A question on quantum mechanics

[GreedyAlgorithm]

A photon is coming in where the arrow is. The red lines are half-silvered mirrors, and the green lines with ticks on their ends are full mirrors. The blue circles are detectors. Here's what I'm getting as the probability of detectors 1, 2, or 3 going off given that a detector goes off:

D1=1/6, D2=1/6, D3=4/6
This is since all of the amplitude that you'd think was coming from the path where the photon goes down first, bounces around, and hits 1 or 2 is in fact canceled out.

Is this right? If not, can someone explain what the answer should be?

[69dodge]

I agree with your explanation, but not with your numbers.

For detectors 1, 2, and 3, I get probabilities of 1/4, 1/4, and 1/2.

The photon is equally likely to go through the first mirror or to be reflected from it. If it goes through, it will certainly end up at detector 3. If it is reflected, it will end up at 1 or 2 with equal probability.

So to speak.

I think.

[vidiviciveni]

I get a different result

For detectors 1, 2, and 3, I get probabilities of 3/8, 3/8, and 1/4.

[69dodge]

Originally Posted by vidiviciveni

I get a different result

For detectors 1, 2, and 3, I get probabilities of 3/8, 3/8, and 1/4.

Did you consider the possibility of interference?

[sol invictus]

The question is ill-posed. The answer depends on both the thickness of the mirrors (in units of the wavelength of the light) and their orientation.

Where did you get this from?

[69dodge]

I assumed for each mirror that the reflected wave is delayed in phase by pi/2 radians relative to the transmitted wave and that both have equal intensity, regardless of which side of the mirror the incoming wave hits.

Apparently this is reasonable? (See the end of section IV.A.)

[sol invictus]

Originally Posted by 69dodge

I assumed for each mirror that the reflected wave is delayed in phase by pi/2 radians relative to the transmitted wave and that both have equal intensity, regardless of which side of the mirror the incoming wave hits.

That depends on the type of beam splitter and on the wavelength of the light. Half-silvered mirrors (one type of beam splitter) aren't symmetrical like that.

[GreedyAlgorithm]

Originally Posted by sol invictus

The question is ill-posed. The answer depends on both the thickness of the mirrors (in units of the wavelength of the light) and their orientation.

Where did you get this from?

I'm trying to expand on this series of blog posts. I've ordered Feynman's QED but it hasn't arrived yet.

Originally Posted by 69dodge

I assumed for each mirror that the reflected wave is delayed in phase by pi/2 radians relative to the transmitted wave and that both have equal intensity, regardless of which side of the mirror the incoming wave hits.

That's what I was assuming.

Originally Posted by sol invictus

That depends on the type of beam splitter and on the wavelength of the light. Half-silvered mirrors (one type of beam splitter) aren't symmetrical like that.

Suppose instead I said it was the symmetrical type. In fact, if such a thing exists, suppose it's a device where the two dominant flows are: 1) the phase is not changed nor the angle; 2) the phase is delayed pi/2 and the flow is reflected.

[sol invictus]

Originally Posted by GreedyAlgorithm

Suppose instead I said it was the symmetrical type. In fact, if such a thing exists, suppose it's a device where the two dominant flows are: 1) the phase is not changed nor the angle; 2) the phase is delayed pi/2 and the flow is reflected.

OK - in that case, I get:

1/3,1/3,1/3

[Dilb]

Originally Posted by 69dodge

I agree with your explanation, but not with your numbers.

For detectors 1, 2, and 3, I get probabilities of 1/4, 1/4, and 1/2.

The photon is equally likely to go through the first mirror or to be reflected from it. If it goes through, it will certainly end up at detector 3. If it is reflected, it will end up at 1 or 2 with equal probability.

So to speak.

I think.

I agree with these numbers.

[Beerina]

Originally Posted by GreedyAlgorithm

A photon is coming in where the arrow is. The red lines are half-silvered mirrors, and the green lines with ticks on their ends are full mirrors. The blue circles are detectors. Here's what I'm getting as the probability of detectors 1, 2, or 3 going off given that a detector goes off:

D1=1/6, D2=1/6, D3=4/6
This is since all of the amplitude that you'd think was coming from the path where the photon goes down first, bounces around, and hits 1 or 2 is in fact canceled out.

Is this right? If not, can someone explain what the answer should be?

I get:

#      %       x/y
1     37.5     3/8
2     37.5     3/8
3     25       1/4

The discrete breakdown is this:

1/2 up at mirror A
	1/4 up at mirror D, to 1
	1/4 down at mirror D, to 2
1/2 down
	1/4 up at mirror B
		1/8 up at mirror C
			1/16 up at mirror D, to 1
			1/16 down at mirror D, to 2
		1/8 down at mirror C, to 3
	1/4 down at mirror B
		1/8 up at mirror C
			1/16 up at mirror D, to 1
			1/16 down at mirror D, to 2
		1/8 down at mirror C, to 3

[sol invictus]

Originally Posted by Beerina

I get:

#      %       x/y
1     37.5     3/8
2     37.5     3/8
3     25       1/4

The discrete breakdown is this:

1/2 up at mirror A
	1/4 up at mirror D, to 1
	1/4 down at mirror D, to 2
1/2 down
	1/4 up at mirror B
		1/8 up at mirror C
			1/16 up at mirror D, to 1
			1/16 down at mirror D, to 2
		1/8 down at mirror C, to 3
	1/4 down at mirror B
		1/8 up at mirror C
			1/16 up at mirror D, to 1
			1/16 down at mirror D, to 2
		1/8 down at mirror C, to 3

That doesn't take interference into account.

I think my answer is correct (with the assumptions spelled out above).

[sol invictus]

Originally Posted by 69dodge

I agree with your explanation, but not with your numbers.

For detectors 1, 2, and 3, I get probabilities of 1/4, 1/4, and 1/2.

The photon is equally likely to go through the first mirror or to be reflected from it. If it goes through, it will certainly end up at detector 3. If it is reflected, it will end up at 1 or 2 with equal probability.

So to speak.

I think.

By the way, the problem with this logic is that the intensity of the light that ends up at 3 is 1/4, not 1/2 (because there are two paths, each passing through three beam splitters, for a total of 1/8+1/8=1/4). So I think the answer is

1/4,1/4,1/4, which (since we were asked for probabilities, not intensities) is 1/3,1/3,1/3

.

[69dodge]

Originally Posted by sol invictus

By the way, the problem with this logic is that the intensity of the light that ends up at 3 is 1/4, not 1/2 (because there are two paths, each passing through three beam splitters, for a total of 1/8+1/8=1/4).

But don't those two paths interfere constructively, because they're in phase at the end, each containing two transmissions and one reflection, so that the resulting intensity is more than just the sum of the individual intensities?

[sol invictus]

You're saying the amplitude gets decreased at each splitter by a factor of the square root of 2, not 2... yes, you're right.

I was wrong. I agree with your numbers.

[GreedyAlgorithm]

Originally Posted by sol invictus

You're saying the amplitude gets decreased at each splitter by a factor of the square root of 2, not 2... yes, you're right.

I was wrong. I agree with your numbers.

Ah! Now this finally makes sense to me. Let's see if I've got it: After the first splitter, we've i/sqrt(2) going NE and 1/sqrt(2) going SE. That top i/sqrt(2) will bounce off of the top mirror to become -1/sqrt(2), then hit detectors 1 and 2 with -i/2 and -1/2 respectively. Meanwhile the bottom splits again into i/2 going NE and 1/2 going SE, bouncing off mirrors to become -1/2 SE and i/2 NE. Then comes another splitter - going NE we get (-i/2sqrt(2))+(i/2sqrt(2))=0, and going SE we get (-1/2sqrt(2))+(-1/2sqrt(2))=-1/sqrt(2) at detector 3.

Probabilities come from the squared abs of amplitude which give us 1/4, 1/4, and 1/2.