# $\nabla$ and non-locality in simple relativistic model of quantum mechanics

In Wavefunction in quantum mechanics and locality, wavefunction is constrained by $H = \sqrt{m^2 - \hbar^2 \nabla^2}$, and taylor-expanding $H$ results in:

$$H = \dots = m\sqrt{1 - \hbar^2/m^2 \cdot \nabla^2} = m(1-\dots)$$

While the person who asked this question accepted the answer, I was not able to understand fully.

Why would $\nabla^{200}$ in the taylor-expansion of $H$ be so problematic (200 can be replaced by any arbitrary number) - resulting in non-locality? Isn't this just some exponentiation of dot product of gradient?

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–  Qmechanic Oct 9 '12 at 18:01

Non-locality comes from presence of infinite many terms in that expansion. To see that, lets assume we are applying the non-polynomial function $f(\vec{z})$ of $i\nabla$ on any function $\psi(\vec{x})$: $f(i\nabla) \psi(\vec{x})$. Assuming that $f(z)$ is "nice enough" to have the Fourier representation $f(\vec{z}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot \vec{z}}$ for suitable transform function $F(\vec{k})$. Then:

$f(i \nabla) \psi(x) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot (i \nabla)} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{- k\cdot \nabla} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) \psi(\vec{x}-\vec{k})$

This is the superposition of values of $g$ calculated at points different than $\vec{x}$. that is the non-locality.

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It is non-local because the time evolution of the wave function depends instantaneously on parts of the field which are arbitrary far away. In the case of a linearization using the square root you get.

$$i \hbar \frac{\partial}{\partial t}\, \psi ~~=~~ \tilde{H}\,\psi ~~=\ \sqrt{\mathbf{\tilde{p}}^2 c^2 + m^2c^4}\ \psi\ =\ \nonumber$$

Where the tildes on the $\tilde{H}$ and $\tilde{p}$ defines them as operator. We can use here the series.

$$\sqrt{1 + \tilde{p}^2} ~~=$$ $$\mbox{ 1 + \frac{1}{2}\tilde{p}^2 - \frac{1}{8}\tilde{p}^4 + \frac{1}{16}\tilde{p}^6 - \frac{5}{128}\tilde{p}^8 + \frac{7}{256}\tilde{p}^{10} - \frac{21}{1024}\tilde{p}^{12} + \frac{33}{2048}\tilde{p}^{14} - \frac{429}{32768}\tilde{p}^{16} + ....}\nonumber$$

This series is found to be represented by a finite expression and we can define $\tilde{H}$ as following.

$$i \hbar \frac{\partial}{\partial t}\, \psi ~~=~~ \tilde{H}\,\psi ~~=\ \sqrt{\mathbf{\tilde{p}}^2 c^2 + m^2c^4}\ \psi\ =\ \nonumber$$ $$=\ \pm \ mc^2\left\{ 1 + \sum_{n=1}^\infty \frac{(-1)^{n+1}\ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\mathbf{\tilde{p}}^2}{m^2c^2}\right)^{n} \right\} \psi$$

Or written out as a differential operator:

$$i \hbar \frac{\partial}{\partial t} \, \psi ~~=~~ \tilde{H}\,\psi ~~=$$ $$\pm \ mc^2\left\{ 1 - \sum_{n=1}^\infty \frac{ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\hbar^2}{m^2 c^2}\ \right)^{n} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right)^{n}\right\} \psi \nonumber$$

Applying this type of operator series on a wave function $\psi$ amounts to a convolution with a Bessel K function. For instance in the one dimensional case one can show, using the Fourier transform, that:

\begin{aligned} & \sqrt{ \partial_x^2 - m^2}^{\,-1} \psi &=~~ &~~\, K_0(mx) ~*~ \psi \\ \end{aligned}

Where $*$ denotes convolution, From this result then one can derive:

\begin{aligned} & \sqrt{ \partial_x^2 - m^2}~~ \psi &=~~ &\tfrac{m}{x} K_1(mx) ~*~ \psi \\ \end{aligned}

This convolution means that $\partial\psi/\partial t$ depends on non-local values of $\psi$

Hans

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