On Wednesday, January 30, 2019 at 5:16:05 AM UTC-7, Bruno Marchal wrote: > > > On 30 Jan 2019, at 02:59, [email protected] <javascript:> 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 * > > > > > > The probabilities dos not depend on the choice of the base, but they are >> different when the components are different, given that the probabilities >> are given by the quake of those components, in the base corresponding to >> what you decide to measure, and this in all base used to describe the wave. >> > > *OK, AG* > >> >> Bruno >> >> >> >> >> >>> >>> >>> >>> >>>> >>>>> 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. >>> 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. >> >> >> > -- > 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.
