breitenlohner freedman stability condition I am looking for a simple way to derive the breitenlohner-freedman bound. Actually I can't understand why we have stability above the BF bound and instability below the BF bound,while both have negative energy so the time dependent part of the filed grow exponentially and it leads to instability?
 A: For a quick (and somewhat dirty) way of deriving the bound, do the following: Recall, that in Poincaré coordinates, the metric of $AdS_{d+1}$ is
$$
ds=\frac{1}{z^2}\left(dz^2+\eta^{\mu\nu}dx^\mu dx^\nu\right).
$$
The equation of motion for a scalar is
$$
(\Box-m^2)\,\phi(z,x)=0,
$$
where $\Box=\frac{1}{\sqrt{-g}}\partial_a \sqrt{-g}g^{ab}\partial_b$. Plugging in the AdS metric you get
$$
\partial_z^2\phi(z,x)-\frac{d-1}{z}\partial_z \phi(z,x)+z^2\eta^{\mu\nu}\partial_\mu\partial_\nu\phi(z,x)-\frac{m^2}{z^2}\phi(z,x)=0.
$$
Now, use a plane wave basis for to express the $x$-dependence of the solutions, i.e.
$$
\phi(z,x)=\varphi(z)e^{ik_\mu x^\mu}
$$
Then, the equations of motion become
$$
\partial_z^2\varphi(z)-\frac{d-1}{z}\partial_z\varphi(z)-k_\mu k^\mu \varphi(z)-\frac{m^2}{z^2}\varphi(z)=0
$$
Make the replacement $\varphi(z)\rightarrow z^{\frac{-d+1}{2}}\varphi(z)$
to arrive at
$$
(-\partial_z^2 +V(z))\varphi(z)=\omega^2\varphi(z).
$$
where
$$
V(z)=\vec{k}^2+\frac{1}{z^2}\left(m^2+\frac{d^2-1}{4}\right).
$$
Now, you can use the well-known (?) result that a time dependent Schrödinger equation only admits a stable solution if $V>-\frac{1}{4}$, i.e.
$$
m^2>-\frac{d^2}{4}.
$$
Find details about this last arguement here and references therein. Another source is this solution to a problem set from a class by Gary Horowitz: web.physics.ucsb.edu/~phys230B/Solution2.pdf
The reason why negative mass$^2$ solutions are allowed is that AdS includes a gravitational potential that gives the negative mass$^2$ eigenstates an overall positive energy.
