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

```> On 18 Nov 2018, at 14:00, agrayson2...@gmail.com 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 <javascript:> 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”.

> 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.

>
>> 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 about
measurement.

> 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
>>>> 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
>>> 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
>> 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
> To post to this group, send email to everything-list@googlegroups.com
> Visit this group at https://groups.google.com/group/everything-list
> For more options, visit https://groups.google.com/d/optout