6
$\begingroup$

I was going through Mark Srednicki's book on QFT. It says in the relativistic limit the Schrodinger equation becomes something like :

$$ i\hbar\frac{\partial}{\partial t} \psi(\vec x,t) = \sqrt{-\hbar^2c^2\nabla^2+m^2c^4}\psi(\vec x,t) $$ Now he says that if I expand the square root (say binomially) it will have infinite no. of spatial derivatives acting on $\psi(x,t)$; this implies that equation is not local in space.

What exactly does it mean to say the equation is not local in space?

$\endgroup$
  • 2
    $\begingroup$ Possibly a duplicate of physics.stackexchange.com/q/13624 or more directly physics.stackexchange.com/q/18762, but this approaches from a different enough angle that I'm not personally inclined to close. $\endgroup$ – user10851 Feb 5 '14 at 4:31
  • $\begingroup$ Is there something you're wondering about that is not covered in Grumiller's answer to the question in the first link? $\endgroup$ – joshphysics Feb 5 '14 at 4:35
  • $\begingroup$ I am just not very sure how it works in a more general manner. Say you don't consider the discretisation scheme to look at what is happening !! $\endgroup$ – user35952 Feb 5 '14 at 4:53
4
$\begingroup$

If you consider a standard differential operator $B$ working on functions defined in $\mathbb R^n$, like $\partial/\partial x_i$ or a polynomial of partial derivatives, and pick out a sufficiently smooth function $f$ vanishing in a neighbourhood $\Omega$, you see that also $Bf$ vanishes therein. This is the relevant notion of locality for operators.

In the RHS of the equation you wrote down an operator shows up which does not fulfil locality in the sense I said. That equation is, in fact, the equation satisfied by the positive energy solutions of Klein-Gordon equation.

The operator in the RHS cannot be defined by formal Taylor expansion (it works only formally), but one has to use spectral theory. In the considered case it is equivalent to translate that equation in Fourier transform.

Non locality arises here due to a known property of the operator $A:= \sqrt{-\Delta + aI}$ and, more generally, for $(\Delta + aI)^\nu$ with $\nu \not \in \mathbb Z$. This property is called anti locality (I.E. Segal, R.W. Goodman, J. Math. Mech. 14 (1965) 629) and is related to the famous Reeh and Schlieder property in QFT.

Anti locality means that If both $f$ and $Af$ vanish in a bounded region $\Omega \subset \mathbb R^3$ then $f$ is everywhere zero.

If $f$ has support included in a bounded open set $\Omega$, then, remarkably and very differently from what happens for standard differential operators, $Af$ does not identically vanish outside $\Omega$ otherwise $f$ would be the everywhere zero function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.