[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*}[/step]

[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: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*}[/step]

[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: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*}[/step]

[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: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.[/step]

[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]