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19th April 2008, 12:11 PM | #1 |
Muse
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A question on quantum mechanics
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:
Is this right? If not, can someone explain what the answer should be? |
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20th April 2008, 12:33 AM | #2 |
Illuminator
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I agree with your explanation, but not with your numbers.
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20th April 2008, 01:38 AM | #3 |
Thinker
Join Date: Aug 2007
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I get a different result
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20th April 2008, 02:12 AM | #4 |
Illuminator
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20th April 2008, 09:20 AM | #5 |
Philosopher
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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? |
20th April 2008, 01:04 PM | #6 |
Illuminator
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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.) |
20th April 2008, 08:37 PM | #7 |
Philosopher
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20th April 2008, 10:02 PM | #8 |
Muse
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I'm trying to expand on this series of blog posts. I've ordered Feynman's QED but it hasn't arrived yet.
That's what I was assuming. 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. |
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21st April 2008, 07:10 AM | #9 |
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21st April 2008, 09:46 AM | #10 |
Muse
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21st April 2008, 10:48 AM | #11 |
Sarcastic Conqueror of Notions
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"Great innovations should not be forced [by way of] slender majorities." - Thomas Jefferson The government should nationalize it! Socialized, single-payer video game development and sales now! More, cheaper, better games, right? Right? |
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21st April 2008, 11:03 AM | #12 |
Philosopher
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21st April 2008, 11:19 AM | #13 |
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21st April 2008, 07:07 PM | #14 |
Illuminator
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21st April 2008, 08:11 PM | #15 |
Philosopher
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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. |
21st April 2008, 09:16 PM | #16 |
Muse
Join Date: Aug 2005
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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. |
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