Why is a laserpointer able to "erase" a glow-in-the-dark sticker? a while ago I tried to charge a glow-in-the-dark sticker using a simple red laser pointer. It was a large sticker, of the type used to mark emergency exits and fire extinguishers here in Germany. I thought I could draw a picture this way, with glowing strokes.
To my surprise, the exact opposite happened - the sticker was already glowing a little bit and the areas I illuminated with my laser turned dark instantly, as if the charge of the sticker was removed by the laserlight.
The effect was reversible, the darkened areas would charge and glow as usual if exposed to normal light.
So I was able to draw a picture after all - however, it was a negative.
Can anyone explain why the sticker was discharged instead of charged by the laserlight?
 A: "dmckee" is probably right about the mechanism. In our lab, we often work with high-power 1064 nm NIR laser beams, and an obvious issue when dealing with them is that the light is invisible to the eye. For visualization purposes, many companies like ThorLabs and Newlight Photonics sell IR visualization cards that consist of a pink material bonded to a laminated card. The cards need to be "charged" by holding them to a light source, but after charging they do not emit any light. However, when they are struck by NIR light, they glow red brightly, and the spot that has been struck by NIR light will slowly be "depleted" until it no longer responds to NIR exposure. I strongly suspect that the material is excited to a metastable triplet state by visible light, and that NIR light promotes it to a state which has large decay channels.
In any case, if your glow-in-the-dark sticker is one of the usual glowing green types, you probably already know that they can't be excited using light which is of a longer wavelength than they emit. That's why red light will not charge glow stickers, but blue and UV will.
A: How interesting. 
Presumably the "glow-in-the-dark" effect comes from the decay of a meta-stable excited state. It gets charged by sufficiently energetic photons, and decays slowly because some selection rule prevents a direct transition without an external influence.
If this is the case, we can guess that the laser is exciting the meta-stable state to a still higher energy state which is not subject to the selection rule so that it promptly falls to the ground state1, from which the low energy photons in the laser don't have the energy to push back into the meta-stable state. 
I think I'll try to make a demo out of that. Thanks.

1 Presumably you could see the light of this decay if you filtered out the intense red from the laser. Try a green or blue filter. Or a spectrograph.
