The Heisenberg uncertainty principle establishes a robust lower bound on the product of variances of Hermitian operators, which depends on their commutator. We established this bound for any arbitrary, normalizable state \(|\psiangle\); however, can we determine the state (or states) for which the Heisenberg uncertainty principle is saturated? That is, on what state do we have the equality

\[\begin{equation*}\sigma_{x} \sigma_{p}=\frac{\hbar}{2} ? \tag{4.162}\end{equation*}\]

(a) In our derivation of the uncertainty principle, we exploited the CauchySchwarz inequality at an early stage. In particular, Cauchy-Schwarz implies that

\[\begin{equation*}\left\langle(\hat{x}-\langle\hat{x}angle)^{2}\rightangle\left\langle(\hat{p}-\langle\hat{p}angle)^{2}\rightangle \geq|\langle(\hat{x}-\langle\hat{x}angle)(\hat{p}\langle\hat{p}angle)angle|^{2}, \tag{4.163}\end{equation*}\]

where all expectation values are taken on some state \(|\psiangle\). Argue that if this inequality is saturated and becomes an equality, then

\[\begin{equation*}(\hat{x}-\langle\hat{x}angle)|\psiangle=-\frac{2 i \sigma_{x}^{2}}{\hbar}(\hat{p}-\langle\hat{p}angle)|\psiangle \tag{4.164}\end{equation*}\]

Don't forget to use the saturated uncertainty principle, too.

(b) Now, express the relationship in Eq. (4.164) as a differential equation for the wavefunction \(\psi(x)=\langle x \mid \psiangle\) in position space. Solve the differential equation for \(\psi(x)\) for general \(\langle\hat{x}angle\) and \(\langle\hat{p}angle\). Don't worry about an overall normalization.

(c) Now, Fourier transform \(\psi(x)\) to momentum space. How does the functional form of its Fourier transform compare to that of \(\psi(x)\) in position space?