Exploring "Not Local in Space" in Relativistic Quantum Mechanics

In summary: So given that we know something about the mathematics of higher-derivative equations, and the requirements of quantum mechanics, we can conclude that the square root Hamiltonian is not appropriate for our equation. One might think that the square root Hamiltonian is just some 'new thing' that we need to get used to, but it is actually just the Dirac Hamiltonian for spin 1/2 particles (and the square root operator is the Dirac operator), which has been around for some time (though its interpretation has changed). So the square root Hamiltonian is not new, and it is not that we are not used to it, it is just a matter of the mathematics not
  • #1
asub
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One approach to get a relativistic QM might be to take hamiltonian as sqrt(p2c2 +m2c4). But apparently this does not work because expanding H in inverse powers of c2 will lead to higher order derivative in space. Apparently, one reason this is bad is because higher derivatives mean that the equation is not local in space.

My question: what does "not local in space" mean? Why does higher derivatives of x make it not local in space?
 
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  • #2
asub said:
One approach to get a relativistic QM might be to take hamiltonian as sqrt(p2c2 +m2c4). But apparently this does not work because expanding H in inverse powers of c2 will lead to higher order derivative in space. Apparently, one reason this is bad is because higher derivatives mean that the equation is not local in space.

I've never understood this argument. I think that sqrt(p2c2 +m2c4) is a perfect 1-particle Hamiltonian.

Eugene.
 
  • #4
asub said:
One approach to get a relativistic QM might be to take hamiltonian as sqrt(p2c2 +m2c4). But apparently this does not work because expanding H in inverse powers of c2 will lead to higher order derivative in space. Apparently, one reason this is bad is because higher derivatives mean that the equation is not local in space.

My question: what does "not local in space" mean? Why does higher derivatives of x make it not local in space?

I can offer some off-hand remarks, but you should check the details of what I say. First, to quickly respond to your question about "not local in space": ultimately, interactions between objects should satisfy causality...objects not within each others light cones should not interact, even in quantum mechanics where "spooky action at a distance" occurs. Look up Bell's Inequalities for a critical look at locality in quantum mechanics. In quantum field theory, the idea of local interactions comes under the guise of the "cluster decomposition principle" (Weinberg mentions it in this talk printed here http://arxiv.org/abs/hep-th/9702027). To get wavefunctions and interaction amplitudes that satisfy locality, we write our Hamiltonians and Lagrangians using "local operators", which we get to now.

To make it explicit, in quantum mechanics the operator equation we're looking at is
[itex]\sqrt{\partial^{2}/\partial x^{\mu}\partial x_{\mu}+\cdots}\Psi=E\Psi[/itex]
where [itex]x_{\mu}[/itex] are spacetime coordinates. In mathematics, a local operator is one that consists of polynomials of x and p(x) acting at the same point x in spacetime (as position space operators acting on wavefunctions these are x and d/dx); the square root in our operator makes it non-local, and there is a whole branch of mathematics that talks about the "spectra" of operators (solutions to differential equations e.g.).
But what does this mean for us? As you mentioned, we can expand our operator so that it is a polynomial (though an infinite series), but there are two ways of doing this. Expanding in powers of [itex]p^{2}/(mc)^{2} [/itex] leads to an infinite series of higher and higher powers of p (and therefore derivatives in position space). We could truncate this series if [itex]p<<mc [/itex], which amounts to a non-relativistic limit reproducing the NR Schrodinger eqn with a constant potential term. But we are interested in the relativistic arena so that the full expansion must be kept. The problem is, higher derivative equations (above 2nd order) have solutions that generally have instabilities (see mathematics literature on this subject), and stability of solutions is one of the requirements in quantum mechanics (see the boundedness, or square-integrability, issue in its axioms). Related to this is the existence of negative norm states (violating the unitarity axiom), which will necessarily be involved in transititions (interaction amplitudes); the presence of these states can be seen as "negative mass poles" in the propagators used to calculate scattering amplitudes. Furthermore, in quantum field theory, higher derivative terms will appear in the Lagrangian density for the fields, and are necessarily accompanied by larger and larger negative mass-dimension coupling "constants", which is troublesome for renormalizability. None of this is to say that higher derivative theories are un-useful, or don't appear in nature, or that no sense can be made of them. But for the relativistic S.E., these issues are too sophisticated.
If you instead made an expansion in [itex](mc)^{2}/p^{2}[/itex], we can truncate to the first two terms in the relativistic limit, but the second one involves 1/p, which in position space is [itex](\partial/\partial x)^{-1} [/itex]; this is a non-local operator in position space (so this is the expansion that leads to manifestly non-local terms). Wavefunctions, as solutions of this equation, would have causality problems. Non-locally-interacting theories also show up in physics, but again, these are more sophisticated situations than the relativistic S.E.
 

FAQ: Exploring "Not Local in Space" in Relativistic Quantum Mechanics

What is the concept of "Not Local in Space" in Relativistic Quantum Mechanics?

"Not Local in Space" refers to the idea that particles or systems in quantum mechanics can be entangled, meaning that they are connected in a way that their properties are dependent on each other regardless of the distance between them. This challenges the classical notion of locality, where objects can only affect each other if they are in close proximity.

How does Relativistic Quantum Mechanics explain "Not Local in Space"?

In Relativistic Quantum Mechanics, the concept of "Not Local in Space" is explained through the theory of entanglement. This theory suggests that particles can become entangled through interactions, and their properties and behaviors will remain connected even if they are separated in space. This phenomenon is supported by numerous experiments and has been a key aspect of understanding the quantum world.

What implications does "Not Local in Space" have for our understanding of the universe?

The idea of "Not Local in Space" challenges our traditional understanding of cause and effect, as entangled particles can have instantaneous influences on each other regardless of distance. This has implications for our understanding of space, time, and the fundamental nature of reality. It also has potential applications in quantum communication and computing technologies.

Can "Not Local in Space" be observed in everyday life?

While the phenomenon of "Not Local in Space" has been observed in various experiments, it is not something that can be observed in everyday life. This is because it occurs at the quantum level, which is incredibly small and not observable by the naked eye. However, its effects can be seen in certain technologies, such as quantum encryption.

How is "Not Local in Space" related to Einstein's theory of relativity?

Einstein's theory of relativity and the concept of "Not Local in Space" are related in that they both challenge our traditional understanding of space and time. While relativity introduced the idea of spacetime and the relativity of simultaneity, "Not Local in Space" suggests that particles can have instantaneous connections regardless of their position in spacetime. Both theories have been crucial in shaping our understanding of the universe.

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