From: *smitra* <smi...@zonnet.nl <mailto:smi...@zonnet.nl>>
On 09-08-2019 07:54, Bruce Kellett wrote:
> On Fri, Aug 9, 2019 at 3:16 PM smitra <smi...@zonnet.nl
<mailto:smi...@zonnet.nl>> wrote:
>
>> On 09-08-2019 05:35, Bruce Kellett wrote:
>>
>>> It is really quite simple. If a state is a sum of two components,
>>> |psi> = (|A> + |B>), then we measure <psi|psi> = (<A| + <B|)(|A>
>> +
>>> |B>) = <A|A> + <B|B> + 2<A|B>. If <A|B> does not vanish (the
>>> components are not orthogonal), then there is interference. For
>>> orthogonal components <A|B> = 0, and there is no interference.
>>> Introducing separate states for the two slits does not aid
>>> comprehension here.
>>
>> Nonsense.
>
> What is nonsense? The fact that separate states for the two slits does
> not aid comprehension? Or the fact that orthogonal states do not
> interfere?
>
Your statement that orthogonal states don't interfere is plain nonsense.
Huh?? Did you not understand my example above of the state (|A> + |B>)?
There is interference in the norm only if |A> and |B> are not
orthogonal. This is elementary text book stuff.
>> What we observe at a point x on the screen is the expectation
>> value of the projection operator |x><x|.
>
> No, we don't observe an expectation value, which is a weighted average
> over possible outcomes. We measure a particular outcome at each point
> on the screen.
>
And that's precisely given by the expectation value of the projection
operator |x><x|, which is
<psi|x><x|psi> = psi*(x) psi(x) = |psi(x)|^2
That is the expectation value over all possible results. We only observe
one spot on the screen for each photon through the slits -- we do not
directly observe expectation values. The states <x|0> and <x|1> are not
orthogonal.
Bruce
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