On 10-07-2021 07:58, Bruce Kellett wrote:
On Sat, Jul 10, 2021 at 10:21 AM smitra <smi...@zonnet.nl> wrote:

On 08-07-2021 01:51, Bruce Kellett wrote:

Do you dispute that that is what the paper by Hornberger et al.
says?

Bruce

I don't dispute these results. The buckyballs are coming from a
thermal
reservoir at some finite temperature. We can avoid working with
mixed
states by simply considering the interference pattern for each pure
state separately and then summing over the probability distribution
over
the pure states. But for this discussion we want to focus on what
the
interference pattern will be if all the buckyballs are in the same
exited state. If we put the entire system ina finite volume then we
have
a countable set of allowed k-values for the photon momenta. We then
have
a set of allowed states for the photons defined by the allowed
momenta
and a polarization. We can then label these photon states using a
number
and then specify an arbitrary state for the photons by specifying
how
many photons we have in each state, and these numbers can then be
equal
to zero. If they are all zero then no photons are present.

If only the right slit is open, the state of the buckyball and the
photons just before the screen is hit can be denoted as:

|Right> = sum over n1, n2,n3,...|R(n1,n2,n3,...)>|n1,n2,n3,......>

where |R(n1,n2,...)> denotes the quantum state of the buckyball if
it
emits n1 photons in state 1, n2 photons in state 2 etc. The state of
the
photons is then denoted as |n1,n2,n3,......>

If only the left slit is open, the state of the buckyball and the
photons just before the screen is hit can be denoted as:

|Left> = sum over n1, n2,n3,...|L(n1,n2,n3,...)>|n1,n2,n3,......>

where |R(n1,n2,...)> denotes the quantum state of the buckyball.
Here we
note that the state of the photons will pick up a phase factor
relative
to the case of only the right slit being open, but we can then
absorb
this phase factor in |L(n1,n2,n3,...)>.

With both slits open, we'll then have a state of the form:

|psi> = 1/sqrt(2) [|Right> + |Left>]

The inner product of |psi> with some position eigenstate |x>,
<x|psi> is
then a state vector for the photon states, the squared norm of that
state vector is the probability of finding the buckyball at position
x,
because this is the sum over the probabilities for photons over all
possible photon states. So, the probability is then:

P(x) = ||<x|psi>||^2 = 1/2 [||<x|Right>||^2 + ||<x|Left>||^2] +
Re[<Left|x><x|Right>]

We can then evaluate the interference term as follows:

Re[<Left|x><x|Right>] = Re sum over n1, n2,n3,...m1,m2,m3

<x|L(m1,m2,m3,...)>*<x|R(n1,n2,n3,...)><m1,m2,m3,...|n1,n2,n3,......>

Using that <m1,m2,m3,...|n1,n2,n3,......> = 0 unless m_j = n_j for
all
j, in which case this inner product equals 1, we then have:

Re[<Left|x><x|Right>] = Re sum over n1, n2,n3
<x|L(n1,n2,n3,...)>*<x|R(n1,n2,n3,...)> =

Re[<x|L(0,0,0...)>*<x|R(0,0,0,...>] +
Re[<x|L(1,0,0...)>*<x|R(1,0,0,...>] + .....  (1)

As explained above when there are photons present then we've
absorbed
the phase factor due to translation of the photon states in the
states
|L(n1,n2,n3...)>. For each wave vector k there is a factor for each
photon with that wavevector of exp(i k dot r) where r is the
position of
the left slit w.r.t. the right slit. So, the total phase factor will
be:

Product over j of exp(i kj nj dot r)

In the experiment there is then an additional summation over the
pure
states of the buckyballs. If the temperature is low then the
summation
will consist of states for which |R(0,0,0,..> and |L(0,0,0,..> are
the
dominant terms, as most of the time no photons will be emitted. At
higher temperatures the typical states there will be contributions
from
different numbers pf photons, so the interference pattern will be a
sum
of many different terms in (1) with comparable norms, they come with

different phase factors due to the different numbers of photons with

different momenta. So, the interference pattern will be washed out.

This analysis contradicts what you said in your first analysis. In the
first analysis, you claimed that no interference would be seen if the
IR photons were not detected.

In that case I considered a simple model where one photon would be emitted such that the state of that photon emitted from the ball moving through one slit would be orthogonal to that of the state of the photon emitted by the ball moving through the opther slit.


You seem to have dropped this notion in
the above.

In the derivation above I work in the momentum basis for the photons. Obviously, exp(i k dot r) and exp(i k dot (r+u)) are not orthogonal, but then one needs to consider a superposition of such momentum eigenstates as I do above.

 You now say that there is interference when no photons are
present, but this is washed out when there are different numbers of
photons.


Yes, the above derivation captures more of the details.

I think you are making the same basic mistake that you made when we
previously discussed  two-slit interference: You are analysing the two
slit case as a sum over single slit patterns, and assuming the
emission of independent photons from the ball through each slit. The
trouble is that there are not two balls, one for each slit. The same
ball goes through both slits and there is only one ball emitting (or
not emitting) photons.

I agree that there are no two balls for each slit, the derivation involves only one ball and treats the superposition in the usual way.


Neither of your analyses actually explain the observed behaviour.

It does so, otherwise the experiment would have proven that QM is invalid.

Saibal

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