On Sat, Jul 10, 2021, 1:58 AM Bruce Kellett <[email protected]> wrote:
> 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. > This conversation brought to mind Feynman's question: what happens in the limit of going from 2 slits to infinite slits, such that the wall disappears? Jason -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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