= Solution
For $\psi(x)=Ce^{-\alpha x^2}$,
$$
\psi''=(4\alpha^2x^2-2\alpha)\psi.
$$
The stationary <Schrödinger equation> becomes
$$
\left[
\frac{\hbar^2\alpha}{m}
+\left(k-\frac{2\hbar^2\alpha^2}{m}\right)x^2
\right]\psi=E\psi.
$$
Hence, choosing the positive root and a positive normalization constant,
$$
\alpha=\frac1\hbar\sqrt{\frac{mk}{2}},
\qquad
E=\hbar\sqrt{\frac{k}{2m}},
\qquad
C=\left(\frac{2\alpha}{\pi}\right)^{1/4}.
$$
The state is even, so $\langle x\rangle=\langle p\rangle=0$. The Gaussian <integrals> give
$$
(\Delta x)^2=\frac1{4\alpha},
\qquad
(\Delta p)^2=\hbar^2\alpha,
$$
and therefore $\Delta x\Delta p=\hbar/2$. These values are collected in <gaussian eigenstate for a quadratic potential>.
Finally, equality in the derivation of the uncertainty relation requires the centred <vectors> to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,
$$
(\widehat p-p_0)\psi=is(\widehat x-x_0)\psi.
$$
In position space this says
$$
\psi'=\left(\frac{ip_0}{\hbar}-\frac{s(x-x_0)}{\hbar}\right)\psi,
$$
whose normalizable solutions are
$$
\psi(x)=C_0\exp\left(\frac{ip_0x}{\hbar}-\frac{s(x-x_0)^2}{2\hbar}\right).
$$
\b[Thus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the <equality case of the Heisenberg uncertainty relation>.]
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