On Sunday, November 11, 2018 at 7:52:00 AM UTC, Bruno Marchal wrote: > > > On 10 Nov 2018, at 01:27, [email protected] <javascript:> wrote: > > > > On Friday, November 9, 2018 at 12:26:52 PM UTC, Bruno Marchal wrote: >> >> >> On 8 Nov 2018, at 18:25, [email protected] wrote: >> >> >> >> On Thursday, November 8, 2018 at 11:04:20 AM UTC, Bruno Marchal wrote: >>> >>> >>> On 6 Nov 2018, at 12:22, [email protected] wrote: >>> >>> >>> >>> On Tuesday, November 6, 2018 at 9:27:31 AM UTC, Bruno Marchal wrote: >>>> >>>> >>>> On 4 Nov 2018, at 22:02, [email protected] wrote: >>>> >>>> >>>> >>>> On Sunday, November 4, 2018 at 8:33:10 PM UTC, jessem wrote: >>>>> >>>>> >>>>> >>>>> On Wed, Oct 31, 2018 at 7:30 AM Bruno Marchal <[email protected]> >>>>> wrote: >>>>> >>>>>> >>>>>> On 30 Oct 2018, at 14:21, [email protected] wrote: >>>>>> >>>>>> >>>>>> >>>>>> On Tuesday, October 30, 2018 at 8:58:30 AM UTC, Bruno Marchal wrote: >>>>>>> >>>>>>> >>>>>>> On 29 Oct 2018, at 13:55, [email protected] wrote: >>>>>>> >>>>>>> >>>>>>> >>>>>>> On Monday, October 29, 2018 at 10:22:02 AM UTC, Bruno Marchal wrote: >>>>>>>> >>>>>>>> >>>>>>>> On 28 Oct 2018, at 13:21, [email protected] wrote: >>>>>>>> >>>>>>>> >>>>>>>> >>>>>>>> On Sunday, October 28, 2018 at 9:27:56 AM UTC, Bruno Marchal wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> On 25 Oct 2018, at 17:12, [email protected] wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> On Tuesday, October 23, 2018 at 10:39:11 PM UTC, agrays...@ >>>>>>>>> gmail.com wrote: >>>>>>>>>> >>>>>>>>>> If a system is in a superposition of states, whatever value >>>>>>>>>> measured, will be repeated if the same system is repeatedly >>>>>>>>>> measured. But >>>>>>>>>> what happens if the system is in a mixed state? TIA, AG >>>>>>>>>> >>>>>>>>> >>>>>>>>> If you think about it, whatever value you get on a single trial >>>>>>>>> for a mixed state, repeated on the same system, will result in the >>>>>>>>> same >>>>>>>>> value measured repeatedly. If this is true, how does measurement >>>>>>>>> distinguish superposition of states, with mixed states? AG >>>>>>>>> >>>>>>>>> >>>>>>>>> That is not correct. You can distinguish a mixture of particles in >>>>>>>>> the up or down states with a set of 1/sqrt(2)(up+down) by measuring >>>>>>>>> them >>>>>>>>> with the {1/sqrt(2)(up+down), 1/sqrt(2)(up-down}) discriminating >>>>>>>>> apparatus. >>>>>>>>> With the mixture, half the particles will be defected in one >>>>>>>>> direction, >>>>>>>>> with the pure state, they will all pass in the same direction. >>>>>>>>> Superposition would not have been discovered if that was not the case. >>>>>>>>> >>>>>>>> >>>>>>>> >>>>>>>> *And someone will supply the apparatus measuring (up + down), and >>>>>>>> (up - down)? No such apparatuses are possible since those states are >>>>>>>> inherently contradictory. We can only measure up / down. AG* >>>>>>>> >>>>>>>> >>>>>>>> You can do the experience by yourself using a simple crystal of >>>>>>>> calcium (CaCO3, Island Spath), or with polarising glass. Or with >>>>>>>> Stern-Gerlach devices and electron spin. Just rotating (90° or 180°) >>>>>>>> an >>>>>>>> app/down apparatus, gives you an (up + down)/(up - down) apparatus. >>>>>>>> >>>>>>> >>>>>>> *I don't understand. With SG one can change the up/down axis by >>>>>>> rotation, but that doesn't result in an (up + down), or (up - down) >>>>>>> measurement. If that were the case, what is the operator for which >>>>>>> those >>>>>>> states are eigenstates? Which book by Albert? AG * >>>>>>> >>>>>>> >>>>>>> David Z Albert, Quantum Mechanics and Experience, Harvard University >>>>>>> Press, 1992. >>>>>>> >>>>>>> https://www.amazon.com/Quantum-Mechanics-Experience-David-Albert/dp/0674741137 >>>>>>> >>>>>>> Another very good books is >>>>>>> >>>>>>> D’Espagnat B. Conceptual foundations of Quantum mechanics, I see >>>>>>> there is a new edition here: >>>>>>> >>>>>>> https://www.amazon.com/Conceptual-Foundations-Quantum-Mechanics-Advanced/dp/0738201049/ref=sr_1_1?s=books&ie=UTF8&qid=1540889778&sr=1-1&keywords=d%27espagnat+conceptual+foundation+of+quantum+mechanics&dpID=41NcluHD6fL&preST=_SY291_BO1,204,203,200_QL40_&dpSrc=srch >>>>>>> >>>>>>> It explains very well the difference between mixtures and pure >>>>>>> states. >>>>>>> >>>>>>> Bruno >>>>>>> >>>>>> >>>>>> *Thanks for the references. I think I have a reasonable decent >>>>>> understanding of mixed states. Say a system is in a mixed state of phi1 >>>>>> and >>>>>> phi2 with some probability for each. IIUC, a measurement will always >>>>>> result >>>>>> in an eigenstate of either phi1 or phi2 (with relative probabilities >>>>>> applying). * >>>>>> >>>>>> >>>>>> If the measurement is done with a phi1/phi2 discriminating apparatus. >>>>>> Keep in mind that any state can be seen as a superposition of other >>>>>> oblique >>>>>> or orthogonal states. >>>>>> >>>>> >>>>> I don't know if you're restricting the definition of phi1 and phi2 to >>>>> some particular type of eigenstate or not, but in general aren't there >>>>> pure >>>>> states that are not eigenstates of any physically possible measurement >>>>> apparatus, so there is no way to directly measure that a system is in >>>>> such >>>>> a state? >>>>> >>>> >>>> *Yes, such states exist IIUC. That's why I don't understand Bruno's >>>> claim that Up + Dn and Up - Dn can be measured with any apparatus, * >>>> >>>> >>>> Not *any*¨apparatus, but a precise one, which in this case is the same >>>> apparatus than for up and down, except that it has been rotated. >>>> >>>> >>>> >>>> >>>> *since they're not eigenstates of the spin operator, or any operator. * >>>> >>>> >>>> This is were you are wrong. That are eigenstates of the spin operator >>>> when measured in some direction. >>>> >>> >>> *If what you claim is true, then write down the operator for which up + >>> dn (or up - dn) is an eigenstate? AG * >>> >>> >>> >>> It is the operator corresponding to the same device, just rotated from >>> pi/2, or pi (it is different for spin and photon). When I have more time, I >>> might do the calculation, but this is rather elementary quantum mechanics. >>> (I am ultra-busy up to the 15 November, sorry). It will have the same shape >>> as the one for up and down, in the base up’ and down’, so if you know a bit >>> of linear algebra, you should be able to do it by yourself. >>> >>> Bruno >>> >> >> *You don't have to do any calculation. Just write down the operator >> which, you allege, has up + dn or up - dn as an eigenstate. I don't think >> you can do it, because IMO it doesn't exist. AG * >> >> >> >> If up and down are represented by the column (1 0) and (0 1) the >> corresponding observable is given by the diagonal matrix >> >> 1 0 >> 0 -1 >> >> Then the up’ = 1/sqrt(2) (1 1), and down’ = 1/sqrt(2) (1 -1), >> >> So the operator, written in the base up down, will be >> >> 0 1 >> 1 0 >> >> Here the eigenvalue +1 and -1 correspond to up (up’) or down (down’). >> >> I have no clue why you think that such operator would not exist. >> > > *Because the measured spin state is Up or Dn along some axis, never > anything in between. Up + Dn or Up - Dn is not physically realizable in > unprimed basis. AG* > > > > If the measured spin state is Up or Dn along some axis, the measured spin > state will be Up + Dn or Up - Down along the axis obtained by rotating the > measuring apparatus adequately. >
*But NOT along the original spin axis! You can't measure Up + Dn or Up - Dn along any particular spin axis that you choose. That was my point. If you rotate the axis, the same situation exists. AG * That is physically realisable with spin (by just rotating the Stern-Gerlach > apparatus) of with light polarisation (rotating the polariser or the CaCO3 > crystal). > > Bruno > > > > > > All pure state can be seen as a superposition, in the rotated base, and >> you can always build an operator having them as eigenvalues. >> >> Bruno >> >> >> >> >> >> >> >> >> >>> >>> >>> >>> >>> >>> >>>> Julian Swinger (and Townsend) showed that the formalism of (discrete, >>>> spin, qubit) quantum mechanics is derivable from 4 Stern-Gerlach >>>> experiments, using only real numbers, but for a last fifth one, you need >>>> the complex amplitudes, and you get the whole core of the formalism. >>>> >>>> Bruno >>>> >>>> >>>> >>>> >>>> *Do you understand Bruno's argument in a previous post on this topic? >>>> AG * >>>> >>>> -- >>>> 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]. >> 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.
