# Double Slit Experiment and Electron Spin

1. Feb 20, 2009

### Wallin

Can anyone direct me to any version of the double-slit experiment which used only electrons with aligned spin axes? If you can't direct me to an experiment, can you postulate how or if the double-slit experiment might be different using such electrons?

Thanks!

2. Feb 20, 2009

### sirchasm

Unfortunately 'aligning' the spins of electrons, is exactly what a double slit does.

So if you align the spin of an electron and send it through, it gets re-aligned.
If the electrons have a global spin-phase, the slit preserves it, but still creates a divided wavefunction, which evolves in 'probability-space'.

3. Feb 20, 2009

### Wallin

I don't understand how going past the biprism would re-align the electron. Can you explain that? Are you saying that there is no way to do the double slit experiment using eletrons with aligned angular moment and preserving that angular moment throughout the course of the experiment?

4. Feb 21, 2009

### sirchasm

"If the electrons have a global spin-phase, the slit preserves it"; so if you align the electrons by giving them a global spin(-phase angle), it will be preserved - except you won't be able to measure it with a double-slit since that will give each electron another phase angle.

The second angle, with the path through the 'slit interferometer' that introduces it, can be determined (say by a screen, in the usual way); the first is a global phase which the screen won't 'see'.

"Relative phase is detectable; global phase isn't":- QM 101.

5. Feb 21, 2009

### sirchasm

Of course, you can build a circuit that exploits spin-phase, you just need to connect the spins and the paths they take, with 'gates' that preserve and differentiate global (overall) and local (relative) phase angles. This is much easier with laser optics and mirrors and polarizers.

Mirrors and filters are the gates, beams of coherent light are the signals. Electrons can be used in another basis - you could have gates 'inside' the slit-interferometer. The gates or slits (or magnets, say) rotate phases, so if you can mix two signals that are each 1/2 the output of 2 gates, that gate the same initial input(s), you can measure the angles. Measuring destroys or collapses any local phase - QM 101.1

6. Feb 21, 2009

### per.sundqvist

Where have you got this from? The only way you could coupe the spinor components in a double slit (which is non-magnetic) is by the spin-orbit interaction.

$$\hat {H}_{so}\propto (\vec{E}\times\vec{p})\cdot\vec{\sigma}$$

acting on the spinor: $$\vec{\Psi}=(\Psi_{\uparrow},\Psi_{\downarrow})^T$$

Having an incoming wave with only spin up, my strong guess is that you would get a verys small component of spin down both as transmitted and reflected waves. Say probability of spin up is of order 1, then transmitted would be of order 1e-5 or something like that. Haven't done the calculation but this is my guess. Have never heard of double slits used to obtain spin polarization, if so it would be great break through in spintronics!

7. Feb 21, 2009

### ZapperZ

Staff Emeritus
I would like to hear this as well since I have never heard of such a thing being done already. Sirchasm, can you please cite published papers that have done exactly what you wrote in this thread?

The clearest demonstration of a "double slit" experiment for electrons are using SQUIDs. Even in there, there is no discrimination of the spins of the individual electrons.

Zz.

8. Feb 21, 2009

### Wallin

Thanks to all of you. If you find anything out, I would appreciate any additional information. I have been told that if the spin allignment is retained that there would be no interference pattern even though no attempt is made to detect which slit the electrons pass through. I can't figure out why. Does the uncertainty principle come into play here? I was thinking that it wouldn't since spin is the only information I "know".

Sirchasm, could you further explain what you mean by global and relative phase angle? I don't think I follow you...which isn't a surprise given my background. Please keep the explanation relatively simple.

Regards!!

9. Feb 21, 2009

### sirchasm

What happens in a Stern-Gerlach apparatus? What happens in the AB effect?
The only way you could couple spinor components, is to use electrons with spinor components.

An electron diffraction experiment (AB, Bragg scattering, ...) divides electrons according to their spin. If you "process" electron spin, using diffraction, it separates electrons, or sorts them. Momentum and charge sort the electrons by selecting a spin for them (or, they already have one).
The only other thing you need to sort electrons, is some matter that looks like a double slit.

I think there's a bit of misunderstanding; you don't use this apparatus to spin-polarize them, you use it to create a phase-difference.
If you send spin-up electrons into a sample of graphite, what happens? Do you get one ring on the screen?

10. Feb 21, 2009

### ZapperZ

Staff Emeritus
But a stern-gerlach is NOT a "double slit" experiment. The spin components are not made to interfere.

Bragg scattering divides electrons according to their spin? Since when? I've done several Bragg scattering experiments and in none of these are their spins separated. The same goes with LEED experiments. The interference effects are not due to any spin components at all.

This is getting to be highly dubious.

Zz.

11. Feb 22, 2009

### sirchasm

What's dubious about accelerating thermionic electrons? Or scattering a beam of such electrons through a sample of ordinary graphite?

Or the two circles that light up on a phosphor screen after the electrons have 'sampled' the graphite? The double circle is because of the geometry of the graphite interatomic scale, which is flat-looking?

Why does it make the beam divide into two, and why are these beams circular?

A single inhomogenous magnetic field isn't a double slit, obviously, it's one thing; the electron's momentum diverges because spin is diverging it.

In neutron interferometry, a magnetic field aligns one-half of a neutron beam sorted by a Si lattice. What's a phase gate in QIS?

12. Feb 22, 2009

### ZapperZ

Staff Emeritus
You seem to have totally forgotten the OP.

And what does the spin properties have anything to do with the interference effect that's due to the spin, which is what the whole question is all about. All you seem to be doing here, it seems, is to show the existence of the spin bipartite state, which is NOT what is being questioned here.

Instead, you're bringing a bunch of experiments which has nothing to do with the question.

Zz.

13. Feb 23, 2009

### sirchasm

Are you implying that fermionic spin should not be discussed if someone asks about electron interferometry, i.e. a double-slit experiment?

I can't decipher what anyone (that's you, pal) might intend or mean with (a): "The spin properties" then (b): "to do with the interference effect that's due to the spin", by connecting both phrases with: "what does the first have to do with the second?"; you can't personally see any?

Spin has to do with interference in fermions; this is well-understood. Are you wanting to know "what it has to do" with interference? Don't you know?
You can't see any connection between neutron and electron spin experiments?

A sample of graphite isn't a double-slit? The two circular images you see are separated because of Bragg scattering, nothing to do with waves or Planck?

Sorry, it's just I can still remember being told it was something to do with it. I don't know how the faculty could have made such a big mistake...

But that's what I learned "the spin properties have to do with". OK?

14. Feb 23, 2009

### Staff: Mentor

I assume you're referring to the experiment described in the following PDF, which is commonly done in undergraduate modern-physics labs:

www.uow.edu.au/eng/phys/200labs/phys205/ediff.pdf[/URL]

The radius of a diffraction ring in this experiment depends on the spacing between planes of atoms in the crystals that make up the target. You have two rings because the atoms can be grouped into two different sets of parallel planes, with different spacings.

You get a ring rather than a single spot because the target isn't a single crystal, but instead, a collection of many randomly-oriented crystals. If a crystal happens to have the right "radial" orientation angle with respect to the beam, part of the beam is reflected, due to constructive interference. Different such crystals have different "azimuthal" orientation angles with respect to the beam, so they reflect the beam in different directions, but always with the same radial angle.

The two rings are formed by reflections off of different sets of crystals. The spin of the electrons has nothing to do with it.

Last edited by a moderator: Apr 24, 2017
15. Feb 23, 2009

### sirchasm

I see.
Does spin have anything to do with Planck's constant, you know, h?

If spin is something that two rings on a phosphor screen has nothing to do with, then a Bragg diffraction cannot yield any measurement of electron spin? Is this correct?
(just so I get this right, and don't make the same mistake again)

16. Feb 23, 2009

### ZapperZ

Staff Emeritus
Electrons exhibit interference effects NOT because of its spin! In fact, many electron interference effects have NOTHING to do with its spin (example: look at LEED, SQUIDs, etc). So it is a puzzle why you are invoking spins to explain the observation of such interference effects. THAT is what I do not understand. I can do x-ray Bragg scattering with with photons as well. What's the "fermionic" component here that are so required here?

When polarized particles are used, it has more to do with wanting to probe the spin states of the material. This is done, for example, in electron spin resonance experiment and neutron scattering experiment. However, the interference effects do NOT require that these be done with such spin polarized particles. I can get various interference patterns with unpolarized particles as well!

Zz.

17. Feb 23, 2009

### sirchasm

When you do Bragg diffraction with electrons, you boost them off a thermionic cathode through an accelerating voltage. When they arrive on a phosphor screen there are two circular areas.

This has absolutely nothing to do with fermionic spin.

Is this correct? (note this is simply asking if the previous statement has any basis in physical reality, is all)

P.S. I think you will find that photons are spin-1 bosons, whereas electrons are spin-1/2 fermions.
(but I might have that wrong too)

18. Feb 23, 2009

### ZapperZ

Staff Emeritus

http://www.upscale.utoronto.ca/IYearLab/elecdiff.pdf

You will notice that (i) no mention of electron spin is being given as the source of the diffraction pattern, and (ii) that you get different things depending on the nature of the MATERIAL. So how could the electron spin be the source of such a thing? I could have a well-ordered single crystals that get no "rings".

These are standard experiments with standard explanations, done in universities advanced undergraduate labs.

They are, but they are certainly not fermions the way you were insisting in your earlier posts. And the spins of these photons have no bearing on the results, since I could get almost the same time of bragg scattering using them. The spins are NOT the origin of such interference.

Zz.

19. Feb 23, 2009

### DeShark

As has probably been stated (although not explicitly that I can see), using aligned spin electrons has no effect on the final result of interference (i.e. probability to find an electron at a certain position). This is analogous to using polarised light in a double-slit experiment. Sending the electrons through a slit has no effect on the spin of the electron.

20. Feb 23, 2009

### sirchasm

So therefore, there is "no spin" and you cannot get a result from Bragg diffraction that means you can calculate Planck's constant. OK, that's what I wanted to know.

So it's impossible to do a scattering experiment if you use Bragg angles, to calculate h? You need to do another kind, and use different formulas - Bragg scattering only gives an angle.
If there are no circular rings on the phosphor, is there no angle to measure?

P.S. just a heads-up, I'm trying really hard, to get you to admit "you cannot calculate Planck's constant with Bragg scattering experiments".

So are you going to?

Last edited: Feb 23, 2009