[proofplan]
We compute the first and second moments of $X$ and $P$ by rewriting them in terms of the annihilation and creation operators $a$ and $a^*$. The coherent-state identity $a\psi_\alpha=\alpha\psi_\alpha$ gives the expectations of $a$, $a^*$, $a^2$, and $(a^*)^2$, while the commutation relation $aa^*=a^*a+I$ supplies the single extra contribution in the mixed second moments. After subtracting the squares of the means, all terms depending on $\alpha$ cancel, leaving the ground-state variances $\hbar/(2m\omega)$ and $m\hbar\omega/2$.
[/proofplan]
[step:Compute the first moments from the coherent state eigenvalue equation]
We work with the [Hilbert space](/page/Hilbert%20Space) $H$, dense invariant domain $D \subset H$, operators $a: D \to D$ and $a^*: D \to D$, constants $m>0$, $\omega>0$, and $\hbar>0$, and normalized coherent state $\psi_\alpha \in D$ from the theorem statement. Let $I: D \to D$ denote the identity operator on $D$. The density of $D$ is part of the operator-theoretic setup for the adjoint; the computation below uses the adjoint identity on $D$, invariance of $D$ under $a$ and $a^*$, the commutation relation, and normalization. Define the constants
\begin{align*}
c_X := \sqrt{\frac{\hbar}{2m\omega}}.
\end{align*}
\begin{align*}
c_P := \sqrt{\frac{m\hbar\omega}{2}}.
\end{align*}
The position and momentum operators are the linear maps $X: D \to H$ and $P: D \to H$ given by
\begin{align*}
X\psi := c_X(a\psi+a^*\psi)
\end{align*}
and
\begin{align*}
P\psi := -ic_P(a\psi-a^*\psi)
\end{align*}
for each $\psi \in D$.
Since $a\psi_\alpha=\alpha\psi_\alpha$ and $\|\psi_\alpha\|_H=1$, linearity of the first inner-product argument gives
\begin{align*}
(a\psi_\alpha,\psi_\alpha)_H = \alpha.
\end{align*}
Using the adjoint relation and conjugate symmetry,
\begin{align*}
(a^*\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,a^*\psi_\alpha)_H} = \overline{(a\psi_\alpha,\psi_\alpha)_H} = \bar{\alpha}.
\end{align*}
Therefore
\begin{align*}
\mathbb E_{\psi_\alpha}[X] = (X\psi_\alpha,\psi_\alpha)_H = c_X(\alpha+\bar{\alpha}) = 2c_X\operatorname{Re}\alpha = \sqrt{\frac{2\hbar}{m\omega}}\operatorname{Re}\alpha.
\end{align*}
Similarly,
\begin{align*}
\mathbb E_{\psi_\alpha}[P] = (P\psi_\alpha,\psi_\alpha)_H = -ic_P(\alpha-\bar{\alpha}) = 2c_P\operatorname{Im}\alpha = \sqrt{2m\hbar\omega}\operatorname{Im}\alpha.
\end{align*}
[guided]
The first moments are expectations of $X$ and $P$, so we first record the exact setting in which these expectations are defined. The theorem gives a complex Hilbert space $H$, a dense invariant domain $D \subset H$, linear maps $a: D \to D$ and $a^*: D \to D$, positive constants $m$, $\omega$, and $\hbar$, and a normalized vector $\psi_\alpha \in D$ satisfying $a\psi_\alpha=\alpha\psi_\alpha$. Let $I: D \to D$ denote the identity operator on $D$. The density of $D$ is what makes the adjoint terminology natural; the moment computation itself uses the adjoint identity on vectors in $D$, the invariance of $D$, the commutation relation, and normalization. Define
\begin{align*}
c_X := \sqrt{\frac{\hbar}{2m\omega}}.
\end{align*}
\begin{align*}
c_P := \sqrt{\frac{m\hbar\omega}{2}}.
\end{align*}
Then $X: D \to H$ and $P: D \to H$ are the maps
\begin{align*}
X\psi := c_X(a\psi+a^*\psi)
\end{align*}
and
\begin{align*}
P\psi := -ic_P(a\psi-a^*\psi)
\end{align*}
for $\psi \in D$.
The coherent state condition gives the expectation of $a$ immediately. Since the [inner product](/page/Inner%20Product) is linear in the first argument and $\|\psi_\alpha\|_H=1$,
\begin{align*}
(a\psi_\alpha,\psi_\alpha)_H = (\alpha\psi_\alpha,\psi_\alpha)_H = \alpha(\psi_\alpha,\psi_\alpha)_H = \alpha.
\end{align*}
To compute the expectation of $a^*$, we use the definition of the adjoint. The identity
\begin{align*}
(a\psi_\alpha,\psi_\alpha)_H = (\psi_\alpha,a^*\psi_\alpha)_H
\end{align*}
holds because $a^*$ is the Hilbert-space adjoint of $a$. Taking complex conjugates gives
\begin{align*}
(a^*\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,a^*\psi_\alpha)_H} = \overline{(a\psi_\alpha,\psi_\alpha)_H} = \bar{\alpha}.
\end{align*}
Now substitute these two scalar identities into the definitions of $X$ and $P$:
\begin{align*}
\mathbb E_{\psi_\alpha}[X] = (X\psi_\alpha,\psi_\alpha)_H = c_X\bigl((a\psi_\alpha,\psi_\alpha)_H+(a^*\psi_\alpha,\psi_\alpha)_H\bigr) = c_X(\alpha+\bar{\alpha}).
\end{align*}
Since $\alpha+\bar{\alpha}=2\operatorname{Re}\alpha$, this becomes
\begin{align*}
\mathbb E_{\psi_\alpha}[X] = 2c_X\operatorname{Re}\alpha = \sqrt{\frac{2\hbar}{m\omega}}\operatorname{Re}\alpha.
\end{align*}
For momentum,
\begin{align*}
\mathbb E_{\psi_\alpha}[P] = (P\psi_\alpha,\psi_\alpha)_H = -ic_P\bigl((a\psi_\alpha,\psi_\alpha)_H-(a^*\psi_\alpha,\psi_\alpha)_H\bigr) = -ic_P(\alpha-\bar{\alpha}).
\end{align*}
Since $\alpha-\bar{\alpha}=2i\operatorname{Im}\alpha$, we get
\begin{align*}
\mathbb E_{\psi_\alpha}[P] = 2c_P\operatorname{Im}\alpha = \sqrt{2m\hbar\omega}\operatorname{Im}\alpha.
\end{align*}
[/guided]
[/step]
[step:Compute the second moment of position and subtract the square of its mean]
Because $D$ is a linear subspace and $a,a^*:D\to D$, the formula $X\psi=c_X(a\psi+a^*\psi)$ shows that $X\psi\in D$ for every $\psi\in D$. Hence $X^2\psi_\alpha$ is well-defined. Expanding $X^2$ on $D$ gives
\begin{align*}
X^2 = c_X^2(a+a^*)^2 = c_X^2(a^2+aa^*+a^*a+(a^*)^2).
\end{align*}
The coherent-state equation gives
\begin{align*}
(a^2\psi_\alpha,\psi_\alpha)_H = \alpha^2.
\end{align*}
Indeed, $a^2\psi_\alpha=a(\alpha\psi_\alpha)=\alpha^2\psi_\alpha$, and $\|\psi_\alpha\|_H=1$. Since $\psi_\alpha \in D$ and $D$ is invariant under $a$ and $a^*$, the vectors $a\psi_\alpha$, $a^*\psi_\alpha$, $a^2\psi_\alpha$, and $(a^*)^2\psi_\alpha$ are all defined. Applying the adjoint relation twice gives
\begin{align*}
(a^2\psi_\alpha,\psi_\alpha)_H=(a\psi_\alpha,a^*\psi_\alpha)_H=(\psi_\alpha,(a^*)^2\psi_\alpha)_H.
\end{align*}
Hence conjugate symmetry gives
\begin{align*}
((a^*)^2\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,(a^*)^2\psi_\alpha)_H} = \overline{(a^2\psi_\alpha,\psi_\alpha)_H} = \bar{\alpha}^2.
\end{align*}
Finally, the adjoint relation gives
\begin{align*}
(\psi_\alpha,a^*a\psi_\alpha)_H = (a\psi_\alpha,a\psi_\alpha)_H = (\alpha\psi_\alpha,\alpha\psi_\alpha)_H = |\alpha|^2.
\end{align*}
Since $|\alpha|^2$ is real, conjugate symmetry gives
\begin{align*}
(a^*a\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,a^*a\psi_\alpha)_H} = |\alpha|^2.
\end{align*}
Using the commutation relation $aa^*=a^*a+I$ on $D$,
\begin{align*}
(aa^*\psi_\alpha,\psi_\alpha)_H = (a^*a\psi_\alpha,\psi_\alpha)_H+(\psi_\alpha,\psi_\alpha)_H = |\alpha|^2+1.
\end{align*}
Hence
\begin{align*}
\mathbb E_{\psi_\alpha}[X^2] = c_X^2\bigl(\alpha^2+\bar{\alpha}^2+2|\alpha|^2+1\bigr).
\end{align*}
Since
\begin{align*}
\mathbb E_{\psi_\alpha}[X]^2 = c_X^2(\alpha+\bar{\alpha})^2 = c_X^2\bigl(\alpha^2+\bar{\alpha}^2+2|\alpha|^2\bigr),
\end{align*}
we obtain
\begin{align*}
(\Delta_{\psi_\alpha}X)^2 = \mathbb E_{\psi_\alpha}[X^2]-\mathbb E_{\psi_\alpha}[X]^2 = c_X^2 = \frac{\hbar}{2m\omega}.
\end{align*}
[guided]
The goal is to compute the variance of position, so we need the second moment and then subtract the square of the first moment. Before expanding $X^2$, we check that this expression is defined on the coherent state. Since $D$ is a linear subspace and $a,a^*:D\to D$, the vector $a\psi+a^*\psi$ lies in $D$ for every $\psi\in D$, and therefore $X\psi=c_X(a\psi+a^*\psi)$ also lies in $D$. In particular $X^2\psi_\alpha$ is well-defined. Expanding $X^2$ on $D$ gives
\begin{align*}
X^2 = c_X^2(a+a^*)^2 = c_X^2(a^2+aa^*+a^*a+(a^*)^2).
\end{align*}
The coherent-state eigenvalue equation gives the pure $a$ term:
\begin{align*}
a^2\psi_\alpha = a(\alpha\psi_\alpha)=\alpha^2\psi_\alpha.
\end{align*}
Because $\|\psi_\alpha\|_H=1$, this implies
\begin{align*}
(a^2\psi_\alpha,\psi_\alpha)_H = \alpha^2.
\end{align*}
For the pure $a^*$ term, we must use the adjoint relation carefully because $a$ and $a^*$ are unbounded operators. The theorem supplies the needed domain information: $\psi_\alpha \in D$, and $D$ is invariant under both $a$ and $a^*$. Therefore $a\psi_\alpha$, $a^*\psi_\alpha$, $a^2\psi_\alpha$, and $(a^*)^2\psi_\alpha$ are all defined. Applying the adjoint relation first to $a\psi_\alpha$ and $\psi_\alpha$, and then to $\psi_\alpha$ and $a^*\psi_\alpha$, gives
\begin{align*}
(a^2\psi_\alpha,\psi_\alpha)_H=(a\psi_\alpha,a^*\psi_\alpha)_H=(\psi_\alpha,(a^*)^2\psi_\alpha)_H.
\end{align*}
Taking complex conjugates then gives
\begin{align*}
((a^*)^2\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,(a^*)^2\psi_\alpha)_H} = \overline{(a^2\psi_\alpha,\psi_\alpha)_H} = \bar{\alpha}^2.
\end{align*}
For the normally ordered mixed term, the adjoint relation first gives
\begin{align*}
(\psi_\alpha,a^*a\psi_\alpha)_H = (a\psi_\alpha,a\psi_\alpha)_H = (\alpha\psi_\alpha,\alpha\psi_\alpha)_H = |\alpha|^2.
\end{align*}
Because the inner product is linear in the first argument, this is the adjoint identity in the correct direction. To obtain the expectation with $a^*a\psi_\alpha$ in the first argument, we use conjugate symmetry and the fact that $|\alpha|^2$ is real:
\begin{align*}
(a^*a\psi_\alpha,\psi_\alpha)_H = \overline{(\psi_\alpha,a^*a\psi_\alpha)_H} = |\alpha|^2.
\end{align*}
The remaining mixed term is not normally ordered. Using the commutation relation $aa^*=a^*a+I$ and $\|\psi_\alpha\|_H=1$, we get
\begin{align*}
(aa^*\psi_\alpha,\psi_\alpha)_H = (a^*a\psi_\alpha,\psi_\alpha)_H+(\psi_\alpha,\psi_\alpha)_H = |\alpha|^2+1.
\end{align*}
Substituting these four scalar expectations into the expansion of $X^2$ gives
\begin{align*}
\mathbb E_{\psi_\alpha}[X^2] = c_X^2\bigl(\alpha^2+\bar{\alpha}^2+2|\alpha|^2+1\bigr).
\end{align*}
The square of the first moment is
\begin{align*}
\mathbb E_{\psi_\alpha}[X]^2 = c_X^2(\alpha+\bar{\alpha})^2 = c_X^2\bigl(\alpha^2+\bar{\alpha}^2+2|\alpha|^2\bigr).
\end{align*}
Subtracting this from the second moment cancels every term depending on $\alpha$:
\begin{align*}
(\Delta_{\psi_\alpha}X)^2 = \mathbb E_{\psi_\alpha}[X^2]-\mathbb E_{\psi_\alpha}[X]^2 = c_X^2 = \frac{\hbar}{2m\omega}.
\end{align*}
[/guided]
[/step]
[step:Compute the second moment of momentum and subtract the square of its mean]
Because $D$ is a linear subspace and $a,a^*:D\to D$, the formula $P\psi=-ic_P(a\psi-a^*\psi)$ shows that $P\psi\in D$ for every $\psi\in D$. Hence $P^2\psi_\alpha$ is well-defined. Expanding $P^2$ on $D$ gives
\begin{align*}
P^2 = (-ic_P)^2(a-a^*)^2 = -c_P^2(a^2-aa^*-a^*a+(a^*)^2).
\end{align*}
The identities for $(a^2\psi_\alpha,\psi_\alpha)_H$, $((a^*)^2\psi_\alpha,\psi_\alpha)_H$, and $(a^*a\psi_\alpha,\psi_\alpha)_H$ were derived above from the coherent-state equation, the invariant-domain hypothesis, and the adjoint relation. The remaining mixed term follows from the canonical commutation relation $aa^*u=a^*au+u$ applied to $u=\psi_\alpha$. Therefore
\begin{align*}
\mathbb E_{\psi_\alpha}[P^2] = -c_P^2\bigl(\alpha^2-(|\alpha|^2+1)-|\alpha|^2+\bar{\alpha}^2\bigr) = c_P^2\bigl(-\alpha^2-\bar{\alpha}^2+2|\alpha|^2+1\bigr).
\end{align*}
Also,
\begin{align*}
\mathbb E_{\psi_\alpha}[P]^2 = \bigl(-ic_P(\alpha-\bar{\alpha})\bigr)^2 = -c_P^2(\alpha-\bar{\alpha})^2 = c_P^2\bigl(-\alpha^2-\bar{\alpha}^2+2|\alpha|^2\bigr).
\end{align*}
Therefore
\begin{align*}
(\Delta_{\psi_\alpha}P)^2 = \mathbb E_{\psi_\alpha}[P^2]-\mathbb E_{\psi_\alpha}[P]^2 = c_P^2 = \frac{m\hbar\omega}{2}.
\end{align*}
[guided]
The momentum computation is parallel to the position computation, but the sign from $P=-ic_P(a-a^*)$ changes which terms cancel. Before expanding the square, we check the domain. Since $D$ is a linear subspace and $a,a^*:D\to D$, the vector $a\psi-a^*\psi$ lies in $D$ for every $\psi\in D$, and therefore $P\psi=-ic_P(a\psi-a^*\psi)$ lies in $D$. In particular $P^2\psi_\alpha$ is well-defined. Expanding $P^2$ on $D$ gives
\begin{align*}
P^2 = (-ic_P)^2(a-a^*)^2 = -c_P^2(a^2-aa^*-a^*a+(a^*)^2).
\end{align*}
We now restate the scalar identities needed for this expansion. The coherent-state equation gives $a^2\psi_\alpha=\alpha^2\psi_\alpha$, so normalization gives
\begin{align*}
(a^2\psi_\alpha,\psi_\alpha)_H = \alpha^2.
\end{align*}
Applying the adjoint relation twice and then conjugate symmetry gives
\begin{align*}
((a^*)^2\psi_\alpha,\psi_\alpha)_H = \bar{\alpha}^2.
\end{align*}
For the normally ordered mixed term, the adjoint relation first gives $(\psi_\alpha,a^*a\psi_\alpha)_H=(a\psi_\alpha,a\psi_\alpha)_H=|\alpha|^2$, and conjugate symmetry then gives
\begin{align*}
(a^*a\psi_\alpha,\psi_\alpha)_H = |\alpha|^2.
\end{align*}
The remaining mixed term uses the canonical commutation relation from the theorem statement. Applying $aa^*u=a^*au+u$ to $u=\psi_\alpha \in D$ and using $\|\psi_\alpha\|_H=1$ gives
\begin{align*}
(aa^*\psi_\alpha,\psi_\alpha)_H = (a^*a\psi_\alpha,\psi_\alpha)_H+(\psi_\alpha,\psi_\alpha)_H = |\alpha|^2+1.
\end{align*}
Substituting these four values into the expansion of $P^2$ yields
\begin{align*}
\mathbb E_{\psi_\alpha}[P^2] = -c_P^2\bigl(\alpha^2-(|\alpha|^2+1)-|\alpha|^2+\bar{\alpha}^2\bigr) = c_P^2\bigl(-\alpha^2-\bar{\alpha}^2+2|\alpha|^2+1\bigr).
\end{align*}
The first moment of $P$ was $-ic_P(\alpha-\bar{\alpha})$, so its square is
\begin{align*}
\mathbb E_{\psi_\alpha}[P]^2 = \bigl(-ic_P(\alpha-\bar{\alpha})\bigr)^2 = -c_P^2(\alpha-\bar{\alpha})^2 = c_P^2\bigl(-\alpha^2-\bar{\alpha}^2+2|\alpha|^2\bigr).
\end{align*}
Subtracting again cancels all $\alpha$-dependent terms and leaves only the commutator contribution:
\begin{align*}
(\Delta_{\psi_\alpha}P)^2 = \mathbb E_{\psi_\alpha}[P^2]-\mathbb E_{\psi_\alpha}[P]^2 = c_P^2 = \frac{m\hbar\omega}{2}.
\end{align*}
[/guided]
[/step]
[step:Multiply the standard deviations to obtain equality in the uncertainty bound]
The constants $c_X$ and $c_P$ are positive because $m>0$, $\omega>0$, and $\hbar>0$. Hence
\begin{align*}
\Delta_{\psi_\alpha}X = \sqrt{\frac{\hbar}{2m\omega}}, \qquad \Delta_{\psi_\alpha}P = \sqrt{\frac{m\hbar\omega}{2}}.
\end{align*}
Multiplying these two positive quantities gives
\begin{align*}
(\Delta_{\psi_\alpha}X)(\Delta_{\psi_\alpha}P) = \sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{m\hbar\omega}{2}} = \frac{\hbar}{2}.
\end{align*}
This proves the stated expectation formulas and the coherent-state uncertainty minimization identity.
[guided]
The standard deviations are the non-negative square roots of the variances. Since $m>0$, $\omega>0$, and $\hbar>0$, both constants $c_X$ and $c_P$ are positive. From the two variance computations,
\begin{align*}
\Delta_{\psi_\alpha}X = \sqrt{\frac{\hbar}{2m\omega}}.
\end{align*}
\begin{align*}
\Delta_{\psi_\alpha}P = \sqrt{\frac{m\hbar\omega}{2}}.
\end{align*}
Multiplying these positive quantities gives
\begin{align*}
(\Delta_{\psi_\alpha}X)(\Delta_{\psi_\alpha}P) = \sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{m\hbar\omega}{2}} = \frac{\hbar}{2}.
\end{align*}
Together with the first-moment formulas already computed, this proves both expectation identities and the coherent-state uncertainty minimization identity.
[/guided]
[/step]
