On Sat, Jul 10, 2021 at 10:21 AM smitra <[email protected]> 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. You seem to have dropped this notion in the 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. 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. Neither of your analyses actually explain the observed behaviour. Bruce -- 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/CAFxXSLSo8YmXPD89d7yuX_fQD8PQJFT%2BU-6%2BUrAZQg0%3DDsbeeA%40mail.gmail.com.
