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2 votes
0 answers
96 views

Relativistic Schrödinger Equation: How is it relativistic and can it be useful? [duplicate]

As is well known, the usual Schrödinger equation, $$\mathrm{i}\hbar\frac{\partial}{\partial t}\psi=-\frac{\hbar^2}{2m}\Delta\psi+V\psi,$$ is not relativistic. It can be derived formally by applying ...
Caesar.tcl's user avatar
1 vote
2 answers
162 views

How is the non-locality of a theory 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{\...
NeonGabu's user avatar
  • 229
6 votes
1 answer
440 views

Functional Analytic Square Root of Hamiltonian Alternative to Dirac

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^...
Ivan Burbano's user avatar
  • 3,915
3 votes
2 answers
3k views

Non-local potential

I read that the Schrodinger equation for a non-local potential is given by $$-\frac{\hbar^2}{2m}\nabla^2\psi(x)+\int V(x,x')\psi(x')dx'=E\psi(x).$$ In case of a local potential, $$ V(x,x')=V(x)\delta(...
Thuliyan's user avatar
  • 158
10 votes
2 answers
2k views

What's wrong with the square root version of the Klein-Gordon equation?

The Wikipedia article has a derivation of the Klein-Gordon equation. It gets to this step: $$\sqrt{\textbf{p}^2 c^2 + m^2 c^4} = E$$ and inserts the QM operators to get $$\left( \sqrt{ (-i \hbar \...
Nick's user avatar
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