 Solution

Identify the nonrelativistic mass with the rest mass, $m=m_0$. Expanding the relativistic kinetic energy in $p^2/(m^2{c}^2)$ givesFirst-order perturbation theory requires $\Delta E_0=⟨0|H^{\prime }|0⟩$. Put $\alpha =\sqrt{m\hslash \omega /2}$, so $p=i\alpha (a^{†}-a)$. The annihilation relations are $a|0⟩=0$ and $⟨0|a^{†}=0$. In particular, $a^{†}|0⟩$ does not vanish. A short calculation retaining the operator ordering givesEquivalently, of the ordered words in $(a^{†}-a{)}^4$, only $a{a}^{†}a{a}^{†}$ and $aa{a}^{†}{a}^{†}$ have nonzero ground-state expectation values, respectively $1$ and $2$. Words with a creation operator on the far left or an annihilation operator on the far right vanish; the remaining unbalanced words also vanish. ThusThe correction is negative because the relativistic kinetic energy is smaller than $p^2/(2m)$ at fixed nonzero momentum. This relativistic oscillator ground-state kinetic correction requires $\hslash \omega \ll m{c}^2$.