> On 3 Feb 2019, at 00:03, [email protected] wrote: > > > > On Saturday, February 2, 2019 at 2:59:30 PM UTC-7, [email protected] wrote: > > > On Saturday, February 2, 2019 at 1:40:29 AM UTC-7, Bruno Marchal wrote: > >> 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] <> 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). > > Wrong! Not equal to one. I see you like to talk the talk, but refuse to walk > the walk. Just read the 4th paragraph of the Wiki link. I correctly > calculated the probability density for orthonormal eigenstates! AG
I understand that if you have a plane wave, say, psi(x,t) = Ae^i(kx-wt), the probability density if given by the square of the square of the modulus of A. I am not sure why you need this, and why this should be problematic with Bruce’s (or mine) explanation. > > > >> 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). > > Right. I was mistaken. AG OK. Good. > > 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. > > Wiki shows I correctly calculated the probability density. Also I agree with > Phil, and noted the error I made (not in any calculation, but in > interpretation). Didn't you read it? AG That was not enough clear, sorry. The wiki is also rather unclear on the Born Rule. I mean that there are clearer exposition. > > 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. > > If you put in the projection operator, you're calculating the probability of > getting some eigenvalue, not the probability density of the position. AG You lost me. I was just explaining why the relative phase does play a role for the probability of finding specific values. Bruce was correct, and I still don’t know if you agree on this or not. I am not sure that I understand what is your problem. Bruno > > You could help yourself by reading plain English. SEE PARAGRAPH 4 OF WIKI > LINK. THEY CALCULATED THE PROBABILITY DENSITY AND DIDN'T PUT IN THE > PROJECTION OPERATOR! AG > > 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] <>. >>> To post to this group, send email to [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. >> 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 >> <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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