 |
|
|
|
|
|
| |
| |
|
|
|
|
| |
| |
|
|
What if instead of having a double-slit, you have a triple-slit?
I suppose the interference pattern would be that which can be expected
from a wave which goes through the three slits.
However, what happens if you put a detector in one of the slits?
--
- Warp
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Warp wrote:
> What if instead of having a double-slit, you have a triple-slit?
> I suppose the interference pattern would be that which can be expected
> from a wave which goes through the three slits.
>
> However, what happens if you put a detector in one of the slits?
As I understand it, the only way to detect a photon is to destroy it.
Then again, I'm hardly an expert...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Orchid XP v8 <voi### [at] dev null> wrote:
> Warp wrote:
> > What if instead of having a double-slit, you have a triple-slit?
> > I suppose the interference pattern would be that which can be expected
> > from a wave which goes through the three slits.
> >
> > However, what happens if you put a detector in one of the slits?
> As I understand it, the only way to detect a photon is to destroy it.
That's not what I was asking.
--
- Warp
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Warp wrote:
> What if instead of having a double-slit, you have a triple-slit?
> I suppose the interference pattern would be that which can be expected
> from a wave which goes through the three slits.
I think that's right, yes.
> However, what happens if you put a detector in one of the slits?
I *think* what would happen (by analogy with other experiments I'm familiar
with) is that each time you detected an electron, it would land on the
screen in the probability it would have if you only had the one slit. I.e.,
you know what slit it went through, so you'll have a fuzzy dot behind that
slit from those electrons. If you don't see it go through that slit, it'll
land in the pattern you'd get if you had the two open slits and not the
monitored slit.
Or, in other words, as I understand it, you'll get (detected plus slit A)
overlapped on (not detected plus slit B or C) patterns, so some interference
and some not interference. An interference pattern with a blob behind the
slit with the detector. And if you actually note which dot on the screen
went with the detector's triggering and which dot didn't (assuming they're
coming slowly enough to sort out), you can separate the picture into a pure
blob and a pure interference pattern.
(And the "delayed quantum erasure" experiment does essentially this, except
they don't put the slit-detector in place until the electron has already hit
the screen. And that's the results they get. Freaky, huh?)
Of course, there is some probability that it goes thru the slit with the
detector and you don't detect it, so you'll get a faint pattern that falls
into that probability distribution as well, giving you a triple-slit pattern.
Even with a double-slit pattern with both slits monitored, you get
"interference patterns" between electrons that manage to go through without
being detected at the slits at all.
The way the probability works is the same addition and multiplication rules
as regular probability, except with complex numbers. So the probability of a
particle leaving the detector, being detected going thru slit A, and hitting
spot X on the screen is the product of those probabilities. The probability
of leaving the detector, going through slit B or slit C, then hitting spot X
on the screen is the sum (in some real+imaginary sense) of going through
slit B and hitting spot X or thru slit C and hitting spot X. The
"interference pattern" is caused by the fact that when you add two complex
numbers, even if they're both unit length, you can wind up with any
resultant length between 0 and 2.
--
Darren New, San Diego CA, USA (PST)
The question in today's corporate environment is not
so much "what color is your parachute?" as it is
"what color is your nose?"
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Darren New wrote:
> familiar with) is that each time you detected an electron, it would land
> on the screen in the probability it would have if you only had the one
> slit.
... if it was detected. Otherwise, you get a 2-slit interference pattern.
Sorry I didn't finish that sentence.
--
Darren New, San Diego CA, USA (PST)
The question in today's corporate environment is not
so much "what color is your parachute?" as it is
"what color is your nose?"
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Warp <war### [at] tagpovrayorg> wrote:
> What if instead of having a double-slit, you have a triple-slit?
> I suppose the interference pattern would be that which can be expected
> from a wave which goes through the three slits.
>
> However, what happens if you put a detector in one of the slits?
>
It measures a wave.
Post a reply to this message
|
|
| |
| |
|
|
From: Phil Cook v2
Subject: Re: Question about the double-slit experiment
Date: 5 Mar 2010 05:38:13
Message: <4b90df15@news.povray.org>
|
|
|
| |
| |
|
|
On 22/02/2010 21:00, Warp wrote:
> What if instead of having a double-slit, you have a triple-slit?
> I suppose the interference pattern would be that which can be expected
> from a wave which goes through the three slits.
>
> However, what happens if you put a detector in one of the slits?
As I understand it is that with two slits and no detector the particle
passes through slits 1 and 2, with a detector it passes through 1 or 2.
So with three slits and no detector you'd get an interference pattern as
it passes through slits 1, 2, and 3; with one detector on 3 it'd pass
through slits 1 and 2, or 3.
--
Phil Cook
--
I once tried to be apathetic, but I just couldn't be bothered
http://flipc.blogspot.com
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
|
|
