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Why the particle velocity in Dirac theory is equal to c?

  1. Jul 17, 2016 #1
    In Dirac theory the electron velocity is equal to the speed of light. Why should that appear? Why should we try to solve this problem outside quantum mechanics hypothesis? Should we look for an alternative theory?
  2. jcsd
  3. Jul 17, 2016 #2


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    When we treat Dirac’s theory as a single-particle theory, the free Dirac particle undergoes a rapidly oscillating motion (Zitterbewegung) around the classical trajectory. This “motion” is caused by the interference between the positive and the negative energy components of the wave packet representing the free Dirac particle. However, the Zitterbewegung vanishes if wave packets with either positive or negative energy are used. This shows that a relativistic single particle theory is not possible, it can only be approximately considered when the corresponding wave packets is restricted to either positive or negative energy range.
    If the above is not clear enough for you, let me know and I will do the math for you. Basically, paradoxical results are obtained if the “velocity operator” is calculated according to
    [tex]\frac{d \hat{x}_{i}}{dt} = \frac{1}{i \hbar} [ \hat{x}_{i} , H_{D}] = c \alpha_{i} .[/tex]
    This shows that
    1) the absolute value of the electron velocity is equal to the speed of light (the einenvalues of [itex]\mathbf{\alpha}[/itex] are [itex]\pm 1[/itex]),
    2) the components of the velocity cannot be measured simultaneously (the [itex]\alpha_{i}[/itex]’s don’t commute with each other).
    Of course, this is just rubbish because, according to Ehrenfest’s theorem, QM is meant to reproduce the classical results for the average values. Indeed, we can show that the classical results are obtained only if wave packets with only positive, or only negative, energy are used. To see that, first integrate the equation
    [tex]\frac{d \alpha_{i}}{dt} = \frac{1}{i \hbar} [ \alpha_{i} , H_{D} ] = \frac{2i}{\hbar} (c \hat{p}_{i} - \alpha_{i}H_{D}) ,[/tex] then average over positive or negative energy wave packet to kill the Zitterbewegung and obtain the relativistic velocity
    [tex]\vec{v} = c \langle \vec{\alpha} \rangle = \frac{c^{2} \vec{p}}{E_{p}} .[/tex]
  4. Jul 19, 2016 #3


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    @samalkhaiat, are there Lorentz covariant versions of the equations of motion you have written above?
  5. Jul 20, 2016 #4


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    No. The whole point of my previous post is to stress the fact that single-particle QM is not compatible with special relativity. Clearly you get into troubles if you write the Heisenberg equation [itex]dX^{j}/dt = [iH , X^{j}][/itex] in the form [tex][ i P^{\mu} , X^{\nu} ] = \partial^{\mu}X^{\nu}=\eta^{\mu\nu}.[/tex] If [itex]X^{j}[/itex] is to represent the (3) position operators, then [itex]P^{0}[/itex] cannot represent the Hamiltonian, [itex]H[/itex], of the single particle theory. Also, what "operator " does [itex]X^{0}[/itex] represent?
  6. Jul 20, 2016 #5
    Time in a co-moving frame?
  7. Jul 21, 2016 #6


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