How can the annihilation operator for photons be realized experimentally?

[jeremyfiennes]

TL;DR Summary

How does a beam of unpolarized light carry energy?

Light consists of transversely varying electric and magnetic fields. A point charge in the path of a vertically polarized light beam experiences a vertically varying force from which energy can be extracted. In an unpolarized beam, the electric fields of individual photons act randomly in all directions. Resulting in no net force on a point charge. So how does an unpolarized light beam carry energy?

 

[ZapperZ]

Summary:: How does a beam of unpolarized light carry energy?

Light consists of transversely varying electric and magnetic fields. A point charge in the path of a vertically polarized light beam experiences a vertically varying force from which energy can be extracted. In an unpolarized beam, the electric fields of individual photons act randomly in all directions. Resulting in no net force on a point charge. So how does an unpolarized light beam carry energy?

Why does it need to be polarized in the first place for it to carry energy?

In a photoelectric effect experiment, for example, the light source is monochromatic, but NOT polarized. Yet, it can still excite the electrons in the conduction band of the photocathode. It carries quanta of energy regardless of its polarity.

Zz.

 

[jeremyfiennes]

"Why does it need to be polarized in the first place for it to carry energy?"
This is not my point. I'm saying that I can see how a polarized beam carries energy. A point charge in its path experiences an oscillating force. But this explanation does not apply to an unpolarized beam where the electric fields of individual photons are randomly oriented.

 

[ZapperZ]

"Why does it need to be polarized in the first place for it to carry energy?"
This is not my point. I'm saying that I can see how a polarized beam carries energy. A point charge in its path experiences an oscillating force. But this explanation does not apply to an unpolarized beam where the electric fields of individual photons are randomly oriented.

Again, why does it need to be polarized to be able to transfer its energy?

Say you have a free-electron gas. A stream of unpolarized light passes through. A photon can bump into an electron, causing a Compton-like scattering.

So what's the issue here?

Zz.

 

[Nugatory]

I'm saying that I can see how a polarized beam carries energy. A point charge in its path experiences an oscillating force. But this explanation does not apply to an unpolarized beam where the electric fields of individual photons are randomly oriented.

You are confusing yourself by trying to apply the classical wave model when the scale is such that quantum effects matter so you have to use quantum electrodynamics. Your model of a point charge experiencing an oscillating force (the way a floating object is pushed up and down by water waves) feels reasonable, but it's not an accurate description of the interaction between a photon and a charged particle.

 

[jeremyfiennes]

Ok. Thanks.

 

[tech99]

From a classical standpoint, I am uncomfortable in saying a wave is unpolarised. I would rather say it is randomly polarised. For instance, if we receive radio noise from a hot object we might say it is unpolarised. But we can observe received power from it using a linearly polarised antenna, so it is clearly making electrons move. One way to create an "unpolarised" source is to have a pair of cross-polarised antennas, each being driven by a separate noise generator. If this wave passes a charged particle, as in a metal, I would expect it to execute a random motion.

 

[Cthugha]

Summary:: How does a beam of unpolarized light carry energy?

Light consists of transversely varying electric and magnetic fields. A point charge in the path of a vertically polarized light beam experiences a vertically varying force from which energy can be extracted. In an unpolarized beam, the electric fields of individual photons act randomly in all directions. Resulting in no net force on a point charge. So how does an unpolarized light beam carry energy?

How do people go through the wall in the middle, when they take the left door 50% of the time and the right door 50% of the time?
It is important to note that (for a single polarization-degenerate mode of the light field) the term "unpolarized" always refers to an average, usually a time average. If you trace the polarization of a such a "mode" over time, you will find some direction of polarization at every instant, but it randomizes very quickly, giving rise to no net polarization in the average.

 

[jonbovi]

Despite I think the issue is solved. I would like to add: A single photon is always polarized. When we talk about unpolarized light, what we want to says is a beam of light. That beam is composed by many photons randomly polarized. Note that by definition, each of those photons is polarized, but from a macroscopic point of view, the beam has no polarization. In this sense, when we talk about polarized light, we mean that all photons present the same or almost the same polarization state. In the same way, when we talk about quasi-polarized light, most of the photons are around a polarization state.

I hope this help

 

[PeterDonis]

Light consists of transversely varying electric and magnetic fields.

Since this post is in the Quantum Physics forum, this statement is not correct. It would be correct for a thread in the Classical Physics forum, but in classical physics there is no such thing as a "photon".

In an unpolarized beam, the electric fields of individual photons act randomly in all directions.

In classical physics, as above, there is no such thing as a "photon". In quantum physics, individual photons don't have electric fields.

 

[PeterDonis]

It is important to note that (for a single polarization-degenerate mode of the light field) the term "unpolarized" always refers to an average, usually a time average.

I can see how this would be true in classical physics, but I don't think it's true if we are using QED to model the light. Individual photons don't need to be in an eigenstate of any polarization operator, do they?

 

[Cthugha]

I can see how this would be true in classical physics, but I don't think it's true if we are using QED to model the light. Individual photons don't need to be in an eigenstate of any polarization operator, do they?

Well, photons are spin 1 particles, so you will find them with spin (more exactly: helicity) +/-1 in experiments, which is in turn tied to circular polarization. What the polarization state of individual photons is, depends on what you mean by "individual photons". A single photon state as an ensemble average of many identically prepared individually emitted photons may indeed be unpolarized, linearly polarized or take on many interesting polarization states. However, any individual photon out of that ensemble will still transfer its spin to whatever absorbs it.

In terms of quantum optics, unpolarized light would be given by a 2x2 density matrix in helicity space, where the diagonal elements are 0.5 each. Instead, take a pure superposition of the two helicities with a fixed relative phase and you get linear polarization. Take only one helicity and you get circular polarization.

 

[PeterDonis]

photons are spin 1 particles, so you will find them with spin (more exactly: helicity) +/-1 in experiments

I was asking about the photon states (or more generally the states of the quantum electromagnetic field, not all of which are usefully described as "photon" states) before any measurement is made. Of course any spin measurement will yield a definite value, but that's not the same as the field having been in an eigenstate of the operator corresponding to that spin measurement before the measurement was made.

The question I was asking can be rephrased as: given a quantum EM field state, must there always be some photon spin operator of which that state is an eigenstate (which would mean, in terms of measurements, that the measurement corresponding to that spin operator will always yield the same result with 100% probability if the field is in the given state)?

 

[vanhees71]

Despite I think the issue is solved. I would like to add: A single photon is always polarized. When we talk about unpolarized light, what we want to says is a beam of light. That beam is composed by many photons randomly polarized. Note that by definition, each of those photons is polarized, but from a macroscopic point of view, the beam has no polarization. In this sense, when we talk about polarized light, we mean that all photons present the same or almost the same polarization state. In the same way, when we talk about quasi-polarized light, most of the photons are around a polarization state.

I hope this help

Why should a single photon be always polarized? Whether some property (observable) in quantum theory is determined depends on the preparation of the state it is described with.

E.g., if you look at the polarization of a single photon in an entangled polarization-singlet photon pair, it's completely unpolarized, i.e., it's polarization state is maximally undertermined (in the sense of von Neumann entropy), but it's for sure a single photon (even heralded when using the other photon in the pair as "trigger").

 

[Cthugha]

I was asking about the photon states (or more generally the states of the quantum electromagnetic field, not all of which are usefully described as "photon" states) before any measurement is made. Of course any spin measurement will yield a definite value, but that's not the same as the field having been in an eigenstate of the operator corresponding to that spin measurement before the measurement was made.

The question I was asking can be rephrased as: given a quantum EM field state, must there always be some photon spin operator of which that state is an eigenstate (which would mean, in terms of measurements, that the measurement corresponding to that spin operator will always yield the same result with 100% probability if the field is in the given state)?

As a sidenote: I tend to avoid talking about eigenstates and the like for photons. As there are obviously no position eigenstates of photons defined in the standard way, this may open up a lot of pit traps. In my opinion, the Wigner function or maybe the Husimi Q function is a better description. As an added bonus, they also tell you something about the fields associated with the states.

With respect to the other question: No, of course there is no need for any light field to have some well-defined polarization. If you lack knowledge about the system and you have to describe it using a density matrix, you may describe unpolarized states. Whether this is also the case if you have all information about the system, which one might potentially have, is a different question that usually leads into interpretation teritory and I have no intention to go there. ;)
To briefly summarize what I mean: Considering the helicity basis, every fixed-phase superposition of the two helicity states gives you some well-defined polarization. Depending on the relative amplitude and phase of the two components, this might be a pretty odd polarization, but it is well-defined and you will always measure this polarization if you can up with a measurement that is appropriate for measuring exactly this polarization. You need to randomize this relative phase in order to arrive at completely unpolarized light. If you now consider the question whether this randomization is a consequence of lack of knowledge or whether it is fundamental, this is the path towards interpretation territory. As correctly mentioned by @vanhees71, this is what happens as well in experiments on polarization entanglement,where any individual photon of the pair will be found to be completely unpolarized. This pretty much extends to every optical process, where the vacuum state needs to be taken into account in a thorough treatment.

 

[vanhees71]

Sure, but of course photons have a well-defined Fock space and thus also well-defined single-photon (pure or mixed) state as any quantum(-field) theoretical description of an entity. In this case it's the description of the quantized electromagnetic field.

I guess what you are referring to is, how to get observable facts from such a state, and of course the first thing you think about is the description of detection of photons. Many photodetectors are based on the photoeffect, i.e., the interaction of the em. field with atoms/molecules in a solid (e.g., a CCD cam), and in this way indeed you get some "spatio-temporal information", but it's information on a detection event. Using the dipole approximation it turns out that it's described by the two-point electric-field current correlation function, and of course you can use the Wigner transform of this to evaluate all kinds of observable quantities. The most simple one is the detection probability for a photon for a given photon state, i.e., you get the probability to detect a photon at some location at a given time, where the location is determined by the place you put the detector (or the pixels of the CCD cam). Nowhere in this description an observable like "photon position" has been used, and this is indeed problematic, because a photon has no position observable to begin with. That's why one should not think about photons as little localized bullets. What's however well-defined are detection events, and if you have a single photon, it's always detected once or not at a point defined by the location of the detector (or which pixel "fired" in your CCD cam).

 

[Cthugha]

Maybe this follow-up discussion should be moved to a separate topic, as it is somewhat different from the initial question. Also in the following I will try to keep the language such that everybody can follow it. Please let me know if it gets too simplified somewhere.

Well, the position issue is one issue, but although it is the one most commonly raised, I do not think it is the most relevant one. Indeed, the question how to relate the formalism to experiments is relevant to me. To me the most relevant pit trap that many people are not aware of is that even in theory the photon number operator cannot be applied to all states and is not really an operator that can be realized in a measurement (in photon number measurements one usually destroys all photons, creates other excitations of the same number and counts these excitations, which is something very different) and even the annihilation operator may only be realized approximately if one has a rough idea about the mean photon number of the light field to investigate.

However, even in that case one clearly finds a significant difference between the photon number operator and standard measurement operators like position or momentum operators acting on e.g. a massive particle. The latter operators may always be applied, while there are situations, where the photon number operator and the photon annihilation operators simply cannot be applied. This is why I would loosely call them conditional operators. The reason for that is trivial and well known: $\hat{a}^\dagger \hat{a} |n\rangle=n|n\rangle$ does not apply to every n. It does not apply to the vacuum. $\hat{a}^\dagger\hat{a}|0\rangle=0$, not $0|0\rangle$. Due to the notation, people sometimes miss that 0 represents the empty set instead of the number 0 or a state with zero photons. This is the mathematical equivalent of "you cannot perform this operation". This is why I loosely called the annihilation operator a conditional operator from the experimental point of view. One may only apply it in a conditional manner, provided there is at least one photon present within the light field.

This has significant consequences. Consider any arbitrary light field $|l\rangle$. Due to the fact mentioned before, the operations $\hat{a}^\dagger|l\rangle$ and $\hat{a}^\dagger \hat{a}^\dagger\hat{a}|l\rangle$ will in many cases result in different final states - the latter state cannot have any contributions of photon number 1, while the first one can. This also allows one to play other fancy tricks. Prepare a thermal state of some well-defined photon number. Annihilate a photon and measure the photon number again. You will find that annihilating a photon actually double the mean photon number of any thermal state of the light field (which also reflects in $g^{(2)}(0)=2$ for the thermal light field). The reason is simply that 0 is the most probable photon number for any thermal state and applying the annihilation operator fully removes the vacuum contribution. Even simpler if you assume a state that contains 1000 photons with a probability of 1/1000 and 0 photons with a probability of 999/1000, you can increase the mean photon number of this state from 1 to 999 just by applying the annihilation operator. This is of course fully consistent and in line with theory. It is just non-intuitive and often overlooked.

This conditional nature of the operators also results in common misunderstandings about coherent states, which are eigenstates of the annihilation operator. I frequently hear the statement that coherent states in experiments are only approximations of coherent states in theory because the latter a eigenstates of the annihilation operator and as such one could annihilate an infinite amount of photons from them, which would be unphysical. This is of course not how the formalism should be interpreted. A coherent state has a finite probability to find zero photons and the meaning of coherent states being eigenstates of the annihilation operator is that if one is able to apply the annihilation operator, the probabilities to find each photon number from 0 to n do not change due to the absorption process. However, in any repetition there is a finite probability that the annihilation operator cannot be applied any further.

All of this is well-defined and fine, but one should be aware that these operators and states are pretty special and many people often neglect that and jump to wrong conclusions.

In order to return to the question why I prefer Wigner functions: Standrard experiments on photocount statistics usually measure the second-order correlation function of a light field. Loosely speaking, this is the conditional probability to detect a photon in mode b at time $\tau$ if one was already detected in mode a at time 0 (where in most cases a=b), normalized to light fields of the same mean intensity that are statistically independent of each other. This quantity is not even defined for the vacuum state as it is not possible to detect the first photon in the first place. From the theory point of view, both coherent and thermal light fields converge towards a vacuum state if the photon number goes to 0, which also shows that this is not a quantity that makes sense for a vacuum state. Calculating a Wigner function of the vacuum is, however, trivial and it can be measured as well by using homodyne detection as one needs to measure the field directly (and not the intensity/photon number) in order to get to the Wigner function.

 

[vanhees71]

Just to begin: I didn't understand your remark about the photon-number operator and the vacuum. Since $\hat{a}(p,\lambda)|\Omega \rangle =0$ for $\hat{n}(p,\lambda)|\Omega \rangle =0$, i.e., the vacuum state is the (unique) state with photon number 0, i.e., it's an eigen vector of all photon-number operators with eigenvalue 0. Also 0 does not represent the empty set but the number zero, and the vacuum state $|\Omega \rangle$ is not the zero-vector of course.

Then I'm a bit puzzled, what the problem is. Of course it's difficult to prepare a state with a definite number of photons. I'm not an expert in quantum optics. So I'm sure I'm missing something, but afaik an example for a single-photon preparation is to use an entangled photon pair from parametric down conversion and meaure one of them (which then indeed usually gets absorbed by the detector) and then being sure to have the other photon and in that way have prepared a single-photon Fock state. In the literature they call that "a heralded photon".

The last paragraph I think to fully understand, and I agree with it. That's what one can find in standard Quantum Optics textbooks like my favorite

J. Garrison, R. Chiao, Quantum optics, Oxford University Press, New York (2008).

 

[Cthugha]

Just to begin: I didn't understand your remark about the photon-number operator and the vacuum. Since $\hat{a}(p,\lambda)|\Omega \rangle =0$ for $\hat{n}(p,\lambda)|\Omega \rangle =0$, i.e., the vacuum state is the (unique) state with photon number 0, i.e., it's an eigen vector of all photon-number operators with eigenvalue 0. Also 0 does not represent the empty set but the number zero, and the vacuum state $|\Omega \rangle$ is not the zero-vector of course.

The annihilation operator is a ladder operator that performs the following operation: $\hat{a}|n\rangle=\sqrt{n}|n-1\rangle$. The vacuum state is the obvious exception as you do not get $\hat{a}|0\rangle=0|-1\rangle$, but $\hat{a}|0\rangle=0$. The ladder operator does not lead to any new state as there are obviously no excitations to annihilate in this state. Or put differently: The set of all states that one may get to by removing one photon from the vacuum state is the empty set.

Edit: Your statement that 0 denotes the number 0 still puzzles me. Usually, it is quite commonly accepted that in $\hat{a}|0\rangle=0$, the last 0 represents the zero vector of the vector space (iirc Sakurai called it the null ket in his book) and not the scalar 0. Or did this statement focus on the photon number 0? That one is of course indeed a scalar 0.

The problem is not that it is difficult to prepare single photons. That is not even that difficult unless you require them to be non-heralded. The "problem" is the very special role of the vacuum, which does not allow one to interpret the vacuum as an eigenstate of the photon number in the same manner as a Fock state containing a fixed number of excitations may be considered an eigenstate of the photon number operator. This problem does not show up for Fock states, but for most naturally occurring light fields, e.g. thermal states. Maybe the most thorough demonstration of the problem has been given in Science 317, 1890 (2007) by Parigi et al. You can still find the paper on the homepage of the group (https://sites.unimi.it/aqm/wp-content/uploads/Probing-Quantum-Commutation-Rules-Science.pdf). There, they clearly show that applying the experimental equivalent of the photon number operator to the thermal state increases the photon number significantly, simply because the vacuum contributions present initially are cut off.

If you do not see where the problem is, just try to apply the two sets of operators $\hat{a}^\dagger \hat{a}^\dagger \hat{a}$ and $\hat{a}^\dagger$ to any relevant state containing vacuum contributions, e.g. a thermal state. Do you agree that the results differ?

 

[PeterDonis]

The set of all states that one may get to by removing one photon from the vacuum state is the empty set.

Which in turn means that the set of all states that one may get to by applying the number operator to the vacuum state is also the empty set, correct? (Since applying the number operator means applying the annihilation operator, then applying the creation operator.) Which is why you are saying that it's not really true that the vacuum state is an eigenstate of the number operator with eigenvalue zero: that would require that you get the same state back when you apply the number operator, which can't be the case if the set of states you can get by applying the number operator is empty.

 

[Cthugha]

Which in turn means that the set of all states that one may get to by applying the number operator to the vacuum state is also the empty set, correct? (Since applying the number operator means applying the annihilation operator, then applying the creation operator.) Which is why you are saying that it's not really true that the vacuum state is an eigenstate of the number operator with eigenvalue zero: that would require that you get the same state back when you apply the number operator, which can't be the case if the set of states you can get by applying the number operator is empty.

Well, theory is of course consistent and you get an eigenstate with eigenvalue 0. However, as this is a zero ket, it will have exactly 0 weight in any experiment and whatever follow-up operation (such as photon creation) you perform, all of this processes will have 0 weight.

I just want to point out that photon number measurements usually do not implement the photon number operator. Instead one rather applies an operator that does something like $(\hat{c}^\dagger a)^n$. You destroy all of the n photons you have and instead create different excitations - usually photoelectrons or something like that, so strictly speaking the creation operator $\hat{c}^\dagger$ will create a photoelectron in a slightly different mode for every photon. This is obviously not an implementation of the photon number operator as the photons are gone afterwards.

If one wants to implement the photon number operator, this is nontrivial and one has to do it in the manner as outlined in the manuscript above. However, from the experimental point of view, it is hard to implement the annihilation operator. You want to annihilate exactly one photon, not 2 photons or all of them, but exactly one. The usual way - as in the manuscript I linked - to implement the operator is a beam splitter, usually a glass plate, that splits off only a tiny fraction of the beam followed by a single-photon sensitive detector that verifies the absorption process. If the photon number in the original field is low enough, the probability that two or more photons were reflected by the beam splitter simultaneously may be made negligible.

Assuming for simplicity that the state could be described by a diagonal density matrix before, where the probability to have n photons was given by $p_n$, the probabilities to find n' photons in the photon-subtracted light field will be given by $p_{n'}=p_n|\langle n'|\hat{a}|n\rangle|^2$. This obviously only yields contributions for n'=n-1 and you get $p_{n'=n-1}=p_n n$. So, the weight of finding n photons in the photon-subtracted state is given by the probability that you had n+1 photons in the initial state times the number of photons. This makes sense as the operator you applied is the single photon annihilation operator and when placing a single beam splitter even if you had the same initial probability to find e.g. 1 or 5 photons inside your beam, it will happen 5 times as often that a single photon gets reflected from the n=5 state as compared to the n=1 state. Thus, the detector will also click 5 times as often. This rescaling of probabilities is at the core of many non-intuitive effects in quantum optics, such as doubling of the photon number inside a thermal beam if you subtract a photon. However, it also means that the vacuum contribution of the initial state will not contribute anything here. The probability that one photon gets reflected from the beam splitter resulting in a detector click is of course exactly 0.

This was the implementation of the annihilation operator. Implementing the creation operator now would be "simple". One would just have to add a single photon source that adds a photon to be the beam as soon as the detector clicked. In the manuscript I linked before this was done in a stochastic manner using heralding an an entangled photon pair. When this is done, you of course get the correct photon numbers out of it. However, in terms of finding eigenstates in a measurement and therefore in terms of state preparation, the behavior is somewhat different. If I consider a standard experiment such as a spin measurement or a position measurement on some mixed state, I expect the different eigenvalues to show up with a relative probability given by their weights. This is not as easy for the photon number operator. Here, for example, the probability to find the system in the vacuum state after applying the photon number operator is a constant 0. This is of course fully consistent, but has important consequences for experiments and e.g. preparation of states.

 

[vanhees71]

No, the vacuum state is the unique state with eigenvalue 0 for all photon-number operators. Of course
$$\hat{a}^{\dagger} \hat{a} |\Omega \rangle=0 |\Omega \rangle=0.$$
An eigenvector $\hat{O}$ with eigenvalue 0 is mapped to the zero-vector since the eigenvalue equation reads
$$\hat{O} |\lambda \rangle=\lambda |\lambda \rangle.$$
If $\lambda=0$ it's maped to the 0-vector and this equation is true. On the other hand, by definition, the zero-vector itself is not an eigenvector of any operator.

I've never seen any textbook, where this is defined otherwise, and anything else wouldn't make sense. You don't map anything to a set either. The operators in a free QFT operate in the corresponding Fock space.

 

[vanhees71]

Well, theory is of course consistent and you get an eigenstate with eigenvalue 0. However, as this is a zero ket, it will have exactly 0 weight in any experiment and whatever follow-up operation (such as photon creation) you perform, all of this processes will have 0 weight.

I just want to point out that photon number measurements usually do not implement the photon number operator. Instead one rather applies an operator that does something like $(\hat{c}^\dagger a)^n$. You destroy all of the n photons you have and instead create different excitations - usually photoelectrons or something like that, so strictly speaking the creation operator $\hat{c}^\dagger$ will create a photoelectron in a slightly different mode for every photon. This is obviously not an implementation of the photon number operator as the photons are gone afterwards.

If one wants to implement the photon number operator, this is nontrivial and one has to do it in the manner as outlined in the manuscript above. However, from the experimental point of view, it is hard to implement the annihilation operator. You want to annihilate exactly one photon, not 2 photons or all of them, but exactly one. The usual way - as in the manuscript I linked - to implement the operator is a beam splitter, usually a glass plate, that splits off only a tiny fraction of the beam followed by a single-photon sensitive detector that verifies the absorption process. If the photon number in the original field is low enough, the probability that two or more photons were reflected by the beam splitter simultaneously may be made negligible.

Assuming for simplicity that the state could be described by a diagonal density matrix before, where the probability to have n photons was given by $p_n$, the probabilities to find n' photons in the photon-subtracted light field will be given by $p_{n'}=p_n|\langle n'|\hat{a}|n\rangle|^2$. This obviously only yields contributions for n'=n-1 and you get $p_{n'=n-1}=p_n n$. So, the weight of finding n photons in the photon-subtracted state is given by the probability that you had n+1 photons in the initial state times the number of photons. This makes sense as the operator you applied is the single photon annihilation operator and when placing a single beam splitter even if you had the same initial probability to find e.g. 1 or 5 photons inside your beam, it will happen 5 times as often that a single photon gets reflected from the n=5 state as compared to the n=1 state. Thus, the detector will also click 5 times as often. This rescaling of probabilities is at the core of many non-intuitive effects in quantum optics, such as doubling of the photon number inside a thermal beam if you subtract a photon. However, it also means that the vacuum contribution of the initial state will not contribute anything here. The probability that one photon gets reflected from the beam splitter resulting in a detector click is of course exactly 0.

This was the implementation of the annihilation operator. Implementing the creation operator now would be "simple". One would just have to add a single photon source that adds a photon to be the beam as soon as the detector clicked. In the manuscript I linked before this was done in a stochastic manner using heralding an an entangled photon pair. When this is done, you of course get the correct photon numbers out of it. However, in terms of finding eigenstates in a measurement and therefore in terms of state preparation, the behavior is somewhat different. If I consider a standard experiment such as a spin measurement or a position measurement on some mixed state, I expect the different eigenvalues to show up with a relative probability given by their weights. This is not as easy for the photon number operator. Here, for example, the probability to find the system in the vacuum state after applying the photon number operator is a constant 0. This is of course fully consistent, but has important consequences for experiments and e.g. preparation of states.

Again: The vacuum state is not the zero-vector! The notation $|0 \rangle$ is misleading, which is why I always write $|\Omega \rangle$ for it.

It's also not true that the vacuum has 0 weight in any experiment. E.g., if you have a coherent state $|\alpha \}$, then there's always the possibility that you don't detect a photon, because in the coherent state the vacuum state is always in the representation in terms of Fock states:
$$\langle \Omega|\alpha \rangle=\exp(-|\alpha|^2/2),$$
i.e., the probability to detect no photon is $\exp(-|\alpha|^2/2) \neq 0$.

 

[Cthugha]

Again: The vacuum state is not the zero-vector! The notation $|0 \rangle$ is misleading, which is why I always write $|\Omega \rangle$ for it.

You do not seem to read my posts carefully. My claim is not that the vacuum state is the zero vector, but that you get the zero vector from applying the annihilation operator to the vacuum. I am a bit puzzled because this is pretty trivial. It is explicitly mentioned in quantum theory: concepts and methods by Peres on page 138 and this is also what Sakurai denotes as the null ket in his book on modern quantum mechanics. Unfortunately, I am obviously not really supposed to visit the library of our chair at current. Otherwise I could dig up the exact wording in this and other books.

It's also not true that the vacuum has 0 weight in any experiment. E.g., if you have a coherent state $|\alpha \}$, then there's always the possibility that you don't detect a photon, because in the coherent state the vacuum state is always in the representation in terms of Fock states:
$$\langle \Omega|\alpha \rangle=\exp(-|\alpha|^2/2),$$
i.e., the probability to detect no photon is $\exp(-|\alpha|^2/2) \neq 0$.

Again, this is absolutely not related to my post. Note that the discussion started from the complex meaning of what can be considered an eigenstate for light fields. What you describe is the overlap of a coherent state with the vacuum state, which is given by $\langle 0|\alpha\rangle$. This is related to the probability to have a zero photon Fock state present inside the coherent state. You can measure it using any destructive operator that annihilates any photons present, e.g. a CCD. This is obviously not an experiment that leaves the state of the light field unchanged if you have a Fock state arriving at the CCD and it is not what applying the photon number operator gives you. What I describe is NOT the overlap between the two states, but the actual result of $\langle 0|\hat{a}^\dagger\hat{a}|\alpha\rangle$, which is 0. Or simply speaking: When applying the photon number operator to an arbitrary light field, there is no way to herald the vacuum. This is expressed in the paper I linked by "When the initial number of particles is precisely known, the quantum and the classical cases give exactly the same results for arbitrary sequences of additions and subtractions. However, the situation changes completely for general superpositions or mixtures of Fock states. After the operation of particle subtraction, the average number of quantum particles in certain states may unexpectedly grow instead of diminishing, as one would be naturally tempted to expect."

I am a bit puzzled and somehow we must talk past each other because this is really elementary quantum optics stuff. Let me emphasize once more that the focus is on preparing (eigen)states, not measuring photon numbers. These are different animals for light fields. When talking about the problem to relate theory to experiment with respect to photons as eigenstates of the photon number operator, I am not talking about difficulties in creating faithful Fock states experimentally, but about difficulties in implementing the photon number operator faithfully. Any implementation will end up in a conditional measurement. In case you disagree: In your opinion, what would a faithful experimental implementation of an unconditional measurement that corresponds to applying the photon number operator (or even just the annihilation operator) look like?

 

[PeterDonis]

from the experimental point of view, it is hard to implement the annihilation operator.

Isn't the annihilation operator non-Hermitian? Wouldn't that mean it's impossible to have an exact physical realization of it? (Or perhaps by "implement" here you only mean "implement the best possible approximation to"?)

 

[Cthugha]

Isn't the annihilation operator non-Hermitian? Wouldn't that mean it's impossible to have an exact physical realization of it? (Or perhaps by "implement" here you only mean "implement the best possible approximation to"?)

Yup, unlike the photon number operator, the annihilation operator is non-Hermitian. Still, it has eigenstates. Coherent states are eigenstates of the annihilation operator with complex-valued eigenvalues corresponding to the complex amplitude of a light field in the classical limit. This results in some not-so-nice features. The basis of coherent states is not orthogonal and overcomplete. So you can expand every coherent state in terms of different coherent states.
And: sure, as a consequence only approximations may be realized in the lab. The process becomes necessarily non-deterministic, which is at the heart of the problem. People have actually quantified how well this works. For example quantum process tomography of photon addition and subtraction has been performed in the group of Alex Lvovsky when he was still back in Calgary. See the following PhD thesis: https://pdfs.semanticscholar.org/0d5d/1b7ee43b0fc7298e3efc13cdf2c73f67d2b7.pdf

 

[vanhees71]

You do not seem to read my posts carefully. My claim is not that the vacuum state is the zero vector, but that you get the zero vector from applying the annihilation operator to the vacuum. I am a bit puzzled because this is pretty trivial. It is explicitly mentioned in quantum theory: concepts and methods by Peres on page 138 and this is also what Sakurai denotes as the null ket in his book on modern quantum mechanics. Unfortunately, I am obviously not really supposed to visit the library of our chair at current. Otherwise I could dig up the exact wording in this and other books.

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).

 

[vanhees71]

Isn't the annihilation operator non-Hermitian? Wouldn't that mean it's impossible to have an exact physical realization of it? (Or perhaps by "implement" here you only mean "implement the best possible approximation to"?)

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

 

[PeterDonis]

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

I can see how it's easy to annihilate an entire light beam. But that's not the same as annihilating exactly one photon from a light beam that contains a lot of them, or from a coherent state that's not even an eigenstate of the photon number operator.

 

[vanhees71]

True, 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.

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. Another example is black-body radiation. Of course it's very difficult (if not impossible) to just absorb one and only one photon with certainty from such a state.

 

[PeterDonis]

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.

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.

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

Exactly.

 

[Cthugha]

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?

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).

 

[vanhees71]

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.

 

[Cthugha]

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.