Stern-Gerlach Question: Can Interference be Seen?

[msumm21]

Quoting from Wikipedia (Stern-Gerlach experiment):
as the particles pass through the Stern-Gerlach device, they are "being observed."

Is that true? For example let's say we have spin 1/2 particles, can you not create an appropriate double-slit (DS) setup on the other side of the Stern-Gerlach (SG) device and see an interference pattern? I would have thought you could... if you don't "observe" the particles and their spin is not determined they effectively get deflected in all directions by the SG device and then pass through both slits of the DS and display interference. However, the above-mentioned quote seems to imply that as a particle enters the SG device it picks a particular spin state and then goes through a specific path through the DS, no interference.

Thanks

 

[Vanadium 50]

A Stern-Gerlach apparatus makes a measurement. A measurement is an "observation".

 

[msumm21]

So putting a particle in a magnetic field (SG device) constitutes an observation? If so, how are two electrons kept entangled? Each sees a magnetic field due to the spin of its counterpart, and therefore is observed?

 

[Vanadium 50]

What entanglement?

 

[msumm21]

What I'm saying is that, if putting a particle in a magnetic field (SG device) constitutes an observation, then I wonder how some other things work (like certain entanglement experiments). For example, when two electrons are entangled each sees a magnetic field due to the spin of the other, and therefore they are continuously being "observed" right? But then they wouldn't display the properties of entanglement, right?

 

[Vanadium 50]

I think you may be trying to try and do science by putting streams of words together, rather than understanding QM quantitatively, and I think that's why you're getting tangled up. The fact that spins can be entangled doesn't mean spins are always entangled. The fact that a Stern-Gerlach experiment uses a magnetic field doesn't mean any magnetic field is a Stern-Gerlach experiment.

 

[James R]

Merely passing through a Stern-Gerlach apparatus does not constitute an "observation", in my opinion. More generally, passing through a magnetic field does not constitute an observation in and of itself.

 

[msumm21]

I think you may be trying to try and do science by putting streams of words together, rather than understanding QM quantitatively, and I think that's why you're getting tangled up. The fact that spins can be entangled doesn't mean spins are always entangled. The fact that a Stern-Gerlach experiment uses a magnetic field doesn't mean any magnetic field is a Stern-Gerlach experiment.

No, I'm not saying those things... I never said spins were always entangled. I am doubting (along with JamesR) that letting a particle pass through a SG device constitutes an observation because it seems to imply some strange things. The Wikipedia article and you are saying that it is an observation.

Could you be more specific about what (when passing a particle through a SG device) causes it to be "observed." The SG device is just a magnetic field, right? When you said that passing a particle through an SG device constituted an observation of the particle I thought you were saying that placing a particle in a magnetic field constituted an observation, is that not true? If not, what constitutes the observation of the particle?

 

[Phrak]

Quoting from Wikipedia (Stern-Gerlach experiment):
as the particles pass through the Stern-Gerlach device, they are "being observed."

hmm.. looks like the Wikipedia writer wants to ambiguate entanglement with observation. Best not to, i'd think. It makes them more difficult to tell apart.

 

[strangerep]

Could you be more specific about what (when passing a particle through a SG device) causes it to be "observed." The SG device is just a magnetic field, right? When you said that passing a particle through an SG device constituted an observation of the particle I thought you were saying that placing a particle in a magnetic field constituted an observation, is that not true? If not, what constitutes the observation of the particle?

I offer my $0.02 worth...

A measurement resulting in a particlar outcome is modeled by a projection operator
of the form [itex]P := |\psi\rangle\langle\psi|[/itex] and is such that [itex]P^2 = P[/itex].
The output from a Stern Gerlach magnet is 2 beams. Roughly speaking, the beam going
"up" consists only of spin-up particles, and the beam going down consists of spin-down
particles. So far, there is no "observation" nor "measurement". You've merely operated
on the original beam with a magnetic field, resulting in two beams.

However, if you now select one of the beams (the "up" beam, say) for subsequent
manipulation, then you have performed a measurement, because you've acted
on the combined pair of output beams with the projection operator [itex]|\uparrow\rangle\langle \uparrow|[/itex] .

As to whether this affects a subsequent double-slit experiment, the action of the slits
can be modeled as yet another operator (ideally a sum of delta functions in position
space, or an equivalent convolution in momentum space), so it depends on whether the
[itex]J_z[/itex] (spin orientation) operator commutes with the 2-slit operator.

 

[Fredrik]

If you send an electron through the inhomogeneous magnetic field produced by a Stern-Gerlach apparatus, its state expressed as |spin>|position> changes to |+>|left>+|->|right>. This isn't a measurement, but a subsequent position measurement is also a spin measurement, because the spin and position states are correlated. If we're talking about a device that doesn't measure the position, I don't think it's correct to call it a Stern-Gerlach apparatus.

 

[msumm21]

Thanks for the info, one more question. So if we pass an electron down through an inhomogeneous magnetic field its state becomes |+>|left> + |->|right> (assuming the appropriate orientation of the m field), so its spin remains unmeasured. At this point measurement of either spin or position is effectively a measurement of both, which makes sense. However, if neither are measured then the particle effectively passes through both slits of the double slit, and interference is seen right? Or are you saying that even after passing through the double slit the subsequent position measurement constitutes a spin measurement, giving which-way info, and no interference?

Thanks

 

[Phrak]

I offer my $0.02 worth...

how about a fiver? well done!