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Units in QM

  1. Nov 25, 2013 #1
    Greetings,

    Just checking if i'm getting this ... please correct me if I'm wrong.

    The value of the wavefunction is 'probability amplitude' in discrete case and 'probability amplitude density' in continuous case. The former is a dimensionless complex number and the latter is the same multiplied by the inverse of the units of coordinare vectors in space on which the distribution is defined. Eg. Ψ(x,t) would have units ##L^{-3}##. The actual scale factor is irrelevant.

    State operator ρ is dimensionless with Tr{ρ}=1. However for the operator R representing observable the eigenvalues must have proper units if we want the measurements to come out right. That is,λ in ##R|ψ\rangle=λ|ψ\rangle## has to have the right units. Now what does it mean in terms of domain and range of operator R? I thought it was supposed to be defined over some vector space ##\Omega^\times##, but how can ##| ψ\rangle## and ##(5kg)|ψ\rangle## belong to the same space? I guess these are in fact two different spaces with the same structrure but as you never need to add elements from different spaces and the units always cancel out nicely in the end, the distinction is quietly ignored.

    Looking at the generators of Galilei group, the units of P are necesserily ##L^{-1}##, H is ##T^{-1}##, J is 1 and G is ##L^{-1}T## (so we can have eg. ##e^{iPx}##, ##e^{iHt}## etc. But in order to connect them to familiar notions, we introduce this funny coefficient ##\hbar## which has units of action ##J \cdot s##. But I 'm just wondering, is there some deeper meaning whereby measuring energy in ##s^{-1}## and momentum in ##m^{-1}## would actually make sense? Let's say we introduce a unit of action , ##\hbar=1 fubar##.Then we'll be measuring energy in fubars per second and momentum in fubars per meter. Mass will come out as fubar-second/meter² (hmm, not sure about this one). What would the quantity of 1 fubar signify?

    Regards, DK
     
  2. jcsd
  3. Nov 25, 2013 #2

    atyy

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    Planck's constant has units of angular momentum or "action". In old quantum physics, Bohr-Sommerfeld quantization was quantization of action.
     
  4. Nov 25, 2013 #3

    bhobba

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    The best way of looking at Planks constant IMHO is the path integral formalism:
    http://hitoshi.berkeley.edu/221a/pathintegral.pdf

    The wavefunction is dimensionless so the action in the path integral, which of course has units of action, must be divided by a constant that also has units of action, so its dimensionless. That constant is Planks constant - its required by dimensional analysis, and of course its exact value is an experimental matter - but can, again of course, be set to 1 for theoretical convenience by a simple rescaling of units.

    Thanks
    Bill
     
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