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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Foldy-Wouthusien velocity operator

**Physics Forums | Science Articles, Homework Help, Discussion**