06 - Sunday

Continuing from yesterday's notes, the Bogoliubov coefficients can be calculated using the Klein-Gordon symplectic form. It's more commonly referred to as the Klein-Gordon “inner product” but this is a misnomer as it is not positive definite. The full form is a relatively complicated integral:

\[\langle f_i,f_j\rangle=-i\int_{\Sigma}\mathrm{d} x\sqrt{-g}g^{00}\left[f_if^*_{j;\mu}-f^*_jf_{i;\mu}\right]\mathrm{d}\Sigma^{\mu}=\delta_{ij}\]

Here we are taking an integral over a Cauchy surface \(\Sigma\), essentially just a surface on which we can completely define boundary conditions of the field (there is a lot more going on here especially around what makes this surface suitable that I'm skipping over), and the measure \(\mathrm{d}\Sigma^\mu\) is the measure over this surface multiplied by a normal vector on the surface \(n^\mu\). For null fields we can take the surface to be a null surface, which is particularly well suited when using null/light cone coordinates.

\(g\) in the square root is the determinant of the metric, though some people write the product above in terms of the determinant of the induced metric on the surface \(\Sigma\), which may be easier to calculate.

We can obtain the Bogoliubov coefficients using the following:

\[\alpha_{ij}=\langle f_i,p_j\rangle\]\[\beta_{ij}=\langle f_i^*,p_j\rangle\]

This is all very nice and provides a way forward for a lot of problems, but there is another method that doesn't require as much messing around with the background geometry. While we can write the field as a sum over discrete frequency modes, we can also write it as an integral over the range of frequencies. I'm not sure if we can always do this, I haven't come across a rigorous explanation yet. Often I see the sum expansion given in the theoretical background while an integral is used in the actual body of the work, I suspect it's either how one writes down the wave solutions or it's just something we can always do. Regardless:

\[\phi=\int\mathrm{d}\omega\left[f_{\omega}\hat{a}_{\omega}+f_{\omega}^*\hat{a}_{\omega}^{\dagger}\right]\]\[\phi=\int\mathrm{d}k\left[p_{k}\hat{b}_{k}+p_{k}^*\hat{b}_{k}^{\dagger}\right]\]

And the Bogoliubov transform takes a similar form as before:

\[f_{\omega}=\int\mathrm{d}k\left[\alpha_{\omega k}p_k+\beta_{\omega k}p_k^*\right]\]

Let's be a bit more explicit and write down our wave solutions as 2D plane waves, to do this we choose some coordinates. For the \(p_k\) waves we will use the coordinates \((u,v)\) and for the  \(f_{\omega}\) waves we will use the coordinates \((x,y)\). Thus we have:

\[f_\omega=\frac{1}{\sqrt{4\pi\omega}}e^{-i\omega x}\]\[p_k=\frac{1}{\sqrt{4\pi k}}e^{-iku}\]

It's important to stress that these two sets of coordinates describe the same 2D plane, and we can usually write one in terms of the other which is essentially a requirement for calculating the Bogoliubov coefficients. As I mentioned on yesterday's notes, these coordinates can be thought of as belonging to different observers in the same universe, but each sees things from a different perspective. 

As a quick aside, when we work with 2D toy models we often have a single spatial coordinate and a single time coordinate, and we then choose a suitable combination of these to form null coordinates like the pairs above, technically time is still an axis on this 2D plane but we've absorbed it into a reparameterization of the space.

Let's continue with the maths. If we write out the integral explicitly, it almost looks like a Fourier transform:

 

\[f_{\omega}=\int\mathrm{d}k\frac{1}{\sqrt{4\pi k}}\left[\alpha_{\omega k}e^{-iku}+\beta_{\omega k}e^{iku}\right]\]

We can thus, extract the the Bogoliubov coefficients by way of an inverse Fourier transform:

\[\begin{aligned} \frac{1}{2\pi}\int_{-\infty}^\infty\mathrm{d}u f_{\omega}(x)e^{i\Omega u}=&\int\mathrm{d}k\frac{1}{\sqrt{4\pi k}}\int_{-\infty}^\infty\mathrm{d}u\frac{1}{2\pi}e^{i\Omega u}\left[\alpha_{\omega k}e^{-iku}+\beta_{\omega k}e^{iku}\right]\\ =&\int\mathrm{d}k\frac{1}{\sqrt{4\pi k}}\int_{-\infty}^\infty\mathrm{d}u\frac{1}{2\pi}\left[\alpha_{\omega k}e^{iu(\Omega-k)}+\beta_{\omega k}e^{iu(\Omega+k)}\right]\\ =&\int\mathrm{d}k\frac{1}{\sqrt{4\pi k}}\left[\alpha_{\omega k}\delta(\Omega-k)+\beta_{\omega k}\delta(\Omega+k)\right]\\ \end{aligned}\]

Because we required our plane waves to have positive frequencies, i.e. \(k>0\) we can select one of the components at a time by choosing the sign of \(\Omega\). If we choose \(\Omega>0\), the Dirac delta enforces \(k=\Omega\)

\[\begin{aligned}     \alpha_{\omega k}=&\frac{\sqrt{4\pi k}}{2\pi}\int_{-\infty}^\infty\mathrm{d}u f_{\omega}(x)e^{ik u}\\     =&\frac{1}{2\pi}\sqrt{\frac{k}{\omega}}\int_{-\infty}^\infty\mathrm{d}u e^{-i\omega x}e^{ik u} \end{aligned}\]

If we have a relation between \(x\) and \(u\):

\[x=f(u)\]

 

\[\alpha_{\omega k}=\frac{1}{2\pi}\sqrt{\frac{k}{\omega}}\int_{-\infty}^\infty\mathrm{d}u e^{-i\omega f(u)}e^{ik u}\]

This method is really nice, and often simpler to evaluate than using the Klein-Gordon symplectic form. If the relation between the two coordinates is simple enough then this can be done by hand, I have not had such luck. I'm working with some very bizarre coordinates which require switching between an intermediary set of coordinates to make the transform between the two field expansions, things are very messy.