Re: Quantum Rebel

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Sat, 14 Aug 2004 10:40:08 -0400

Russell Standish wrote:
>
>Let |i> refer to the state where the photon travels on path i. Then
>one can write down a few relations, such as:
>
>|1> = 1/sqrt{2}|3> + 1/sqrt{2}|4> = |5>
>|2> = 1/sqrt{2}|3> - 1/sqrt{2}|4> = |6>
>
>If a photon is detected on path 5, then the probability it travelled
>along path i is <5|i>. Since <5|1>=1 and <5|2>=0, we have "which way"
>information.
>
>Now inserting an absorber on path 4 is mathematically equivalent to
>inserting a projection operator |3><3| in the middle of the
>propagator. The the probability of a photon detected at path 5 taking
>path i becomes <5|3><3|i>. Computing these values by the above
>formulae gives:
>
> <5|3><3|1>=1/2 and <5|3><3|2>=1/2

Thanks for the elaboration, it's been a while since I studied QM. A
question: I had thought the notion of "probability" only makes sense when
talking about actual measured outcomes, and that paths in a path integral
can only be assigned a probability amplitude, not a probability, since if
you tried to talk about the "probability" of each path (just by squaring the
path's amplitude, I guess) the probabilities would not necessarily add
together classically. Is my memory wrong, or when you talk about the
"probability" that a photon took a path i do you really mean the probability
amplitude?

Jesse Mazer
Received on Sat Aug 14 2004 - 10:46:41 PDT