Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mind experiment

  1. Jul 8, 2005 #1
    I've been thinking about something for quite a long time now. A muon can decay into two electrons, right? Since the spin of the muon is zero, the total spin of the elctrons must also be zero. That means one of the electron has up spin and the other one has down spin. According to the Copenhagen model the electron actually has neither (or both) before we measure it. ok, so long all is fine. If we measure one electron to be up, we know the other one must be down. (Which by itself is a bit odd). Now to the tricky part. What if, we measure the spin of both electrons with a small time discrepancy. The time between the measurement must not be longer than the time it takes to travel from one of the electrons to the other with the speed of light. Every time we will find, that the electrons has opposite spin, but there is no chance of the electrons ever "interacting". How can the second electron "know" which spin to apply if both electrons, before the measurement, was in the same indeterminate mode?


    Ah, i just found the name of the setup
    EPR (Einstein Poldalsky Rosen) -experiment
    i'll read some more on the subject
    Last edited: Jul 8, 2005
  2. jcsd
  3. Jul 8, 2005 #2


    User Avatar
    Science Advisor
    Gold Member

    Muons don't have a direct decay path into 2 electrons, as far as I know. EPR tests do not use muons for entangled pair production.
  4. Jul 8, 2005 #3
    Muon (either plus, minus or zero) cannot decay into 2 electrons, because muon has the same charge as single electron. The simplest decay would be probably pion at rest to 2 gammas:

    [tex]\pi^0 \to \gamma + \gamma[/tex]

    where gamma's are created in singlet state (conservation of ang. momentum):

    [tex]\Psi_{12} = \frac{1}{\sqrt{2}}\left( \Psi_1(\mathrm{up})\Psi_2(\mathrm{down}) - \Psi_1(\mathrm{down})\Psi_2(\mathrm{up})\right)[/tex] .

    We cannot write this function as a product: [tex]\Psi_{12} \neq \Psi_1 \times \Psi_2[/tex], which we could if the 2 photons (gammas) would be separate entities (I'm not sure if this is the right word). If I understand correctly, in this sense, the state [tex]\Psi_{12}[/tex] can be called "entangled" state (meaning: "connected"). According to QM, it's correct only to refer them as a single entity or as a one system, not as 2 separate photons. Of course, this doesn't really solve the problem of instant interaction. A little comfort to Einstein's theory of relativity is that this weird interaction still doesn't allow to transfer information at instant (so causality is not affected).
    Last edited: Jul 8, 2005
  5. Jul 8, 2005 #4


    User Avatar
    Science Advisor

    Note to igor s - you described pion decay, not muon. Muons have spin 1/2 and decay into one electron + gamma ray. The electron has spin 1/2. Muon and electron each have charge -1.
  6. Jul 8, 2005 #5
    Yes, I made some corrections to my post that may have confused you. Hope it's clear now. My intention wasn't to describe muon decay.
  7. Jul 10, 2005 #6
    ok, sorry. muon, pion, remembered it the wrong way 'round.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook