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] >> 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] >>>>>> 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? 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* > > 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. > For more options, visit 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]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
