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Qmechanic
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I am reading Relativistic Quantum Mechanics by Bjorken and Drell and on page 5 they present the following attempt at a relativistic Hamiltonian for a free particle

\begin{equation} i\hbar\frac{\partial\psi}{\partial t} = \sqrt{-\hbar^2 c^2\nabla^2+m^2c^4}\psi. \end{equation}

About this equation they say "If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory". They also mentioned how the equation's time and space components are asymmetrical and would transform differently under Lorentz transformations.

I can see the asymmetry of the equation as a problem for its relativistic ambitions, but how does it relate to non-locality. If feels they are equating being un-transformable under Lorentz transformations with non-locality but I can't quite see the connection. Even if a covariant equation preserves the theory's locality, does it mean a non-covariant equation is necessarily non-local?

Thank you very much.

I am reading Relativistic Quantum Mechanics by Bjorken and Drell and on page 5 they present the following attempt at a relativistic Hamiltonian for a free particle

\begin{equation} i\hbar\frac{\partial\psi}{\partial t} = \sqrt{-\hbar^2 c^2\nabla^2+m^2c^4}\psi. \end{equation}

About this equation they say "If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory". They also mentioned how the equation's time and space components are asymmetrical and would transform differently under Lorentz transformations.

I can see the asymmetry of the equation as a problem for its relativistic ambitions, but how does it relate to non-locality. If feels they are equating being un-transformable under Lorentz transformations with non-locality but I can't quite see the connection. Even if a covariant equation preserves the theory's locality, does it mean a non-covariant equation is necessarily non-local?

Thank you very much.

I am reading Relativistic Quantum Mechanics by Bjorken and Drell and on page 5 they present the following attempt at a relativistic Hamiltonian for a free particle

\begin{equation} i\hbar\frac{\partial\psi}{\partial t} = \sqrt{-\hbar^2 c^2\nabla^2+m^2c^4}\psi. \end{equation}

About this equation they say "If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory". They also mentioned how the equation's time and space components are asymmetrical and would transform differently under Lorentz transformations.

I can see the asymmetry of the equation as a problem for its relativistic ambitions, but how does it relate to non-locality. If feels they are equating being un-transformable under Lorentz transformations with non-locality but I can't quite see the connection. Even if a covariant equation preserves the theory's locality, does it mean a non-covariant equation is necessarily non-local?

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NeonGabu
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How doesis the non-locality of a theory is apparent from its mathematical form?

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NeonGabu
  • 229
  • 1
  • 7

How does the non-locality of a theory is apparent from its mathematical form?

I am reading Relativistic Quantum Mechanics by Bjorken and Drell and on page 5 they present the following attempt at a relativistic Hamiltonian for a free particle

\begin{equation} i\hbar\frac{\partial\psi}{\partial t} = \sqrt{-\hbar^2 c^2\nabla^2+m^2c^4}\psi. \end{equation}

About this equation they say "If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory". They also mentioned how the equation's time and space components are asymmetrical and would transform differently under Lorentz transformations.

I can see the asymmetry of the equation as a problem for its relativistic ambitions, but how does it relate to non-locality. If feels they are equating being un-transformable under Lorentz transformations with non-locality but I can't quite see the connection. Even if a covariant equation preserves the theory's locality, does it mean a non-covariant equation is necessarily non-local?

Thank you very much.