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?


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:


  • 5
    $\begingroup$ 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? $\endgroup$ – Sklivvz Apr 7 '11 at 9:27
  • $\begingroup$ 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 physics.stackexchange.com/questions/8191/… $\endgroup$ – Georg Apr 7 '11 at 9:57
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Apr 7 '11 at 11:20
  • 3
    $\begingroup$ If you want Stern-Gerlach experiments with individual electrons, be sure that they have also been done - see e.g. 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. $\endgroup$ – Luboš Motl Apr 7 '11 at 11:23
  • $\begingroup$ Thanks. It's much clearer now, for the sake of the next user. :-) $\endgroup$ – Sklivvz Apr 7 '11 at 18:18

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.

  • $\begingroup$ Do you have references to such an experiment (measure the spin of free electrons)? $\endgroup$ – verdelite Aug 19 at 18:18

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

  • $\begingroup$ 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 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. $\endgroup$ – Marty Green Dec 23 '11 at 16:40

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

  • 1
    $\begingroup$ ""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! $\endgroup$ – Georg May 21 '11 at 9:52
  • 2
    $\begingroup$ 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. $\endgroup$ – Marty Green May 21 '11 at 17:42
  • $\begingroup$ 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. $\endgroup$ – Peter Shor Dec 23 '11 at 18:37
  • $\begingroup$ 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. $\endgroup$ – user24209 May 8 '13 at 17:20

protected by Emilio Pisanty Oct 4 '17 at 12:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.