6

I was thinking about the history of the Dirac equation and asked myself, what happens if one simply considers the Schrödinger equation $$i\hbar\frac{\partial\phi}{\partial t}=\sqrt{-c^2\hbar^2\Delta+m^2c^4}\phi~?$$ The literature seems to suggest that the square root is troublesome. However, in spectral theory it is well known how to take the square root of positive self-adjoint operators. So, what goes wrong with this?

Qmechanic
  • 201,751
Ivan Burbano
  • 3,875
  • It is not clear what benefit we have from using $\sqrt{…}$? We must consider two roots (particles, antiparticles) as $\pm \sqrt{…}$ and consider $\phi$ as a spinor. In Dirac's formulation, these are 4 linear equations. In this formulation, these are 4 equations with a fractional degree of operator. – Alex Trounev Apr 04 '20 at 19:58
  • Just a comment. The Klein-Gordon-Fock-Schrödinger equation (today known only the first two names) was around by the end of 1926. The probabilistic interpretation came around in 1927 and in 1928 Dirac ignores the square root possibility and comes up with his equations. The mathematics allowing us to speak about a sqrt of a linear oporator on a Hilbert space was just under creation by Von Neumann in Germany and M.Stone in the US by that time. So by the time it was ready, the success of Dirac equations (reinterpreted in terms of QFT) made returning to the sqrt option uninteresting. – DanielC Apr 05 '20 at 00:11
  • 2
    Related: What's wrong with the square root version of the Klein-Gordon equation? and Delocalization in the square root version of Klein-Gordon equation – Chiral Anomaly Apr 05 '20 at 15:40

1 Answers1

7

I think that the problem is that the square root of the Laplacian is a non-local operator and non-locality is usully regarded as a bad thing in physics. The long range nature shows up in the general expression
$$ (-\nabla^2)^s f(x)\equiv \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \int_{{\mathbb R}^n} d^ny \frac{\left\{f(x)-f(y)\right\}}{|x-y|^{2s+n}}, \quad 0<s<1, $$ for fractional powers of the Laplacian as a convolution integral. (The limit on the range of $s$ is to ensure that no further subtractions are required to define $(-\nabla^2)^s$ as a distribution.)

mike stone
  • 52,996