On 12-07-2021 03:42, Bruce Kellett wrote:
On Mon, Jul 12, 2021 at 8:43 AM Bruce Kellett <[email protected]>
wrote:
On Mon, Jul 12, 2021 at 3:44 AM smitra <[email protected]> wrote:
On 11-07-2021 02:46, Bruce Kellett wrote:
This is where you go seriously wrong. Simply recording where the
photons land does not quantum erase the information they
carried. Once
the photons carry off the which way information, the
interference
pattern is restored only if the information carried by the
photons is
quantum erased. Simply running the photons into a screen (or the
wall), even if you record where they land, is not quantum
erasure.
See, for example, the paper arXiv:1206.6578 on quantum erasure.
In
this paper they say "the presence of path information anywhere
in the
universe is sufficient to prohibit any possibility of
interference. In
other words, the atoms' path states alone are not in a coherent
superposition due to the atom-photon entanglement." This
transfers
directly to the buckyball experiment under discussion. Running
the
photons into a screen, or the wall, does not destroy the
ball-photon
entanglement.
Recording where the photons land on the screen is enough, this is
very
easy to see. Let's consider an interference experiment where the
particle gets entangled with another particle that carries away
the
which way information. If we work in the position representation
then we
have a wavefunction:
psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)] (1)
where x is the position of particle 1 just before it hits the
screen and
y the position of the other particle at that time, and psi_1(x,y)
is the
wavefunction when only slit 1 is open while psi_2(x,y) the
wavefunction
with only slit 2 open. We also assume that particle 2 carries the
which
way information, this means that psi_1(x,y) and psi_2(x,y) for x
kept
fixed and considered as a function of y are eigenfunctions with
different eigenvalues of an observable that corresponds to
extracting
the which way information from the second particle. This then
implies
that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:
Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)
This is the same mistake that you have routinely made in all of
these discussions. You calculate for particles through each slit
independently. But if that is the case, then no interference can
ever be seen. The possibility of interference relies on coherence
between the amplitudes at the two slits. In your calculations, you
have never allowed for this coherence -- the amplitudes at the two
slits are not independent.
To see this in more detail, we can go through your original argument.
You write : denote the buckyball moving through the left slit by |L>
and the right slit by |R> and the photons by |PL> and |PR>. Now the
states |R> and |L> are coherent; thus correlated; and the overlap
<L|R> does not vanish. Writing
|psi> = (|L>|PL> + |R>|PR>)
is the same as assuming that the photons are emitted from the slits if
they detect a ball passing through that slit. This is a direct
"which-way" measurement, and since the photon states are independent,
they are, as you say, orthogonal, and there can be no interference if
such measurements are made at the slits. This is the essential loss of
coherence.
Your claim then is that if we observe the photons on another screen
and keep track of the ball positions on their screen for photons that
arrive at the same point on this other screen, then the interference
pattern is restored. This is wrong, because the photons emitted from
the slits if and when they detect a ball passing through the
corresponding slit are still independent. The fact that they might
arrive at the same point on a further screen is irrelevant. The photon
states are still independent. In fact, there can only ever be one
photon for each buckyball, so there can be no case in which two
photons arrive at the photon screen at the same time. So the photon
states are necessarily always independent and orthogonal, and your
overlap function
<y|PL_t><y|PR_t> = 0, always.
No, we perform a detection at a point y on the photon screen, the
product of these amplitudes are not zero, it only becomes zero if you
integrate this over y. The rest of your argument does not apply
Saibal
Consequently, your joint detection of photons does not restore the
interference. Decoherence, once present, destroys coherence. And this
coherence cannot be restored simply by observing the which-way
photons. You have to quantum erase the which-way information. And this
is not easy for this particular set-up. Quantum erasure essentially
requires a measurement of a conjugate variable, measurement of which
means that the original state cannot be recovered. For example,
measurement of photon polarization at 45 degrees will quantum erase a
horizontal/vertical polarization state.
Going back to the original buckyball experiment, the balls are laser
heated and this puts them in an excited state. The state then decays
by IR emission. This happens after the balls have passed the initial
slits, so it is not a direct which-way measurement. It is only as the
temperature increases, and the emitted photons have shorter
wavelengths, that the wavelength is such that the photon can
discriminate between the slits, that which-way information can be
recovered from the IR photons. Letting the photons go undisturbed to
infinity, or detecting them on some other screen, does not destroy the
which-way information that they carry. As long as these photons are
created, the interference pattern of the balls is lost. The photon
information could be quantum-erased, I suppose, but I have no idea how
that might be achieved.
Your analyses, neither the first nor subsequent analyses, come
anywhere near explaining this observed behaviour.
Bruce
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