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Degrees of Freedom in QM

  1. Apr 14, 2012 #1


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    Consider one particle traveling at a relativistic velocity in 3-space. Then the configuration space of the system consisting of that one particle would have 3 degrees of freedom – 1 particle times 3 dimensions.

    Because of its high energy, the particle decays into, say, 2 particles. Now the configuration space of the system has 6 degrees of freedom.

    Is the above correct?

    But, the 2 particles are now entangled in position (as well as momentum). So does the configuration space truly have 6 degrees of freedom, or something less than that due to position correlation between the 2 particles?

    Thanks in advance.
  2. jcsd
  3. Apr 14, 2012 #2


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    For a relativistic particle, there are two invariants: the mass and the (total) spin. For simplicity, let's suppose we're talking about a spinless particle. Then, although it might seem like there's 4 degrees of freedom (1 energy and 3 momenta), these are constrained by the relativistic mass relation:
    M^2 ~=~ E^2 - |{\mathbf P}|^2
    so there's really only 3 degrees of freedom. The equation above defines the "mass hyperboloid", or "mass shell" for the particle.

    Total energy and total momentum are both conserved. This places further constraints on any particles that might emerge from a decay, thus reducing the degrees of freedom applicable to such situations.
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