On Wed, Aug 14, 2019 at 8:28 AM smitra <[email protected]> wrote:
> On 13-08-2019 13:33, Bruce Kellett wrote:
> >
> > Of course A(x) and B(x) refer to the same point on the screen. That is
> > not a collapse, that is just what the notation means.
>
> A(x) and B(x) considered as the representations of |A> and |B> in the
> position basis, i.e. A(x) = <x|A> and B(x) = <x|B> are still orthogonal
> states, as they represent the orthogonal states |A> and |B>:
>
> 0 = <A|B> = Integral over x of <A|x><x|B>d^3x = Integral over x of
> A*(x)B(x) d^3x
>
I don't think this really works out. You are claiming that the integral of
the interference terms over the whole screen vanishes. If we look at the
usual derivation of the interference from two slits, we get something like
Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))
where the term involving the angle beta is the superposed diffraction
pattern from the finite width of the slits. The cos (delta) term is the
interference, but it has this form only in a small angle approximation, and
the phase difference delta is, of course, limited by the separation of the
slits. So, although the cos(delta) term may integrate to zero over small
angles, the presence of the diffraction envelope, and the limitations of
the small angle approximation, mean that is almost certainly will not
vanish when integrated over the whole screen.
So <A|B> will not vanish in general. Which is what I would have thought
because the paths through the separate slits are not independent -- each
particle essentially has to see both slits (go through both slits) in order
to maintain coherence. So they cannot be orthogonal (independent).
In practice, to see the interference pattern you need coherent illumination
over both slits. This is easy these days with lasers, but in older books,
coherence was ensured by having a preparatory single slit followed by
suitable condenser lenses. If the slits could be treated as independent
entities, this would not have been necessary.
But you interpret this as the total counts of particles on the entire
> screen not changing which you call an absence of interference. However
> interference is what we detect locally on each point on the screen. You
> can't say that for each point x0 on the screen, A(x0) and B(x0) are
> particle states. These values are not the quantum states of the particle
> before it hits the screen,
No, they are the amplitudes of the wave function at each point on the
screen. This is what give the probability of the particle being detected
(by the screen) at this point.
Bruce
unless you would have done a measurement
> localizing the particle near x0.
>
> So, your argument only makes sense if you invoke collapse via a position
> measurement.
>
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/everything-list/CAFxXSLR_LU5MzJxwxxx2xw8ZGRNQKtP_MP1mjMhQku%3DGm4XjJA%40mail.gmail.com.
