On 19-08-2019 07:56, Bruce Kellett wrote:
On Mon, Aug 19, 2019 at 2:55 PM smitra <[email protected]> wrote:
On 19-08-2019 03:52, Bruce Kellett wrote:
On Sat, Aug 17, 2019 at 10:28 AM smitra <[email protected]> wrote:
On 16-08-2019 09:01, Bruce Kellett wrote:
On Fri, Aug 16, 2019 at 4:43 PM smitra <[email protected]>
wrote:
I think you need to prove that. In my understanding, A(x) =
<x|A>
is
to be interpreted as the amplitude for a wave through slit A to
get to
the screen at x. There is nothing 3-dimensional about this. The
'x' is
just the distance from the centre of the screen in the plane of
the
screen. Nothing else is relevant. You do not have to integrate
over
all space because you use a complete set of states in the
x-direction:
int |x><x| dx = 1.
A quantum state is defined in the position representation by
assigning
an amplitude to all points in space. It can be the case that the
amplitude is zero outside of a narrow volume surrounding the the
screen,
in which case the integration can be approximated as an integral
over
the screen's surface.
The orthogonality can be rigorously proved as follows. If we
have a
single particle incident on the two slits described by a time
dependent
wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that
at
A(x,0)
is nonzero at one slit and B(x,0) at the other slit, then A(x,0)
and
B(x,0) are obviously orthogonal. Since time evolution will
preserve
inner products, A(x,t) and B(x,t) will remain orthogonal as a
function
of t. One can then describe the interaction with the screen as
an
effective collapse that will happen with the largest probability
when
the peak of the wavefunction has arrived at the screen.
The slits are orthogonal only if they are eigensatates of the
position
operator (in the x direction). That is the case only if you have
a
measurement that gives which-way information. Then there is no
interference, as advertised. The amplitudes at the slits aren't
orthogonal just because you say so.
The particle will be in one of the two eigenstates if such a
measurement
is made, but if we don't make that measurement then the particle
ends up
in a superposition of the two eigenstates.
Come on, Saibal. You are just blowing smoke. The particle is not in a
superposition of the two orthogonal eigenstates -- it is in a
superposition of an indefinite number of eigenstates after passing the
slits, and a different superposition at each slit, so there is a
non-zero overlap (It spreads out as a cylindrical wave after each
slit.). Integrating over all 3-space is not going to help you here. In
fact, that is a meaningless operation in the context. Integration over
the direction along the screen is, at most, all that is required for
the complete set of states. And that does not integrate to zero.
If there is no overlap immediately after the wavepacket spreads out of
the slits, then there can't be any overlap later. Immediately after
leaving the slits the amplitudes for left slit is zero at the right slit
and vice versa, so that there is no overlap immediately after the
wavepacket spreads out the slits is trivially true. The overlap is then
zero at all later times even when the wavepackets emanating from the two
slits cross and "overlap" in the colloquial sense, due to unitary time
evolution: If U(t) is the time evolution operator and |psi_j(t)> is the
state the particle would be in, if only slit j were open, then
<psi_1(t)|psi_2(t)> = <psi_1(0)|U^{\dagger}(t)U(t)|psi_2(0)> =
<psi_1(0)|psi_2(0)>
Therefore, if both slits are open and the initial state is some
superposition of the two states, then it will remain a superposition of
two orthogonal states, even though the two orthogonal components will
"overlap" in the colloquial sense. Orthogonality involves overlap in the
mathematical sense: <psi_1|psi_2> = 0
And we may evaluate this in the position representation by inserting a
complete set of position eigenstates:
Int d^3x |x><x| = 1, therefore:
0 = <psi_1|psi_2> = int d^3x <psi_1|x><x|psi_2> = int d^3x psi_1(x)*
psi_2(x)
So, an integral over all space will yield zero. This means that if an
integral over only the screen would not yield zero, then that
integration doesn't capture all the regions where the wavefunctions have
an overlap in the colloquial sense.
The point raised about there being many different possible states for a
particle when only one slit is open doesn't change the above argument.
What matters is that after some initial wavepacket moves through both
slits, it can be written as a superposition of the two states it would
be in corresponding to only 1 of the two slits being open. And these two
states are orthogonal per the above argument.
Saibal
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