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.

  • $\begingroup$ might be relevant: the on-shell hermitial form $(f,g)$ is independent of $\Sigma$. In other words, for $f,g$ solutions of the KG equation you have the freedom to choose any surface $\Sigma$ over which to integrate. It seems to me that this freedom allows you to choose $n\propto k$ regardless of whether the metric is static or not. $\endgroup$ Commented Feb 1, 2017 at 20:07
  • $\begingroup$ Thanks for the comment. I think that there is some restriction on $\Sigma$ in fact. I believe the proof that $\mathcal{L}_k$ is antiHermitian (which we need to be the case in order to talk about positive frequency in the first place) with respect to the KG form actually requires the Killing field to correspond to translations in the same 'time' $t$ as $\Sigma$ is a level surface of. I then agree that whichever level surface of $t$ we choose is inconsequential, but this prevents us from freely choosing $\Sigma$ so that its normal is proportional to $k$. $\endgroup$
    – gj255
    Commented Feb 1, 2017 at 22:51

1 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.