If one takes the derivative of the position operator in the Dirac Hamiltonian, the result is [itex]\dot{\vec{x}} = c \vec{\alpha}[/itex]. This, however, disagrees with the classical limit in which [itex]\dot{\vec{x}}\sim \dot{\vec{p}}/m[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to show that the time derivative of the position operator [itex]\vec{X} = U^\dag \vec{x} U[/itex] is [itex]\frac{c^2\vec{p}}{E^2}H[/itex], and thus satisfies the non-relativistic approximation.

[itex]U = \frac{\beta \vec{\alpha}\cdot \vec{p} c + mc^2 + E}{\sqrt{2E(mc^2+E)}}[/itex] is the FW transformation (Relativistic quantum mechanics by Greiner p. 277 or Advanced Quantum Mechanics by Sakurai p. 176) .

I tried using the Heisenberg equation of motion

[itex]\dot{\vec{X}} = \frac{1}{i\hbar}[X,H][/itex]

but the expression for X is long and complicated so I wasn't able to get to the result using this method.

I note that [itex]\frac{1}{i\hbar}[\vec{x},E] = (\vec{p}c^2)/E[/itex] which looks a bit like the answer.

I'm also aware that [itex]U^\dag H U = \beta E[/itex]. So I want to show that

[itex]\frac{1}{i\hbar}[U^\dag \vec{x} U, H ] = \frac{1}{i\hbar}[\vec{x}, \beta U^\dag H U ][/itex] (since [itex]\beta^2=1[/itex]) but I can't figure out how to show this.

Any help would be greatly appreciated!

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# Foldy-Wouthusien velocity operator

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