> On 10 Nov 2018, at 01:27, [email protected] wrote: > > > > On Friday, November 9, 2018 at 12:26:52 PM UTC, Bruno Marchal wrote: > >> On 8 Nov 2018, at 18:25, [email protected] <javascript:> 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. 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] <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.
