Under the null hypothesis, if we have $$\sqrt{n} \vec{x} \, \rightarrow_d \, N(0, I_p),$$ the test statistic can be construct as: $$\hat{\Psi} = n \vec{x}^{\top} \vec{x} \, \rightarrow_d \,\chi^2_p.$$ And we reject the null hypothesis if $\hat{\Psi} > \chi^2_{p, 1 - \alpha}$ under level $\alpha$.

Now, if under the alternative hypothesis $H_1$, $$\sqrt{n} \left( \vec{x} - \vec{\mu} \right)\, \rightarrow_d \, N(0, \Sigma),$$ I want to know the power of using test statistic $\hat{\Psi}$, i.e. $$\mathrm{P} \left( \left. \hat{\Psi} > \chi^2_{p, 1 - \alpha} \right| H_1 \right) = ?$$

I know under $H_1$, $\hat{\Psi}$ can be decomposed as: $$\hat{\Psi} \, \rightarrow_d \, \sum_{j = 1}^p \xi_j^2, \quad \left( \xi_1, \cdots, \xi_p \right)^{\top} \, \sim \, N(\vec{\mu}, \Sigma),$$ but I don't know how to deal with this square sum. Can anyone help me? Thanks a lot!!