• Quick note - the problem with Youtube videos not embedding on the forum appears to have been fixed, thanks to ZiprHead. If you do still see problems let me know.

A question on quantum mechanics

GreedyAlgorithm

Muse
GreedyAlgorithm
Joined
Aug 29, 2005
Messages
569
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?
 

Attachments

  • qm.jpg
    qm.jpg
    5.1 KB · Views: 14
I agree with your explanation, but not with your numbers. :confused:

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 get a different result

For detectors 1, 2, and 3, I get probabilities of 3/8, 3/8, and 1/4.
 
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 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.)
 
69dodge said:
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.
Click to expand...

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.
 
sol invictus said:
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?
Click to expand...
I'm trying to expand on this series of blog posts. I've ordered Feynman's QED but it hasn't arrived yet. :)

69dodge said:
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.
Click to expand...
That's what I was assuming.

sol invictus said:
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.
Click to expand...
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.
 
GreedyAlgorithm said:
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.
Click to expand...

OK - in that case, I get:

1/3,1/3,1/3
 
69dodge said:
I agree with your explanation, but not with your numbers. :confused:

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.
Click to expand...

I agree with these numbers.
 
GreedyAlgorithm said:
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?
Click to expand...


I get:
Code: 
#      %       x/y
1     37.5     3/8
2     37.5     3/8
3     25       1/4

The discrete breakdown is this:

Code: 
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

 
Last edited:
Beerina said:

I get:
Code: 
#      %       x/y
1     37.5     3/8
2     37.5     3/8
3     25       1/4

The discrete breakdown is this:

Code: 
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

Click to expand...

That doesn't take interference into account.

I think my answer is correct (with the assumptions spelled out above).
 
69dodge said:
I agree with your explanation, but not with your numbers. :confused:

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.
Click to expand...

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
.
 
sol invictus said:
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).
Click to expand...

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?
 
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.
 
sol invictus said:
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.
Click to expand...

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.
 
Last edited:
You must log in or register to reply here.

Back
Top Bottom