Positive frequency solutions in a stationary spacetime

Question

I'm studying the basics of QFT in curved spacetime, and for the simplest case of real scalar fields we can define the Klein-Gordon Hermitian form by $$ (f,g) = i \int_\Sigma \mathrm{d}^3x \sqrt{h} n^\mu ( \overline{f} \partial_\mu g - g \partial_\mu \overline{f}) $$ With $\Sigma$ a spatial hypersurface, $n$ its unit normal and $h$ the determinant of the spatial metric. In a stationary spacetime, we say that a solution $f$ of the Klein-Gordon equation is positive frequency if, for Killing field $k$, we have that $$ \mathcal{L}_k f = - i \omega f \qquad \omega > 0$$ Question: how can we prove that positive frequency solutions have positive Klein-Gordon 'norm' – that is to say, $(f,f) > 0$?

It's clear to me that if the spacetime is static, or equivalently the metric has no cross-terms between space and time, then the unit normal $n^\mu$ is proportional to the Killing field $k^\mu$ and the Klein-Gordon norm is positive since we we have $n^\mu \partial_\mu \propto \mathcal{L}_k$. However, if this is not the case, then there is no easy way to relate $n$ to $k$, and I cannot see how to proceed.

If staticity is in fact a requirement for this statement to hold, is there an obvious counterexample? That is to say, is there a solution $f$ to the Klein-Gordon equation in a stationary (but not static) spacetime with both $k^\mu \partial_\mu f = -i\omega f$ ($\omega > 0$) and $(f,f)<0$? Thanks for the help.

Answer 1

For a stationary spacetime we can rewrite the metric as \begin{equation} g=-\alpha^{2}dt^{2}+h_{ij}( dx^{i}+\beta^{i}dt)( dx^{j}+\beta^{j}dt) \end{equation} Now the one-form normal to $\Sigma$ will read $n^{\flat}=-\alpha dt$ and the corresponding vector field $n^{\sharp}=\frac{1}{\alpha}(k-\beta)$, where $k$ is the stationary Killing vector field.

The KG norm of a positive freq solution $f$ is \begin{equation} (f,f)=2\omega \int_{\Sigma} d^{3}x \frac{\sqrt{\det h}}{\alpha} |f|^{2} + I \end{equation} where \begin{equation} I=-i \int_{\Sigma} d^{3}x \frac{\sqrt{\det h}}{\alpha} (\bar{f}\langle \beta,df\rangle - f \langle \beta, d\bar{f}\rangle ). \end{equation} We now need to estimate I. Note that since $f$ is a solution to the KG equation we have \begin{equation} \label{intkg}\tag{1} 0=\int_{\Sigma} d^{3}x \sqrt{\det h}\;[\bar{f}(\square - m^{2})f]\alpha. \end{equation} Rewriting \begin{align} \square f & = \frac{1}{\sqrt{-\det g}}\partial_{\mu}(\sqrt{-\det g} g^{\mu\nu}\partial_{\nu} f ) \\ & = - \frac{1}{\alpha^{2}}\partial^{2}_{t} f + \frac{1}{\alpha^{2}}\beta^{i}\partial_{i}\partial_{t}f+\frac{1}{\alpha\sqrt{\det h}}\partial_{i}\left(\beta^{i}\frac{\sqrt{\det h}}{\alpha} \partial_{t}f\right)+\frac{1}{\alpha\sqrt{\det h}}\partial_{i}\left(\alpha\sqrt{\det h}g^{ij}\partial_{j}f\right) \end{align} and integrating \eqref{intkg} by parts we get \begin{align} 0=\int d^{3}x \sqrt{\det h} \alpha \left[ \frac{\omega^{2}}{\alpha^2}|f|^{2}-\frac{i\omega}{\alpha^{2}}(\bar{f}\langle \beta,df\rangle - f \langle \beta, d\bar{f}\rangle )-g^{ij}\partial_{i}f\partial_{j}\bar{f}-m^{2}|f|^{2}\right] \end{align} and thus \begin{align} I=-\omega \int \frac{\det h}{\alpha}|f|^{2}+ \frac{1}{\omega}\int \sqrt{\det h}\alpha \left(m^{2}|f|^2 + g^{ij}\partial_{i}f \partial_{j} \bar{f}\right). \end{align} We can substitute this in our original expression for $(f,f)$ \begin{align} (f,f)=\omega \int d^{3}x \frac{\sqrt{\det h}}{\alpha} |f|^{2}+\frac{1}{\omega}\int d^{3}x \sqrt{\det h} \alpha \left(m^{2} |f|^{2} + g^{ij} \partial_{i}f\partial_{j}\bar{f} \right) \ge 0 \end{align}