B Destructive interference and conservation of momentum

[Quantum of Solace]

If two photons traveling in the same direction but out of phase cancel each other out, what happens to the energy and momentum?

 

[BvU]

They do not cancel each other out.
If they have a lot of energy, they can collide and form other particles (with the right quantum numbers). Energy and momentum are conserved.

 

[haael]

If they cancel in one region of space, they must enhance in other.

 

[BvU]

Who says 'they' cancel? What exactly cancels ?

 

[stevendaryl]

Who says 'they' cancel? What exactly cancels ?

I think the OP is thinking about interference of light, as is demonstrated by the two-slit experiment, for example. I would split his question up into three parts:

1. Is it possible to arrange two sources of light so that you get total destructive interference? He seems to think the answer is "yes", but I'm pretty sure the answer is "no"; if the two sources interfere destructively in some regions, then they will interfere constructively in other regions.
2. Assuming that the answer to the first question is "yes", does that imply that two photons can interfere so that there is zero probability of detecting any photon anywhere? I'm pretty sure the answer is "no".
3. Assuming that the answer to the second question is "yes", how do you explain where the energy went?

Since the answer to the first question is "no", the other two questions are moot.

 

[BvU]

OP ? What's your status now ?

 

[Quantum of Solace]

Stephendaryl covered it pretty well. Thanks for replying.

 

[Quantum of Solace]

Thread necromancer, here. I watched an excellent video by YouTuber "Applied Science", where he discusses and experiments with Rugate filters used as lens coatings. I urge you to check it out if that sounds interesting. In the video, he explains how you can create an anti-reflective coating on a lens by creating a layer at a specific depth, with a different index of refraction. So instead of one reflected ray/photon, you get two, at such a phase difference that they cancel out. Instead, manifesting as having entered the lens, improving the transmissivity of it. OK, you can see how this discussion is so far related to my original question.

My problem now, however, is that you could set this scenario up whereby the reflected photons could be directed to interfere and cancel each other, or NOT, based on a delayed choice. You could also set up a sensor to measure the transmissivity of the lens based on the intensity of light received from a fixed-intensity light source transmitting through it.

If your delayed choice to throw the switch and send those reflected photons in opposite directions, or have them cancel each other was significantly later in time. What would happen to the detection of modulation of light intensity at the receiver? More importantly *when* would the detection occur?

Delayed choice experiments on the wave function of individual photons seems to imply retrocausality in a classical sense, but causality is preserved because the correlation cannot be checked without reference to the choice-maker's data, which is separated in spacetime. Hence no information can violate the speed of causality. However, the transmissivity of a lens can be measured almost instantaneously, whereby the choice to make those reflected photons interfere destructively could theoretically be done light-years away.

How can causality be preserved if:
1) The destructive interference of two reflected photons gives rise to a transmitted (not reflected) photon(s)
2) The choice in whether to destructively interfere those photons can be done in distant spacetime, such that any change in transmissivity could be detected "before" the choice was made. Effectively violating the speed of causality.

Or, which of 1) and 2) is wrong?

 

[BvU]

Sorry I missed the 'same direction' in post #2 (spent a lot of time studying $\gamma\gamma$). -- made me shut up and listen.

I get the impression you mix up photons and (plane) waves now and then. Anywhere from same thing to altogether different beast. Perhaps you want to be introduced to QED gently: Feynman on photons

 

[Quantum of Solace]

Yes, I admit I'm out of my comfort zone and this particular question mixes classical theory (destructive wave interference) with implications from quantum mechanics (wave function interference). I can't find any references to delayed choice light wave interference except in the context of single-photon self-interference of the wave function. Thanks for your kind reply.

 

[Demystifier]

2. Assuming that the answer to the first question is "yes", does that imply that two photons can interfere so that there is zero probability of detecting any photon anywhere? I'm pretty sure the answer is "no".

It's illuminating to try to find a counterexample to see why exactly is that impossible. So let us consider non-relativistic free Schrodinger equation in 1 spatial dimension. Suppose that we have two packets, $\psi_1(x,t)$ and $\psi_2(x,t)$, one moving from the left to the right and the other moving from the right to the left, such that at some time $t_0$ they "collide" and satisfy
$$ \psi_1(x,t_0)=f(x), \;\;\psi_2(x,t_0)=-f(x) \;\;\; (1) $$
for some arbitrary function $f(x)$. Hence at $t=t_0$ we have
$$\psi_1(x,t_0)+\psi_2(x,t_0)=0 \;\;\; (2)$$
for all $x$, so we have a total destructive interference at $t=t_0$. But before that, at $t<t_0$, we had two separated wave packets so
$$\psi_1(x,t)+\psi_2(x,t)\neq 0 \;\;\; (3) $$
for at least some $x$. But that should be impossible, because it contradicts the continuity equation. What went wrong?

The answer is the following. A priori, there is nothing wrong with (1). We can take $t_0$ to be the initial time, so (1) can be chosen to be the initial condition. But the Schrodinger equation is a first-order equation in time, so the initial condition $\psi(x,t_0)$ determines $\psi(x,t)$ uniquely, for all $t$. Furthermore, the initial condition (1) says that the two packets are identical (up to the global phase factor -1) at the initial time, so by the Schrodinger equation they must be identical at all times. So contrary to the initial assumption, it cannot be the case that one wave packet travels to the left and the other to the right. Instead, they must travel identically for all times $t$, depending on the function $f(x)$. In particular, if $f(x)$ is a symmetric Gaussian $f(x)=e^{-x^2/\sigma^2}$, then $\psi_1(x,t)$ and $\psi_2(x,t)$ travel neither to the right nor to the left. Instead, their centers are at rest at $x=0$, while only their widths change with time. In other words, if (2) is right, then Schrodinger equation implies that (3) is wrong. If two packets cancel each other at one time $t_0$, then they cancel each other for all times $t$.