In Weinberg's QFT Vol. 1 he says the Dirac equation is not a true generalization of Schrodinger's equation, that it does not stand up to inspection when viewed in this light. He says it should be viewed as an approximation to a true relativistic quantum field theory of photons and electrons.

  • I do not understand what this means, would someone mind filling me in? (Assuming he's saying something more general & subtle than merely saying Dirac is not a one-particle equation)

One of Dirac's motivations for his derivation was that Klein-Gordon was not first order in time like the non-relativistic Schrodinger equation is.

  • Since the 2'nd order in time Klein-Gordon equation does actually describe something physical, does this mean Dirac's point about not being first-order in time is actually flawed, and that Dirac's equation worked for some other reason?

From browsing Cartan's Spinor's book it seems the Dirac equation holds for any spinor, it apparently relates left & right representations of a spinor or something, thus it holds in GR etc... There is also this great quote from Atiyah that a spinor is a square root of a geometry.

  • What is the Dirac equation & how does this explain why Dirac's derivation worked, why it relates representations of a spinor & explains this square root of a geometry business?
  • $\begingroup$ You should consider splitting up this question, which may be somewhat too broad as of right now, into separate question, each covering one (maybe 2) subquestions of the current question. $\endgroup$
    – Danu
    Oct 18, 2014 at 17:07
  • 3
    $\begingroup$ Weinberg also says on the Introduction of his Lectures on QM: "there is one topic I was not sorry to skip: the relativistic wave eq. of Dirac. It seems to me that the way this is usually presented in books on QM is profoundly mistaken". Dirac thought his eq. was a relativisti generalization of Schrödinger's [...]. The right way to combine relativity and QM is through QFT, in which the Dirac wave function appears as the matrix element of a quantum field between a one-particle state and the vacuum [...]." I never understood what he meant, maybe by answering your question this can be explained. $\endgroup$
    – Drarp
    Oct 18, 2014 at 17:09
  • $\begingroup$ Closely related: physics.stackexchange.com/questions/261893/… , physics.stackexchange.com/questions/356323/… $\endgroup$
    – Name YYY
    Sep 9, 2017 at 15:27

1 Answer 1


'Square root of geometry'

A Dirac spinor field $\psi^\alpha(x)$ under Lorentz transformations behaves as,

$$\psi^\alpha(x)\to S[\Lambda]^\alpha_\beta \psi^\beta(\Lambda^{-1}x)$$

where $\Lambda = \exp(\frac{1}{2} \Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$ and $S[\Lambda] = \exp(\frac{1}{2}\Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$, where $\mathcal{M}$ are the generators of the Lorentz transformations; they are anti-symmetric and obey the Lorenz Lie algebra. The matrices $S$ satisfy the spinor representation of the Lorenz group (exercise!), and are defined by,

$$S^{\mu\nu} = \frac{1}{4}[\gamma^\mu,\gamma^\nu]$$

A general rotation matrix acting on a spinor may be written as,

$$\left( \begin{array}{cc} e^{i\varphi \cdot \sigma /2} && 0\\ 0 && e^{i\varphi \cdot \sigma /2} \end{array} \right)$$

A rotation by $2\pi$, i.e. an entire turn, around the $x^3$ axis is given by the choice $\varphi = (0,0,2\pi)$, and the rotation matrix becomes,

$$\left( \begin{array}{cc} e^{i\pi \sigma^3} && 0\\ 0 && e^{i\pi \sigma^3} \end{array} \right) = -1$$

Hence, under a full turn, a spinor transforms as, $\psi^\alpha(x) \to -\psi^\alpha(x)$, in other words it does not return to its original state. That's why we call them the 'square root of geometry.' You in fact have to rotate them fully twice rather than once.

  • $\begingroup$ Is that all it is? In Atiyah's talk "What is a Spinor" on youtube youtube.com/watch?v=SBdW978Ii_E I got the impression it was far more than that, what you're saying is more related to spin-statistics ala Finkelstein books.google.ie/… (mentioned around minute 50 in this Susskind video in terms of belts :p theoreticalminimum.com/courses/advanced-quantum-mechanics/2013/… ) which is another question I may put up. $\endgroup$
    – bolbteppa
    Oct 18, 2014 at 17:44
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    $\begingroup$ @bolbteppa: There's more to it, but I can't answer at the moment on my phone. I'll try get back to you some time tonight. $\endgroup$
    – JamalS
    Oct 18, 2014 at 17:49
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    $\begingroup$ @JamalS: I'd be most interested in what more you might have said. $\endgroup$
    – iSeeker
    Oct 5, 2016 at 0:34

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