# How can energy be conserved in photon absorption at a detector?

This question is somewhat of an elaboration on the discussion here, which didn't really get a satisfactory answer. A related (but not identical) question of mine is found here, which also was not resolved.

Let us imagine an experiment composed of three parts. The first part is a system that emits a single photon, at distant times. It could, for example, be a machine that puts, one by one, single excited atoms in the room of the experiment, and waits for a time much longer than than the half-life of the excited state, before bringing in the next atom. The upshot is that we get with a very high probability, a single photon in the room at any time.

The second part of the experiment is simply some space for the photon to travel in, before it reaches the third part which is a photodetector. The photodetector is a metal plate, and when a photoelectron is ejected from it, it causes a current pulse and we hear a click.

Now quantum electrodynamics predicts that after emission, the electromagnetic field around the emitter becomes excited and an electromagnetic wave with the shape of dipole radiation is produced around it, as discussed more thoroughly here in Andrew Steane and G Smith's answers. The excitation travels away from the emitting atom as it expands, and it carries with it an amount of energy $$\hbar\omega$$ - this is the crucial point.

At a later time this electromagnetic wave reaches our detector. Standard quantum calculations on the photoelectric effect show that the probability of photoejection (and thus a click) is proportional to the amplitude (squared) of the electric field near the absorber. So with any photon emitted, there's a finite probability (which we can calculate) for a click, and a finite probability for not getting a click. Overall, we can see that on average, the amount of clicks that we should get after a long time corresponds exactly to an amount of energy absorbed that cancels the amount of energy emitted by the emitter (together with the energy that escaped to directions other than that of the detector). This is mysterious: why does energy seem to be conserved here only on the average?

We can now elaborate on the experiment and add a second detector spatially separated from the first one, at a different direction of the emitter. This detector should have the same probability of absorption - so there's a finite probability that an atom will emit a single photon (energy $$\hbar\omega$$) but both detectors will click, that is energy of $$2\hbar\omega$$ will be absorbed. How is it possible energy seems to not be conserved in this experimet - what is wrong in my understanding?

• Photons are detected as discrete particles. The probability of the photoelectric effect occurring is not proportional to the square of an electric field, it depends on the photon energy. There is no probability that a single photon cause two photoelectrons in different detectors. – Rob Jeffries Jan 6 at 13:53
• @RobJeffries You can find a derivation of the photoejection probability, based on a semiclassical approximation, showing that the probability is indeed proportional to the electric field square amplitude, in this paper by Lamb: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19680009569.pdf – roymend Jan 6 at 15:07
• You are correct that the probability of detection at each detector is nonzero and equal. However, you must ALSO consider the joint probability of detection two photons. This can be calculated using Glaubers higher order coherence functions. You will see for. A single photon state the probability of joint detection is zero. In the language of probability theory we say the two statistical events are NOT independent. (We cant get the joint probability from the individual probabilities) – jgerber Jan 6 at 16:16
• @jgerber This is more of an answer. Comments are for asking for clarification of the question or raising thought-provoking single points. – Bill N Jan 6 at 16:50
• Thanks for that. The Lamb paper certainly looks interesting. It suggests the transition probability is proportional to $E^2$, which is the same I suppose as saying it's proportional to the number of photons. However, I don't see where it predicts the energy dependence of the cross-section; but then I don't really understand what the $\sigma$ is in Lamb's paper. – Rob Jeffries Jan 6 at 18:48

This is a great question. It can sometimes be hard to find satisfactory answers to these kinds of questions, but I will do my best. The key to understanding this issue is to understand the nature of the quantum states of light and detector involved here.

The quantum state of light after spontaneous emission

"Now quantum electrodynamics predicts that after emission, the electromagnetic field around the emitter becomes excited and an electromagnetic wave with the shape of dipole radiation is produced around it..." "The excitation travels away from the emitting atom as it expands, and it carries with it an amount of energy ℏω - this is the crucial point."

This is in fact a crucial point, though for my discussion, I am going to advocate for a slightly more exact description. When spontaneous emission occurs, there are many possible modes that light can be emitted into. The important thing about these modes is that they are eigenstates of the quantum electromagnetic field Hamiltonian. We can write something like $$| 1_k \rangle$$ to denote a state of the electromagnetic field which has a single photon in mode $$k$$. Since there are many modes that can be emitted, the actual state of the radiation field $$|\psi_{photon}\rangle$$ is a superposition over all of these modes. With this description, we can now write something like $$|\psi_{photon}\rangle = \sum_k a_k |1_k\rangle,$$ where $$a_k$$ is a coefficient which depends on the mode. In practice, the coefficient $$a_k$$ enforces two critical things

1. The modes represented have the correct energy $$\hbar\omega$$ which is consistent with energy conservation from whatever transition caused this light to be produced.
2. The modes carry the dipole emission pattern (or whatever the appropriate radiation pattern is if the transition is not a dipole transition). So to be concrete, $$a_k = 0$$ for any $$k$$ which point along the dipole moment, and $$a_k$$ is maximal for $$k$$ which are orthogonal to the dipole moment.

Quantum Optics by Marlan Scully has graphic discussions of these things, for anybody interested in the details of such states. But the important thing is that the electromagnetic field is in a superposition of many quantum states. With this established, we can now move on to...

The quantum states of two-level systems

In your post, you use the example of photoelectric detection. While this is completely valid, I'm going say that the detector is a two level system (perhaps another atom), which has the ability to absorb light at the same frequency we are considering. The state of a single two level system can be described as either "down" $$|\downarrow\rangle$$ or "up" $$|\uparrow\rangle$$, which correspond to "not detected" and "detected" states respectively. In a semiclassical picture, we can think of the detector as some atomic with a dipole moment which acts as an antenna to receive light.

If we have a single detector at some location, then we can use quantum electrodynamics to calculate the probability that the detector flips its state from down to up, given the state $$|\psi_{photon}\rangle$$ that we started with. One will find that the probability of detection of course depends on the location of the detector, and that detection will be best in the region of maximal "classical dipole radiation", and worst in regions where the "atomic antenna" cannot emit. Our detector can now interact with its environment, eventually providing us with information that the photon was detected by that detector. Measurement and decoherence are very tricky things, but I think this description should be sufficient. Now we get to the final, and most interesting issue, of...

More than one detector

Say we now have two of these 2-level system detectors which are some distance away from each other. For now, let's assume they are actually quite close to each other (we will move them far apart shortly). If we have two atoms next to each other, and a single incident photon incoming, it is very tempting to think that either one atom OR the other atom must absorb the light. But this is actually not true, because quantum superposition is still at play. If we do a QED calculation, we will find that the final state of the two detectors can be described as $$|\psi_{detectors}\rangle = \alpha|\uparrow\downarrow\rangle + \beta|\downarrow\uparrow\rangle,$$ which means that we have a superposition of the photon being absorbed into either of the two detectors. The coefficients $$\alpha$$ and $$\beta$$ will depend on the locations of the detectors. This kind of state generalizes to many atoms, and these states are called Dicke states. Importantly, a state like $$|\uparrow\uparrow\rangle$$ is not included in the possible final states, because this would not conserve energy, and QED calculations such as these only permit final states which conserve energy. (Technically, a state where neither of the detectors are activated, and the photon goes undisturbed, or is scattered into a different single photon state, is also permitted, though not needed for this discussion.) Then, we can take a measurement on the detectors. Since the two level systems are in entangled state, only two outcomes are possible

1. The measurement indicates the system is in state $$|\uparrow\downarrow\rangle$$ with probability $$|\alpha|^2$$, and the first detector clicks.
2. The measurement indicates the system is in state $$|\downarrow\uparrow\rangle$$ with probability $$|\beta|^2$$, and the second detector clicks.

If the detectors are extremely close to each other, then these probabilities may be quite similar, and the question of which detector beeps may be determined by 50/50 chance. However, if the detectors are far apart (many wavelengths), then the probabilities of detection will be determined by the dipole radiation pattern. The key is that even though a single photon can put two detectors in a superposition, the states will always be entangled such that when measurements are made, only one detector can click.