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It is a well-known fact that Klein-Gordon scalar $\Psi(x)$, $$ (\partial^{2} + m^2) \Psi (x) = 0 $$ as well as 4-vector $A_{\mu}(x)$, $$ (\partial^{2} + m^{2})A_{\mu} = 0,\quad \partial_{\mu}A^{\mu} = 0, $$ (and even function of an arbitrary integer spin) describe the field: first, there aren't positive definite norm (with Lorentz invariant fullspace integral) for this functions, and the second, the free solutions are represented in a form of independent harmonic oscillators, like for case of classical electromagnetic field. So we naturally assume commutation relations for amplitude operators of these fields.

Then let's have the Dirac equation and corresponding function (in general - let's see the function of arbitrary half-integer spin). Let's also assume, that we don't know that it describes some particle. We can build positive definite norm (with Lorentz invariant fullspace integral), and the solution for field also looks like harmonic oscillator. But for positive definite of energy we must assume anticommutation relations.

So, the question: why do we assume that Dirac spinor $\Psi$ (or, in general, tensors of an arbitrary spin) describes only the particle, not the field? In my opinion, the fact about positive definite norm leaves the possibility for the description of the field by this spinor (not the particle).

My question is not about formal definition of these functions. Of course, all of them are relativistic fields. But they describe different physical objects in classical limit - fields and particles correspondingly. Maxwell function $A_{\mu}$ describes the EM field even in classical limit, but the Dirac spinor $\Psi$ describes the electron only in the quantum case (when QM postulates work).

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  • $\begingroup$ Correct me if I am mistaken, but is not the Dirac spinor $\Psi(\mathbf x,t)$ a field function defined on spacetime coordinates? This function does not give probability of position of particle or particles in the classical meaning of the word (as in Born's interpretation of Schroedinger's non-relativistic equation). In quantum field theory, it is an abstract operator field. $\endgroup$ – Ján Lalinský Mar 22 '14 at 0:24
  • $\begingroup$ @JánLalinský : your comment is very useful. I think that the answer on it is following. Yes, according to the definition of the relativistic field as function which determined on minkowskian space your first statement is true. But my question is about what physical object this function describes, not about the mathematical status of the function. As for the next statements we can assume free fields, so we don't even need to quantize field, and so do not assume the quantum field theory (operates only with relativistic QM). $\endgroup$ – Andrew McAddams Mar 22 '14 at 0:47
  • $\begingroup$ I think two frameworks are mixed in your question, both the KG and the Dirac solutions were first used as an extension of the first quantization framework, and both describe particles/probability-waves in this framework: bosons for KG and fermions for Dirac. Second quantization is a different mathematical framework/view turning the solutions into creation and annihilation operators. It works in calculating crossections etc but is not particularly useful in visualizing/fitting "particles-in/particles-out". We tend to keep the framework of first quantization in describing specific interactions. $\endgroup$ – anna v Mar 22 '14 at 4:14
  • $\begingroup$ "But my question is about what physical object this function describes, not about the mathematical status of the function." That is a very good question! Perhaps it would help if you could add it to the original question. I'm curious about answers too. $\endgroup$ – Ján Lalinský Mar 22 '14 at 9:21
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In QFT, the Dirac spinor will also be promoted to a field, whose oscillation mode coefficients are creation and annihilation operators.

BUT: For the Dirac spinor it is possible to well-define a probablility density and current:

$$\rho^\mu \propto \bar\psi \gamma^\mu \psi$$

This current's zero component is positive definite and using the Dirac equation one can show that it is conserved, i.e. $\partial_\mu \rho^\mu \equiv 0$.

Therefore, besides being interpreted as a quantum field, the Dirac spinor can be interpreted as a particle wavefunction in regular QM.

Let me remind you, though, that the energy eigenvalues of the Dirac operator are not bounded from below. This is not as problematic, if one agrees to the concept of the Dirac sea of electrons already occupying all negative energy states. While the construction of the Dirac sea is very hand-waving, it provides a key prediction: particle-antiparticle pair creation from 'pure energy' (i.e. a photon).

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  • $\begingroup$ "...the Dirac spinor can be interpreted as a particle wavefunction in regular QM...", - but may it be interpreted as field wavefunction in regular QM, like $A_{\mu}$? $\endgroup$ – Andrew McAddams Mar 22 '14 at 12:22
  • $\begingroup$ I am not sure what you mean by "field wavefunction" in regular QM. Either you have a quantum field theory (which is not regular QM) or you have quantum particles and classical fields (where there is no concept like a "field wavefunction"). $\endgroup$ – Neuneck Mar 22 '14 at 12:53
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    $\begingroup$ @Neuneck Your formula for $\rho^\mu$ is that of KG field! The one for Dirac field involves $\gamma^\mu$ matrices! Please correct. Actually, the situation is very similar to that of complex KG equation. In that case energy is bounded below while the conserved charge is not positive (with definite sign). However if considering only solutions which are superposition of positive frequency modes, the charge is positive and the energy is bounded below. For Dirac equation, considering only positive frequency solutions, both energy and charge are positive (with definite sign). $\endgroup$ – Valter Moretti Mar 22 '14 at 16:50
  • $\begingroup$ Thank you, I corrected. For the KG field no physical reason for just looking at the positive frequency modes is available in regular QM. For the Dirac equation - as we are dealing with fermions - once the negative energy states have been occupied, there is no way a particle can reduce its energy by decaying into an every lower lying mode. For bosons this exclusion does not exist. $\endgroup$ – Neuneck Mar 23 '14 at 9:12
  • $\begingroup$ So, do I understand correctly the: Dirac equation outside of QFT can describe a particle, whereas the Klein-Gordon equation cannot because of the undefined sign of "norm" of its solutions? (I am not the OP) $\endgroup$ – Lurco Jul 2 '14 at 18:30

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