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A Probability density of dirac spinors

  1. Aug 11, 2017 #1
    The probability density of the dirac spinor is known to be ∑(Ψ)2 and I know how it is derived. However, I'm just wondering why it should be positive definite. Since the lower two components represent antiparticles, so shouldn't the probability density contribution of those two components be negative? Or actually what does that probability density represent? Thanks!
  2. jcsd
  3. Aug 11, 2017 #2


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    I'm not sure about your notation. In fact the Dirac-field operators are fermionic operators, and that's why the conserved current due to invariance of the action under global gauge transformations ##\psi \rightarrow \exp(\mathrm{i} \alpha)##, ##\bar{\psi} \rightarrow \exp(-\mathrm{i} \alpha) \bar{\psi}##, defined by the Lagrangian
    $$\mathcal{L}=\bar{\psi}(\mathrm{i} \partial_{\mu} \gamma^{\mu}-m) \psi$$
    for the free field, is
    $$j^{\mu} = \overline{\psi} \gamma^{\mu} \psi.$$
    In order to make this finite in the quantized version you have to normal order this expression, and due to the fermionic nature of the fields you get positive (negative) contributions to the density ##\rho=j^{0}## from particles (antiparticles). Thus, here the conserved Noether current can not be interpreted as a probability current, as in the case of Schrödinger fields, because ##\rho## is not positive definite.
  4. Aug 11, 2017 #3
    Hi vanhees, thanks for the reply but sorry I think I don't really get your point. What I mean is essentially this:


    I see quite a few textbooks interpret this as the probability density, and I didn't really thought about the lagrangian.
  5. Aug 11, 2017 #4


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    In which textbook have you found this? It's highly misleading at best!

    There is no proper interpretation of a 1st-quantization approach to relativistic QT or only in the non-relativistic limit. The reason is that at relativistic energies you have always the possibility of creating and destroying particles, and that's why relativistic QT is formulated as QFT, which is the most convenient way to describe situations, where the particle number is not conserved, and in my opinion that's the way relativistic QT should be introduced in the 21st century!
  6. Aug 11, 2017 #5
    Thanks for the clarification vanhees. So I guess you mean, combined with your previous post, the conserved current represents the density and current of some charge where antiparticles will give negative contribution?
  7. Aug 11, 2017 #6


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    Yes, the most simple example is QED. You just minimally couple the electromagnetic field to it, and then the above derived conserved current is interpreted as the electromagnetic current:
    $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \bar{\psi}(\mathrm{i} \gamma^{\mu} \partial_{\mu} + e \gamma^{\mu} A_{\mu}),$$
    $$F_{\mu \nu}=\partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}.$$
    Note that now the action is invariant under local gauge transformations!
    Last edited: Aug 11, 2017
  8. Aug 11, 2017 #7
    Thanks vanhee! Your replies are very insightful!
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