On 07-07-2021 02:45, 'Brent Meeker' via Everything List wrote:
On 7/6/2021 12:21 AM, smitra wrote:
On 05-07-2021 12:18, Bruce Kellett wrote:
On Mon, Jul 5, 2021 at 7:39 PM smitra <[email protected]> wrote:
On 05-07-2021 09:00, Bruce Kellett wrote:
On Mon, Jul 5, 2021 at 2:23 PM smitra <[email protected]> wrote:
I don't think this is actually done in the experiment. What is
observed is the presence or absence of the interference pattern on
the
screen where the balls hit. The photons are not detected. But if,
in
principle, they are of suitable wavelength to resolve the slit
difference, then the interference pattern vanishes. The experiment
is
convincing in that they start wil cold buckyballs which show a
clear
interference pattern. They then gradually heat the balls so that
the
typical wavelength of the photons decreases. This gradually washes
out
the interference pattern. (Because at lower temperatures, the
wavelength distribution of the IR photons is such that a few of
them
have shorter wavelengths.) As the temperature is increased so that
most IR photons have short enough wavelengths, the interference
pattern disappears completely. The paper by Hornberger et al. is
at
arXiv:quant-ph/0412003v2
This is then what I said previously, what you denied, i.e. that you
are
only considering part of the system which is defined by the reduced
density matrix. The complete system of buckyball plus photons will
show
interference, even if the wavelength is small enough to resolve the
slits provided you perform the right sort of measurement on the
balls
and photons.
That is false.
This is easy to see. Denote the buckyball state of a buckball moving
through the left slit by |L> and moving through the right slit by |R>.
Suppose that a photon is emitted by the by the buckyballs such that
the ball moving through the left slit emits a photon in a state |PL>
that will be orthogonal to the state |PR> of the photon emitted by the
ball moving through the right slit . The state of the system after the
ball passes the slits is then:
|psi> = 1/sqrt(2) [|L>|PL> + |R>|PR>]
This state then evolves under unitary time evolution, we can write the
state just before the ball hits the screen as:
|psi_s> = 1/sqrt(2) [|L_s>|PL_s> + |R_s>|PR_s>]
There is then no interference patter on the screen for the buckyballs
because |PL_s> and |PR_s> are orthogonal,
In the Bucky Ball experiment there's no interference pattern when the
photons have long wave length. So it's not just a question of the
states being orthogonal.
If there are many photons, then you have large number of inner products
between the unobserved photon states and the interference pattern then
gets suppressed as well.
the unitary time evolution preserves the orthogonality of the initial
states. The probability to observe a buckyball on position x on the
screen is:
P(x) = ||<x|psi_s>||^2 = 1/2 [|<x|L_s>|^2 + |<x|R_s>|^2] + Re[<x|L_s>
<x|R_s>* <PR_s|PL_s>]
And the last interference term is zero because <PR_s|PL_s> = 0
But if we also observe the photon on another screen and keep the joint
count for buckyballs landing on spot x on the buckyball screen and for
photons landing on spot y on the photon screen as a function of x and
y, then we do have an interference pattern as a function of x for
fixed y. If we de note by U the unitary time evolution for the photons
until they hit their screen, and put |PL_t> =U|PL_s> and |PR_t> =
U|PR_s>, then the probability distribution is:
P(x,y) = |<x,y|U|psi_s>|^2 = 1/2 [|<x|L_s>|^2|<y|PL_t>|^2 +
|<x|R_s>|^2|<y|PR_t>|^2] +Re[<x|L_s> <x|R_s>* <y|PL_t><y|PR_t>*]
The interference term Re[<x|L_s> <x|R_s>* <y|PL_t><y|PR_t>*] does not
vanish as it involves evaluating the components of the buckyball and
photon states in the position basis and so there is no inner product
involved anymore. For fixed y the quantity <y|PL_t><y|PR_t>* will have
some value that will be nonzero in general, so if we keep y fixed then
there will be an interference term.
The position operators are projections and include decoherence. I
don't think having fixed y will recover the interference pattern, but
I'm not clear on what y is measuring? Are there just two spots on the
y-screen corresponding to L and R slits?
Y is measuring the position at which the photon is landing on the screen
for the photon. When you measure y and x then there is an interference
in the joint probability distribution. For every fixed y, the function
P(x,y) will show interference as a function of x. You can also say that
measuring y destroys the which way information present in the photon
states.
So, we can conclude that invoking escaping IR photons does not male
any sense in this discussion because all it does is it scrambles the
interference pattern to make it invisible in a way that allows it to
be recovered in principle using measurements on those IR photons. You
can, of course, erase the interference patter by measuring the
observable for the photons that has |PR> and |PL> as its eigenstates.
But even in that case the information will still be there in the state
of all the atoms of the measurement apparatus for the photons. But if
you don't perform any measurement then the information will simply
continue to exists in the escaping photons.
And per the first sentence of the paragraph the interference will be
eliminated by the escaping IR photons. Are you contradicting this in
the last sentence? The experiment showed the interference
disappeared as the IR photon wave length decreased. It said nothing
about them being observed or escaping into space.
Brent
I don't disagree with that, if you don't measure y then there is no
interference pattern. And even if you do measure y, you would then have
to count the number of buckyballs on the screen only when the photon
lands on some specific position. If you just consider the total numbers
of buckyballs irrespective of where the photon is detected then the
interference pattern will vanish again, because you are then integrating
over the term <y|PL_t><y|PR_t>* which yields the inner product between
the photon states which was assumed to be zero.
Saibal
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