Shooting a single photon at a 1/2 transparent 1/2 mirror 45 degree piece of glass

Question

I have very limited knowledge of any of this (read: none), but have wondered this for a number of years:

What would happen if one were to shoot a single photon at a piece of half transparent / half mirror glass that was angled at a 45 degree angle.

Assuming everything was done in vacume like conditions, the glass was EXACTLY 1/2 transparent, the angle was exactly 45 degrees, etc.

Would the photon pass through or would it be deflected?

I appoligze if this is not 'how things work', but you can see where I am going with this.

Answer 1

The answer depends on what you mean by "what happens?" It seems simple at first, but gets weird quickly.

To start out, let's put two detectors around the half-silvered mirror: one that catches the photons that are transmitted, one that catches photons that are reflected.

Only one photon is ever flying through the system. It seems impossible to predict which detector will pick up each photon, but we can observe that half will hit Detector 1 and half will hit Detector 2. Each photon only gets detected at one detector, never both. I've labeled the paths the photons take to each detector as 1 and 2.

We could say that when the photon hits the half-silvered mirror, it simply chooses path 1 or 2 at random with equal probability.

Let's complicate the setup a bit with fully reflecting mirrors and see what happens when the two paths are brought back together (L = Light source, HSM = Half-Silvered Mirror, D1/D2 = Detector 1/2).

The two paths cross, but nothing seems to have changed. It is still possible that each photon picks path 1 or 2 at random. The fact that the paths cross each other doesn't matter.

Now, let's add another half-silvered mirror at the crossing point (M = Mirror (fully reflecting)).

What?! You would think that no matter which path the photon took after the first half-silvered mirror (HSM), it would again have a 50-50 chance of either passing through or reflecting off the second half-silvered mirror, leading to each detector getting 50% of the photons. Somehow, mixing the beams with the second half-silvered mirror has resulted in all of the photons going into Detector 2.

We can conclude that the half-silvered mirror does not simply randomly reflect or transmit each photon. Each photon must somehow be present in both path 1 and path 2. We can see this by blocking one of the paths.

Blocking the upper path changes how many photons each detector receives. Half the photons are blocked, and the rest are split evenly between the two detectors. The presence or absence of path 1 affects how photons that go along path 2 react to the second half-silvered mirror. Remember, there is only ever one photon in the system, so each photon has to be present in both path 1 and path 2 at the same time (whatever "present" means in this case). The two paths interfere with each other, which would not be the case if the photon only took one path (since there would be nothing on the other path to interfere with it).

We can also imagine that, instead of blocking a path, we put a detector there that detects the photon and lets it through.

This new detector confirms that half the photons go along path 1, and therefore half go along path 2 (if it measures energy, it will also confirm that the photons have the same energy as the photons emitted from the light source, so half-silvered mirrors do not split photons in half). However, this detector also seems to destroy the weird effect when we first put in the second half-silvered mirror and all the photons ended up in Detector 2. We now have a situation where it seems like each photon randomly picks path 1 or path 2 and randomly goes into Detector 1 or Detector 2.

Without detecting which path a photon takes, the photon seems to go both ways simultaneously. Detecting which path collapses the possibilities into path 1 or path 2, but not both.

This is a start to the weirdness that is quantum mechanics.

Answer 2

It depends on how you measure. If you have two detectors, then only one would detect the photon. Each photon you shoot would randomly either pass through or reflect, but not both. Your detectors would never trigger simultaneously measuring a single photon or two "halves" of it in two places at once. The photon is an elementary particle that can be detected only once in one place.

However, a photon in flight also is a wave. So here comes the weird part... If instead of two separate detectors you would use another mirror to send the photon eventually to the same screen regardless of whether it passed through or reflected, then each photon would take both paths at the same time and interfere with itself on the scrreen.

Each single photon would still produce a single dot on the screen. However, given enough photons and the proper geometry of your setup, you may see the dots eventually bundle together to resemble lines of interference. There would be "dark lines" with a destructive interference where no photon "wants to go" and "bright lines" with a constructive interference where each photon "prefers to go".

In essence, this setup would be a version of the famous double-slit experiment. And of course, if you still use the detectors to see which path each photon takes, you would detect each photon only in one place and lose the interference.

Answer 3

One photon is splitting into two. Photon is not the the least energy Inseparable unit. This is proved by a lot experiments. For example Compton effect, in which a X ray photon collides to electron and lost energy, scattered photon is splitted from incoming one, its energy is part of the incoming one. Electron absorbs another part of the incoming photon and changes its velocity.