I Two single-photon Fock states vs one two-photon Fock state

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1. Oct 11, 2016

maxverywell

What is the difference between two single-photon Fock states $|1\rangle |1\rangle$ and one two-photon Fock state $|2\rangle$ (all in the same mode)? In both cases the mean photon number is 2. How do we distinguish them experimentally?

2. Oct 11, 2016

A. Neumaier

The first is two photons at two different times, the second is a pair of simultaneous photons. They give different statistics when observed sufficiently often.

3. Oct 11, 2016

vanhees71

I don't understand, what you are saying. If you have the number basis (aka Fock basis), the correct notation is
$$|\{N(\vec{k},\lambda)\}_{\vec{k} \in\mathbb{R}^3,\lambda \in \{-1,1 \}}.$$
This basis is time-independent (in the Heisenberg picture of free photons), because the $\hat{N}(\vec{k},\lambda)$ are conserved, i.e., commute with $\hat{H}$. Then there's one two-photon single-mode Fock state of the form
$$|\{N(\vec{k}_1,\lambda_1)=2, \quad N(\vec{k},\lambda)=0 \rangle=\frac{1}{\sqrt{2}} \hat{a}^{\dagger 2}(\vec{k}_1,\lambda_1)|\Omega \rangle$$
for all $\vec{k} \neq \vec{k}_1$, $\lambda \neq \lambda_1$ and where $|\Omega \rangle$ is the vacuum state, defined by $$N(\vec{k},\lambda)=0$$ for all $\vec{k}$, $\lambda$.

4. Oct 11, 2016

A. Neumaier

5. Oct 11, 2016

vanhees71

I see, but then the notation is very misleading.

6. Oct 11, 2016

maxverywell

Thank you all for the clarifications.

So it would be more correct to ask the question: what is the difference between the states $|1\rangle |1\rangle$ and $|2\rangle |0\rangle$ (or $|0\rangle |2\rangle$ equivalently)? And now the answer is more obvious. However I have a difficulty in understanding the Fock states for $n>1$.

The state $|2\rangle$ is not just two photons at two different times, but a pair of simultaneous photons, as A. Neumaier said. I guess this is due to their bosonic nature.
But what exactly "a pair of simultaneous photons" means? What a photo-counting detector will see? How to distinguish these two photons? (beamsplitter?) As I understand it, "two simultaneous photons" would mean that they are completely "bunched", but the second order correlation function of the state $|2\rangle$ is 1/2, so it is anti-bunched.

Last edited: Oct 11, 2016
7. Oct 11, 2016

A. Neumaier

Think of the vacuum (the lightless state of the e/m field) as a system of harmonic oscillators in which (under the given conditions) only one mode (monochromatic with given momentum) can be excited. Then |1> is the first excited state and |2> the second excited state. That these states are called 1-photon and 2-photon states is historical baggage that may more mislead than help. In a defendable sense, photons exist only in the brief moment of their detection.

If a pulse of a beam containing a 2-photon state passes a beam splitter it will be localized in the two resulting entangled beams; the detectors at the end of the two beams will fire simultaneously (assuming 100% efficiency, which is an idealization). If you split each of these beams again you get 4 beams, and of the 4 detectors at the end, just two will fire simultaneously. This is actually being used in the concept of ''heralded photons''.

8. Oct 11, 2016

DrChinese

I have seen variations on this statement previously (and various answers provided), and have a question for you.

Doesn't this imply there is no such things as a truly "free" photon? And if so, how are photons detected as CMB? Only those photons that are ultimately detected (absorbed) later could exist.

9. Oct 12, 2016

vanhees71

Photons are damn complicated beasts. First of all as any "particle" in relativistic QFT you can define them properly in a particle sense only concerning the free fields. Transient states in the interacting theory are hard to make physical sense of at least if you like to interpret them in the sense of "countable" "particle-like" entities. So what's really defined at all are S-matrix elements, i.e., transition-probability amplitudes for going from one asymptotic free state to another asymptotic free state involving particles and photons. Of course, Arnold is right saying that in principle what we observer are just detection events of photons, i.e., their interaction with some photodetector (photoplate, CCD, etc.).

Now in the interacting theory there's also some quibbles concerning asymptotic free states, which manifest themselves in IR and collinear divergences of S-matrix elements, which is due to the masslessness of the photon, i.e., the long-range nature of the em. interaction. Usually you solve this by resumming adequately (keyword: Bloch&Nordsieck; for analogous problems in non-Abelien gauge theories Kinoshita-Lee-Nauenberg theorem) certain "ladder diagrams". The physics of this becomes clear in analyzing the true asymptotic-state behavior, which turn out to be coherent states (and matter particles surrounded by a coherent-state cloud). For a readable introduction, see

P. Kulish and L. Faddeev, Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys., 4 (1970), p. 745.

10. Oct 12, 2016

A. Neumaier

There are oscillations in a free electromagnetic field (to a good approximation a beam in the lab is such a free field). This is what exists (i.e., is describable by states in quantum field theory) at the fundamental level. The intensity of the field determines the average frequency of detector events. That's the QED picture. Individual photons that one could talk of don't appear at all, the state is typically a thermal mixture of complicated superpositions of multiphoton states with arbitrarily many photons without any direct interpretation, and not even the mean photon number is well-defined since there is an unbounded number of very soft (low energy) photons. Photons are just building blocks to set up the theory - nothing more.

However, to obtain some kind of intuitive picture, these oscillations are somehow portioned by means of a somewhat vague concept of a photon. However, the process of portioning is not really well-defined, especially when the beam is pulsed, as in photon on demand experiment. One has a time-dependent signal in the unobserved part of the beam, which must be short time Fourier transformed within each pulse to obtain its frequency content and hence the photon content. The length of the chosen window affects the result. If a beam contains many photons, the window of the transform is not really determined by the signal; so the result is ambiguous. If a beam contains very few photons (dim laser) they would appear at random times (in the usual way of speaking) and again, it is not clear how to window it. Only if the preparation guarantees that each pulse contains exactly one photon, the beam content is predictable. But in this case losses mean that the pulse will contain less energy than prepared, which again interferes with the notion of a photon. Moreover, this is a description of the unobserved beam. If we observe it we interfere and change everything. Measuring the full state is impossible (unless one knows it in advance and can do a non-demolition measurement). What is measured is only an event count that tells very little about the state except after sufficient time averaging, assuming the beam or the pulse sequence is stationary.

None of these problems appear when working statistically with a large number of photons. This is why sufficiently long statistics of photon events is the only thing about photons that has scientific content, i.e., is repeatable and hence testable. Therefore all talk about single photons or photon pairs must be understood as inadequate but partially intuitive language for dealing with a complex, indivisible process happening in the electromagnetic field.

Note that the above problematic is exactly the same as what one has classically when interpreting 1-dimensional acoustic signals (phonons in place of photons) in traditional signal analysis. We can resolve music in time and frequency only because music is not arbitrary sound but structured with a fixed rhythm that allows one to recognize where one should place the windows. The musical notes correspond to prepared single phonons, a chord to prepared phonon pairs, triples, etc. But speech or acoustic noise is already very different - no sensible division into frequencies, hence phonons is possible. Instead one divides it into ''formants'' - complex entities adapted to human sound recognition. This analogy between light and sound should be kept in mind when thinking about photons.

11. Oct 12, 2016

zonde

Detection of downconverted photon pairs within the same time window is experimentally observed and adequately described by particle type models. Describing them as oscillations in electromagnetic field won't improve this description but rather will make it much more complicated and ambiguous.

12. Oct 12, 2016

A. Neumaier

It is adequately described by ''particles'' with very strange nonlocal properties.

Stopping to interpret nonlocal 2-photon states in terms of particles (''two photons'') doesn't invalidate the experimental results and their theoretical explanation in terms of quantum stochastic models (Lindblad equations etc.). Nothing gets ambiguous, on the contrary. And if one looks at realistic predictions of loss rates etc. the simple pure entangled state picture is already very inadequate, and things get more complicated anyway. Realistic quantum optics is much more about the electromagnetic field than you may imagine.

13. Oct 12, 2016

maxverywell

Can you elaborate on that please? Sending the state $|2\rangle_a |0\rangle_b$ to a 50:50 beam splitter will result on the entangled state $\propto |2\rangle_c |0\rangle_d-|0\rangle_c |2\rangle_d+2i|1\rangle_c |1\rangle_d$. What will the two detectors at $c$ and $d$ register?
On the other hand, sending the state $|1\rangle_a |1\rangle_b$ will result on $\propto |2\rangle_c |0\rangle_d+|0\rangle_c |2\rangle_d$.
In the first case there is probability that the two detectors will register two simultaneous counts (because of the state $|1\rangle_c |1\rangle_d$ in the superposition), whereas on the second case they will never register simultaneous counts.
But what the detectors register for the state e.g. $|2\rangle_c |0\rangle_d$? One click at one detector and 0 clicks at the other? I.e. the detector at $c$ sees the state $|2\rangle_c$ as one photon?

Last edited: Oct 12, 2016
14. Oct 12, 2016

A. Neumaier

I don't understand what you mean by your notation.

One cannot form a meaningful superposition of states following each other in a pulse train,
i.e., two photons in two different pulses, as opposed to a 2-photon state in a single pulse.

15. Oct 12, 2016

maxverywell

What I meant by $|2\rangle_a |0\rangle_b$ is that the state $|2\rangle$ falls onto the beam splitter's port $a$, while by and $|0\rangle_b$ I represent that at the second port, $b$, there is nothing.

16. Oct 12, 2016

DrChinese

So you have an entangled pair of photons: you route Alice to a detector so you have 1 photon (Bob) somewhere else (it's heralded). Bob is routed to outer space. Can that be considered a free photon, never being absorbed anywhere? Of course it could be eventually absorbed, if eventually is long enough. Or maybe not. That's my question, and I have heard somewhat different answers. Thought I would ask your considered opinion, as well as anyone else who would like to tackle it. Obviously this is off-topic a bit.

17. Oct 12, 2016

A. Neumaier

After Alice's detector clicked, the original prepared 2-photon state no longer exists. Instead the experimenter conventionally updates the model (aka collapses the wave function) to that of a single photon state heading to outer space. This is equivalent to having prepared a single photon on demand and sending it to outer space. In both cases you have an excitation of the e/m field of the given frequency whose intensity integrates to that needed for effecting a single click, and in this sense it can be taken to be a single photon. That's what the preparation procedure tells.

On the other hand, it is impossible to check this. Indeed, a beam cannot be arbitrarily narrow else it can no longer support a photon. Moreover, even a narrow beam broadens tremendously when it travels an interstellar distance; thus the signal weakens so much that one cannot tell the later impact of this presumed photon from that of a stray photon, which exist in abundance. For lack of checkability it is questionable to uphold the claim that a single photon is traveling. This is why I said that the traveling photon picture makes sense only if you have many photons in a well-defined experimental environment where a sufficient amount of statistics can be collected. The science is in the statistics of detection events, not in the intuitive picture of traveling photons.

18. Oct 12, 2016

A. Neumaier

Then the beam splitter will turn |2> into a superposition of |2>_c|0>_d and |0>_c|2>_d. It never changes a (number or polarization) state but only redirects it. Thus there is an enhanced probability of recording 2 simultaneous photons. Note that this is different from Bell-like experiments that generate superpositions of different |1>_c|1>_d states!

19. Oct 12, 2016

A. Neumaier

A detector can click only once in a sufficiently short interval; hence it will mistake the 2-photon state for a single photon.

Last edited: Oct 12, 2016
20. Oct 12, 2016

A. Neumaier

No.

First of all, it seems to me impossible to send your input state. We cannot time photons to pass at exactly the same time (so that they would interqct in the beam splitter) unless the photons were produced in an entangled way and sent through a symmetric arrangement. But in that case they will be in a superposition of different states of the form $|1\rangle_a |1\rangle_b$.

Second, after passing the beam splitter, the state will again be a (usually different) superposition of different states of the form $|1\rangle_a |1\rangle_b$, and not what you proposed. 1 and 2 are not spin labels (which would behave closer to like you said) but Fock number labels!

21. Oct 16, 2016

maxverywell

From a textbook on quantum optics:

$|1\rangle_a |1\rangle_b \xrightarrow{BS} \frac{1}{2}\left(\hat{a}^{\dagger}_{c}+i\hat{a}^{\dagger}_{d}\right)\left(i\hat{a}^{\dagger}_{c}+\hat{a}^{\dagger}_{d}\right)|0\rangle_c |0\rangle_d =\frac{i}{\sqrt{2}}\left(|2\rangle_c |0\rangle_d+|0\rangle_c |2\rangle_d \right)$

Last edited: Oct 16, 2016
22. Oct 16, 2016

A. Neumaier

Please add reference and page. Maybe we are talking about different things because of different notation.

23. Oct 16, 2016

maxverywell

Introductory Quantum Optics, C.Gerry and L.Knight, chapter 6.2 (Quantum mechanics of beam splitters), page 141.

24. Oct 16, 2016

A. Neumaier

I am a bit surprised by this calculation; had never done it before - but yes, you are right, the individual number of photons in a pure photon state is not conserved in a beam splitter, only the total number.

Sorry for my mistake. This also affects the answers I gave in #14 and #18. Instead of what I said there:
For a detector that counts clicks, it is as you say. For a detector that measures currents, the current produced by |2> should be twice that of the current produced by |1>, so that there would be a difference.

25. Oct 16, 2016

vanhees71

Sure, but also the original state, $|1 \rangle_c |1 \rangle_b$ is a two-photon state. So the beam (ideal) splitter doesn't change the total number of photons.

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