I 2-photon interference question

[Killtech]

I don't see how what you wrote is the same mathematically as what I wrote. What you wrote doesn't make sense to me.

I think I got the signs right but it's possible I flipped some. I'll have to check when I have time.

Your matrix has a zero eigenvalue with eigenvectors $\begin{pmatrix}1 & 0 & 0 & -1\end{pmatrix}^T$ and $\begin{pmatrix}0 & 1 & -1 & 0\end{pmatrix}^T$ i.e. that's a total absorption for both polarizations. Therefore I assume you meant:
$$
\frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & -1 & 0 \\
1 & 0 & 0 & -1
\end{pmatrix}
$$
which is what I used in the calculations in my previous post. So that matrix would corresponds to a $\frac \pi 2$ phase shift on reflection. Indeed my mirror used a $\frac \pi 4$ phase shift, so sorry, you are right it's not mathematically the same. However projecting it down to a single polarization subspace doesn't change anything mathematically - because the spaces are independent and can be cleanly separated.

 

[vanhees71]

I've no clue what your $4 \times 4$ matrices mean. There are two incoming photon modes and two outgoing ones. So you need only a $2 \times 2$ matrix as explained in any textbook of quantum optics, or are you referring to something else?

 

[Killtech]

I've no clue what your $4 \times 4$ matrices mean. There are two incoming photon modes and two outgoing ones. So you need only a $2 \times 2$ matrix as explained in any textbook of quantum optics, or are you referring to something else?

each photon beam is composed of two independent components, that makes 4 degrees of freedom for 2 beams. See photon polarization. I haven't seen a 4x4 matrix for beam splitters that often on the internet but it makes sense to me when one has to deal with different incoming polarization states in an experiment. In the 2x2 matrix form I suppose you would have trouble explaining why two orthogonal polarizations don't interfere even though their phases are right.

 

[vanhees71]

Hm, in the usual formalism that's all included in the annihilation operators used to describe the beam splitter for arbitrary modes. That's why in any description of a beam splitter usually there's only a $2 \times 2$ matrix. See my summary of the standard treatment here:

https://www.physicsforums.com/threa...nce-in-an-interferometer.975742/#post-6216820

for more details see

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.40.1371

 

[Killtech]

Hm, in the usual formalism that's all included in the annihilation operators used to describe the beam splitter for arbitrary modes. That's why in any description of a beam splitter usually there's only a $2 \times 2$ matrix. See my summary of the standard treatment here:

Yes, well, this is just because I am great at confusing people. The 4x4 matrix probably came up because I tried to write the total time evolution operator for my setup - just so the presence of other beams outside the beam splitter is not forgotten... and I guess this confused PerterDonis what those additional dimensions were supposed to mean - and the degrees of freedom that come with it. Hence I guess he wrote it down in a way to point out which degrees of freedom he sees there. Technically though the additional ones in my matrix are those of the total experiment setup which has a few more then just those needed for the interaction at the beam splitter. But I also made an honest mistake: I (intuitively) chose an identity matrix to say that the beam splitter does affect anything about beams that aren't its input... which would however mean the photons amplitude in those beams would be trapped in place and could "interfere" with anything coming up in the next time step.

Anyhow let's skip this and get back to discussing the experiment.

 

[PeterDonis]

I've no clue what your $4 \times 4$ matrices mean.

As @Killtech said, I was attempting to capture two polarization degrees of freedom for each input/output arm of the beam splitter. However, I agree that if you don't measure polarizations in the experiment you can ignore that degree of freedom and just use a $2 \times 2$ matrix to represent the beam splitter operator.

 

[vanhees71]

I still don't get it. Do you have a paper/book, where this convention is used? The formalism, I summarized in

https://www.physicsforums.com/threa...nce-in-an-interferometer.975742/#post-6216820

includes the polarization in the symbol of the annihilation operators, i.e., $\hat{a}_1$ stands for $\hat{a}(\vec{k}_1,\lambda_1)$, where $\lambda_1 \in \{1,2 \}$. It refers to a mode with wave number (momentum) $\vec{k}_1$ and a polarization state labelled by $\lambda_1$ and $\lambda_2$ (e.g., you can take some linear polarization states or the circular or elliptical ones). Of course the transfer matrix depends on the specific polarization states, determined by the specifically used beam splitter.

 

[PeterDonis]

Do you have a paper/book, where this convention is used?

Who are you asking? If you are asking me, I'm no longer sure that the $4 \times 4$ matrix I gave correctly captures the polarization degrees of freedom, but anyway I've already agreed we can just ignore them for this discussion.

Of course the transfer matrix depends on the specific polarization states, determined by the specifically used beam splitter.

Can you give an example of a transfer matrix that changes the polarization states? For example, that takes some of the amplitude coming in on an input arm in the "right circular polarization" degree of freedom and transfers it to the "left circular polarization" degree of freedom in an output arm?

That kind of transfer matrix was what I was attempting to capture (quite possibly incorrectly). I don't see how you can capture that in a $2 \times 2$ matrix, since each input arm now has to have two degrees of freedom, not one (the momentum is fixed on each input arm, but now you have two polarization basis states).

 

[Killtech]

In addition to all the correct responses that have been given already, just a short additional one: Splitting a coherent light beam and performing measurements on one of the outputs will do absolutely nothing to the state of the light field in the other output beam. This is the evry definition of a coherent beam. A coherent beam follows a Poissonian phootn number distribution, which is the distribution for independent events. Accordingly, if you have a photon detection event, also the conditional mean photon number of the remaining beam does not change at all. Even for a beam that contains only one photon per minute on average, there is still a finite probability for having two photons. For a coherent beam, the probability for that is exactly so large that the conditional photon number for the remaining beam after one detection event stays exactly at one photon per minute. All detection events are statistically completely independent foor coherent beams.

I really want to know the mathematical model for this. So how do you calculate that it is Poisson distributed?

With my own understanding I would say it should work like this: now a single photon (partial) beam with probability amplitude $a= \sqrt p$ i.e. $\sqrt p |1\rangle$ will be simply Bernoulli distributed with parameter $p$ when it runs into a detector. Now if i have a beam with multiple independent identically distributed photons I get a Binomial distribution and I would write the beam down as $a |n\rangle$ $=\sqrt {\frac p n} \prod_{i=0}^n |A_i>$ with each $|A_i\rangle>$ still Bernoulli distributed with parameter $\frac p n$ and independent of the other $|A_i\rangle$. The beam is tuned down to an amplitude of $\sqrt {\frac p n}$ such that the expected photon number is the same as for the single photon. With this using the Poisson limit theorem it follows that $$\lim_{n \rightarrow +\infty} \sqrt {\frac p n} \prod_{i=0}^n |A_i>$$
is Poisson distributed with $\lambda = lim_{n \rightarrow +\infty} n \frac p n$ $=p$ and with this it is easy to verify that it yields exactly the behavior you said with a beam splitter. That is: detection events with such beams will be always statistically independent due to $n$ being infinite (since taking one particle out of an infinite pool via measurement leaves the pool unchanged) and therefore conditional particle numbers stay unaffected by detections. This will however only hold for as long $n$ is a large number though I guess this is probably always true for a laser beam?

This makes a lot of sense to write down a coherent laser beam in such a way as it enables me to properly calculate its behavior at beam splitters and other devices but I wager this isn't a standard way to denote and calculate such things. So how do you do it normally?

 

[vanhees71]

It's sufficient to consider this for a single mode. Then you have basically just a harmonic oscillator with one annihilation and one creation operator $\hat{a}$ and $\hat{a}^{\dagger}$, obeying the commutation relation
$$[\hat{a},\hat{a}^{\dagger}]=\hat{1}.$$
One complete orthonomal set are the eigenvectors of the number operator
$$\hat{N}=\hat{a}^{\dagger} \hat{a}.$$
These Fock states are $|n \rangle$ with the eigenvalues $n \in \mathbb{N}_0$ and
$$\langle n|n' \rangle=\delta_{nn'}.$$
The annihilation operator acts as
$$\hat{a} |n \rangle=\sqrt{n} |n-1 \rangle.$$
Now a coherent state is an eigenstate of the (NOT self-adjoint!) annihilation operator with complex eigenvalues $\alpha \in \mathbb{C}$. To evaluate it we set
$$|C_{\alpha} \rangle=\sum_{n=0}^{\infty} c_n |n \rangle.$$
Then we have
$$\hat{a} |C_{\alpha} \rangle=\alpha |C_{\alpha} \rangle=\sum_{n=1}^{\infty} \sqrt{n} c_n |n-1 \rangle=\sum_{n=0}^{\infty} \sqrt{n+1} c_{n+1} |n \rangle = \sum_{n=0}^{\infty} \alpha c_n |n \rangle.$$
This leads to the recursion
$$c_{n+1} = \frac{\alpha}{\sqrt{n+1}} c_n \; \Rightarrow \; c_{n} = \frac{c_0}{\sqrt{n!}} \alpha^{n}.$$
So we have
$$|C_{\alpha} \rangle=c_0 \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle.$$
For the norm we get
$$\langle C_{\alpha}|C_{\alpha} \rangle=|c_0|^2 \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!}=|c_0|^2 \exp(|\alpha|^2) \; \Rightarrow\; c_0=\exp(-|\alpha|^2/2).$$
So finally
$$|C_{\alpha} \rangle=\exp(-|\alpha|^2/2) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle.$$
What's now the probability to find exactly $n$ photons? Just use Born's rule:
$$P(n)=|\langle n|C_{\alpha} \rangle|^2 = \exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!},$$
which is a Poisson distribution with mean $\langle N \rangle=|\alpha|^2$.

 

[Killtech]

It's sufficient to consider this for a single mode. [...]

Thanks, that's easy enough. Okay, now I understand how to read a state like $|n\rangle$ in terms of measurement. What I still don't get though is how do you calculate conditional probabilities within this formalism? Or in other words how do you note any kind of entanglement between the sates? Well, in this case of a coherent beam everything is assumed to be independent so there is no entanglement anyway.

But still still let's take it as a simple example anyway to proof the statement of @Cthugha that

All detection events are statistically completely independent foor coherent beams.

So the simplest case would be to run the state $|C_\alpha\rangle$ through a beam splitter and calculate the conditional probabilities for measuring of beam given the results of measurement on the other. After the splitter can I write the state as
$$e^{-|\alpha_A^2|} \sum_{n=0}^{\infty} \frac {\alpha_A^n} {\sqrt{n!}} |A,n \rangle + e^{-|\alpha_B^2|} \sum_{n=0}^{\infty} \frac {\alpha_B^n} {\sqrt{n!}} |B,n \rangle$$
where $A$,$B$ are supposed to denote the two outgoing arms of the splitter? So again the Born rule would give me the probabilities and given that all vectors $|A,n\rangle$, $|B,n\rangle$ are independent (i.e. orthogonal) the statement is perhaps trivial. Still how do I properly and formally calculate a conditional probability here? And how would I denote a state where the outgoing beams $A$,$B$ were statistically dependent, like for a most simple case of a single photon after passing the splitter?

 

[Killtech]

Ah, I think I was just stupid. writing the state via a product rather then a sum makes a lot more sense.

$$(e^{-\frac 1 2 |\alpha_A^2|} \sum_{n=0}^{\infty} \frac {\alpha_A^n} {\sqrt{n!}} |n \rangle_A)(e^{-\frac 1 2|\alpha_B^2|} \sum_{n=0}^{\infty} \frac {\alpha_B^n} {\sqrt{n!}} |n \rangle_B = e^{-\frac 1 2(|\alpha_A^2|-|\alpha_B^2|)}\sum_{n,m=0}^{\infty} \frac {\alpha_A^n \alpha_B^m} {\sqrt{n! m!}} |n\rangle_A |m\rangle_B$$

It that about right? The product form of the sum coefficients $c_{n,m} = \frac {\alpha_A^n \alpha_B^m} {\sqrt{n! m!}}$ makes it immediate obvious that the beams are independent in this case and more generally i can easily express any possible dependency between $A$ and $B$ via the coefficients $c_{n,m}$. In this form it's kind natural how to conditional expectations/distributions can be expressed/calculated. So is this actually correct?

 

[vanhees71]

Yes, if you have two independent coherent states, that's the correct description.

 

[Killtech]

Yes, if you have two independent coherent states, that's the correct description.

[...]All detection events are statistically completely independent foor coherent beams.

That said, it would seem to me that a laser beams that are statistically dependent could be constructed? I was just looking at the form of state of two independent beams while also having the HOM effect open in another tab...

So let's say both beams have an expected photon number of 0.01 per some time interval, which means for a Poission distribution that $|\alpha_A|^2=|\alpha_B|^2=0.01$. Hence writing only the dominant terms of the beams i'd get
$$ e^{-0.01}( |0,0\rangle_{A,B} + 0.1(|0,1\rangle_{A,B} + |1,0\rangle_{A,B}) + 0.01|1,1\rangle_{A,B} + \frac {0.01} {\sqrt 2} (|2,0\rangle_{A,B} + |0,2\rangle_{A,B}) + ...) $$
and assuming these beams are the input to a beam splitter in a HOM-setup I would be let to believe the term $|1,1\rangle_{A,B}$ in this expansion would participate in the HOM-effect. While this wouldn't change the outgoing beams in a visible way, it should ever so slightly alter their statistics: for one they should deviate from a true Poisson statistic in that in the case of detecting one photon in one of the beams has a slightly increased chance of detecting at least another one there too. On the other hand the chance of detecting a photon at all would somewhat drop. That increased probability of a 0-detection event in one beam should however be compensated by the increase chance for a two photon detection in the other rendering the two outgoing beams a little bit statistically dependent. This effect would be quite small but recording a sufficiently large statistic it should be enough to make stand out over the statistic error.

Hmm, for my purpose strengthening the $|1,1\rangle$ term above all else is what i want while this would actually diminishing it. Still, it might be salvageable.

 

[Cthugha]

It is not that easy as even for weak coherent light fields, you still have off-diagonal elements of the density matrix (which means that the coherent state has a phase), while the phase for single photon Fock states is completely undefined. So in order to provide some meaningful calculations, you need to define what kind of relative phase between the beams you are talking about. Usually people consider phase randomized states, which means that the relative phase varies quickly, but has some well defined fluctuating value for short instants of time. If you then consider all the classical interference patterns you can get, the results will depend on where you place your detectors. You can get some classical equivalent of the HOM dip that goes down to 50% of the visibility of a Fock state HOM dip that way, but it depends on a lot of details.

The math for the standard spatial interference pattern in the classical phase-randomized case was presented by Mandel already in 1983 https://journals.aps.org/pra/abstract/10.1103/PhysRevA.28.929 .
It is not too different to construct the equivalent for a beam splitter geometry along similar lines.

 

[Killtech]

It is not that easy as even for weak coherent light fields, you still have off-diagonal elements of the density matrix (which means that the coherent state has a phase), while the phase for single photon Fock states is completely undefined. So in order to provide some meaningful calculations, you need to define what kind of relative phase between the beams you are talking about. Usually people consider phase randomized states, which means that the relative phase varies quickly, but has some well defined fluctuating value for short instants of time. If you then consider all the classical interference patterns you can get, the results will depend on where you place your detectors. You can get some classical equivalent of the HOM dip that goes down to 50% of the visibility of a Fock state HOM dip that way, but it depends on a lot of details.

The math for the standard spatial interference pattern in the classical phase-randomized case was presented by Mandel already in 1983 https://journals.aps.org/pra/abstract/10.1103/PhysRevA.28.929 .
It is not too different to construct the equivalent for a beam splitter geometry along similar lines.

Unfortunately I have no access to that paper but I figured it isn't that trivial. After I posted that I have tried to google a bit myself and am currently reading this here: https://www.researchgate.net/publication/45921675_Interference_of_dissimilar_photon_sources - though there is a lot to do at work which is slowing me down. Anyhow, it's using somewhat similar approach albeit simplified to just observe the HOM effect whereas I ultimately want to understand if its possible to filter out/detect an exact probability amplitude value (its modulus more then its phase) of a single photon state. And I am more inclined to understanding the capabilities of the underlying mathematical model here rather then the physics, hence I am a bit lenient in overlooking the practical difficulties in preparing the assumptions I sometimes use :D... well, unless they are already theoretically invalid.

So in the case of my latest posts I was just trying to understand how laser beams are properly described mathematically and then using that to check if they could be made usable for my intentions. The first question for that was to answer whether it was possible to construct statistically dependent beams from laser sources at all - and using the HOM effect was just an idea. The exact form of that dependence is perhaps less of an issue for now and therefore the HOM effect needs to be present just so it doesn't effectively cancel out in the end - just so any statistical dependency is achieved. If it roughly worked though I wonder what would happen if one of the the HOM output beams were interfered destructively with one of the original beams - if there were any difference in statistics and dependence between those beams the destructive leg at such a splitter should work as a filter for that dependent component of the beam which might be what I am looking for. Hmm, should I draw a sketch of what I have in mind or does everyone(anyone?) still follow?

 

[Killtech]

Also, as I am not so familiar with optics, can anyone give me a simple link about optical couplers - like the ones used in the paper I linked before? Can't find a wikipedia article on them like there is one about beam splitters. Google gives me a lot of manufacturer sites... and some papers on how to produce even better and more compact ones. I get the concept of what they do, but would feel better when I saw how they are treated in the calculus.