# The average value of the the square of Dirac velocity operator

Let's have Dirac velocity operator (the case of the free particle: $$\hat {\mathbf v} = i [\hat {H}, \hat {\mathbf r}] = \hat {\alpha}, \quad \hat {H} = (\hat {\alpha} \cdot \hat {\mathbf p}) + \hat {\beta} m,$$ where $\hat {\alpha}$ is $\hat {\gamma}_{0}\hat {\gamma}$.

I need to find $\sqrt{\langle (\Delta \hat {\mathbf v})^{2}\rangle} = \sqrt{\langle \hat {\mathbf v}^{2}\rangle - \langle \hat {\mathbf v}\rangle^{2}}$. The second summand is equal to $\frac{\mathbf p^{2}}{p_{0}^{2}}$, and the first (maybe) $$\langle \hat {\mathbf v}^{2}\rangle = \langle \hat {\alpha}^{2} \rangle = 3 \langle \rangle = 3.$$ So $$\sqrt{\langle (\Delta \hat {\mathbf v})^{2}\rangle} = \sqrt{3 - \frac{\mathbf p^{2}}{p_{0}^{2}}}.$$ But it seems that this result is incorrect, because it is unnatural. Can anyone help me? What is the correct answer?

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