Do Two Photons in a Radiative Cascade Exhibit Directional Correlation?

[Clovis]

TL;DR Summary

Is there any directional correlation between the two photons emitted in a two-photon atomic cascade?

Hello. I have a question about the two photons emitted from a radiative atomic cascade (such as the calcium radiative cascade used by Aspect et al. in their tests of Bell's theorem).

The short version of my question is this: Do the two photons have any directional correlation (any correlation between their directions of propagation outward from the source atom)? For example, if the first photon happens to be detected along the Z-axis does that determine in any way the direction of travel of the second photon?

The longer version of my question is this: In one of Aspect's papers the author's cite a 1973 paper by Edward Fry in which the correlation rates of the two-photons emitted from a two-photon radiative cascade are analyzed at a theoretical level. The Edward Fry paper can be found here:

https://oaktrust.library.tamu.edu/b...29/PhysRevA.8.1219.pdf?sequence=1&isAllowed=y

In figure 1 of the above Fry paper the experimental set up for Fry's analysis is shown. Two detectors face the source of the two-photon radiative cascade, with an angle of about 45 degrees between the two detectors. The angle between the two detectors is labeled in the figure as ΘS. The fact that Fry included and labeled the angle between the two detectors implied (to me) that the angle between the detectors plays some role in the probability of detecting both photons, which would in turn imply that there is some directional correlation between the two photons.

But in equation 1 (which, I must admit, I find mathematically overwhelming) I cannot see ΘS anywhere, which implied to me that the angle between the two detectors does not affect the probability of detecting the two photons, which would seem to imply that there is no directional correlation between the two photons.

But in figure 2b of the paper ΘS reappears.

Although my ability to understand this paper is limited one possibility is that the angle between the detectors only plays a role in the coincidence rate because the detectors have polarizing filters, and so changing the angle between the detectors changes the orientation of the polarizing filters relative to one another. And given that there is a correlation between the polarizations of the two photons, that would explain why the angle between the detectors plays a role in the coincidence rate.

So if one were to remove the polarizing filters from the detectors in figure 1, would there no longer be any effect of the angle between the detectors on the coincidence rate?

Thank you in advance for any help you can give me. My math is not very strong so qualitative answers will be more helpful.

 

[anuttarasammyak]

TL;DR Summary: Is there any directional correlation between the two photons emitted in a two-photon atomic cascade?

The short version of my question is this: Do the two photons have any directional correlation (any correlation between their directions of propagation outward from the source atom)? For example, if the first photon happens to be detected along the Z-axis does that determine in any way the direction of travel of the second photon?

Another photon is also along the Z-axis but in opposite direction. This is my thought from conservation of momentum of the system which is also applied to a pair of emitted photons in PET.

 

[Clovis]

Thank you, Anuttarasammyak. A similar thought had occurred to me at first but then I thought that the momentum of the system would still be conserved if, for example, both photons were emitted along the positive z-axis and the calcium atom recoiled along the negative z-axis. So it seems to me that conservation of momentum by itself does not dictate that the photons must be emitted in opposite directions. Does that seem correct?

 

[anuttarasammyak]

A similar thought had occurred to me at first but then I thought that the momentum of the system would still be conserved if, for example, both photons were emitted along the positive z-axis and the calcium atom recoiled along the negative z-axis. So it seems to me that conservation of momentum by itself does not dictate that the photons must be emitted in opposite directions. Does that seem correct?

Say a calsium atom has been at rest in a IFR.
Momentum of the system = 0
Energy of the system = $mc^2$

After emission of two photons
Momentum of the system = $\mathbf{p}+\hbar \mathbf{k_1}+\hbar \mathbf{k_2}$ = 0
Energy of the system = $\sqrt{m'^2c^4+p^2c^2}+\hbar ck_1+\hbar ck_2 = mc^2$

We may be able to deduce $\mathbf{p}=0, \mathbf{k_1}+ \mathbf{k_2}=0$ from these relations.

 

[Clovis]

Allow me to test the formmatting before I post a reply:

mc
².

 

[Clovis]

Oops, my formatting test didn't work. Here is my reply with imperfect formatting. Apologies.

Hello again, Anuttarasammyak. I very much appreciate you taking the time to consider my question in detail and write a thoughtful response. But I am having trouble understanding why the two equations that you wrote in your last post (conservation of momentum for the system and conservation of energy for the system) must necessarily lead to the conclusion that p (the momentum of the calcium atom after emission of the two photons) must be zero and also that the sum of the momentum of the two photons must be zero.Please allow me to make a counter argument why I believe that p cannot be zero. And again, please know that I am not counter arguing out of any disrespect for you.The energy of the excited calcium atom before emission of the two photons is mc2. After emission of the photon pair the energy of calcium atom is m'c2. By conservation of energy m'c2 must be less than mc2.Given that m'c2 must be less than mc2 in the conservation of energy equation in your post, then it seems to me that it is impossible to also set p = zero and k1 + k2 = zero. Phrased in a different way, we were to set p = zero and k1 + k2 = zero then the equation produces m'c2 = mc2, which cannot be correct by conservation of energy.

 

[PeterDonis]

we were to set p = zero and k1 + k2 = zero then the equation produces m'c2 = mc2

No, it doesn't. The $k_1$ and $k_2$ in the energy equation are magnitudes, not vectors. Even if the vector sum of the photon momenta is zero (as it must be if $p = 0$), the sum of their magnitudes is not zero, it's positive.

 

[Clovis]

Hi again, Anuttarasammyak. Again, thank you. I see now that I made a mistake in the way I formulated my previous reply, so please ignore it.What I should have said is this: In your first reply, I believe that you concluded that p (the momentum of the calcium atom after emission of the two photons) must be zero and also that the vector sum of the 2 photons' momentums must also be zero. I agree that such a situation would indeed satisfy the two conservation equations (the conservation of momentum and conservation of energy equations) in your reply. But I believe that the two equations also have solutions in which (a) p is non-zero and (b) the momentum sum of the two photons is non-zero. And if there are indeed such solutions to the two equations then conservation of energy and momentum do not, in and of themselves, show that p must equal zero and the two photons' momentums must sum to zero.

 

[PeterDonis]

I believe that the two equations also have solutions in which (a) p is non-zero and (b) the momentum sum of the two photons is non-zero.

If you believe this, you should try to find such a solution.

 

[Clovis]

Hello anuttarasammyak and PeterDonis. I want to thank you both for making me think about my original posted question in more depth. After reviewing your comments I have analyzed, to the best of my abilities, the conservation of energy and conservation of momentum of the calcium atom emitting the two photons. Based on my analysis I believe still that the atom's final momentum does not have to be zero and I believe that the sum of the photon momentums do not have to sum to zero.

If you would be so kind please review my analysis below. Again, I apologize for the lack of formatting (I have yet to master the LaTeX formatting).

============

(a) According to an article by Aspect et al, the two photons emitted from the excited calcium atom have frequencies of 551.3 nm and 422.7 nm. Converting these in wave number k values, we have 1.81E9 and 2.37E9.

(b) Before the excited calcium atom emits the photon pair, we will define it as at rest. So total momentum initially is zero. The excited atom's total energy is mc(2). By conservation of energy the total energy of the excited atom is the energy of the atom in it ground state (which we will denote as m'c(2)) plus an amount of energy equal to the energy of the two photons that it will emit when it moves from its excited state to it ground state.

(c) Using the formula that the energy of a photon is E(photon) = hck, the two photon energies are (in Joules) 3.6E-16 and 4.71E-16, respectively. These sum to a total energy of 8.31E-16. Using the reasoning in step (b) we have mc(2) = m'c(2) + 8.31E-16. In words, the total energy of the excited atom is equal to the total energy of the atom in its ground state atom plus 8.31E-16 Joules. Rearranging this we have m'c(2) = mc(2) - 8.31E-16.

(d) We imagine a scenario where the excited calcium atom emits both photons along the positive z-axis. The momentum of each photon is p(photon) = hk. From the k values of the two photons this gives momentums of 1.20E-24 and 1.57E-24, respectively. The total momentum of the two photons together is 2.77E-24.

(e) The excited atom was initially at rest and so the total momentum of the system before the photons were emitted was zero. By conservation of momentum the ground state atom must recoil along the negative z-axis with a momentum of -2.77E-24.

(f) In the second reply to my original post, two equations are given. The first equation is the conservation of momentum for the system and the second equation is the conservation of energy for the system.

(g) Into the conservation of momentum equation we substitute p = -2.77E-24 for the ground state atom and 1.20E-24 and 1.57E-24 for the momentums of the two photons. The equation does return zero for the total momentum. So (as far as the conservation of momentum equation is concerned) the momentum values p = -2.77E-24, photon 1 = 1.20E-24, and photon 2 = 1.57E-24 are valid solutions.

(h) Into the conservation of energy equation we first substitute 3.6E-16 and 4.71E-16 for the two photon energies (from step (c)). These add to a total photon energy of 8.31E-16 on the left side of the equation. Before substituting any further values into the conservation of energy equation, we rearrange it by subtracting the total photon energy from both sides of the equation. The right side of the equation becomes mc(2) - 8.31E-16.

(i) Next, under the square root on the left side of the equation, we substitute m'c(2) = mc(2) - 8.31E-16 (from step (e)) and p = -2.77E-24. The m'c(2) term appears as its square under the square root and p term appears as pc squared. After making all of the above substitutions and expanding all squares, we arrive at the following for the complete term under the square root: (mc(2))2 - 1.66E-15(mc(2)) + 6.91E-31.

(j) Next, we square booth sides of the equation. This removes the square root from the left side. In other words, the left side of the equation becomes (mc(2))2 - 1.66E-15(mc(2)) + 6.91E-31 with no square root. Squaring both sides of the equation changes the right side of the equation (which was mc(2) - 8.31E-16 before squaring) into (mc(2))2 - 1.66E-15(mc(2)) + 6.91E-31 after squaring. This is exactly the same as the left side of the equation.

(k) In summary in regard to the conservation of energy equation, by substituting m'c(2) = mc(2) - 8.31E-16, p = -2.77E-24, and (for the two photon energies) 3.6E-16 and 4.71E-16, we arrived at a valid equation (an equation for which the left side is equal to the right side). Thus the values in that were substituted into the conservation of energy equation are valid solutions to the equation.

(l) Combining the results from steps (g) and (k), we conclude that it is possible for the excited calcium atom in this analysis to emit two photons whose combined momentum does not sum to zero and yet conservation of momentum and conservation of energy are still obeyed for the system (due to the recoil of the ground state atom). Stated in the opposite way, momentum and energy conservation do not dictate that the excited calcium atom must necessarily emit its two photons in opposite directions and the conservation laws do not dictate that the atom's initial momentum of zero must be remain zero after it has emitted the two photons. As an analogy, a rifle at rest that then fires two bullets does not have to remain at rest. Conservation laws do not dictate that the bullets must be fired in opposite directions.

 

[Clovis]

Discovered a math error in my analysis above. Please ignore for now.

 

[anuttarasammyak]

(contd. from #4)
All the three vectors are coplanar otherwise cancellation is impossible. We can choose it XY plane. Without loosing generality, non dimensional wave number vectors introducing momentum unit of mc are:
$$\mathbf{\kappa_1}=(\kappa_1,0):=\mathbf{k_1}\frac{\hbar}{mc}$$
$$\mathbf{\kappa_2}=(\kappa_2\cos\theta,\kappa_2\sin\theta):=\mathbf{k_2}\frac{\hbar}{mc}$$
$$\mathbf{K}=(K\cos\phi,K\sin\phi):=\frac{\mathbf{p}}{\hbar}\frac{\hbar}{mc}$$
Momentum conservation
$$\kappa_1+\kappa_2\cos\theta+K\cos\phi=0$$
$$\kappa_2\sin\theta+K\sin\phi=0$$
Energy Conservation
$$\kappa_1+\kappa_2+\sqrt{K^2+(\frac{m'}{m})^2}=1$$

I hope it could help your analysis. Approximation by smallness of
$$0<\epsilon:=1-\frac{m'}{m}<<1$$
would be useful.

 

[Clovis]

Hi again, Anuttarasammyak. This question is much more complicated than I originally thought. Thank you very much for explaining it. I will think more about what you wrote.

 

[anuttarasammyak]

I write down here what I found as momerandum. Please revisit when you are prepared.

Spoiler

K^2 in the energy conservation equation is expressed by $\kappa_1,\kappa_2,\theta$ using momentum conservation.　Making use this expression, the energy conservation equation is reduced to
$$\kappa_1+\kappa_2-\kappa_1\kappa_2(1-\cos\theta)=\epsilon-\frac{\epsilon^2}{2}$$
Comparing quantity of first order of $\epsilon$ in the both sides, we know
$$\kappa_1+\kappa_2=\epsilon$$
as first order approximation. It means all the calcium electron transition energy is transformed to generation of two photons and no transfer to calcium atom kinetic energy. Thus in the same approximation
$$K=0$$
Momentum conservation says
$$\kappa_1=\kappa_2=\frac{\epsilon}{2},\ \ \theta=\pi$$
in the same approximation.

 

[PeterDonis]

I write down here what I found as momerandum

I think you need to give more detail about the expression you get for $K^2$ using momentum conservation.