On Tue, Aug 13, 2019 at 7:41 PM smitra <[email protected]> wrote: > On 13-08-2019 05:14, Bruce Kellett wrote: > > 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. > > The states |A> and |B> have components psi_A(x) = <x|A>, and psi_B(X) = > <x|B>, these components depend on x, not the states |A> and |B>. We can > work with these components (wavefunctions) psi_A(x), and psi_B(x) > instead of |A> and |B> describing the particle just before it interacts > with the screen, but |A> and |B> are still orthogonal, the wavefunctions > psi_A(x), and psi_B(x) are obviously also orthogonal. You can argue that > when restricting the two wavefunctions in some small region near x = x0 > must yield two functions that are not orthogonal, otherwise there cannot > be interference at x = x0. However, what you are then doing is letting > the states collapse by letting them get localized near x0. So, you're > not talking about |A> and |B>, you are instead talking about two states > that are proportional to |x0> > > > > 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. > > As explained above, what you call |A> and |B> are in fact both > proportional to some arbitrarily chosen |x0>. You have effectively > replaced |A> by <x0|A> |x0> and |B> by <x0|B> |x0>, so you are letting > the two states collapse at some point x0 on the screen first which yield > states that depend on x0 and then you argue on the basis that the > squared norm of what you now have should have an interference term, > which then precludes the two states being orthogonal. But you made them > co-linear by letting them collapse at the same position first. >
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. > Before collapse you can write |A> in the position representation as > integral over <x|A>|x> d^3x and similarly B = integral over <y|B>|y> > d^3y, and we not only have that |A> and |B> are orthogonal, but also the > simple fact that |x> and |y> are orthogonal for x not equal to y. > I think you are over-elaborating what was intended as a simple schematic. I am glad that you finally recognize that orthogonal states do not show interference. 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/CAFxXSLRtjg_ZkSqTMzKTM7buRPfRABFZQrjCjpHqSNafqGgpXQ%40mail.gmail.com.
