> On 31 Jan 2019, at 21:10, agrayson2...@gmail.com wrote: > > > > On Thursday, January 31, 2019 at 6:47:12 AM UTC-7, Bruno Marchal wrote: > >> On 31 Jan 2019, at 01:28, agrays...@gmail.com <javascript:> wrote: >> >> >> >> On Wednesday, January 30, 2019 at 2:38:58 PM UTC-7, agrays...@gmail.com >> <http://gmail.com/> wrote: >> >> >> On Wednesday, January 30, 2019 at 5:16:05 AM UTC-7, Bruno Marchal wrote: >> >>> On 30 Jan 2019, at 02:59, agrays...@gmail.com <> wrote: >>> >>> >>> >>> On Tuesday, January 29, 2019 at 4:37:34 AM UTC-7, Bruno Marchal wrote: >>> >>>> On 28 Jan 2019, at 22:50, agrays...@gmail.com <> wrote: >>>> >>>> >>>> >>>> On Friday, January 25, 2019 at 7:33:05 AM UTC-7, Bruno Marchal wrote: >>>> >>>>> On 24 Jan 2019, at 09:29, agrays...@gmail.com <> wrote: >>>>> >>>>> >>>>> >>>>> On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com >>>>> <http://gmail.com/> wrote: >>>>> >>>>> >>>>> On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote: >>>>> >>>>>> On 18 Jan 2019, at 18:50, agrays...@gmail.com <> wrote: >>>>>> >>>>>> >>>>>> >>>>>> On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote: >>>>>> >>>>>>> On 17 Jan 2019, at 14:48, agrays...@gmail.com <> wrote: >>>>>>> >>>>>>> >>>>>>> >>>>>>> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote: >>>>>>> >>>>>>>> On 17 Jan 2019, at 09:33, agrays...@gmail.com <> wrote: >>>>>>>> >>>>>>>> >>>>>>>> >>>>>>>> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote: >>>>>>>> >>>>>>>> >>>>>>>> On 1/16/2019 7:25 PM, agrays...@gmail.com <> wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> On 1/13/2019 9:51 PM, agrays...@gmail.com <> wrote: >>>>>>>>>> This means, to me, that the arbitrary phase angles have absolutely >>>>>>>>>> no effect on the resultant interference pattern which is observed. >>>>>>>>>> But isn't this what the phase angles are supposed to effect? AG >>>>>>>>> >>>>>>>>> The screen pattern is determined by relative phase angles for the >>>>>>>>> different paths that reach the same point on the screen. The >>>>>>>>> relative angles only depend on different path lengths, so the overall >>>>>>>>> phase angle is irrelevant. >>>>>>>>> >>>>>>>>> Brent >>>>>>>>> >>>>>>>>> Sure, except there areTWO forms of phase interference in Wave >>>>>>>>> Mechanics; the one you refer to above, and another discussed in the >>>>>>>>> Stackexchange links I previously posted. In the latter case, the wf >>>>>>>>> is expressed as a superposition, say of two states, where we consider >>>>>>>>> two cases; a multiplicative complex phase shift is included prior to >>>>>>>>> the sum, and different complex phase shifts multiplying each >>>>>>>>> component, all of the form e^i (theta). Easy to show that >>>>>>>>> interference exists in the latter case, but not the former. Now >>>>>>>>> suppose we take the inner product of the wf with the ith eigenstate >>>>>>>>> of the superposition, in order to calculate the probability of >>>>>>>>> measuring the eigenvalue of the ith eigenstate, applying one of the >>>>>>>>> postulates of QM, keeping in mind that each eigenstate is multiplied >>>>>>>>> by a DIFFERENT complex phase shift. If we further assume the >>>>>>>>> eigenstates are mutually orthogonal, the probability of measuring >>>>>>>>> each eigenvalue does NOT depend on the different phase shifts. What >>>>>>>>> happened to the interference demonstrated by the Stackexchange links? >>>>>>>>> TIA, AG >>>>>>>>> >>>>>>>> Your measurement projected it out. It's like measuring which slit the >>>>>>>> photon goes through...it eliminates the interference. >>>>>>>> >>>>>>>> Brent >>>>>>>> >>>>>>>> That's what I suspected; that going to an orthogonal basis, I departed >>>>>>>> from the examples in Stackexchange where an arbitrary superposition is >>>>>>>> used in the analysis of interference. Nevertheless, isn't it possible >>>>>>>> to transform from an arbitrary superposition to one using an >>>>>>>> orthogonal basis? And aren't all bases equivalent from a linear >>>>>>>> algebra pov? If all bases are equivalent, why would transforming to an >>>>>>>> orthogonal basis lose interference, whereas a general superposition >>>>>>>> does not? TIA, AG >>>>>>> >>>>>>> I don’t understand this. All the bases we have used all the time are >>>>>>> supposed to be orthonormal bases. We suppose that the scalar product >>>>>>> (e_i e_j) = delta_i_j, when presenting the Born rule, and the quantum >>>>>>> formalism. >>>>>>> >>>>>>> Bruno >>>>>>> >>>>>>> Generally, bases in a vector space are NOT orthonormal. >>>>>> >>>>>> Right. But we can always build an orthonormal base with a decent scalar >>>>>> product, like in Hilbert space, >>>>>> >>>>>> >>>>>> >>>>>>> For example, in the vector space of vectors in the plane, any pair of >>>>>>> non-parallel vectors form a basis. Same for any general superposition >>>>>>> of states in QM. HOWEVER, eigenfunctions with distinct eigenvalues ARE >>>>>>> orthogonal. >>>>>> >>>>>> Absolutely. And when choosing a non degenerate >>>>>> observable/measuring-device, we work in the base of its eigenvectors. A >>>>>> superposition is better seen as a sum of some eigenvectors of some >>>>>> observable. That is the crazy thing in QM. The same particle can be >>>>>> superposed in the state of being here and there. Two different positions >>>>>> of one particle can be superposed. >>>>>> >>>>>> This is a common misinterpretation. Just because a wf can be expressed >>>>>> in different ways (as a vector in the plane can be expressed in >>>>>> uncountably many different bases), doesn't mean a particle can exist in >>>>>> different positions in space at the same time. AG >>>>> >>>>> It has a non null amplitude of probability of being here and there at the >>>>> same time, like having a non null amplitude of probability of going >>>>> through each slit in the two slits experience. >>>>> >>>>> If not, you can’t explain the inference patterns, especially in the >>>>> photon self-interference. >>>>> >>>>> >>>>> >>>>> >>>>>> >>>>>> Using a non orthonormal base makes only things more complex. >>>>>>> I posted a link to this proof a few months ago. IIRC, it was on its >>>>>>> specifically named thread. AG >>>>>> >>>>>> But all this makes my point. A vector by itself cannot be superposed, >>>>>> but can be seen as the superposition of two other vectors, and if those >>>>>> are orthonormal, that gives by the Born rule the probability to obtain >>>>>> the "Eigen result” corresponding to the measuring apparatus with Eigen >>>>>> vectors given by that orthonormal base. >>>>>> >>>>>> I’m still not sure about what you would be missing. >>>>>> >>>>>> You would be missing the interference! Do the math. Calculate the >>>>>> probability density of a wf expressed as a superposition of orthonormal >>>>>> eigenstates, where each component state has a different phase angle. All >>>>>> cross terms cancel out due to orthogonality, >>>>> >>>>> ? Sin(alpha) up + cos(alpha) down has sin^2(alpha) probability to be fin >>>>> up, and cos^2(alpha) probability to be found down, but has probability >>>>> one being found in the Sin(alpha) up + cos(alpha) down state, which would >>>>> not be the case with a mixture of sin^2(alpha) proportion of up with >>>>> cos^2(alpha) down particles. >>>>> Si, I don’t see what we would loss the interference terms. >>>>> >>>>> >>>>> >>>>>> and the probability density does not depend on the phase differences. >>>>>> What you get seems to be the classical probability density. AG >>>>> >>>>> >>>>> I miss something here. I don’t understand your argument. It seems to >>>>> contradict basic QM (the Born rule). >>>>> >>>>> Suppose we want to calculate the probability density of a superposition >>>>> consisting of orthonormal eigenfunctions, >>>> >>>> Distinct eigenvalue correspond to orthonormal vector, so I tend to always >>>> superpose only orthonormal functions, related to those eigenvalue. >>>> >>>> >>>> >>>> >>>> >>>>> each multiplied by some amplitude and some arbitrary phase shift. >>>> >>>> like (a up + b down), but of course we need a^2 + b^2 = 1. You need to be >>>> sure that you have normalised the superposition to be able to apply the >>>> Born rule. >>>> >>>> >>>> >>>> >>>>> If we take the norm squared using Born's Rule, don't all the cross terms >>>>> zero out due to orthonormality? >>>> >>>> ? >>>> >>>> The Born rule tell you that you will find up with probability a^2, and >>>> down with probability b^2 >>>> >>>> >>>> >>>>> Aren't we just left with the SUM OF NORM SQUARES of each component of the >>>>> superposition? YES or NO? >>>> >>>> If you measure in the base (a up + b down, a up -b down). In that case you >>>> get the probability 1 for the state above. >>>> >>>> >>>> >>>>> If YES, the resultant probability density doesn't depend on any of the >>>>> phase angles. AG >>>>> >>>>> YES or NO? AG >>>> >>>> >>>> Yes, if you measure if the state is a up + b down or a up - b down. >>>> No, if you measure the if the state is just up or down >>>> >>>> Bruno >>>> >>>> I assume orthNORMAL eigenfunctions. I assume the probability densities sum >>>> to unity. Then, using Born's rule, I have shown that multiplying each >>>> component by e^i(theta) where theta is arbitrarily different for each >>>> component, disappears when the probability density is calculated, due to >>>> orthonormality. >>> >>> >>> That seems to violate elementary quantum mechanics. If e^I(theta) is >>> different for each components, Born rule have to give different >>> probabilities for each components---indeed given by the square of >>> e^I(theta). >>> >>> The norm squared of e^i(thetai) is unity, except for the cross terms which >>> is zero due to orthonormality. AG >>> >>>> What you've done, if I understand correctly, is measure the probability >>>> density using different bases, and getting different values. >>> >>> The value of the relative probabilities do not depend on the choice of the >>> base used to describe the wave. Only of the base corresponding to what you >>> decide to measure. >>> >>> >>> >>>> This cannot be correct since the probability density is an objective >>>> value, and doesn't depend on which basis is chosen. AG >>> >>> Just do the math. Or read textbook. >>> >>> Why don't YOU do the math ! It's really simple. Just take the norm squared >>> of a superposition of component eigenfunctions, each multiplied by a >>> probability amplitude, and see what you get ! No need to multiply each >>> component by e^i(thetai). Each amplitude has a phase angle implied. This >>> is Born's rule and the result doesn't depend on phase angles, contracting >>> what Bruce wrote IIUC. If you would just do the simple calculation you will >>> see what I am referring to! AG >> >> >> Bruce is right. Let us do the computation in the simple case where >> e^i(theta) = -1. (Theta = Pi) >> >> Take the superposition (up - down), conveniently renormalised. If I multiply >> the whole wave (up - down) by (-1), that changes really nothing. But if I >> multiply only the second term, I get the orthogonal state up + down, which >> changes everything. (up +down) is orthogonal to (up - down). >> >> Bruno >> >> Fuck it. You refuse to do the simple math to show me exactly where I have >> made an error, IF I have made an error. You talk a lot about Born's rule >> but I seriously doubt you know how to use it for simple superposition. AG >> >> If you take the inner product squared (Born's rule) using an orthonormal set >> of eigenfunctions, you get a sum of the form (a_j)^2 + (b_j) ^2 where A_j >> is the complex probability amplitude for the jth component, A_j = a_j + i * >> b_j. The cross terms drop out due to orthonormality, and the phase angles >> are implicitly determined by the relative values of a_j and b_j for each j. > > If you have prepared the state, so that you know that the state of your > object is given by > > phi = A_1 up + A_2 down, say, then, if you decide to measure the up/down > state, and use the device doing that, you do not need to make the inner > product between phi and phi, but between the base state up and/or down to get > the probability given by the square of phi * up (to get the probability of > up) and the square of phi*down, to get the probability of down. They will > both depend on the value of A_1 and A_2. They are respectively (A_1)^2 and > (A_2)^2. Of course, we suppose that we have renormalised the state so that > (A_1)^2 + (A_2)^2 = 1 (which makes them into probability of getting up and > down). > > > > > > > >> The question then becomes how do we calculate the probability density with >> the phase angles undetermined. Are we assuming they are known given the way >> the system is prepared? AG > > > Yes. The Born rule, written simply, is only that if phi = A_1 up + A_2 down, > (so the state has been prepared in advance) then if you measure if the object > is in up or down, you will find up with a probability given respectively by > (A_1)^2 and (A_2)^2. > All probabilities are relative to the state of the object and the choice of > what you decide to measure. It is always simpler to write the state in the > base corresponding to the measurement, so that the “simple” Born rule above > can be applied immediately. > > Bruno > > For reference I repeat my last comment and add a significant point: > > If you take the inner product squared (Born's rule) using an orthonormal set > of eigenfunctions, you get a sum of the form (a_j)^2 + (b_j) ^2 where A_j is > the complex probability amplitude for the jth component, A_j = a_j + i * b_j. > The cross terms drop out due to orthonormality, and the phase angles are > implicitly determined by the relative values of a_j and b_j for each j. The > question then becomes how do we calculate the probability density with the > phase angles undetermined. Are we assuming they are known given the way the > system is prepared? AG > > The question for me is how the phase angles are related to interference.

But that is explained by may calculation above. You calculation does not make sense to me. You compute an inner product of the wave to itself? I don’t see the relation with your problem. > The calculation above shows that the cross terms drop out due to > orthonormality. Do it again, explicitly. Take the simple state phi = A_1 up + A_2 down. Up and down are orthonormal, but phi is not orthonormal with either up or down. If “up” means go to the left hole, and “down” is go the right hole, the amplitude A_1 and A_2, if not null, will interfere, even if only one photon is sent.The wave go through both silts, and interfere constructively along some direction and destructively along other direction, making it impossible for that photon to lend on those last place, like anyway, by the laws of addition of sinus/wave. > But IIUC these are the terms which account for interference. I am not sure what you say here. The interferences comes only from the fact that we have a superposition of two orthogonal state, and that superposition is a new state, which is not orthogonal to either up or down. > Thus, applying Born's rule to a superposition of states where the components > are orthonormal, leaves open the question of interference. That does no make sense. The Born rule just say that if you measure (up/down) on phi = A_1 up + A_2 down, you get up with probability (A_1)^2 and down with probability (A_2)^2. But if you do any measurement, the state beg-have like a wave, and the amplitudes add up, constructively or destructively. If you don’t understand that, it means you begin to understand quantum mechanics, as nobody understand this, except perhaps the Mechanist Philosophers …(which predicts something at least as weird and counter-intuitive). > Bruce wrote that the phase angles are responsible for interference. I doubt > that result. Am I mistaken? AG Yes, I’m afford you are. The relative phase (in a superposition) angles are responsible for the interference. A global phase angle changes nothing. I really wish you to read the first 60 pages of David Albert’s book. Its exposition of the functioning of the interferometer is crystal clear. I am still not sure if you have a problem with the formalism or with the weirdness related to it. Read that piece of explanation by Albert, and if you still have problem, we can discuss it, but it would be too long (here and now) to do that here. Bruno > > -- > 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 everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to everything-list@googlegroups.com > <mailto:everything-list@googlegroups.com>. > Visit this group at https://groups.google.com/group/everything-list > <https://groups.google.com/group/everything-list>. > For more options, visit https://groups.google.com/d/optout > <https://groups.google.com/d/optout>. -- 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.