Stern-Gerlach Something missing in the analysis?

[geistkiesel]

Below is a schematic of two Stern--Gerlach transition experiments. In experiment A. The T segment is obstructed in the +T and –T channels, allowing 1/3 of the beam to pass through the +- T channel. The particle exiting in the +-T state then enters the S segment where the +S and –S channels are obstructed. Again 1/3 of the beam survives transition through the S segment in the +-S state.

 

In the B experiment the obstructions are removed from the T segment. Quoting Feynman, the “wide open” T segment affects the transition as if the T segment “were not present”. Here the particle exits the T segment and enters the S segment where the +S and –S channels are obstructed. No particles survive the transition through the S segment

 

Symbolically the two transition are,

A. +S -> T + b -> T and T -> S + b -> S.

B. +S -> T -> S and S -> S + b -> 0.

 

Here “+b” indicates the presence of the obstructions or “blocks” in the transitions.

 

Question: What is the physical reason for the difference in the results of the two transitions?

 

Geistkiesel

[Qfwfq]

I can't see why no particles survive the transition through the S segment, but I belive I haven't understood your description. It isn't easy to describe things without diagrams or the like. :) Is there any material that you could link to or supply?

[geistkiesel]

I can't see why no particles survive the transition through the S segment, but I belive I haven't understood your description. It isn't easy to describe things without diagrams or the like. :esheriff: Is there any material that you could link to or supply?

 

Qfwfq, Here is a schematic of transition rules and nomenclature etc. The best Stern-Gerlach reference I can give is Feynman's "

Lectures on Physics" Vol. III Chapter 5

(the most complete source I am aware of, which is self contained. I have analyzed the problem and concluded there is only one straight forward explanation. In one sense the problem here is a remake (analogous),of the two hole difraction problem and how does a single electron go through two holes?

I can start with a "hint" if you like.

The obstructions are placed in locations that the particle would have taken had the particle been so polarized when entering the T segment. RF makes the point that a particle produced with two obstructed channels, say the +-S and -S channels, that when transitioning this particle through another equally arranged S segment that the particle will always take the +S channel

.

 

The channel that are obstructed in the S segment in the "B." transition, all +S particles will take the upper channel and crash into the obstructions there. The fact of the "wide open" T segment, free of obstructions, means the particle transitions through that segment "as if the segment wasn't there". (an RF quote) I take some exception to this as the particle

is affected

when inside the T segment. The particle is necessarily polarized to one of the three possible T states and upon exiting reforms to the +S state in the field free volume outside the T segment. Unlike the comapss needle that returns to North by the force of the earth's magnetic field after being perturbed, the spin=-1 particle returns to the prepolarized state in a field free environment when exiting teh T segment.!!

 

The selection of which channel the particle uses in the T segment is 'particle selected' and is effectively random , or seemingly random.

 

The standard model has the particle states determined from the heat of the "tungsten filament" that boils off the particles used in the experiments. This is a " grasping for solutions" as I view the matter. The only statistical requirement as determined from experimental results is that 1/3 of a pure spin-1 beam source will take one of the three channels equally 1/3 of the time. If this is the case then we may as well place the determination of the particle state when entering the polarizing segment's field/gradient. This implies the particle generates its own states in some rapid fire sequential manner. Lets define the particle state functionally as Y(100) = +S, Y(010) = +-S and Y(001) = -S states. If we used a random model such that the particle would genrate these states effectively evenly in 1/3 distribution we may as well just define the generation of the states as linearly sequential. The time history for the particle generating the states would then be, dropping thje 'Y',

. . .001 100 010 001 100 010 001 100 010 .....

 

And here we lose some randomness in the determination of particle states. Likewise to generalize the function we write it as Y(000f) where it is understood that any '1' observed is the current observed state and that the function changes states at a rate f a where the liftime of the current state is ~ < 1/f.

 

The state the particle is in when coming into first effective contact with the segment polarizing field/gradient (in the S -> T transition) locks the state of the particle to what ever state is "observed" at that instant, (100) for the +S state.. All of this has a moral and an ending.

 

The obstructed channels are those channels the particle would have taken had the partcle been polarized to that particlular state. It would be physically improper to assign any "physical" attrinbutes to the "0" states such as "off" and "on", like a light switch on one's wall. Rather, we should recognize that the "0" states are in fact nonlocal states of the particle that continue to function nonlocally as indicated by the placement of obstructions in the "nonlocal" designated channels. In the presence of the field/gradient the obstructions impose perturbations on the nonlocal elements of the particle such that the pre-polarized +S state is now permannently changed into the +T state in the same manner that the +S state was produced originally.

 

The physics must be the same, do you see this? Both are alien-to-domestic transitions.

 

When the obstructions are removed there are no perturbations to the particle satte and upon exiting the segment and the effect of the field/grasdinet the nonlocal elemebnts are sufficient to reconstruct the pre-polarized +S state.

 

I will follow up with more, but here is how I see the particle described.

 

+S = Y(1 00[+S]) where the 1 refers to the +S observed state and the 00[+S] are the nonlocal, unobserved, elements of the +S state that guarantee the reformation of the +S state in the +S -> T -> +S transition. The 00[+S] nomenclature is purely arbitrary and reflects only the existence of functional nonlocal attributes of the particle state.

 

Here is the schematic I referred to. I will post a more complete set of transition rules derived from Feynman's text, shortly.

 

There will be some "disagreement" I suspect from 'standard model' QM folks.

Geistkiesel