I have read this question:

Conservation of Energy in photon exchange between two atoms

where Kurshal Shah in a comment says:

As per the energy-time uncertainty relation, the emitted photon does not have a definite energy, but a spread. You need to account for that too.

No well-defined frequency for a wave packet?

where Ben Crowell says:

Let's say an isolated atom emits a photon. The excited state in the atom has some lifetime τ. Through the energy-time uncertainty relation, that gives the excited state some uncertainty in energy δE∼h/τ (not the same as ΔE, which is a difference in energy between atomic states). The photon then has the same uncertainty δE in its energy, which corresponds to an uncertainty in frequency. The photon isn't in an eigenstate of energy. Yes, when you measure the energy of the photon, you get a random outcome. However, there is a quantum-mechanical correlation between this energy and the energy of the atom, so that energy is exactly conserved (not just statistically, on an average basis).

Now energy must be conserved in a closed QM system.

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.


But if the energy of the emitted photon has some uncertainty as per QM, but the emitting atom's energy levels are specific energy levels, an excited and a ground state, then the difference between the energy levels must be a specific energy level, which should be the energy level of the photon.

Now if I detect (observe, absorb) this photon, then I will get a specific frequency, an eigenvalue. What I do not understand is, why the source of the photon, that is the emitting atom has a definite energy level difference between excited and ground state.

  1. that gives a specific energy level difference

  2. the photon that is emitted must have this specific energy level, since the electron lost this specific kinetic energy to go from excited to ground level

  3. but the questions say, that the photon does have an uncertainty in its energy level

  4. when I detect (observe, absorb) this photon, I get a certain eigenvalue


  1. Does the photon have a well defined frequency or not?
  • $\begingroup$ All atomic energy levels has an uncertainty ,for consrevation ,maybe my answer here will help physics.stackexchange.com/q/489691 $\endgroup$
    – anna v
    Commented Jul 14, 2019 at 15:04
  • $\begingroup$ You really—*really*—should read Lamb's "Anti-photon" essay. The word "photon" doesn't have one well fixed meaning in all contexts, and there are multiple correct answer to questions like this unless and until you nail down which meaning you want to talk about. $\endgroup$ Commented Jul 14, 2019 at 15:15

1 Answer 1


Not sure about the definition of a photon (that part gets me confused too) but here is what I understand about energy conservation in such scenarios : If the photon is emitted in some superposition of energies distributed according to some $f(e)$ (I'll assume discrete energy) then the joint system of the photon and the emitting atom should be described by the entangled state $$|\psi\rangle=\sum_ef(e)|e\rangle_{photon}|E-e\rangle_{atom}$$ where $E$ is the initial energy of the atom (assuming it was in an energy eigenstate) and the kets in the sum are energy eigenstates. Then, when the photon's energy is measured both the atom and the photon collapse into a consistent energy conserving product state. It does not make sense for the photon to be in energy superposition while the atom is in energy eigenstate.


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