On Tue, Aug 13, 2019 at 12:16 PM smitra <[email protected]> wrote: > On 13-08-2019 02:49, Bruce Kellett wrote: > > On Tue, Aug 13, 2019 at 10:43 AM smitra <[email protected]> wrote: > > > >> On 13-08-2019 01:41, Bruce Kellett wrote: > >>> On Tue, Aug 13, 2019 at 9:09 AM smitra <[email protected]> wrote: > >>> > >>>> On 12-08-2019 08:29, Bruce Kellett wrote: > >>>>> > >>>>> Look at this another way. It is just an illustration of > >>>>> complementarity. Measuring which slit the photon went through > >> is > >>>> a > >>>>> position measurement at the slits. Measuring the interference > >>>> pattern > >>>>> at the screen is equivalent to a momentum measurement at the > >>>> slits. > >>>>> Such measurement operators do not commute -- the measurements > >> are > >>>>> complementary and cannot be performed simultaneously. > >>>>> > >>>> > >>>> It doesn't matter for orthogonality of the states whether or not > >>>> they > >>>> are measured. > >>> > >>> Of course it does. The slits are not orthogonal states unless > >> they are > >>> measured position eigenstates. If they are not measured, they are > >>> individually superpositions of many position eigenstates > >> (including > >>> eigenstates that overlap both slits), so the slits themselves are > >> no > >>> longer orthogonal. Orthogonal states cannot interfere, that is > >> why a > >>> position measurement at the slits makes the interference pattern > >> on > >>> the screen disappear. > >>> > >>> The fact remains, that orthogonal states cannot interfere: > >>> > >>> (<A| + <B|)(|A> + |B>) = <A|A> + <B|B> + 2 <A|B> > >>> > >>> and the interference term <A|B> vanishes if |A> and |B> are > >>> orthogonal. You can't get away from this basic fact about quantum > >>> mechanics. > >>> > >> > >> <A|B> is zero in the two slit experiment, if you integrate the > >> interference term over the screen you get zero. > >> > >> Thing is that the interference we can observe at some position x on > >> the > >> screen is Re[<A|x><x|B>], which for general x is nonzero despite > >> the > >> fact that <A|B> = 0. > > > > So you agree that if the overlap vanishes you do not get interference. > > You go to some lengths to try and avoid this fact. Saying that the > > integral over the screen vanishes is beside the point. > > > <A|B> is the integral over all space of <A|x><x|B> and this is zero, but > the interference we observe at some point x on the screen is > Re[<A|x><x|B>], and that's in general nonzero even for orthogonal |A> > and |B>. >
I don't know what you are trying to prove, but that is not what my formula meant. What I intended was that |A> represents the wave at the screen coming from slit A, and |B> is the wave at the screen from slit B. There is already and implicit dependence on the position along the screen A = A(x), etc. So inserting a complete set of position states, \int dx |x><x| is entirely redundant, and serves only to obscure the point. If we take a general state |A>, which depends on x, the overlap of two such is <A|B>, and this cannot vanish if there is to be interference. Classically, the intensity at x from slit A is |A|^2, and similarly for slit B. The fact that quantum mechanically we add amplitudes, not intensities, means that there is the overlap term <A|B>. If this does not vanish, [ |A> and <B| are NOT orthogonal], then we have interference. There is no more to it than that. The waves at the screen are not orthogonal! It really has nothing to do with whether the slit states are orthogonal or not. In general, they are not. Bruce -- 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/CAFxXSLScH0pCEA1%2BdQm9%3DT6BwujQKz7vCVp2QQF-SNpcnSv0DQ%40mail.gmail.com.
