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.
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.
Saibal
--
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/08af8238a838cb395acc8970b40f5463%40zonnet.nl.
