On Wednesday, January 30, 2019 at 2:38:58 PM UTC-7, agrays...@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 >>>> 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. 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 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 everything-list+unsubscr...@googlegroups.com. >>>> To post to this group, send email to everyth...@googlegroups.com. >>>> 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 everything-li...@googlegroups.com. >>> To post to this group, send email to everyth...@googlegroups.com. >>> 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 everything-li...@googlegroups.com. >> To post to this group, send email to everyth...@googlegroups.com. >> 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 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.
