> On 11 Nov 2018, at 10:56, [email protected] wrote: > > > > 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, [email protected] >>>>>>>>> <http://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 >>>>>> >>>>>> <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 >>>>>> >>>>>> <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!
Of course. But up’ = 1/sqrt(2)(up +down), so you get up or down when you measure that state in the up/down base, but you get up’ with probability 1, in the base/apparatus up’/down'where you would get it only with probability 1/2 in the other base. > 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 ? We know that the state was prepared as up’ = 1/sqrt(2)(up +down). So we get probability one to get up’ when measure in the up’/down’ measuring apparatus. Which is not the case for a mixture of halve up and down, as each up and each down will only have a probability 1/2 to go through up’, and through down’. It is conceptually mysterious, but it is well accounted by the formalism, and by the facts which have necessitated that formalism. Bruno > > 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 >>>>> <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] <>. >>> 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] <javascript:>. >> To post to this group, send email to [email protected] >> <javascript:>. >> 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. 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.
