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

I Identical particles distinguished by their trajectory?

  1. Apr 18, 2016 #1
    Hello.
    It is said that if we exchange two electrons, we can't tell which is which. Identical mass, charge, etc. So if I hold two electrons, one in each hand, and someone switched them, I wouldn't be able to know. But one way to distinguish particles is their trajectory. If I have a very long 1D line with an electron at each end, and I measure their kinetic energies, and a very short while after, I measure their positions again. I can see both of them have moved, but their positions can be related to their velocities times the time interval, i.e d=v*t for each electron... Now I can pretty confidently say that the electron to the right is still to the right, and the electron that was to the left before is still to the left, because let's say their speed was small. Or alternatively, it would break the speed of light if the left electron had moved all the way to the other end of the 1D line in that small time interval and vice versa, and hence I can distinguish the two electrons. Maybe a long time after, the wave functions would overlap due to Schrodinger equation, and then they would be truly indistinguishable.
    So is it possible in situations like these to actually distinguish two electrons, or did I mess up?
     
  2. jcsd
  3. Apr 18, 2016 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    Sure, but having "a thing over here" and "a thing over there" has nothing to do with the indistinguishably of electrons so you are setting up a strawman.
     
  4. Apr 19, 2016 #3
    I think strawman is something people use in a fallacious argument, not a question asking for understanding.

    Bootleg, indistinguishabililty comes into play from the statistics of the states, and is expressed in terms of math. It states that if you exchange the two identical particles with each other, the probability of the state doesn't change. By the Born rule, this means the norm of the wavefunction must not change. Since two exchanges is equivalent to no exchanges, this means that the wavefunction must stay the same or be multiplied by -1 during exchange. It's not something that can be explained by a classical analogy.
     
  5. Apr 19, 2016 #4

    phinds

    User Avatar
    Gold Member
    2016 Award

    which is, I think, exactly what he presented although I do see how you interpret it more as just a question.
     
  6. Apr 19, 2016 #5
    I think I can make it a little clearer. Suppose you have two particles which are indistinguishable, and you hypothetically label them 1 and 2. In an experiment, you have a particle go left and a particle go right. Classically, you would think that either 1 went left and 2 went right, or 1 went right and 2 went left. But the math of indistinguishability says that both situations are equally probable. Before you make a measurement, there is no randomness and therefore the left particle has to be a superposition of 1 and 2, and the right has to also be in a superposition of 1 and 2. You can't say the left particle is 1 and the right particle is 2, because why would 1 be biased to go left and 2 biased to go right, if they are identical? Now since both left and right are both superpositions of 1 and 2, there's no (measurable) change if you swap 1 and 2.
     
  7. Apr 24, 2016 #6
    Thanks guys this helped.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Identical particles distinguished by their trajectory?
  1. Identical particles (Replies: 45)

  2. Identical particles (Replies: 3)

  3. Identical particles (Replies: 1)

Loading...