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Energy-Time Uncertainty Relation

  1. Dec 5, 2012 #1
    I was trying to Go from the uncertainty principle to its energy-time counter part. i know the maths is a bit off,but the idea is correct?

    dx=position
    p=momentum
    e=energy
    [itex]\upsilon[/itex]=frequency
    [itex]\lambda[/itex]=wavelength
    c=velocity of electromagnetic radiations
    dt=time

    now ,
    [itex]\lambda[/itex]=h/p.............(i)
    c=[itex]\upsilon[/itex].[itex]\lambda[/itex].............(ii)

    e=h.[itex]\upsilon[/itex]
    e=(h.c)/[itex]\lambda[/itex]

    replacing [itex]\lambda[/itex]'s value here from (i)

    e=(h.c)/(h/p)
    e=c.p

    now c = velocity of light , it can be written as dx/dt
    e= (dx/dt).p
    multiplying by dt on both sides

    e.dt=(dx/dt).dt.p
    e.dt=dx.p

    Therefore frome this relation if we straight away incorporate this in place of the
    σx.σp≥h/4π

    cannot we get
    σe.σt≥4π
     
  2. jcsd
  3. Dec 5, 2012 #2
    Your derivation only works symbolically (and is a good way to convince youself of the idea), as some of the equations above are meaningful only in a narrow range of applications. The uncertainty principle for position-momentum has a rigorous definition and proof (which is caused by the non-commutative relation of X and P operator), but not the energy-time relation. A better explanation for the energy-time uncertainty I've seen is from Shankar's Principles of Quantum Mechanics, this is due to the fact that particles' life time is only finite (at least on the one end), thus its fourier transform into the frequency domain fails to peak at a single frequency, instead will be spreaded - causing the uncertainty in energy measurement.
     
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