Destructive interference of light

[mbe]

I moved this from a different thread as i thought it would be more appropriate here.

If two photons were to arrive at a single spot by accident such that they were half a lambda out of phase they would destructively interfere. Where would the energy go?

 

[scilover89]

This is what I think:

When two photons destructively interfere, the wave amplitude become zero. However, the two photon still exist, and the energy didn't change, as frequency of light is not affected. (energy of light = hf)

Someone please correct me if I am wrong.

 

[Edgardo]

Hello mbe,

I found the same question in another forum here:
http://www.madsci.org/posts/archives/oct99/941139504.Ph.q.html

and the answer:
http://www.madsci.org/posts/archives/oct99/941139504.Ph.r.html

 

[michael879]

scilover89, I am pretty sure your right, the wave is undectable, but it still exists because it has energy/mass (E=mc^2)

 

[DaveC426913]

Ignore. Can't seem to delete this double post.
OK, edgardo, that answer talks about two beams intersecting, but mbe carefully posed the setup so that they *stayed* that way.

So, if they canceled out but were still there, that means it is theoretically possible to send a beam of light that contains an arbitrary amount of energy, but is undetectable. Surely that's impossible!



You know, I think this could be simulated with water or sound - at least in principle - enough to observe the result.

 

[DaveC426913]

and the answer:
http://www.madsci.org/posts/archives/oct99/941139504.Ph.r.html

That answer talks about two beams intersecting, but mbe carefully posed the setup so that they stayed that way.

scilover89, I am pretty sure your right, the wave is undectable, but it still exists because it has energy/mass (E=mc^2)

So, if they canceled out but were still there, that means it is theoretically possible to send a beam of light that contains an arbitrary large amount of energy, but is undetectable - which also means it can't be received.




You know, I think this could be simulated with water or sound - at least in principle - enough to observe the result.

 

[mbe]

Davec i agree with you but do you have any idea how i might set it up because what i need are beams of light.

 

[mbe]

i don't know how u would demonstrate it using water waves or sound waves because u can't columnate them whereas light beams are columnated.

Could it be that this time energy is not conserved for that brief moment. This shouldn't be that unbelivable since there are other times when energy is not conserved. ex: quantum fluctuation.

 

[vanesch]

If two photons were to arrive at a single spot by accident such that they were half a lambda out of phase they would destructively interfere. Where would the energy go?

In fact, two photons do not interfere. Stated a bit simplistically, a photon only interferes with itself.
Somewhat smarter: each photon can take several paths at once to get somewhere, and it are the different phase contributions of each path (for one and the same photon) that interfere constructively or destructively.
That's in fact the basis of the path integral formalism of quantum theory.

cheers,
Patrick.

 

[ZapperZ]

In fact, two photons do not interfere. Stated a bit simplistically, a photon only interferes with itself.
Somewhat smarter: each photon can take several paths at once to get somewhere, and it are the different phase contributions of each path (for one and the same photon) that interfere constructively or destructively.
That's in fact the basis of the path integral formalism of quantum theory.

cheers,
Patrick.

I just want to make sure that what vanesch has said here is not lost, nor the subtle but absolutely CRUCIAL meaning of what he has said. Keep in mind that Dirac (I think) has said that two photons almost NEVER interfere.

It is important that one has a clear understanding that within the QM picture, the interference pattern that we obtain from 1, 2, 3, 4, etc... multislit experiment is the intereference due to ONE photon.[1] The superposition of paths of that one single photon (or electron, or neutron, etc.) is the crucial ingredient here. 2-photon, 3-photon, etc... are very rare and are considered as higher-order effects that are NOT responsible for the typical interference pattern.[2]

So what vanesch said may appear trivial, but it is ALL of the interference effects.

Zz.

[1] L. Mendel "Quantum effects in one-photon and two-photon interference", Rev. Mod. Phys. v.71, p.274 (1999).
[2] T.B. Pittman et al., "Can two-photon interference be considered the inteference of two photons?", Phys. Rev. Lett. v.77, p.1917 (1996).

 

[DaveC426913]

I just want to make sure that what vanesch has said here is not lost, nor the subtle but absolutely CRUCIAL meaning of what he has said. Keep in mind that Dirac (I think) has said that two photons almost NEVER interfere.

Oh yeah. Duh. Um that's what I meant to say.

Nonetheless, you can still set up the experiment that way (though you'd have to use light, not water or sound). You'd still just split the beam and then put one a half wavelength out of phase.

I suspect though that diffraction would stop it from happening. You just can't line *every* photon up 1/2wv out of phase with itself.

 

[masudr]

Photons are packets of energy, not necessarily concentrated within certain points in space.

 

[michael879]

Ignore. Can't seem to delete this double post.
OK, edgardo, that answer talks about two beams intersecting, but mbe carefully posed the setup so that they *stayed* that way.

So, if they canceled out but were still there, that means it is theoretically possible to send a beam of light that contains an arbitrary amount of energy, but is undetectable. Surely that's impossible!



You know, I think this could be simulated with water or sound - at least in principle - enough to observe the result.

Ive seen this done with both water and sound, the wave becomes undetectable as long as the two waves overlap each other in the right way. You could never use this practically for information, since there is no way to have the wave reappear when you want it to, either it hits the transmitter a another wave faces directly at it, or the intersection is a point, not a line.

 

[El Hombre Invisible]

I was asked this very question not long ago and, after rooting around various books and the net, I got somewhat different answers to those posted above. According to Feynman (so forgive me if I'm decades out of date here), photons can only undergo positive interference, never negative. I assume this means the same as 'constructive' and 'destructive' interference, but bring forth the abuse if I'm wrong (I'm a QED newby). From a particle point of view this makes sense as either the photons occupy the same point in space and time and so would constructively interfere, or they are not and so would not interfere at all. It's certainly a nicer sounding explanation than the EM equivilent I found - that EM waves exactly 180 degs out of synch would 'redistribute' themselves to allow positive interference, a mysterious phenomenon that I have yet to find a mechanical explanation for. I'll butt out now.

 

[ZapperZ]

I was asked this very question not long ago and, after rooting around various books and the net, I got somewhat different answers to those posted above. According to Feynman (so forgive me if I'm decades out of date here), photons can only undergo positive interference, never negative. I assume this means the same as 'constructive' and 'destructive' interference, but bring forth the abuse if I'm wrong (I'm a QED newby). From a particle point of view this makes sense as either the photons occupy the same point in space and time and so would constructively interfere, or they are not and so would not interfere at all. It's certainly a nicer sounding explanation than the EM equivilent I found - that EM waves exactly 180 degs out of synch would 'redistribute' themselves to allow positive interference, a mysterious phenomenon that I have yet to find a mechanical explanation for. I'll butt out now.

No, it is not out of date. Since the pattern that is detected is for when a photon hits the detector, then where it doesn't hit is the "dark" pattern.

Again, if we go by the fact that the interference pattern that we detect came from one and only one photon at a time, the issue of "phase difference" for constructive and destructive interference between two or more photons does not come up.

Zz.