We happen to find that the ground state of the harmonic oscillator (and the coherent states in general) is a minimum uncertainty state, but can we do the converse? That is, can we directly determine what state saturates the Heisenberg uncertainty principle directly? In the following, we will assume that the state \(|\psiangle\) on the Hilbert space is the state that saturates the Heisenberg uncertainty such that

\[\begin{equation*}\sigma_{x}^{2} \sigma_{p}^{2}=\frac{\hbar^{2}}{4} \tag{6.134}\end{equation*}\]

(a) For simplicity, we will first assume that the mean values of both position and momentum on this state are \(0:\langle\psi|\hat{x}| \psiangle=\langle\psi|\hat{p}| \psiangle=0\). Thus this state satisfies
\[\begin{equation*}\left\langle\psi\left|\hat{x}^{2}\right| \psi\rightangle\left\langle\psi\left|\hat{p}^{2}\right| \psi\rightangle=\frac{\hbar^{2}}{4} \tag{6.135}\end{equation*}\]
Because this state minimizes the uncertainty relation, any other state will necessarily have a larger uncertainty. Let's express another state as a deviation from \(|\psiangle\) by some other ket \(|\phiangle\) :
\[\begin{equation*}\left|\psi_{\epsilon}\rightangle=|\psiangle+\epsilon|\phiangle, \tag{6.136}\end{equation*}\]
and \(\epsilon \ll 1\). First, if \(|\psiangle\) is normalized, what must the inner product of \(|\psiangle\) and \(|\phiangle\) be if the state \(\left|\psi_{\epsilon}\rightangle\) is still on the Hilbert space? Just work to linear order in \(\epsilon\).
(b) If \(|\psiangle\) is the state that minimizes the uncertainty, then the derivative of uncertainty with respect to the state vanishes for \(|\psiangle\). We can define this derivative on the product of variances as

\[\begin{equation*}\frac{d}{d|\psiangle}\left\langle\psi\left|\hat{x}^{2}\right| \psi\rightangle\left\langle\psi\left|\hat{p}^{2}\right| \psi\rightangle=\lim _{\epsilon \rightarrow 0} \frac{\left\langle\psi_{\epsilon}\left|\hat{x}^{2}\right|\psi_{\epsilon}\rightangle\left\langle\psi_{\epsilon}\left|\hat{p}^{2}\right|\psi_{\epsilon}\rightangle\left\langle\psi\left|\hat{x}^{2}\right|\psi\rightangle\left\langle\psi\left|\hat{p}^{2}\right| \psi\rightangle}{\epsilon} . \tag{6.137}\end{equation*}\]
Evaluate this derivative for the states defined above.
(c) If this derivative vanishes, determine the linear equation that ket \(|\psiangle\) must satisfy, in terms of the variances \(\sigma_{x}^{2}\) and \(\sigma_{p}^{2}\).
Don't forget the result you derived in part (a).
(d) Using Eq. (6.135), show that \(|\psiangle\) must be the ground state of the harmonic oscillator; that is, it is an eigenstate of the harmonic oscillator Hamiltonian with the same eigenvalue.
(e) Repeat parts (b)-(d) with non-zero values for the expectation values \(\langle\psi|\hat{x}| \psiangle\) and \(\langle\psi|\hat{p}| \psiangle\). What equation must \(|\psiangle\) satisfy now? Have we seen it before?