Can photon loss increase energy?

Consider quantum states in a bosonic system looking like this: $$\sum_{n=1}^\infty \frac{1}{n^p}|n\rangle$$ which is normalizable when $p>\frac{1}{2}$. However, we can easily see that if you apply the annihilation operator $a$ to it, the energy ($a^\dagger a$) of the state will increase. And if you keep applying $a$ to it, at a certain point, the state may even become un-normalizable.

This seems strange to me - Losing photon result in the increase of energy. Is this possible? Or does that mean this kind of state cannot exist in real life?

• What would applying the operator in actual physical situation of photon mean? Can you think of an apparatus that can do this to a photon? – Peaceful Feb 24 '17 at 4:15
• Similarly, when you apply the annihilation operator to a coherent state it doesn't change. So it seems strange, because as say, it is losing a photon but the energy of the system remains the same. But in reality, losing a photon is not just applying the annihilation operator. Between two successive photon detections, that are emitted from your system, there is evolution which exponentialy reduces the energy of the system. – AndyK Mar 5 '17 at 1:17