11

Is it possible to measure the spin of a single electron? What papers have been published on answering this question? Would the measurement require a super sensitive SQUID, Superconductive Quantum Interference Device?

QEntanglement
  • 3,831

4 Answers4

10

The spin of a single electron has been measured since the very first moment when the people understood that every electron possesses a spin. A Stern-Gerlach experiment - a magnetic field - is enough to measure the spin:

http://en.wikipedia.org/wiki/Stern-Gerlach_experiment

Sklivvz
  • 13,499
  • 7
  • 64
  • 87
Luboš Motl
  • 179,018
  • 6
    The Wikipedia page does not say anything about single electrons, and the original paper, "Das magnetische Moment des Silberatoms" seems to refer to silver atoms - not sure about this because the paper is in German and behind a paywall. Can you please clarify your answer? – Sklivvz Apr 07 '11 at 09:27
  • 1
    The Stern-Gerlach experiment is on single electrons, but not on free electrons. This is what QEnt... wanted to ask but was not able to formulate. See http://physics.stackexchange.com/questions/8191/measuring-the-magnitude-of-the-magnetic-field-of-a-single-electron-due-to-its-spi – Georg Apr 07 '11 at 09:57
  • 1
    As even the Wikipedia page above says, in 1927, T.E. Phipps and J.B. Taylor reproduced the effect using hydrogen atoms in their ground state, thereby eliminating any doubts that may have been caused by the use of silver atoms. – Luboš Motl Apr 07 '11 at 11:20
  • 4
    If you want Stern-Gerlach experiments with individual electrons, be sure that they have also been done - see e.g. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC323282/pdf/pnas00312-0017.pdf - I just don't understand why someone would find it interesting. It's obviously the same force acting on the same electron. – Luboš Motl Apr 07 '11 at 11:23
  • 1
    Thanks. It's much clearer now, for the sake of the next user. :-) – Sklivvz Apr 07 '11 at 18:18
  • 1
    Stern-Gerlach experiment gives the magnetic moment. Then one needs the relation between magnetic moment and spin to find the spin. But this relation is not trivial! – richard Feb 07 '23 at 10:56
  • 1
    The relationship is direct proportionality which is pretty much by definition trivial. The coefficient is very complex and requires complex QED calculations to be determined numerically, and it is at most a function of other continuous parameters of QED. But I think that "the measurement of spin" really isn't concerned with this continuous parameter. It is a measurement of an observable with discrete spectrum and it produces a bit of information (or another discrete package if it is more than a spin-1/2 particle). – Luboš Motl Feb 08 '23 at 11:03
  • 1
    Thanks, Luboš. Yes, that is correct. But my problem is if the coefficient is unknown then we cannot determine the value of S. Therefore we need a direct way of measuring the spin for example conservation of total angular momentum. – richard Feb 08 '23 at 13:12
  • 1
    Dear Richard, in the Stern Gerlach experiment, the particle goes up if the spin is up and it goes down if the spin is down. The spin z component is either up or down and the measurement means to determine this single classical bit of information (the measurement picks the classical bit out of the quantum bit). Whatever the magnetic moment is doing, the coefficient is such that the spectrum of j_z of a spin-1/2 particle is either +hbar/2 or -hbar/2. Period. The fact that the components of the spin are this quantized in the simple units is a simple mathematical fact. – Luboš Motl Feb 09 '23 at 14:25
  • 1
    I'm sorry I don't understand. The spin should be a well-defined physical quantity. Now I want to measure its precise value. Suppose I don't know Sz=+hbar/2, -hbar/2. Somebody says go and measure magnetic moment using SG experiment. Then divide it by a constant but I don't know what is the value of that constant. So how do I know Sz=+hbar/2, -hbar/2 ? – richard Feb 11 '23 at 10:25
  • 1
    No, that is not how measurement may be described in QM. When you measure any quantity and you interpret what you are doing as a rational action compatible with science, then you must exactly know what you are measuring and what is the spectrum of the observable. This is what the damn measurement means in QM. You decide what is the observable, represented by a linear Hermitian operator entering various commutation and other relations and algebra, is, and the only unknown thing before the measurement is the right eigenvalue. The measurement gives you the eigenvalue and nothing else. – Luboš Motl Feb 12 '23 at 05:00
  • 1
    You may "measure" things like it is done in classical physics. Indeed, people had to determine the quantum of the angular momentum, which can be proven to be the same (one half) of the reduced Planck constant that appears anywhere. You may send a lot of polarized photons (or atoms), for example, to a thin foil and the foil starts to rotate. But this is a "measurement" in the classical sense. A measurement in QM is a very particular new thing that follows slightly updated rules and the a priori knowledge of the spectrum is a part of it. – Luboš Motl Feb 12 '23 at 05:02
  • 1
    You know, you are still thinking classically, assuming that the angular momentum can be any real continuous number in QM a priori. But this statement is not true in QM. The angular momentum is quantized even a priori. I can algebraically prove that $\exp(4\pi i J_z)=1$ which makes the z-component quantized. Period. You may scream at me and prevent me from proving facts in physics using the axioms of QM but then the action using your hands cannot be interpreted as a useful controllable measurement that can be used in quantum physics. – Luboš Motl Feb 12 '23 at 05:04
  • 1
    Thank you for the explanation. I guess I start to understand (I'm not sure though). We have postulates in QM. From these postulates, we find e.g. the orbital angular momentum operator. By definition, it has hbar: L=r cross p and p=-i hbar grad. That is why the orbital angular momentum is multiples of hbar. In the case of spin, we don't have such an operator. For the spin, we know from the SG experiment that Sz=+s0 , -s0 and we just assume that s0=hbar/2. – richard Feb 12 '23 at 13:08
  • 1
    Sorry, the operator for the spin is a special case of the operator of the angular momentum, spin is one term in the total angular momentum, and the spin and the total angular momentum also obey the same commutation relations and have the same quantization rules except that the spin may also be halved. The spin is not made of X and P but that doesn't mean that it obeys different measurement postulates or quantization. I don't know why you keep on producing so many statements that are so obviously wrong, and seem to be deliberately invented to be the opposite of simple and important truths. – Luboš Motl Feb 13 '23 at 17:50
3

First you need to be assured of a source of single electrons. A good one is from spontaneous decay called conversion electrons. Then you set up a Stern Gerlach magnet setup. The problem is that one would need to cancel out the transverse Lorentz force, and this can be done with a transverse uniform electric field to cancel it out. Then use solid state electron detectors to see the deflected electron event counts.

Eric Reiter
  • 61
1

The ion trap experiments by Hans Dehmelt might be of interest. Though the scientific focus was the precision measurement of the g factor, you can't get far with that without first knowing that your trapped electron has spin 1/2 - or if you don't know that, you'll find out pretty quick when theory doesn't match experiment even to first order.

You might find this a good read: Stern-Gerlach experiments: past, present, and future Jean-Francois Van Huele and Jared Stenson - link to PDF is at http://www.physics.byu.edu/Research/theory/paps.aspx

DarenW
  • 5,869
  • 1
    I stumbled independently on Jared Stenson's master's thesis, and it has some things even more interesting that the paper you've flagged. The Master's Thesis is at http://contentdm.lib.byu.edu/ETD/image/etd908.pdf and I discuss its implications in my blog, which I've linked to in my own answer. – Marty Green Dec 23 '11 at 16:40
-2

I was browsing old questions and noticed this one. I think I ought to take issue with the idea you can measure the spin of a single electron. Suppose I prepare an electron in a definite spin state and send it into another room; I don't think there is any way someone else can tell what state I prepared the electron in. Putting it through a Stern Gerlach apparatus certainly won't do. Isn't saying that you can measure the spin of an electron the same as saying you can measure its position and momentum simultaneously?

EDIT: I notice DarenW referes to a paper by Stenson, and it turns out I stumbled on a related paper on my own and it made a big impression on me. The paper I found is actually Stenson's master's thesis, which I will find a link to once I finish this post, and I will post it in the comment field of Daren's answer. As for my own analysis of Stenson's paper, it spans a number of blogposts beginning here. The conclusion is fascinating: if you put a beam of silver atoms through a Stern Gerlach apparatus, it doesn't split into two paths: it spreads out into a donut! I've sketched the deposition pattern for a polarized beam here, and you can read the analysis on my blog. Stern Gerlach pattern for Polarized Beam

Marty Green
  • 4,079
  • 1
  • 21
  • 27
  • 1
    ""Putting it through a Stern Gerlach apparatus certainly won't do"" What makes You shure? ""Isn't saying that you can measure the spin of an electron the same as saying you can measure its position and momentum simultaneously?"" Totally wrong, look up classical physics on spin. Last not least: You mix up some old question with Your assumptions, this is not an answer! Better You delete this and ask in a new question! – Georg May 21 '11 at 09:52
  • 2
    Perhaps there are two interpretations to this question. Are there experiments that can measure the magnitude of the electron spin? I suppose there are; it is not totally clear to me that Stern-Gerlach happens to be such an experiment. I vaguely suspect that it might only measure the ratio of magnetic moment to angular momentum, similar to experiments that measure the charge-to-mass ratio of the electron without measuring either charge or mass. As for measuring the actual spin, magnitude and direction, of a specific electron...I don't think so. – Marty Green May 21 '11 at 17:42
  • You can measure the spin in a given basis. You can't figure out the axis of the spin the electron was prepared in. But when physicists talk about measuring the spin, they usually mean the spin along some predetermined axis. – Peter Shor Dec 23 '11 at 18:37
  • The donut above is the result of the distribution of atoms not all having their polar axis of rotation facing the same direction. in general, one side of the donut is the distribution of the atoms with their poles facing anywhere from 0 to 90 degrees and spin -1/2, and the other half of the donut being the atoms with their poles aligned 0 to -90 degrees and their spins being +1/2; hence, a donut. –  May 08 '13 at 17:20