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Basic:applying uncertainty principle for confined particle

  1. Apr 24, 2012 #1
    i tried to apply uncertainty principle to an electron confined in a 3d box of size 1fm.

    i got uncertainty in velocity Δv to be of the order 1010 m/sec.so i thought maybe i should have taken relativistic mass instead of rest mass.

    but i realised that for calculating relativistic mass we need to know the velocity of the particle.but since we dont know that,

    i was wondering whether we could ascertain whether its possible for electron to be confined in a 3d box of that size?
  2. jcsd
  3. Apr 24, 2012 #2
    Yes using the minimum position-momentum uncertainty relation I believe your uncertainty for the velocity should be that high. But you shouldn't need the relativistic mass since its uncertainty is of the order of 10^10 and not necessarily its velocity.
  4. Apr 25, 2012 #3
    ok.thank you,
  5. Apr 26, 2012 #4
    I do not understand why you have the uncertainty in velocity of order 10 power of 10m/s,because it is greater than velocity of light?
  6. Apr 27, 2012 #5
    sorry i was a bit late in replying.

    i also was not able to understand the same thing.so i thought since Δp is large then p also must be large.so i thought instead of writing Δp as m0v we should write Δp as Δ(m0v/√(1-(v/c)2)) and then find Δv.

    but since i have never seen anyone applying relativity in heisenberg's uncertainty principle(although my experience is very less) so i was confused whether i am doing right thing or not.

    @grindfreak:i wanted to conclude something about v from Δv.your observation is right but i think if Δv is greater than a particular value there is no possible v below c.i have attached a reasoning.

    Attached Files:

  7. Apr 27, 2012 #6
    The uncertainity relates position with momenta, you just have to use the relativistic expression for momenta in order to get the velocity. With the data you have provided that will be about 0,97c. This obviously confirms that the relativistic expression must be used.

    At first we get Δp and then Δv is derived from Δp.
  8. Apr 27, 2012 #7
    Of course if the space where the electron is confined is so small that the electron total energy becomes much greater that it's rest energy, others effects like pair electron positron creation may occur but that's another matter more complicated
  9. Apr 27, 2012 #8
    Although I am not currently skilled in QFT, I do know that one can have infinite uncertainty of a variable (meaning we are approaching infinite precision for the other). In my mind this does not mean that the particle is moving with a relativistic speed, it just means that the uncertainty is of the very high order of 1010, and since this is greater than the speed of light, we have no real way of measuring its velocity. Basically as Δx -> 0, Δp ->∞, since their product is greater than or equal to hcross/2.
  10. Apr 28, 2012 #9
    thank you for your help guys.
  11. May 2, 2012 #10
    Δp ->∞ means that v -> c. See any basic relativity textbook.
  12. May 3, 2012 #11
    In what world does Δp ->∞ mean that v->c? Δp = √(<p2> - <p>2) where <p> = ∫ψ*pψdp for a wave function in the p basis as seen in any basic quantum physics text. This is the standard deviation, it has nothing to do with the momentum or the velocity approaching a certain value.
  13. May 8, 2012 #12
    You are using a correct definition of Δp but an infinite standard deviation on the momenta means that the momenta is absolutelly undefined and therefore the most probable speeds are those nearer to the speed of light. That's because the "density of states" of the momenta in function of speed is greater for velocities near to c. In the limit v --> c the "density of states" grows infinite.

    If a particle is trapped on a box such small as the one you proposed, then the uncertinity in its momenta will be great and therefore speeds a almost as c are probable. Of coure it's also possible to found the particle v = 0.

    Best regards,
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