B How can the annihilation operator for photons be realized experimentally?

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Light consists of transversely varying electric and magnetic fields, with polarized beams exerting a measurable force on point charges, allowing energy extraction. In contrast, unpolarized light has electric fields that act randomly, resulting in no net force on a charge, raising questions about how it carries energy. The discussion highlights that while individual photons are always polarized, a beam of unpolarized light is composed of many photons with random polarization states, leading to an average that appears unpolarized. Quantum electrodynamics (QED) provides a more accurate framework for understanding these interactions than classical models. Ultimately, the nature of polarization in light is complex, involving both classical and quantum perspectives.
  • #31
vanhees71 said:
I was more thinking of one photon being annihilated by some junk of matter. Here it's the opposite: It's not so easy to prepare a single-photon Fock state to begin with.

Yes.

vanhees71 said:
I'd also think a light beam (i.e., a classical em. field as comes out of a laser pointer) is pretty close to a coherent state.

Yes.

vanhees71 said:
Of course it's very difficult (if not impossible) to just absorb one and only one photon with certainty from such a state.

Exactly.
 
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  • #32
vanhees71 said:
I'm puzzled too, because it's very trivial, and you made it a problem. As I said several times the vacuum state is mapped to the zero-vector by application of the annihilation operator (and thus also the number operator), i.e., it's an eigenstate of the annihilation operator as well as the number operator with eigenvalue 0. You made a complicated issue out of it claiming it's maped to a set or something like this. Of course that's what's stated in all quantum mechanics and quantum field theory books (if they are not wrong).

This is getting repetitive. My claim is that there is no unconditional faithful experimental realization of the annihilation operator (and the photon number operator for the same reason) because the full set of physically realizable states you get by applying the annihilation operator to the vacuum state is the empty set. You claim that the vacuum state is an eigenstate of the photon number operator and that this shows that my claim is wrong. I fully agree with the first half, but not with the second. In order to back up your claim, you would need to show how 0|0\rangle is actually realized physically. How is it normalized? How do you measure it?

vanhees71 said:
Practically it's very easy to implement the annihilation operator. Just put something in the photon's way that absorbs it ;-)).

That is of course completely wrong. The annihilation operator is well defined and maps |6\rangle to \sqrt{6}|5\rangle, |3\rangle to \sqrt{3}|2\rangle and |0\rangle to 0|0\rangle. Your "put something in the way operator" maps |6\rangle to the vacuum state, |3\rangle to the vacuum state and any other photon number to the vacuum state.

We have means to test what an experimental implementation of an operator actually does. That is called quantum process tomography. You put your experimental realization iside a black box, feed it with well defined states, see what comes out and may define a superoperator that way. The easiest way of course would be to feed it with Fock states, but as their preparation is tedious one usually uses coherent states instead as they also form a basis that spans all states. But let us assume just for the sake of the argument that one could feed a black box with Fock states and we want to know whether the black box corresponds to the annihilation operator. We feed it with 6 photons and the experimental fact that we get 5 photons out of it shows that we are on a good way. One may get the prefactor from repeating the experiment as well. We also test all the other finite Fock states and things look good. Now we need to test the vacuum state. We put the vacuum in and you need to get 0|0\rangle on the output side. What would the experimental realization of the zero vector look like?

Of course it is not possible to realize the annihilation operator for this obvious reason. Therefore, the next best thing is to make a non-deterministic operator out of it. You put a beam splitter inside the box that diverts only a tiny amount of the beam and have a detector in the path of the diverted beam that gives you a signal whenever it detects a photon. When the detector clicks, you know that the state you get out of the black box is equivalent to what you get when the annihilation operator does to whatever state you actually just put in. How does this work with the vacuum now? The detector never clicks. We kind of avoid the elephant in the room. However, the operator is of course not trace-preserving. The probability that the detector clicks will of course increase with the number of photons present in the input beam. So you change the relative weights around. The former vacuum state will have no weight at all in the output state and the n=10 state will have way more weight than the n=1 state. So you shuffle the mean photon numbers around. The same holds true for the photon creation operator. The probability to add one more photon to any state will scale with the number of photons already present in that mode. That is essentially just stimulated emission at work. In the end, applying the best experimental approximation of the photon number operator (or as it is rather called: creating the photon-subtracted-then-added state) completely shuffles the photon numbers of the output state in a manner that may be surprising unless the input state is a Fock state.

For that reason indeed nobody implements the photon number operator in experiments (although it would work - the click rate of the detector inside the black box will be proportional to the photon number and one may of course completely ignore the output state), but instead rather uses the "put a block of material there to absorb everything"-method.

The fact that photon annihilation may only be realized in a conditional manner really is not too advanced stuff and I do not see where the problem is (to be precise: perfect absorption is possible in nonlinear cavity QED, but certainly not for all Fock states).
 
  • #33
Obviously there's a mutual misunderstanding here, which I cannot resolve. It's clear that it's very hard to realize the annihilation operator, which takes out precisely one photon from a given photon state, experimentally. I've never claimed otherwise. I guess, we agree on the math though now.
 
  • #34
vanhees71 said:
It's clear that it's very hard to realize the annihilation operator, which takes out precisely one photon from a given photon state, experimentally.

Well, to be precise, it is not just very hard, but fundamentally impossible. That is a small, but important difference.

The standard way of connecting Hilbert space to real-world physics makes use of the fact that all collinear state vectors correspond to the same physical state, so |n\rangle and c|n\rangle represent the same physical state for any non-zero complex c. Accordingly, it is very convenient to work with normalized state vectors.

However, it is equally clear that c|n\rangle and 0|n\rangle are not physically equivalent. The latter is not even realizable physically. Applying the photon annihilation operator to the vacuum would require some process that maps these state vectors onto each other. As this is impossible, it is also not just hard, but impossible to find a precise physical implementation of the annihilation operator.
 

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