Single Photon States: Definition & Real-Life Applications

[LittleSchwinger]

A photon in QFT is defined in the same way as all particles. That is they denote a set of quantum states that transform in the simplest possible way under Poincaré transformations. Properly this is known as an irreducible representation (irrep) of the Poincaré group. You can classify these irreps via their mass and spin, but in the case where the mass is zero you classify by helicity (spin along the direction of momentum).

In real life no state exactly corresponds to these particle states, but they are an extremely useful idealisation for calculations.

 

[vanhees71]

Nowadays the preparation of true one-photon states is standard in any quantum optician's lab. One example is the use of parametric downconversion, where you shoot with a laser at a birefringent crystal like beta-Barium Borat (BBO), from which you get entangled photon pairs. Using one of these photons you know that there's precisely one other photon, which makes a "heralded-photon source". In this sense single-photon states are today as real as any other source of electromagnetic waves.

 

[LittleSchwinger]

Using one of these photons you know that there's precisely one other photon, which makes a "heralded-photon source". In this sense single-photon states are today as real as any other source of electromagnetic waves.

"Strictly" speaking the state is not purely a one-photon state. There are still non-zero contributions from higher number states. However the state can be prepared in such a way that the coefficients on higher order terms can be made so small as to be effectively zero for practical purposes. This is essentially what I meant by an "idealisation" above.

Again very strictly speaking it's a theorem* in QFT that you can't prepare an eigenstate of the photon number operator, just arbitrarily approach one, with errors of the order of an inverse power of the size of the preparation device.

However the spirit of what you're saying is true.

*There are many equivalent formulations of this result. Malament's theorem, the Reeh-Schlieder theorem, etc.

 

[vanhees71]

No these are true one-photon states. See

https://en.wikipedia.org/wiki/Single-photon_source

 

[PeterDonis]

No these are true one-photon states. See

https://en.wikipedia.org/wiki/Single-photon_source

Your reference says that exact single photon states cannot be prepared because of the Heisenberg uncertainty principle. Which I think is the substance of what @LittleSchwinger is saying.

 

[LittleSchwinger]

No these are true one-photon states. See

https://en.wikipedia.org/wiki/Single-photon_source

Basically take this line:

In this context, a single-photon source gives rise to an effectively one-photon number state

So it's effectively one-photon for all intents and purposes, but strictly speaking if you modelled it in QFT there would be a non-zero "tail" to the state that overlaps with higher number states. You can't prepare an exact photon number eigenstate in QFT due to the Reeh-Schleider theorem, which sort of functions as a "step up" from the Heisenberg principle in QFT.

Of course for practical purposes in a lab setting it might as well be a $N = 1$ eigenstate.

 

[vanhees71]

Of course the real single photon state has a finite spectral width, but it's a single-photon Fock state, demonstrated by (almost) perfect anti-bunching in the HOM experiment.

 

[PeterDonis]

it's (almost) a single-photon Fock state, demonstrated by (almost) perfect anti-bunching in the HOM experiment

See the bolded addition above, which corresponds to the same word in parentheses further on in the same sentence. That's all I take @LittleSchwinger to be saying.

 

[LittleSchwinger]

Of course the real single photon state has a finite spectral width, but it's a single-photon Fock state, demonstrated by (almost) perfect anti-bunching in the HOM experiment.

It has a finite spectral width, but it also isn't exactly a single photon state. There's a very small overlap with higher photon number states. See section V.B. of this paper:
https://arxiv.org/abs/quant-ph/0212023
if you want a bit more info and a link to the proofs.

Again for any practical purpose in a lab my point is pure pedantry, but it's just to explain partly why any particle state is an idealisation.

 

[PeterDonis]

It has a finite spectral width, but it also isn't exactly a single photon state.

How can one even have a "single photon state" with "finite spectral width"? The definition of a "single photon state" (Fock state) includes that it is a state with a single specific frequency, or, to put it another way, a state that can be created from the vacuum state by applying a single specific creation operator. "Finite spectral width" implies multiple frequencies, each of which would require a different creation operator applied to the vacuum state.

 

[LittleSchwinger]

How can one even have a "single photon state" with "finite spectral width"? The definition of a "single photon state" (Fock state) includes that it is a state with a single specific frequency, or, to put it another way, a state that can be created from the vacuum state by applying a single specific creation operator. "Finite spectral width" implies multiple frequencies, each of which would require a different creation operator applied to the vacuum state.

It would be a superposition of states with various frequencies, which is still an element of the single-photon state space.

 

[PeterDonis]

It would be a superposition of states with various frequencies, which is still an element of the single-photon state space.

I see what you mean, but I'm not sure how useful this distinction is. If I apply two different creation operators to the vacuum state, it seems like I get two different "photons"; but if I form a linear combination of them, they're now a "single photon with a finite spectral width"? Whereas if I now mix in any state that results from applying more than one creation operator to the vacuum state, I can no longer call it a "single photon", but only "almost" a single photon (if I only mix in a very small amount of states that come from more than one creation operator)?

This might be meaningful if I had a way to prepare states that were purely "one creation operator" states, even if I couldn't control the preparation enough to have the result be 100% coming from one creation operator. But my lack of control over the preparation process affects both types of variation: I can't exactly control either the number of creation operators or their frequency. So any state I can actually prepare is going to have both multiple creation operators and multiple numbers of creation operators. And that being the case, I'm not sure I see the point of reserving the term "single photon" for just states that are entirely "one creation operator" states. IMO ither that term should apply to the states I can actually prepare (with an "almost" qualifier if you like, but with the qualifier applying to both the "number" and "frequency" aspects), or it shouldn't be used at all since it causes more confusion than it solves.

 

[LittleSchwinger]

I see what you mean, but I'm not sure how useful this distinction is

Mathematically it's a fairly clear distinction.

States of the form:
$(\int{f(p)a^{\dagger}(p)dp})\Omega$
have an associated eigenvalue of $1$ for the number operator and would be certain to cause a single click in a detector. In rigorous treatments one says they give a singular response to probes (see Haag's book).
I'm ignoring details concerning the function space $f(p)$ must be drawn from and the exact form of the measure $dp$.

States with two creation operators:
$(\int{f(p,q)a^{\dagger}(p)a^{\dagger}(q)dqdp})\Omega$
give an eigenvalue of $2$ and thus are sure to cause two click events and so forth.

This is entirely separate from spectral width which is the "spread" of frequency/wavelength/etc detections.

So a superposition of single photon states can have various frequencies detected, but always only one "click" in a probe.

 

[PeterDonis]

Mathematically it's a fairly clear distinction.

Of course I understand the mathematical distinction; I described it myself.

My point is that physics is not math.

So a superposition of single photon states can have various frequencies detected, but always only one "click" in a probe.

Yes, but we can never prepare such a state; any state we can actually prepare will have a nonzero probability for zero or two or three or four or... clicks.

Also, your description of the math paid much less attention to the response to a different operator, an operator that distinguishes frequencies. What you are calling a "single photon with a finite spectral width" is not an eigenstate of this operator; but a state formed by applying the same creation operator to the vacuum twice is. If I wanted to focus on the "number of creation operators" aspect of the math instead of the "which creation operator" aspect, I could insist that we call states that only contain a single creation operator "single frequency" states, and refuse to allow that term to be used for what you call a "single photon with a finite spectral width", even if the spectral width was extremely small--you would have to insert an "almost" qualifier, as in "almost a single frequency". And I could call a state that was a superposition of states obtained by applying zero, one, two, three... creation operators of the same frequency to the vacuum a "single frequency state with finite number width". All this would be just as justified by the math as the terminology you are using.

In actual practice, we find it far easier to make "number" detectors than "frequency" detectors, so we tend to focus on the "number" aspect and give it precedence. But the physics itself does not do that, nor does the math; the full Hilbert space accessible from the vacuum includes both kinds of variation, and the states we can actually prepare in practice always contain both kinds of variation.

 

[LittleSchwinger]

I'm not exactly sure of the point you are making, but this is my attempt.

Call frequency $\nu$ and number $N$. All actual preparations have a spread in $\nu$ and $N$. However the idealisation of no spread in $N$, a so called n-photon state is clearly a separate idealisation from no spread in $\nu$.

A single photon state involving a superposition over several frequencies will always have one click in each run, just the frequency will vary between runs. This is clearly different from a state with two detection clicks in each run, i.e. a two photon state.

Take the case of a superposition of two single photon states with two different frequencies. Here we will always have one detection, it will just vary probabilistically what frequency is found. A corresponding two photon state with the same frequencies as above will with certainty result in a detection of two counts, one for each frequency, in a given run.

 

[PeterDonis]

the idealisation of no spread in $N$, a so called n-photon state is clearly a separate idealisation from no spread in $\nu$.

Yes, certainly.

A single photon state involving a superposition over several frequencies will always have one click in each run, just the frequency will vary between runs. This is clearly different from a state with two detection clicks in each run, i.e. a two photon state.

But I can equally well rephrase this as follows: a single frequency state involving a superposition over several numbers will always detect one frequency per run, just the number will vary between runs. This is clearly different from a state with two frequencies detected in each run.

Take the case of a superposition of two single photon states with two different frequencies. Here we will always have one detection, it will just vary probabilistically what frequency is found.

And again I can equally well rephrase this as: take the case of a superposition of two single frequency states with two different numbers. Here we will always detect one frequency, but the number we detect will vary probabilistically.

My point is that the physics and the math are the same either way; all we are doing is changing which aspect (number or frequency) we give precedence to in our descriptions. This might well be driven by the practical fact that, as I said, we can make number detectors far more easily than frequency detectors. But with regard to the original question of this thread, since, as you agree, any actual state we can prepare will contain both multiple frequencies and multiple numbers, any description of such a state as "a single photon" will have to be qualified by having a finite "width" in both aspects.

 

[LittleSchwinger]

But I can equally well rephrase this as follows: a single frequency state involving a superposition over several numbers will always detect one frequency per run, just the number will vary between runs. This is clearly different from a state with two frequencies detected in each run

Yes.

And again I can equally well rephrase this as: take the case of a superposition of two single frequency states with two different numbers. Here we will always detect one frequency, but the number we detect will vary probabilistically.

Yes.

What are you getting at? Sorry I really don't understand. I mean yes a monofrequency state could vary in number, but I'm not sure what the problem or issue is.

 

[PeterDonis]

What are you getting at?

The fact that you answered "yes" to both of the things you quoted from me just now.

I'm not sure what the problem or issue is.

As you are using the term "single photon state", it is impossible for us to actually prepare a single photon state. Any state we prepare will have some nonzero probability for a number of photon detections (clicks) other than one, which by your definition means it isn't a "single photon state". But neither you nor @vanhees71 have been adding any qualifiers about that; you both seem ok with just calling them "single photon states" and ignoring the (very small) probability of some other number, except when pressed into using strict terminology.

Yet when it comes to frequency, both you and @vanhees71 have been adding a qualifier: "finite spectral width". But this doesn't even affect the number of photon detections. So I'm confused about why there has to be a qualifier added for variation in frequency, but no qualifier needs to be added for variation in number. On the face of it it would seem like the opposite should be true. Or (my preference), since any state we can actually prepare will have both kinds of variation, one could simply either add a qualifier for both, or ignore the qualifier for both (unless pressed into using strict terminology).

 

[LittleSchwinger]

you both seem ok with just calling them "single photon states" and ignoring the (very small) probability of some other number

Because that's the standard name for those Fock space states, surely I don't have to justify standard terminology.
Also I don't see how I'm ignoring the tail of very small probabilities, I'm the one who introduced it and mentioned it a few times. Surely the fact that we can't actually prepare a single photon state is conveyed by my calling them an "idealisation". In fact that was my point above, no actual real preparation is a single-photon state:

So it's effectively one-photon for all intents and purposes, but strictly speaking if you modelled it in QFT there would be a non-zero "tail" to the state that overlaps with higher number states. You can't prepare an exact photon number eigenstate in QFT due to the Reeh-Schleider theorem

I really don't see how I'm ignoring this. I'm explicitly talking about it.

have been adding a qualifier: "finite spectral width"

I really don't follow. We are saying the states have a finite spectral width, because there is a non-zero spread of frequencies.

 

[PeterDonis]

We are saying the states have a finite spectral width, because there is a non-zero spread of frequencies.

Yes, but you aren't (unless pressed) saying that the states have a "finite number width" because there is a non-zero spread of numbers. Why not?

 

[LittleSchwinger]

Yes, but you aren't (unless pressed) saying that the states have a "finite number width" because there is a non-zero spread of numbers. Why not?

But I did say that, in post #1:

In real life no state exactly corresponds to these particle states, but they are an extremely useful idealisation for calculations.

and then said more in #3:

"Strictly" speaking the state is not purely a one-photon state. There are still non-zero contributions from higher number states. However the state can be prepared in such a way that the coefficients on higher order terms can be made so small as to be effectively zero for practical purposes. This is essentially what I meant by an "idealisation" above.

I don't know how I'm not saying it, when I'm explicitly the one who introduced the concept of all states having a spread in $N$, mentioned the theorem its a consequence of and gave a reference for it.
Seriously I don't understand what's going on here!

 

[PeterDonis]

I don't understand what's going on here!

You're not the only one posting in this thread. You said you were using standard terminology. Presumably @vanhees71 is too. Here is what he said:

No these are true one-photon states.

Of course the real single photon state has a finite spectral width, but it's a single-photon Fock state

But of course it isn't exactly a "single-photon Fock state", as you pointed out in response to him.

So maybe my confusion is about what "standard terminology" is. Is it "standard terminology" to add a "finite spectral width" qualifier when describing states, but not a "finite number width" qualifier? That seems to be what @vanhees71 was doing.

 

[LittleSchwinger]

I think @vanhees71 just wasn't aware of this particular corollary of the Reeh-Schleider theorem, i.e. that it's not possible to exactly prepare a state with a well-defined photon count.

 

[LittleSchwinger]

Is it "standard terminology" to add a "finite spectral width" qualifier when describing states, but not a "finite number width" qualifier?

Well it's standard terminology to add both, but in a state where the width is extremely small in either case there would be a tendency not to add it, i.e. a mono or single frequency state, a n-photon state and so forth.

However I still don't get this:

Yes, but you aren't (unless pressed) saying that the states have a "finite number width" because there is a non-zero spread of numbers. Why not?

Wasn't I saying that all along?

 

[PeterDonis]

it's standard terminology to add both

I think @vanhees71 just wasn't aware of this particular corollary of the Reeh-Schleider theorem, i.e. that it's not possible to prepare a state with a well-defined photon count.

Hm. I'll let @vanhees71 respond on that.

Wasn't I saying that all along?

I should have clarified that I was referring more to @vanhees71's posts than yours with that particular comment.

 

[LittleSchwinger]

Hm. I'll let @vanhees71 respond on that.

Well a different way of saying it might be that in real life you can prepare a state so close to a single photon state that my point becomes pedantic from a pragmatic perspective and thus most will just say its a "one photon state".

 

[PeterDonis]

a different way of saying it might be that in real life you can prepare a state so close to a single photon state that my point becomes pedantic from a pragmatic perspective and thus most will just say its a "one photon state".

And I would be fine with that: such a state will have both types of variation, but they're both so small that they can be ignored for practical purposes.

 

[LittleSchwinger]

And I would be fine with that: such a state will have both types of variation, but they're both so small that they can be ignored for practical purposes.

Just to be clear "one photon" would be applied if the number variation is small, "single frequency" if the frequency variation is small. If both are small you might see something like a "monochromatic single photon state".

 

[vanhees71]

You're not the only one posting in this thread. You said you were using standard terminology. Presumably @vanhees71 is too. Here is what he said:
But of course it isn't exactly a "single-photon Fock state", as you pointed out in response to him.

So maybe my confusion is about what "standard terminology" is. Is it "standard terminology" to add a "finite spectral width" qualifier when describing states, but not a "finite number width" qualifier? That seems to be what @vanhees71 was doing.

I don't think that the Reeh-Schlieder or Haag's theorem are very important concerning relativistic QFT as a physical theory describing phenomenology. Of course, these mathematical theorems make it clear that in fact relativistic QFT of interacting particles is not mathematically strictly defined, and what's used with great success is renormalized perturbative QFT, where these theorems are irrelevant. They are of course highly relevant in the investigation of the mathematical properties of QFTs in the sense of algebraic or axiomatic approaches.

It's also clear that strictly speaking the "photon number" is not an observable to begin with, because it's not a gauge-covariant quantity. What's observable are detector clicks due to the presence of the electromagnetic field. What can be prepared are photon states, which are as close to single-photon states as you want to get them.

It's as in non-relativistic quantum mechanics with momentum (and energy for free particles) eigenstates: They are no true states, and you cannot strictly speaking prepare any particle in a momentum eigenstate, because the plane wave $\exp(\mathrm{i} \vec{x} \cdot \vec{p})$ is not square integrable and thus no representant of a Hilbert-space vector in position representation. What you can, however, prepare are "wave packets" with an arbitrarily sharp momentum.

The same is true for "single-photon" states. Of course the plane-wave momentum eigenstate (in quantum-optics lingo a "single-mode one-photon Fock state" is not a true state too, but you can build normlizable one-photon Fock states in the form
$$|\Psi \rangle=\int \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \hat{a}^{\dagger}(\vec{p},\lambda) |\Omega \rangle,$$
where $A(\vec{p})$ is square integrable. The creation-annihilation operators obey the (non-covariant) commutation relations $[\hat{a}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'}$; $\lambda \in \{1,-1\}$ denotes the helicity eigenstates . All this is in the Coulomb gauge for free fields which leads to transverse four-potential fields, $A^0=0$, $\vec{\nabla} \cdot \vec{A}=$ (which completely fixes the gauge at the cost of lost manifest Poincare invariance).

In this convention the total number operator for free photons is
$$\hat{N}=\sum_{\lambda} \int_{\mathrm{R}^3} \mathrm{d}^3 k \hat{a}^{\dagger}(\vec{k},\lambda') \hat{a}(\vec{p},\lambda').$$
Now we have
$$\hat{N} |\Psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda'} \hat{a}^{\dagger}(\vec{k},\lambda') \hat{a}(\vec{k},\lambda') \hat{a}^{\dagger} (\vec{p},\lambda) |\Omega \rangle.$$
Since $\hat{\vec{k},\lambda'}|\Omega \rangle=0$ we can as well write
$$\hat{N} |\Psi \rangle= \int \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} A(\vec{p}) \int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda'} \hat{a}^{\dagger}(\vec{k},\lambda') [\hat{a}(\vec{k},\lambda'), \hat{a}^{\dagger} (\vec{p},\lambda)] |\Omega \rangle.$$
and use the commutator relation
$$[\hat{a}(\vec{k},\lambda'), \hat{a}^{\dagger} (\vec{p},\lambda)]=\delta^{(3)}(\vec{p}-\vec{k}) \delta_{\lambda \lambda'}$$
to get
$$\hat{N} |\Psi \rangle = |\Psi \rangle,$$
i.e., you have a single-photon state, which is normalized to 1, $\langle \psi|\psi \rangle$.

Of course there's no contradiction to the Reeh-Schlieder theorem since this is not a "localized state" in any sense. Photons can't be localized in any strict sense since they don't even admit the definition of a position operator.

Of course you can't measure the photon number $N$ but only true observables. AFAIK the first preparation of single-photon states have been realized by Clauser in 1973. Of course there a true observable was measured (what else can you meausure), i.e., the registration of the single photons by photo multipliers and clearly the expected non-classicality of the photo effect for such single-photon states have been demonstrated:

https://doi.org/10.1103/PhysRevD.9.853
https://escholarship.org/content/qt3wm0v847/qt3wm0v847.pdf?t=p0lmks (for a freely downloadable scan of the preprint)

 

[LittleSchwinger]

I don't think that the Reeh-Schlieder or Haag's theorem are very important concerning relativistic QFT as a physical theory describing phenomenology. Of course, these mathematical theorems make it clear that in fact relativistic QFT of interacting particles is not mathematically strictly defined, and what's used with great success is renormalized perturbative QFT, where these theorems are irrelevant

That's not true I think. The Reeh-Schlieder theorem is not the same type of theorem as Haag's theorem.
Haag's theorem shows that the free and interacting theories are not related by a unitary transformation when there is no volume cutoff.
The Reeh-Schlieder theorem however is a theorem about what kinds of states can be prepared in a finite volume. It remains true in renormalized perturbative QFT as well and in fact there are papers calculating the entropy of entanglement in perturbative renormalized QFT that show it has the form the Reeh-Schlieder theorem predicts.

The Reeh-Schlieder theorem is more like the CPT theorem or various analyticity theorems for correlation functions, i.e. something first formally proved in the axiomatic approach but still clearly true in regular QFT. It's not really related to difficulties in rigorously defining QFT like Haag's theorem is.

It's also clear that strictly speaking the "photon number" is not an observable to begin with, because it's not a gauge-covariant quantity

Yeah that's why I mentioned the "Probe" treatment above.

 

[vanhees71]

But what are you making of the above constructed normalizable state which is an eigenstate of eigenvalue 1 of the photon-number operator? I don't think that it contradicts in any way the Reeh-Schlieder theorem, and of course it can't, because it's free-field theory, which is well-defined. I think the Reeh-Schlieder theorem simply doesn't apply here, because it's precisely not a state that describes a "localized photon". I think it's just the mathematical rigorous version of the old gedanken experiment by Rosen and Bohr about the impossibility to localize relativistic quanta: Trying to confine even massive particles you rather create new particle-antiparticle pairs than localizing the single particle to a small volume. That's the more true for photons, which as massless spin-1 quanta don't even admit a position observable.

Also what then do you think are the "single photons" used by Clauser in the above quoted historical experiment or the "heralded single photons" in modern preparations using parametric down conversion, all of which clearly show the "single-photon features" like antibunching in the HOM experiment, etc?

 

[LittleSchwinger]

I don't think that it contradicts in any way the Reeh-Schlieder theorem, and of course it can't, because it's free-field theory, which is well-defined. I think the Reeh-Schlieder theorem simply doesn't apply here, because it's precisely not a state that describes a "localized photon"

I agree.

That state is a perfectly valid Fock state, but no realistic preparation procedure in a finite volume can prepare it, i.e. it's an "idealization".

Just to be clear to others. Single photon states definitely exist mathematically as valid states, it's just that no preparation procedure in a finite volume can prepare them.

One can take the idealized limit where the preparation device is infinitely large and then these states can be prepared, but of course this isn't a real procedure. And of course in the actual de Sitter cosmology of real life this limit can't be taken, so we must also be using the idealization of Minkowski space.

I think it's just the mathematical rigorous version of the old gedanken experiment by Rosen and Bohr about the impossibility to localize relativistic quanta: Trying to confine even massive particles you rather create new particle-antiparticle pairs than localizing the single particle to a small volume. That's the more true for photons, which as massless spin-1 quanta don't even admit a position observable

Correct. And I just want to say, that Bohr-Rosen work is a phenomenal paper. I don't know if you've ever seen Freeman Dyson's thoughts on how it relates to gravity. Interesting food for thought.

Regardless yes, the Reeh-Schlieder theorem is nothing but a "rigorous" version of that work.

Also what then do you think are the "single photons" used by Clauser in the above quoted historical experiment or the "heralded single photons" in modern preparations using parametric down conversion, all of which clearly show the "single-photon features" like antibunching in the HOM experiment, etc?

States arbitrarily close to single photon states, so that they effectively are single photon states FAPP. My point is of course complete pedantry that would have me kicked out of an actual lab

 

[vanhees71]

This I fully agree with. For me the best heuristic introduction to relavistic QFT I've seen yet is in S. Coleman's Lectures:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371