# Re: Measuring a system in a superposition of states vs in a mixed state

```
On Monday, November 19, 2018 at 3:52:37 PM UTC, Bruno Marchal wrote:
>
>
> On 18 Nov 2018, at 14:00, agrays...@gmail.com <javascript:> wrote:
>
>
>
> On Sunday, November 18, 2018 at 12:19:20 PM UTC, Bruno Marchal wrote:
>>
>>
>> On 16 Nov 2018, at 15:38, agrays...@gmail.com wrote:
>>
>>
>>
>> On Friday, November 16, 2018 at 10:14:32 AM UTC, scerir wrote:
>>>
>>>
>>> Il 16 novembre 2018 alle 10.19 agrays...@gmail.com ha scritto:
>>>
>>>
>>>
>>> On Thursday, November 15, 2018 at 2:14:48 PM UTC, scerir wrote:
>>>
>>>
>>> Il 15 novembre 2018 alle 14.29 agrays...@gmail.com ha scritto:
>>>
>>>
>>>
>>> On Thursday, November 15, 2018 at 8:04:53 AM UTC, scerir wrote:
>>>
>>> Imagine a spin-1/2 particle described by the state psi = sqrt(1/2)
>>> [(s+)_z + (s-)_z] .
>>>
>>> If the x-component of spin is measured by passing the spin-1/2 particle
>>> through a Stern-Gerlach with its field oriented along the x-axis, the
>>> particle will ALWAYS emerge 'up'.
>>>
>>>
>>> *Why?  Won't the measured value be along the x axis in both directions,
>>> in effect Up or Dn? AG*
>>>
>>> "Hence we must conclude that the system described by the |+>x state is
>>> not the
>>> same as a mixture of atoms in the |+> and !-> states. This means that
>>> each atom in the
>>> beam is in a state that itself is a combination of the |+> and |->
>>> states. A superposition
>>> state is often called a coherent superposition since the relative phase
>>> of the two terms is
>>> important."
>>>
>>> .see pages 18-19 here *https://tinyurl.com/ybm56whu
>>> <https://tinyurl.com/ybm56whu>*
>>>
>>>
>>> *Try answering in your own words. When the SG device is oriented along
>>> the x axis, now effectively the z-axix IIUC, and we're dealing with
>>> superpositions, the outcomes will be 50-50 plus and minus. Therefore,
>>> unless I am making some error, what you stated above is incorrect. AG *
>>>
>>> sqrt(1/2) [(s+)_z +(s-)_z]  is a superposition, but since sqrt(1/2)
>>> [(s+)_z +(s-)_z]  =  (s+)_x the particle will always emerge 'up'
>>>
>>
>> I'll probably get back to on the foregoing. In the meantime, consider
>> this; I claim one can never MEASURE Up + Dn or Up - Dn with a SG apparatus
>> regardless of how many other instruments one uses to create a composite
>> measuring apparatus (Bruno's claim IIUC). The reason is simple. We know
>> that the spin operator
>>
>>
>> Which one?
>>
>
> *Good question. AG*
>
> There are spin operator for each direction in space. The superposition of
>> up and down is a precise pure state, with precise eigenvalues, when
>> measuring state in the complementary directions.
>>
>
> *As I wrote earlier, based on scerir's superpositions on different axes,
> and simulation, I now think that Up + Dn and Up - Dn can be measured along
> the x axis but not along the z axis (which I was focused on). *
>
>
> All you need to do is a change of base. The operator will be defined
> clearly by the Eigen value on the diagonal in the corresponding base. You
> can prepare any state, and measure them “in any base”.
>```
```

*I'll get back to this issue in my next post. AG *

> *You were probably correct about x axis measurements, but perhaps were not
> clear enough. You were not explicit that measurements along the x axis is a
> different SG experiment from along z axis.*
>
>
> OK. Sorry.
>
> * I thought you meant do them in succession, not as separate experiments.*
>
>
> Ah? OK.
>
>
> * Also introducing an infinity of universes seems extraneous and confusing
> for a solution to this problem. AG *
>
> I are probably different on this. I don’t take the word “universe” too
> much seriously, as with mechanism we know at the start that there is
> “physical universe” at all, just the natural numbers with the laws of
> addition and multiplication. Both the computational and the quantum state
> are relative, and high level, pertaining to what is “observable” for some
> the point of view of some locally finite subject, run by some computation.
>
> The empirical point, though, is that to predict correctly an event in
> quantum mechanics, we have to take into account may simultaneous
> “incompatible path”, like going through each hole in a plane. Quantum
> computations, for example, exploits that seemingly parallelism.
>

*I don't like this approach -- in fact I abhor it -- since it implies
simultaneous interference among a multitude of paths to the same point on
the detection screen. This adds an unnecessary mystery to QM. In the
Hilbert Space representation, the wf is what it is, but can be represented
in a multitude of different bases. It is therefore misleading to claim the
system being analyzed is in a multitude of states; rather it is in one
state, which due to linear algebra, has many representations. AG *

> has exactly two eigenstates, each with probability of .5. We can write
>> them down. We also know that every quantum measurement gives up an
>> eigenvalue of some eigenstate. Therefore, if there existed an Up + Dn or Up
>> - Dn eigenstate, it would have to have probability ZERO since the Up and Dn
>> eigenstates have probabilities which sum to unity. Do you agree or not, and
>> if not, why? TIA, AG
>>
>>
>> You add the probabilities, but you need to add the amplitudes of
>> probabilities instead, and then take their square. You simply dismiss the
>> quantum formalism, it seems to me.
>>
>
> *I did not; an incorrect inference on your part.*
>
> All right. (I was just trying to figure out what you did, to be sure).
>
> *I** never mentioned Born's rule (it wasn't necessary), *
>
>
> You did use the probability 1/2 at some place, with the particle in a
> state 1/sqrt(2)(up + down). We use all the time the Born rule when we talk
>

*I just assumed a probability of .5 for Up and Dn states after application
of Born's rule. AG *

> *from which one cannot infer I am criticizing QM itself. AG *
>
>
> I am just trying to understand what you don’t understand, which is not
> easy in a context where the more we understand the formalism, the less we
> understand what it could mean, even more so if we give sense to a dualist
> wave packet reduction.
>
> I am a logician: it is clear that Copenhagen and Everett are not two
> different interpretations, but two different theories. One is Schroedinger
> equation + wave packet reduction + a dualist theory of mind/observation),
> the other is just Schroedinger equation only + the “usual” mechanist theory
> of mind. There are many possible debate on all his of course.
>
> I urge you to study the treatment of the interferometer in David Albert
> books. It is weird. Bohr is right on this: to understand it means to get
> the point that is hard to figure out how nature could to that, but from the
> mechanist post Gödel view, it is rather natural, as we observe is given by
> a statistics on infinitely many computations/histories.
>
> Bruno
>
>
>
>
>
> The states constituted a vector space: the sum (superposition) of
>> orthogonal states are pure state, after a change of base, and I did give
>> you the corresponding operator. You are not criticising an interpretation
>> of QM, but QM itself.
>>
>
>
>> Bruno
>>
>>
>>
>>
>>
>>>
>>>
>>>
>>> In fact (s+)_z = sqrt(1/2) [(s+)_x + (s-)_x]
>>>
>>> and (s-)_z = sqrt(1/2) [(s+)_x - (s-)_x]
>>>
>>> (where _z, _x, are the z-component and the x-component of spin)
>>>
>>> so that psi = sqrt(1/2)[(s+)_z +(s-)_z] = (s+)_x.   (pure state, not
>>> mixture state)..
>>>
>>> AGrayson2000 asked "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?"
>>>
>>> Does Everett's "relative state interpretation" show how to interpret a
>>> real superposition (like the above, in which the particle will always
>>> emerge 'up') and how to interpret a mixture (in which the particle will
>>> emerge 50% 'up' or 50% 'down')?
>>>
>>>
>>> --
>>> 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-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 <javascript:>.
> To post to this group, send email to everyth...@googlegroups.com
> <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