> On 1 Feb 2019, at 21:29, [email protected] wrote: > > > > On Friday, February 1, 2019 at 5:55:30 AM UTC-7, Bruno Marchal wrote: > >> On 31 Jan 2019, at 21:10, [email protected] <javascript:> wrote: >> >> >> >> On Thursday, January 31, 2019 at 6:47:12 AM UTC-7, Bruno Marchal wrote: >> >>> On 31 Jan 2019, at 01:28, [email protected] <> wrote: >>> >>> >>> >>> On Wednesday, January 30, 2019 at 2:38:58 PM UTC-7, [email protected] >>> <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, [email protected] <> wrote: >>>> >>>> >>>> >>>> On Tuesday, January 29, 2019 at 4:37:34 AM UTC-7, Bruno Marchal wrote: >>>> >>>>> On 28 Jan 2019, at 22:50, [email protected] <> wrote: >>>>> >>>>> >>>>> >>>>> On Friday, January 25, 2019 at 7:33:05 AM UTC-7, Bruno Marchal wrote: >>>>> >>>>>> On 24 Jan 2019, at 09:29, [email protected] <> wrote: >>>>>> >>>>>> >>>>>> >>>>>> On Sunday, January 20, 2019 at 11:54:43 AM UTC, [email protected] >>>>>> <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, [email protected] <> wrote: >>>>>>> >>>>>>> >>>>>>> >>>>>>> On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote: >>>>>>> >>>>>>>> On 17 Jan 2019, at 14:48, [email protected] <> wrote: >>>>>>>> >>>>>>>> >>>>>>>> >>>>>>>> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal wrote: >>>>>>>> >>>>>>>>> On 17 Jan 2019, at 09:33, [email protected] <> wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> On 1/16/2019 7:25 PM, [email protected] <> wrote: >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote: >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> On 1/13/2019 9:51 PM, [email protected] <> 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. > > Obviously, you don't know how to apply the rule you speak so highly of, > Born's rule. To calculate the probability density of wf function psi, you > must calculate <psi, psi>. Do you dispute this?
Yes, you need to put some projection operator (corresponding to some eigenvalue you intend to measure) in between. <psi,psi> is the amplitude of probability to go from the psi state to the psi state, and should be equal to one (psi being normalised). > How the phase angles relate to interference is another issue, which I think > Phil explained. AG > >> 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, > > Up and Dn are NOT orhonormal. AG > > 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. > > If I am wrong, it's just because I assumed all interference comes from the > interactions due to the cross terms -- which cancel out for orthonormal > component states. Also, I never introduced a global phase angle in my > calculation. If you would do my calculation, or at least understand it, you'd > understand Born's rule. I don't need to read Albert's book to understand > Born's rule. AG Once you say that up and down are not orthonormal, I am not sure you have studied the QM formalism correctly. Any two distinguishable eigenstates of any observable are orthogonal (and normalised). I have no clue what you don’t understand in my use of the Born rule. You definitely need to study Albert’s book, I think. In your other post you mention wikipedia. No problem there? Actually you can see that they do put the projection operator at the right place. You can help yourself with a dictionary, but books and papers are better. Bruno > > 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 [email protected] <javascript:>. >> To post to this group, send email to [email protected] >> <javascript:>. >> 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 [email protected] > <mailto:[email protected]>. > To post to this group, send email to [email protected] > <mailto:[email protected]>. > 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. 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