Is information lost when a photon is absorbed?

[wywong]

TL;DR Summary

Quantum information seems to be lost in a thought experiment involving absorption of photons

Suppose I fire a photon with an upward spin towards a hydrogen atom. It is absorbed by the hydrogen atom's electron which subsequently emits a photon when it returns to its ground state. I then fire another photon with a downward spin towards the same hydrogen atom and the same thing happens. Is it possible to deduce the spins of the original photons by measuring the emitted photons and the hydrogen atom? If not, is information lost?

 

[Mentz114]

Summary: Quantum information seems to be lost in a thought experiment involving absorption of photons

Suppose I fire a photon with an upward spin towards a hydrogen atom. It is absorbed by the hydrogen atom's electron which subsequently emits a photon when it returns to its ground state. I then fire another photon with a downward spin towards the same hydrogen atom and the same thing happens. Is it possible to deduce the spins of the original photons by measuring the emitted photons and the hydrogen atom? If not, is information lost?

Photons do not have fermionic quantum spin. But photon polarization orientation may be entangled with spin.
In some cases it may be possible to deduce the polarization of the original photon by measuring the polarization of the emitted photon. I don't think any information is lost.

 

[PeterDonis]

Photons do not have fermionic quantum spin. But photon polarization orientation may be entangled with spin.

It's true that photons are not fermions; they have spin 1, not spin 1/2. However, photon polarization is their spin, it's not "entangled with" their spin. Also, because photons are massless, the Hilbert space of photon spins (polarizations) is that of a qubit, i.e., the same as for the spin of a massive spin 1/2 particle like the electron. In fact, it's much easier to realize qubits with photon polarizations, which is why that's the method used in, for example, quantum computing.

 

[PeterDonis]

Is it possible to deduce the spins of the original photons by measuring the emitted photons and the hydrogen atom?

Why do you need to deduce the photon spins? You already know what they are, because you specified them in your scenario. To realize the scenario as you describe it, you would have to prepare the photons in specific polarization (spin) states, which means you know their spins before they get absorbed.

 

[wywong]

Why do you need to deduce the photon spins? You already know what they are, because you specified them in your scenario. To realize the scenario as you describe it, you would have to prepare the photons in specific polarization (spin) states, which means you know their spins before they get absorbed.

To prove a point. In quantum mechanics, it is said that information cannot be destroyed. I think that is not true whenever an irreversible event has taken place, as in the case of my thought experiment. Here the two bits of information encoded in the spin states seem to be lost.

 

[Mentz114]

It's true that photons are not fermions; they have spin 1, not spin 1/2. However, photon polarization is their spin, it's not "entangled with" their spin. Also, because photons are massless, the Hilbert space of photon spins (polarizations) is that of a qubit, i.e., the same as for the spin of a massive spin 1/2 particle like the electron. In fact, it's much easier to realize qubits with photon polarizations, which is why that's the method used in, for example, quantum computing.

Sure. I was thinking of the spin of a nitrogen-vacancy in a diamond where emission can be induced and the polarization of the photon is entangled with the spin of the emitter. Also in laser cooling the rates of emission and absorption depend on polarization.
If I remember correctly, spin and polarization angle both have su(2) symmetry.

 

[PeterDonis]

To prove a point.

You're missing my point. You specified the photon spins in your scenario, so asking how they can be deduced in your scenario is meaningless. If you want to give a scenario to pose your question, you need to specify a different scenario where the question of how to deduce the initial photon spins is meaningful.

If you want to just pose your general question about whether or not information can be destroyed in QM, we can discuss that (though not resolve it) without a scenario at all. See below.

In quantum mechanics, it is said that information cannot be destroyed.

The more technical and precise way of saying this is that unitary operations are reversible, and quantum mechanics models all time evolutions and interactions as unitary operations--except possibly when a measurement (or more generally an irreversible event) takes place. See below.

I think that is not true whenever an irreversible event has taken place

Of course not, because if an event is irreversible, by definition it is not reversible, and therefore it cannot be modeled by a unitary operation.

So the real question is whether or not events that we think of as irreversible, such as measurement results being recorded, are truly irreversible, or if they are actually reversible and it's only the limitations of our current technology that prevent us from demonstrating that experimentally. This is one way of describing what is called the quantum measurement problem, and there is no generally agreed resolution to it. So we don't know for sure what the answer to your title question is.

 

[wywong]

You're missing my point. You specified the photon spins in your scenario, so asking how they can be deduced in your scenario is meaningless. If you want to give a scenario to pose your question, you need to specify a different scenario where the question of how to deduce the initial photon spins is meaningful.

How about this: the two photons come from an entangled pair, and they arrive at the hydrogen atom at different times because the second one takes a longer route. I do not know their spins before the absorption, and it seems that I will never be able to find out that information after the absorption.

The more technical and precise way of saying this is that unitary operations are reversible, and quantum mechanics models all time evolutions and interactions as unitary operations--except possibly when a measurement (or more generally an irreversible event) takes place.

If Rules of Quantum Mechanics indeed do not apply to irreversible events such as measurement, then a lot of conundrums about quantum mechanics, such as black hole information paradox, Schrodinger's cat, time machines, Wigner's friend, etc, can be easily solved by citing the action of irreversible events involved. And we badly need to extend the current model to include irreversible events as well.

 

[Mentz114]

How about this: the two photons come from an entangled pair, and they arrive at the hydrogen atom at different times because the second one takes a longer route. I do not know their spins before the absorption, and it seems that I will never be able to find out that information after the absorption.

If Rules of Quantum Mechanics indeed do not apply to irreversible events such as measurement, then a lot of conundrums about quantum mechanics, such as black hole information paradox, Schrodinger's cat, time machines, Wigner's friend, etc, can be easily solved by citing the action of irreversible events involved. And we badly need to extend the current model to include irreversible events as well.

If an observer did not know their polarization before absorption and still does not after absorption - what information did the observer lose ?

Information is a human construct not a physical observable. If you actually had a case you could write equations otherwise it is hand waving.

 

[ZapperZ]

Coming back to the original question on whether the information is lost when a photon is absorbed, the answer is not necessarily so.

We have already seen evidence for this in Lene Hau's experiment of "stopping light" in cold sodium atoms. The whole point of that experiment is that the light is completely absorbed, the information stored, and then it is "replayed" at a later time.

Zz.

 

[PeterDonis]

If Rules of Quantum Mechanics indeed do not apply to irreversible events such as measurement

That's not what I said. I said that unitary operations can't model irreversible events. But the Rules of Quantum Mechanics aren't just unitary operations. They also include the collapse postulate and the Born rule, which is how events that are treated as irreversible are handled. The question is whether, at a more fundamental level, the collapse is "real" or only apparent (with the underlying dynamics always being unitary), and whether the Born rule can be derived from underlying unitary dynamics or whether it needs to be imposed as an additional requirement (which then leaves the question of where this additional requirement comes from).

 

[PeterDonis]

the two photons come from an entangled pair, and they arrive at the hydrogen atom at different times because the second one takes a longer route

In this case neither photon originally has a well-defined spin; only the two-photon system does. And the total spin of the two emitted photons will be the same as the total spin of the original two-photon system.

 

[wywong]

In this case neither photon originally has a well-defined spin; only the two-photon system does. And the total spin of the two emitted photons will be the same as the total spin of the original two-photon system.

The total spin of the two original entangled photons is zero. Is the total spin of the emitted photons zero? Are the emitted photons entangled? I think they are totally independent.

 

[PeterDonis]

The total spin of the two original entangled photons is zero.

If that's the way they are entangled, yes. Note that there are other possible ways to entangle photon spins.

Is the total spin of the emitted photons zero?

If there are only two of them and the atom ends up in the same state after each emission, then yes, they have to be, because the electron transition for each emission has to be exactly the inverse of the electron transition for each absorption.

Are the emitted photons entangled?

Given the description above, yes, they would have to be.

 

[Heikki Tuuri]

A "collapse" of a wave function does destroy quantum information. Suppose that you have prepared a spin in a pure state where the probability amplitude of "up" is A and "down" is B.

After the measurement you have lost the values of A and B. You just know that the spin is now "up" or you know it is "down".

Unitarity means that the state of a quantum system at any time can be calculated back from the current state. That requires that the system is isolated. A measurement and the resulting "collapse" of the wave function involves coupling the system to a measurement apparatus. The system is no longer isolated then - it interacts with a macroscopic external apparatus.

 

[PeterDonis]

A "collapse" of a wave function does destroy quantum information.

If you are using a collapse interpretation of QM, yes. Not if you are using a no collapse interpretation like MWI. See my post #7.

 

[wywong]

If there are only two of them and the atom ends up in the same state after each emission, then yes, they have to be, because the electron transition for each emission has to be exactly the inverse of the electron transition for each absorption.

Does the emitted photon always travel in the same direction as the absorbed photon? If not, their angular momentum can't be the same.

 

[PeterDonis]

Does the emitted photon always travel in the same direction as the absorbed photon? If not, their angular momentum can't be the same.

Don't you mean their linear momentum? That's what "direction" would imply.

As far as always traveling in the same direction, no, that wouldn't be required, since the nucleus would not have to recoil in the same direction. However, that doesn't prevent the photons from being entangled; it just means they're entangled in their spin (polarization) degree of freedom, not their momentum degree of freedom.

 

[wywong]

Don't you mean their linear momentum? That's what "direction" would imply.

As far as always traveling in the same direction, no, that wouldn't be required, since the nucleus would not have to recoil in the same direction. However, that doesn't prevent the photons from being entangled; it just means they're entangled in their spin (polarization) degree of freedom, not their momentum degree of freedom.

Suppose the first absorbed photon has spin = +1 and the corresponding emitted photon travels north with spin +1. Why can't the excited electron emit a southward photon with spin -1 instead? That alternative photon has the same energy and angular momentum as the actual one. They differ in helicity but that information is not present in the bound electron.

 

[Mentz114]

Does the emitted photon always travel in the same direction as the absorbed photon? If not, their angular momentum can't be the same.

Emission and absorption result in a change of momentum with the sum of the atom 's momentum and the photons momentum conserved.
[I see PeterDonis already mentions this]

 

[PeterDonis]

That alternative photon has the same energy and angular momentum as the actual one.

Same energy, yes. Same angular momentum, no. What counts for balancing angular momentum in the emission process is the photon's spin, not some combination of its spin and linear momentum.

They differ in helicity but that information is not present in the bound electron.

I don't know what you mean by this.

I strongly suggest that you start using math instead of ordinary language. If you think your claims are correct, write down the math that shows that.

 

[wywong]

Same energy, yes. Same angular momentum, no. What counts for balancing angular momentum in the emission process is the photon's spin, not some combination of its spin and linear momentum.

My rationale is, although angular momentum is to be conserved, it does not follow that spin state is conserved. I do not mean linear momentum in any away affects the angular momentum. From definition, spin is the projection of angular momentum along the direction of propagation. Thus the same angular momentum but opposite direction of propagation will flip the sign of the projection. If a northward photon with +1 spin has a certain angular momentum, a southward photon with the same angular momentum must have -1 spin; no? If my argument is correct, then the spin state depends on the direction of emission, which is random.

I am sorry that I don't know enough QM's maths to express my argument better.

 

[Demystifier]

Summary: Quantum information seems to be lost in a thought experiment involving absorption of photons

Suppose I fire a photon with an upward spin towards a hydrogen atom. It is absorbed by the hydrogen atom's electron which subsequently emits a photon when it returns to its ground state. I then fire another photon with a downward spin towards the same hydrogen atom and the same thing happens. Is it possible to deduce the spins of the original photons by measuring the emitted photons and the hydrogen atom?

It depends, did you measure the spin of the electron in the initial ground state?

If not, is information lost?

Information is not lost by absorption, it is just carried by another carrier. Information carried by a photon before absorption is carried by electron after the absorption.

 

[vanhees71]

Photons are massless particles and don't have a spin. What's meant when you say that "photons have spin 1" it means that they are represented in the massless (1/2,1/2) representation of the Poincare group. Since the field is massless and you want only a discrete set of polarization states you have to represent them necessarily as gauge fields, and then you have only 2 polarization-degrees of freedom and not the expected 3(=2*1+1) polarization degrees of freedom of a massive spin-1 field. The 2 polarization-degrees of freedom of the photon can be chosen either as linear polarization states (or any general elliptic polarization states for that matter) with respect to some reference frame or, in a frame-indpendent way, as the left- and right-handed polarization states of helicities $\pm 1$. These are the left- and right-circular polarization states.

 

[Haelfix]

If Schrodinger's cat dies by a measurement. I don't agree that it means that information was lost... Collapse or no collapse postulate. You can always apply a sequence of unitary operators that undoes every set of biological steps that led to the untimely demise. The whole process is guarenteed to be reversible by the laws of quantum mechanics.

There is no difference here at this level compared to breaking an egg. And just like the case of an egg breaking the imprint of the process is spread to the environment in subtle correlations (in the quantum case, entanglement).

 

[wywong]

It depends, did you measure the spin of the electron in the initial ground state?

To make matter simple, I just assume it to have zero spin. So let's say I did and the measured spin is zero.

Information is not lost by absorption, it is just carried by another carrier. Information carried by a photon before absorption is carried by electron after the absorption.

What puzzles me is how the spin information carried by the electron can be transferred to the emitted photon given that the photon is emitted in a random direction. For example, if the emitted photon has spin +1, can we conclude that the absorbed photon must have spin +1, regardless of the angle between the two paths?

 

[vanhees71]

Photons have no spin!

You can derive the intensities of emitted photons in various transitions using perturbation theory (usually in atomic physics the dipole approximation is enough). You get certain selection rules, telling you which transitions are possible and what are the probabilities to emit photons with specific polarization (not spin!) states.

https://theory.physics.manchester.ac.uk/~judith/AQMI/PHYS30201se19.xhtml

 

[PeterDonis]

To make matter simple, I just assume it to have zero spin.

This is not possible. An electron either needs to have spin + 1/2 or spin - 1/2 when measured.

Photons have no spin!

This is a "B" level thread so the technicalities involved with photon polarization are beyond the scope of this discussion. For our purposes here saying that a photon can have "spin" +1 or -1 should be sufficient.

 

[PeterDonis]

the left- and right-handed polarization states of helicities $\pm 1$. These are the left- and right-circular polarization states.

And for our purposes here, since it's a "B" level thread, calling these states of "spin" +1 and -1 should be sufficient.

 

[wywong]

This is not possible. An electron either needs to have spin + 1/2 or spin - 1/2 when measured.

Oops! Let me correct my scenario by adding that I measure the electron spin before firing the first photon and after the second emission, and the electron spin is found to be unchanged (say both +1/2).

 

[vanhees71]

This is not possible. An electron either needs to have spin + 1/2 or spin - 1/2 when measured.
This is a "B" level thread so the technicalities involved with photon polarization are beyond the scope of this discussion. For our purposes here saying that a photon can have "spin" +1 or -1 should be sufficient.

Also in a B-level thread one should not provide wrong information. Usually it's confusing for students thinking a photon would have a spin rather than polarization states.

 

[Demystifier]

What puzzles me is how the spin information carried by the electron can be transferred to the emitted photon given that the photon is emitted in a random direction. For example, if the emitted photon has spin +1, can we conclude that the absorbed photon must have spin +1, regardless of the angle between the two paths?

Spin is not conserved, so the spin of photon (or polarization, as vanhees prefers to call it for some technical reasons that are not essential here) does not need to be equal to the spin of electron. What is conserved is the total angular momentum, which a sum of all spins and all orbital angular momenta. So when electron absorbs or emits a photon, the electron changes its orbital angular momentum such that the total angular momentum is conserved.

 

[wywong]

Spin is not conserved,

In my scenario, the absorbed photons were entangled and had opposite spins (or polarization). Now that the spins are not conserved, can the emitted photons still be entangled?

 

[Demystifier]

Now that the spins are not conserved, can the emitted photons still be entangled?

The emitted photon is entangled with the electron.

 

[DrClaude]

Depending on the polarization of the photon, different atomic states can be reached. If the photon is in a superposition of states, then after absorption, the atom will be in a superposition of the different atomic states. If the photon was initially entangled with another photon, after absorption, the atom is now entangled with that other photon.