What happens if you repeat stern-gerlach-exp. twice?

[simpatico]

Homework Statement

I mean if you take one of the two beams from the experiment, say spin-up
and feed it again into the machine

is it possible to detect spin up/down in a cathodic-tube beam or in a synchrotron?

Homework Equations

none

The Attempt at a Solution

my uneducated guess is that you get again two beams up/down

 

[BruceW]

Is this a homework question, or just curiosity? In any case:
Think about it in terms of eigenstates. When the beams come out of the first Stern-Gerlach apparatus, they are both in eigenstates of the z-component of spin.
So when one of these beams enters the second Stern-Gerlach apparatus, will it be split?

 

[simpatico]

I can't have any homework problems, considerng I'm pushing 70.

Just a curiosity to deduce if it is a causal distribution or an intrinsic quality.

same for second question do you detect spin in a synchroton or e+beam in a MF?
Here, too, my (uneducated) guess would be YES.
I'd appreciate a clear, unequivocal answer.
Much obliged

 

[BruceW]

If you take one of the beams, from the experiment (say spin-up) and feed it back into the machine, then the beam will not be split again

 

[BruceW]

To give a more comprehensive explanation:
A Stern-Gerlach apparatus measures only the component of spin of the electron in the direction which the Stern-Gerlach is oriented in.
So if the Stern-Gerlach is oriented vertically, it will split the beam into an spin-up beam and a spin-down beam.
And if the Stern-Gerlach is on its side, it will split the beam into a spin-left beam and a spin-right beam.
So if you have a vertical Stern-Gerlach (which gives an up-spin and a down-spin beam), then feed the up-spin beam into a horizontal Stern-Gerlach, then the beam will be split again.
But if you feed the up-spin beam into another vertical Stern-Gerlach, then the beam will not be split a second time.

 

[BruceW]

Just to make it clear: the two beams that come out of a Stern-Gerlach apparatus do not have the the same properties as the beam that went into the Stern-Gerlach apparatus

 

[simpatico]

the two beams that come out of a Stern-Gerlach apparatus do not have the the same properties as the beam that went into the Stern-Gerlach apparatus

(next post came first,probably I moved it editing)

in the first place, I meant the same machine, so I was right, now you tell me you have different machines, I was not prepared to that!

knockdown, that is what is disturbing to me. it is not decisive, in the sense that even an intrinsic property can have casual distribution.
know, to knock me out,
tell me that if you feed a pure e+beam into the machine , there is no split-up!..and likewise in a synchrotron

 

[simpatico]

.
So if you have a vertical Stern-Gerlach (which gives an up-spin and a down-spin beam), then feed the up-spin beam into a horizontal Stern-Gerlach, then the beam will be split again.
.

,that's good news
If I got it right: if you had a fantamachine you would have a circle?
(I mean a circular distribution?)

 

[BruceW]

I'm not sure what you mean by fantamachine and circular distribution.

You said a while back that you were wondering if there was a causal distribution or an intrinsic property.
If we had just one electron, and we knew everything that could possibly be known about that electron, we still wouldn't be able to predict which way it came out of the Stern-Gerlach apparatus. We could calculate the probability that it would come out one way or the other, but we wouldn't be able to predict the actual outcome.
Similarly, if we wrote down the exact state of the universe right now, then the outcome of a measurement we make in the future would still be inherently random.

 

[simpatico]

I'm not sure what you mean by fantamachine and circular distribution.
.

I meant a machine that could simultaneously operate in all directions.(kind of a joke)

I meant also that it doesn't prove or disprove much re IQ(intrinsic quality).
we have a beam of atoms, all equals, spinning (we don't know why spinning) in all, (probabilistic) directions.
When you apply a MF in one direction, you split the beam in that direction.That is the way it has to be!.Why expect a diffuse distribution?.
if you apply a similar MF to the two outgoing beams the up splits into up/ down and the down too,
because there was no reason for them to be up/down in the first place.(just casuality)
I suppose that if we change element (say copper), we get a slightly different radius, is there a formula?
If I got it right this proves something

2) now the second question is decisive, conclusive:
if a beam of electrons doesn't split up, that proves what we are looking for.

 

[BruceW]

If you use the same Stern-Gerlach apparatus (oriented in the same direction) on the 'up-spin' beam, it will not be split. (and neither is the 'down-spin' beam).
Also, the standard interpretation of quantum mechanics rejects the notion of causality.

 

[BruceW]

Causality is rejected in that the probability of a certain outcome of experiment can be calculated, but the actual outcome cannot be predicted, even if you knew everything that is possible to know about the system beforehand.

 

[simpatico]

Causality is rejected in that the probability of a certain outcome of experiment .

I meant that the orientation of spin axis is casual, so that in a beam of atoms, all directions are represented and you would get a circular distribution.Is it correct?

1) Could you please specify what determines the displacement,
( I mentioned copper) is there a formula?

2) And what about a beam of electrons, is it split like the silver atoms, or not, or in a different way?

 

[BruceW]

I think the Stern-Gerlach experiment has been done several different ways.
I think the main idea is that the beam is a beam of neutral atoms, and that the 'spin' is the intrinsic spin of the electrons of these atoms. (intrinsic spin doesn't mean the electrons are literally spinning in space, but they do have some properties similar to angular momentum).

Lets say the Stern-Gerlach apparatus is oriented to measure the z-component of the electron's intrinsic spin. Before the beam went into the apparatus, the z-component of a given electron's spin was not defined. When the two beams come out of the apparatus, the z-component of the electron's spin is defined (up for one beam and down for the other).

If a beam of free electrons was sent through the Stern-Gerlach apparatus, it would be a charged particle moving through the magnetic field, so spin effects would be unnoticeable.

 

[simpatico]

If a beam of free electrons was sent through the Stern-Gerlach apparatus, it would be a charged particle moving through the magnetic field, so spin effects would be unnoticeable.

Great, Bruce! ( )
I'm so-glad for poor-so(ul)d-electron!

That is beyond my wildest..,
exactly what I expected to he hear from you!
as your countryman said " ...a thing of iq is a joy forever'



(if you don't think the thread is exausted, you might answer the side- questions)
(copper?, formula for radius,why expect a different distribution...)

or, if you prefer, we might move on to my thread 'Hydrogen orbits...'


Whatever you choose, Thanks, again!

 

[BruceW]

Not sure what you mean about radius distribution for copper. If we're still talking about the Stern-Gerlach experiment, using different elements, this is what I think happens:
The total angular momentum of all the electrons (intrinsic spin and orbital angular momentum) contribute to the observable which is being measured by the Stern-Gerlach apparatus.
The original experiment was done with silver atoms. In silver atoms, all the electrons except one are in filled orbitals. It is a rule that the most important electrons in such a set-up are the ones not in filled orbitals. So the angular momentum of these silver atoms is solely due to the one outer electron. This one electron was known to have zero orbital momentum, yet the beam of silver atoms was split into two. This was explained by the fact that the electron also possesses intrinsic angular momentum (spin).
Copper has one electron in an unfilled shell, so I'd guess it would work similarly to silver in this experiment.

What did you want to know about hydrogen orbits? They're the best, since they are the simplest :) Did you start a new thread about them?

 

[simpatico]

1) Not sure what you mean about radius distribution for copper.
2)Copper has one electron in an unfilled shell, so I'd guess it would work similarly to silver in this experiment.

What did you want to know about hydrogen orbits? They're the best, since they are the simplest :) 3)Did you start a new thread about them?

1)I mean the displacement od spin/up-spin/down, the distance (up-down) / 2

You said it depends on momentum ( I suppose magnetic).So can you predict the length of the displacement?


------------(beam displaced by SGm) ------ --------- ------------- Up-beam
--original silver beam ---------_>>>[S.G.m.]_------------c center
------------(beam displced by SGm) ------ -------- --------------- Down-beam


2)That is why I choose copper.Distance of up-beam of copper will be greater?

3) yes, I chose them because they are the simplest
the thread is 'Hydrogen orbits, MF, orbit-spin interaction
wish you could help!

 

[BruceW]

The copper and silver atoms both have the same 'outer electron', so they would feel the same force from the external magnetic field. The silver atom has more mass, so I reckon it would displaced less than the copper atom. (I think I have come to the same conclusion as you).