> 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] <javascript:> 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. 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] <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.
