How does infrared light 'erase' phosphorescence on zinc sulfide?

Thanks to @Manishearth for the edit

In normal phosphorescence, the atoms are in a "metastable" state--where electrons are in a higher energy level, but do not immediately come to ground state due to partial stability. The electrons come down slowly, giving rise to the (relatively) long lasting glow.

The IR light frees away the electrons from the shallow metastable trap. It's like the usual chemical reaction, where the IR light provides enough activation energy to overcome the barrier. Basically, the IR light promotes the electrons to a higher, non-metastable "energy level" (virtual state). From here, the electron jumps down nearly immediately--dumping all its energy. So, instead of the electrons trickling down like in normal phosphescence, they all come down in a torrent.

The same experiment was performed by this guy: http://ajp.aapt.org/resource/1/ajpias/v29/i3/pxxv_s2

The effect has been explained by the assumption that the long-wave radiation frees electrons from shallow trapping centers into which they have fallen after being excited to the conduction band by the ultraviolet light. The freed electrons recombine quickly with ionized luminous centers with the emmision of light. More recent investigations indicate that probably other processes also are involved.

I'm not sure of this one, comments appreciated.

Update: See @pcr's answer--it links to the correct explanation. In this, the IR rays, instead of doing stimulated emission as I said, promote the trapped electron to a higher non-metastable state. From there, it comes back down. The rest is similar to my answer--substitute "lower intermediate state" with "higher state", and remove the stimulated emission part.

To me, it looks like stimulated emission is occuring.

Phosphorescence works by exciting electrons to an energy level where they sort of get stuck. They come down to the ground state slowly(as in only a few at a time), and each electron coming down to the ground state emits a photon of charateristic color. Due to the slowness of the whole thing, we get light for quite a while.

Here, you are shining light at it. What's happening is that photons of a certain wavelength are stimulating the "stuck" electron to come to some other state--not the ground state as you stated that the light you shone is IR and ZnS is usually in the visible range. From this other state, it jumps to the ground state. This entire process is a fast one, so basically all the trapped electrons are dumped in one go. I guess this should be accompanied by a flash of light, but it may be too weak compared to the laser to be seen.

If this is the case, then most of the other IR wavelengths will be ineffective. Only a wavelength corresponding to a transition between the "stuck" energy level and some lower level will work.

Note: As @TerryBollinger noted below, the mechanism may also involve a variant of two photon emission. It's not required to explain this phenomenon; the above mechanism seems to cover all bases, but it still is a possibility.