,\left[A(\vec k)\,\mathrm{e}^{-\mathrm{i}\,k\cdot x}+B(\vec k)\,\mathrm{e}^{\mathrm{i}\,k\cdot x}\right]\;. $$ In contrast to what you wrote, this is (manifestly) Lorentz invariant. After all you want to describe a scalar field. The next observation is that $$ \frac{\mathrm{d}^4k}{(2\pi)^4}\,\delta(k^2-m^2)\,\Theta(k^0)= \frac{\mathrm{d}^3k}{(2\pi)^3}\,\frac{1}{2\omega_{\vec k}} \quad\text{with}~\omega_{\vec k}=\sqrt{\vec k^2+m^2}\;.$$ This explains the factor $1/\omega_{\vec k}$. In this argument there is also no need to play with minus signs.

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  • $\begingroup$ This is arguably the most straightforward way to get a manifestly Lorentz invariant expression. $\endgroup$ – user178876 Dec 19 '17 at 14:00
  • $\begingroup$ Why do you have the $\Theta(k^0)$? I mean, why do you restrict the $k^0$ to positive values? $\endgroup$ – user171780 May 27 '18 at 3:34
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    $\begingroup$ @user171780 Well, our conventions are such that energies of physical objects, i.e. particles and antiparticles, are positive. $\endgroup$ – user178876 May 27 '18 at 14:58
  • $\begingroup$ Ok, thanks. And what about the extra $\pi$ that disappears in the denominator? I am doing the math and cannot see when it cancels out. $\endgroup$ – user171780 May 27 '18 at 15:21

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