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Uncertainty principle consist with energy conservation?

  1. Feb 16, 2010 #1
    Hi,all, can anyone help to understand how uncertainty principle consist with energy conservation?

    dEdt >= h/2, then in a short time interval, the energy can be really large, even the previous state has lower energy? I cant really understand this..

    say, particle A appeared at t=0 (at rest) and it has life time 10^-15 s and it has initial energy E1 (use E=mc^2), then is that mean at t>0, its energy could growths up?bigger than E1? Then the energy care not conserved..

    Thanks in advance!
  2. jcsd
  3. Feb 16, 2010 #2


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    The interpretation of this uncertainty is NOT the same as that of x and p, because time is not an operator. See for instance Griffith's QM textbook. The point is, that if you want to talk about energy conservation, you have to know the exact amount of energy available.

    I never really understood how virtual particles are justified with merely this "dE*dt>constant" idea and without quantum field theory. But you see it a lot in textbooks, so if anyone knows how to justify this idea, I'm very curious! :)
  4. Feb 16, 2010 #3
    Thanks haushofer for replay. I can not really understand ''the energy available'', is that based on theoretical or experimental? a bit confusing there, thanks for explan:)
  5. Feb 16, 2010 #4
    Hi luxiaolei!

    Energy conservation can be indeed violated, but one cannot observe such processes experimentally. Processes which violate energy conservation (so-called "off-shell" processes) are "virtual". They appear in the internal structure of Feynman diagrams. Experimentally one can only observe the external lines, which correspond to observable particles.

    Think for example of the double-slit experiment.
    If you measure through which of the holes the particle passes, there's no interference pattern to observe. You destroyed the quantum mechanical uncertainty - your measurement is a physical process interacting with the particle.
    Instead, if you do not measure through which of the holes the particle goes, you observe the interference pattern! In this case there is quantum mechanical uncertainty, and in the theoretical description both possibilties are taken into account. Both ways (particle goes through hole 1 OR goes through hole 2) are weighted in some way with their probability, without taking care what "really happens" or how to interpret it, so far. What "really happens" is virtual, you dannot measure it without destroying the interference pattern. The same happens in those Feynman diagrams. All possible ways (internal virtual structures) are weighted with their quantum mechanical probabilty. And energy conservation has not to be fullfilled by virtual, non-observable processes.

    Best regards,
  6. Feb 16, 2010 #5
    Hi Sunset:

    Thanks alot for the replay, very good explanation, got what I wanted! although I would die for knowing why observer can affect experiment result.
  7. Feb 16, 2010 #6
    Hi luxiaolei!

    You're welcome.

    Maybe a word on that issue. Of course one cannot observe the particles by staring at them with one's eyes. Nobody saw an electron with his bare eye. In order to observe an electron one has to be more rude: one has to shoot other particles on it. From the scattering one obtains informations about the electron. And here we are - only if we let something interact with the electron we can observe it. No need to die so early ;)

    Best regards,
  8. Feb 16, 2010 #7
    Hi Sunset:

    actually I havent done double slits experiment, I am wonder even if I did not affect the experiment, just ''observe'' it(the electron goes which hole), say I am a superman and I can track how electron move with my eye(I really hope so), will the pattern still appear? or more abstractly, mean by ''observe'' can be even just known it.
    I think this question can trace back to schrodinger's cat problem, any kind of observe will affect the wavefunction collapse.
    I am sorry, I might just be silly, I am a level3 undergraduate student, confusing about everything...
  9. Feb 16, 2010 #8

    That's the point. You cannot track with your human eye how the electron moves. We humans only know a probabilty (the square of the electron wave function) where to find the electron in a position measurement. You cannot observe momentum and position simultaneously, so even superman could not track how the electron moves (at least according to quantum mechanics). Assume you measure the position of the electron exactly. Then you are left with total uncertainty for the momentum of the electron. Now you say: ok, I measure it directly afterwards. The point is, if you determine afterwards momentum exactly, you are left with total uncertainty for the position after your momentum-measurement.

    Maybe the "Q" from Star Trek know about a more "microscopic" theory than quantum mechanics, which allows them to track electrons. But all theories we know so far do not allow it.

    Speaking of superman: there's a nice movie on the double slit experiment (although a bit silly), where you can see that even superman has to obey the laws of physics ;) Unfortunately it is in german:
    Last edited by a moderator: Sep 25, 2014
  10. Feb 16, 2010 #9


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    Please read Post #5 in the FAQ thread in the General Physics forum.

  11. Feb 16, 2010 #10
    Thanks:) nice video:) like it..quantum world is so interesting and so difficult to understand it with my limited classical experiences. Yeah, thanks again, and gonna sleep now, and see u around. Good night.
  12. Feb 17, 2010 #11


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    I wouldn't call it violation, then. But maybe that's a matter of taste.
  13. Feb 18, 2010 #12
    And more unfortunately the electrons glint in the light as they whizz past so you can see them and yet they still make an interference pattern. It's a shame the fancy animators didn't get more advice from physicists.
    Last edited by a moderator: Sep 25, 2014
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